\(\dfrac{x}{y+2+1}=\dfrac{y}{x+2+1}=\dfrac{n}{x+y+1}=x+y+2\)
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Những câu hỏi liên quan
Tìm x, y, z
dfrac{x+y+2}{z}dfrac{y+z+1}{x}dfrac{z+x-3}{y}dfrac{1}{x+y+z}
Áp dụng tích chất của dãy tỉ số bằng nhau, ta có
dfrac{x+y+2}{z}dfrac{y+z+1}{x}dfrac{z+x-3}{y}
dfrac{x+y+2+y+z+1+z+x-3}{z+x+y}dfrac{2left(x+y+zright)+left(1+2-3right)}{z+x+y}2
Vìdfrac{x+y+2}{z}dfrac{y+z+1}{x}dfrac{z+x-3}{y}dfrac{1}{x+y+z}
2dfrac{1}{x+y+z}2left(x+y+zright)1x+y+zdfrac{1}{2}
dfrac{x+y+2}{z}2x+y+22z
dfrac{y+z+1}{x}2y+z+12x
dfrac{z+x-3}{y}2z+x-32y
dfrac{1}{x+y+z}2x+y+zdfrac{1}{2}
+) x+y+z dfrac{1}{2}y+zdfra...
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Tìm x, y, z
\(\dfrac{x+y+2}{z}=\dfrac{y+z+1}{x}=\dfrac{z+x-3}{y}=\dfrac{1}{x+y+z}\)
Áp dụng tích chất của dãy tỉ số bằng nhau, ta có
\(\dfrac{x+y+2}{z}=\dfrac{y+z+1}{x}=\dfrac{z+x-3}{y}\\ =\dfrac{x+y+2+y+z+1+z+x-3}{z+x+y}=\dfrac{2\left(x+y+z\right)+\left(1+2-3\right)}{z+x+y}=2\\ Vì\dfrac{x+y+2}{z}=\dfrac{y+z+1}{x}=\dfrac{z+x-3}{y}=\dfrac{1}{x+y+z}\\ =>2=\dfrac{1}{x+y+z}=>2\left(x+y+z\right)=1=>x+y+z=\dfrac{1}{2}\\ =>\dfrac{x+y+2}{z}=2=>x+y+2=2z\\ \dfrac{y+z+1}{x}=2=>y+z+1=2x\\ \dfrac{z+x-3}{y}=2=>z+x-3=2y\\ \dfrac{1}{x+y+z}=2=>x+y+z=\dfrac{1}{2}\)
+) x+y+z = \(\dfrac{1}{2}=>y+z=\dfrac{1}{2}-x=>\dfrac{1}{2}-x+1=2x=>3x=\dfrac{3}{2}=>x=\dfrac{1}{2}\)
+)\(x+y+z=\dfrac{1}{2}=>x+y=\dfrac{1}{2}-z=>\dfrac{1}{2}-z+2=2z=>3z=\dfrac{5}{2}=>z=\dfrac{5}{6}\)
\(=>x+y+z=\dfrac{1}{2}+\dfrac{5}{6}+y=\dfrac{1}{2}=>\dfrac{4}{3}+y=\dfrac{1}{2}=>y=\dfrac{-5}{6}\)
Vậy \(x=\dfrac{1}{2}\\ y=\dfrac{-5}{6}\\ z=\dfrac{5}{6}\)
Ê mấy bọn 7B Nguyễn Lương Bằng ơi bài 2 Toán chiều làm thế này đúng chưa! Góp ý nha!
