\(\frac{x}{y+z+1}=\frac{y}{z+x+1}=\frac{z}{x+y-2}=z+y+z\)
Tính x y z
HH
Những câu hỏi liên quan
Cho ba số x , y , z khác 0 thỏa mãn $\frac{y+z-x}{x}$ = $\frac{z+x-y}{y}$ = $\frac{x+y-z}{z}$
Tính giá trị biểu thức P = ( 1+$\frac{x}{y}$ )( 1+$\frac{y}{z}$ )( 1+$\frac{z}{x}$ )
\(\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}=\dfrac{x+y-z}{z}\\ \Rightarrow\dfrac{y+z-x}{x}+2=\dfrac{z+x-y}{y}+2=\dfrac{x+y-z}{z}+2\\ \Rightarrow\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}=\dfrac{x+y+z}{z}\\ \Rightarrow x=y=z\\ \Rightarrow A=\left(1+1\right).\left(1+1\right).\left(1+1\right)=8\)
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\(x+y+z=1,\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=1\)
tính giá trị của biểu thức A=\(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\)
Tính:a) \(A=\frac{2}{x-y}+\frac{2}{y-z}+\frac{2}{z-x}+\frac{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
b) Cho \(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=1\) . Tính \(A=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\)
a) \(A=\frac{2}{x-y}+\frac{2}{y-z}+\frac{2}{z-x}+\frac{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{2\left(y-z\right)\left(z-x\right)+2\left(x-y\right)\left(z-x\right)+2\left(x-y\right)\left(y-z\right)+\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{\left[\left(x-y\right)+\left(y-z\right)+\left(z-x\right)\right]^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=\frac{\left(x-y+y-z+z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=0\)
Áp dụng: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
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b)Ta có: \(\frac{x^2}{y+z}+x=\frac{x^2+x\left(y+z\right)}{y+z}=\frac{x^2+xy+xz}{y+z}=\frac{x\left(x+y+z\right)}{y+z}\)
Tương tự: \(\frac{y^2}{x+z}+y=\frac{y^2+xy+zy}{x+z}=\frac{y\left(x+y+z\right)}{x+z}\)
\(\frac{z^2}{x+y}+z=\frac{z^2+xz+zy}{x+y}=\frac{z\left(x+y+z\right)}{x+y}\)
Suy ra: \(A+\left(x+y+z\right)\)
\(=\frac{x\left(x+y+z\right)}{y+z}+\frac{y\left(x+y+z\right)}{z+x}+\frac{z\left(x+y+z\right)}{x+y}+\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}+1\right)\)
\(=2.\left(x+y+z\right)\)
Nên \(A=2.\left(x+y+z\right)-\left(x+y+z\right)=x+y+z\)
Mình có sai chỗ nào không nhỉ?
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Cho \(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=1.\)
Tính S =\(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\)
\(\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)\left(x+y+z_{ }\right)=x+y+z\)+z
\(\frac{x^2}{y+z}+x+\frac{y^2}{x+z}+y+\frac{z^2}{x+y}+z=x+y+z\)
suy ra S=0
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Cho \(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=1\) . Tính S=\(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\)
\(\Rightarrow\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right).\left(x+y+z\right)=x+y+z\)
\(\Rightarrow\frac{x^2+x\left(z+x\right)}{y+z}+\frac{y^2+y\left(x+z\right)}{x+z}+\frac{z^2+z\left(x+y\right)}{x+y}=x+y+z\)
\(\Rightarrow\frac{x^2}{y+z}+x+\frac{y^2}{x+z}+y+\frac{z^2}{x+y}+z=x+y+z\)
\(\Rightarrow\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}=0\)
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thực hiện phép tính
a,x^3+left[frac{xleft(2y^3-x^3right)}{x^3+y^3}right]^3-left[frac{yleft(2x^3-y^3right)}{x^3+y^3}right]^3
b,frac{frac{xleft(x+yright)}{x-y}+frac{xleft(x+zright)}{x-z}}{1+frac{left(y-zright)^2}{left(x-yright)left(x-zright)}}+frac{frac{yleft(y+zright)}{y-z}+frac{yleft(y+xright)}{y-x}}{1+frac{left(z-xright)^2}{left(y-zright)left(y-xright)}}+frac{frac{zleft(z+xright)}{z-x}+frac{zleft(z+yright)}{z-y}}{1+frac{left(x-yright)^2}{left(z-xright)left(z-yright)}}
c,left[frac{y+z-2x}{fra...
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thực hiện phép tính
a,\(x^3+\left[\frac{x\left(2y^3-x^3\right)}{x^3+y^3}\right]^3-\left[\frac{y\left(2x^3-y^3\right)}{x^3+y^3}\right]^3\)
b,\(\frac{\frac{x\left(x+y\right)}{x-y}+\frac{x\left(x+z\right)}{x-z}}{1+\frac{\left(y-z\right)^2}{\left(x-y\right)\left(x-z\right)}}+\frac{\frac{y\left(y+z\right)}{y-z}+\frac{y\left(y+x\right)}{y-x}}{1+\frac{\left(z-x\right)^2}{\left(y-z\right)\left(y-x\right)}}+\frac{\frac{z\left(z+x\right)}{z-x}+\frac{z\left(z+y\right)}{z-y}}{1+\frac{\left(x-y\right)^2}{\left(z-x\right)\left(z-y\right)}}\)
c,\(\left[\frac{y+z-2x}{\frac{\left(y-z\right)^3}{y^3-z^3}+\frac{\left(x-y\right)\left(x-z\right)}{y^2+yz+z^2}}+\frac{z+x-2y}{\frac{\left(z-x\right)^3}{z^3-x^3}+\frac{\left(y-z\right)\left(y-x\right)}{z^2+xz+x^2}}+\frac{x+y-2z}{\frac{\left(x-y\right)^3}{x^3-y^3}+\frac{\left(z-x\right)\left(z-y\right)}{x^2+xy+y^2}}\right]:\frac{1}{x+