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MK
18 tháng 2 2022 lúc 20:36
Thực hiện các phép tính sau:
a,(\(\dfrac{x}{x+1}\)+\(\dfrac{x-1}{x}\)):(\(\dfrac{x}{x+1}\)-\(\dfrac{x-1}{x}\))
b,(1+\(\dfrac{x}{y}\)+\(\dfrac{x^2}{y^2}\)).(1-\(\dfrac{x}{y}\)).\(\dfrac{y^2}{x^3-y^3}\)
\(\left(\dfrac{x}{x+1}+\dfrac{x-1}{x}\right):\left(\dfrac{x}{x+1}-\dfrac{x-1}{x}\right)\) \(\left(đk:x\ne0;-1\right)\)
\(=\dfrac{x^2+\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}:\left(\dfrac{x^2-\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}\right)\)
\(=\dfrac{x^2+x^2-1}{x\left(x+1\right)}.\dfrac{x\left(x+1\right)}{x^2-x^2+1}\)
\(=\dfrac{\left(2x^2-1\right)x\left(x+1\right)}{x\left(x+1\right)}=2x^2-1\)
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giải các hệ phương trìnha dfrac{5}{x-1}+dfrac{1}{y-1}10dfrac{1}{x-1}-dfrac{3}{y-1}18b dfrac{5}{x+y-3}-dfrac{2}{x-y+1}8dfrac{3}{x+y-3}+dfrac{1}{x-y+1}dfrac{3}{2}c sqrt{x-1}-3sqrt{y+2}22sqrt{x-1}+5sqrt{y+2}15d dfrac{7}{sqrt{x-7}}-dfrac{4}{sqrt{y+6}}dfrac{5}{3}dfrac{5}{sqrt{x-7}}+dfrac{3}{sqrt{y+6}}dfrac{13}{6}e 7x^2+13y-395x^2-11y33f 2left(x-1right)^2-3y^375left(x-1right)^2+6y^34
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giải các hệ phương trình
a \(\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\)
\(\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\)
b \(\dfrac{5}{x+y-3}-\dfrac{2}{x-y+1}=8\)
\(\dfrac{3}{x+y-3}+\dfrac{1}{x-y+1}=\dfrac{3}{2}\)
c \(\sqrt{x-1}-3\sqrt{y+2}=2\)
\(2\sqrt{x-1}+5\sqrt{y+2}=15\)
d \(\dfrac{7}{\sqrt{x-7}}-\dfrac{4}{\sqrt{y+6}}=\dfrac{5}{3}\)
\(\dfrac{5}{\sqrt{x-7}}+\dfrac{3}{\sqrt{y+6}}=\dfrac{13}{6}\)
e \(7x^2+13y=-39\)
\(5x^2-11y=33\)
f \(2\left(x-1\right)^2-3y^3=7\)
\(5\left(x-1\right)^2+6y^3=4\)
a) Ta có: \(\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{5}{x-1}-\dfrac{15}{y-1}=90\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{16}{y-1}=-80\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y-1=\dfrac{-1}{5}\\\dfrac{1}{x-1}=18+\dfrac{3}{y-1}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{4}{5}\\x-1=\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=\dfrac{4}{5}\end{matrix}\right.\)
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Cho x, y, z 0 thoả mãn x+y+z1. Chứng minh rằng:
a) sqrt{x^2+dfrac{1}{x^2}}+sqrt{y^2+dfrac{1}{y^2}}+sqrt{z^2+dfrac{1}{z^2}}gesqrt{82}
b) sqrt{x^2+dfrac{1}{x^2}+dfrac{1}{y^2}}+sqrt{y^2+dfrac{1}{y^2}+dfrac{1}{z^2}}+sqrt{z^2+dfrac{1}{z^2}+dfrac{1}{x^2}}gesqrt{163}
c)sqrt{x^2+dfrac{2}{y^2}+dfrac{3}{z^2}}+sqrt{y^2+dfrac{2}{z^2}+dfrac{3}{x^2}}+sqrt{z^2+dfrac{2}{z^2}+dfrac{3}{y^2}}gesqrt{406}
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Cho x, y, z > 0 thoả mãn x+y+z=1. Chứng minh rằng:
a) \(\sqrt{x^2+\dfrac{1}{x^2}}+\sqrt{y^2+\dfrac{1}{y^2}}+\sqrt{z^2+\dfrac{1}{z^2}}\ge\sqrt{82}\)
b) \(\sqrt{x^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}}+\sqrt{y^2+\dfrac{1}{y^2}+\dfrac{1}{z^2}}+\sqrt{z^2+\dfrac{1}{z^2}+\dfrac{1}{x^2}}\ge\sqrt{163}\)
c)\(\sqrt{x^2+\dfrac{2}{y^2}+\dfrac{3}{z^2}}+\sqrt{y^2+\dfrac{2}{z^2}+\dfrac{3}{x^2}}+\sqrt{z^2+\dfrac{2}{z^2}+\dfrac{3}{y^2}}\ge\sqrt{406}\)
Rút gọn các biểu thức:a) {dfrac{1}{x^2} + dfrac{1}{y^2} + dfrac{2}{x+y}(dfrac{1}{x} + dfrac{1}{y})} : dfrac{x^3+y^3}{x^2y^2}b) {dfrac{1}{left(2x-yright)^2} + dfrac{2}{4x^2-y^2} + dfrac{1}{left(2x+yright)^2}} . dfrac{4x^2+4xy+y^2}{16x}c) (dfrac{x^2-xy}{x^2y+y^3} - dfrac{2x^2}{y^3-xy^2+x^2y-x^3})(1 - dfrac{y-1}{x} - dfrac{y}{x^2})
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Rút gọn các biểu thức:
a) {\(\dfrac{1}{x^2}\) + \(\dfrac{1}{y^2}\) + \(\dfrac{2}{x+y}\)(\(\dfrac{1}{x}\) + \(\dfrac{1}{y}\))} : \(\dfrac{x^3+y^3}{x^2y^2}\)
b) {\(\dfrac{1}{\left(2x-y\right)^2}\) + \(\dfrac{2}{4x^2-y^2}\) + \(\dfrac{1}{\left(2x+y\right)^2}\)} . \(\dfrac{4x^2+4xy+y^2}{16x}\)
c) (\(\dfrac{x^2-xy}{x^2y+y^3}\) - \(\dfrac{2x^2}{y^3-xy^2+x^2y-x^3}\))(1 - \(\dfrac{y-1}{x}\) - \(\dfrac{y}{x^2}\))
VD
2 tháng 4 2022 lúc 21:30
1, x,y,z∈N*. CMR x+3z-y là hợp số biết `x^2+y^2=z^2`
2,Tìm n∈N* để \(\left(4n^3+n+3\right)⋮\left(2n^2+n+1\right)\)
3, CMR:\(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{4}{xy}\forall x\ne y,xy\ne0\)
2.
\(4n^3+n+3=4n^3+2n^2+2n-2n^2-n-1+4=2n\left(2n^2+n+1\right)-\left(2n^2+n+1\right)+4\)-Để \(\left(4n^3+n+3\right)⋮\left(2n^2+n+1\right)\) thì \(4⋮\left(2n^2+n+1\right)\)
\(\Leftrightarrow2n^2+n+1\in\left\{1;-1;2;-2;4;-4\right\}\) (do n là số nguyên)
*\(2n^2+n+1=1\Leftrightarrow n\left(2n+1\right)=0\Leftrightarrow n=0\) (loại) hay \(n=\dfrac{-1}{2}\) (loại)
*\(2n^2+n+1=-1\Leftrightarrow2n^2+n+2=0\) (phương trình vô nghiệm)
\(2n^2+n+1=2\Leftrightarrow2n^2+n-1=0\Leftrightarrow n^2+n+n^2-1=0\Leftrightarrow n\left(n+1\right)+\left(n+1\right)\left(n-1\right)=0\Leftrightarrow\left(n+1\right)\left(2n-1\right)=0\)
\(\Leftrightarrow n=-1\) (loại) hay \(n=\dfrac{1}{2}\) (loại)
\(2n^2+n+1=-2\Leftrightarrow2n^2+n+3=0\) (phương trình vô nghiệm)
\(2n^2+n+1=4\Leftrightarrow2n^2+n-3=0\Leftrightarrow2n^2-2n+3n-3=0\Leftrightarrow2n\left(n-1\right)+3\left(n-1\right)=0\Leftrightarrow\left(n-1\right)\left(2n+3\right)=0\)\(\Leftrightarrow n=1\left(nhận\right)\) hay \(n=\dfrac{-3}{2}\left(loại\right)\)
-Vậy \(n=1\)
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1. \(x^2+y^2=z^2\)
\(\Rightarrow x^2+y^2-z^2=0\)
\(\Rightarrow\left(x-z\right)\left(x+z\right)+y^2=0\)
-TH1: y lẻ \(\Rightarrow x-z;x+z\) đều lẻ.
\(x+3z-y=x+z-y+2x\) chia hết cho 2. \(\Rightarrow\)Hợp số.
-TH2: y chẵn \(\Rightarrow\)1 trong hai biểu thức \(x-z;x+z\) chia hết cho 2.
*Xét \(\left(x-z\right)⋮2\):
\(x+3z-y=x-z+4z-y\) chia hết cho 2. \(\Rightarrow\)Hợp số.
*Xét \(\left(x+z\right)⋮2\):
\(x+3z-y=x+z+2z-y\) chia hết cho 2 \(\Rightarrow\)Hợp số.
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Đặt $ X a - b; Y b - c; Z c - a Rightarrow X + Y + Z 0$Với X + Y + Z 0, ta chứng minh được :$ ( dfrac{1}{X} + dfrac{1}{Y} + dfrac{1}{Z} )^2 dfrac{1}{X^2} + dfrac{1}{Y^2} + dfrac{1}{Z^2}$Thật vậy, ta có :$ ( dfrac{1}{X} + dfrac{1}{Y} + dfrac{1}{Z} )^2 dfrac{1}{X^2} + dfrac{1}{Y^2} + dfrac{1}{Z^2} + dfrac{2}{XY} + dfrac{2}{YZ} + dfrac{2}{ZX}$$ dfrac{1}{X^2} + dfrac{1}{Y^2} + dfrac{1}{Z^2} + 2.dfrac{X + Y + Z}{XYZ}$$ dfrac{1}{X^2} + dfrac{1}{Y^2} + dfrac{1}{Z^2}$ ( do X + Y + Z 0)$ Righta...
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Đặt $ X = a - b; Y = b - c; Z = c - a \Rightarrow X + Y + Z = 0$
Với X + Y + Z = 0, ta chứng minh được :
$ ( \dfrac{1}{X} + \dfrac{1}{Y} + \dfrac{1}{Z} )^2 = \dfrac{1}{X^2} + \dfrac{1}{Y^2} + \dfrac{1}{Z^2}$
Thật vậy, ta có :
$ ( \dfrac{1}{X} + \dfrac{1}{Y} + \dfrac{1}{Z} )^2 = \dfrac{1}{X^2} + \dfrac{1}{Y^2} + \dfrac{1}{Z^2} + \dfrac{2}{XY} + \dfrac{2}{YZ} + \dfrac{2}{ZX}$
$ = \dfrac{1}{X^2} + \dfrac{1}{Y^2} + \dfrac{1}{Z^2} + 2.\dfrac{X + Y + Z}{XYZ}$
$ = \dfrac{1}{X^2} + \dfrac{1}{Y^2} + \dfrac{1}{Z^2}$ ( do X + Y + Z = 0)
$ \Rightarrow \sqrt{\dfrac{1}{X^2} + \dfrac{1}{Y^2} + \dfrac{1}{Z^2}} = \sqrt{( \dfrac{1}{X} + \dfrac{1}{Y} + \dfrac{1}{Z} )^2} = |\dfrac{1}{X} + \dfrac{1}{Y} + \dfrac{1}{Z}|$
Suy ra : $ \sqrt{\dfrac{1}{(a - b)^2} + \dfrac{1}{(b - c)^2} +\dfrac{1}{( c - a)^2}} = |\dfrac{1}{a - b} + \dfrac{1}{b - c} + \dfrac{1}{c - a}|$
Do a, b, c là số hữu tỷ nên $|\dfrac{1}{a - b} + \dfrac{1}{b - c} + \dfrac{1}{c - a}|$ cũng là số hữu tỷ. Ta có điều phải chứng minh.
Rút gọn các biểu thức:
a) (x - \(\dfrac{1}{1-x}\)) : \(\dfrac{x^2-x+1}{x^2-2x+1}\)
b) (1 + \(\dfrac{x}{y}\) + \(\dfrac{x^2}{y^2}\))(1 - \(\dfrac{x}{y}\))\(\dfrac{y^2}{x^3-y^3}\)
a) Ta có: \(\left(x-\dfrac{1}{1-x}\right):\dfrac{x^2-x+1}{x^2-2x+1}\)
\(=\left(x+\dfrac{1}{x-1}\right):\dfrac{x^2-x+1}{\left(x-1\right)^2}\)
\(=\dfrac{x^2-x+1}{x-1}\cdot\dfrac{\left(x-1\right)^2}{x^2-x+1}\)
\(=x-1\)
b) Ta có: \(\left(1+\dfrac{x}{y}+\dfrac{x^2}{y^2}\right)\left(1-\dfrac{x}{y}\right)\cdot\dfrac{y^2}{x^3-y^3}\)
\(=\left(\dfrac{y^2}{y^2}+\dfrac{xy}{y^2}+\dfrac{x^2}{y^2}\right)\cdot\left(\dfrac{y-x}{y}\right)\cdot\dfrac{y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{x^2+xy+y^2}{y^2}\cdot\dfrac{-\left(x-y\right)}{y}\cdot\dfrac{y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{-1}{y}\)
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Cho x, y, z 0 thoả mãn x+y+z2. Tìm GTNN của các biểu thức:
a) Asqrt{x^2+dfrac{1}{x^2}}+sqrt{y^2+dfrac{1}{y^2}}+sqrt{z^2+dfrac{1}{z^2}}
b) Bsqrt{x^2+dfrac{1}{y^2}+dfrac{1}{z^2}}+sqrt{y^2+dfrac{1}{z^2}+dfrac{1}{x^2}}+sqrt{z^2+dfrac{1}{x^2}+dfrac{1}{y^2}}
c) Csqrt{2x^2+dfrac{3}{y^2}+dfrac{4}{z}}+sqrt{2y^2+dfrac{3}{z^2}+dfrac{4}{x^2}}+sqrt{2z^2+dfrac{3}{x^2}+dfrac{4}{y^2}}
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Cho x, y, z > 0 thoả mãn x+y+z=2. Tìm GTNN của các biểu thức:
a) \(A=\sqrt{x^2+\dfrac{1}{x^2}}+\sqrt{y^2+\dfrac{1}{y^2}}+\sqrt{z^2+\dfrac{1}{z^2}}\)
b) \(B=\sqrt{x^2+\dfrac{1}{y^2}+\dfrac{1}{z^2}}+\sqrt{y^2+\dfrac{1}{z^2}+\dfrac{1}{x^2}}+\sqrt{z^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}}\)
c) \(C=\sqrt{2x^2+\dfrac{3}{y^2}+\dfrac{4}{z}}+\sqrt{2y^2+\dfrac{3}{z^2}+\dfrac{4}{x^2}}+\sqrt{2z^2+\dfrac{3}{x^2}+\dfrac{4}{y^2}}\)
Áp dụng liên tiếp bất đẳng thức Mincopxki và bất đẳng thức Cauchy-Schwarz:
\(A=\sqrt{x^2+\dfrac{1}{x^2}}+\sqrt{y^2+\dfrac{1}{y^2}}+\sqrt{z^2+\dfrac{1}{z^2}}\)
\(A\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(A\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{\left(1+1+1\right)^2}{x+y+z}\right)^2}\)
\(A\ge\sqrt{4+\dfrac{81}{4}}=\sqrt{\dfrac{97}{4}}\)
Dấu "=" xảy ra khi: \(x=y=z=\dfrac{2}{3}\)
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\(B=\sqrt{x^2+\dfrac{1}{y^2}+\dfrac{1}{z^2}}+\sqrt{y^2+\dfrac{1}{z^2}+\dfrac{1}{x^2}}+\sqrt{z^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(B=\sqrt{\left(x+y+z\right)^2+2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+2\left(\dfrac{\left(1+1+1\right)^2}{x+y+z}\right)^2}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+\dfrac{162}{\left(x+y+z\right)^2}}\)
\(B\ge\sqrt{4+\dfrac{162}{4}}=\sqrt{\dfrac{89}{2}}\)
Dấu "=" xảy ra khi: \(x=y=z=\dfrac{2}{3}\)
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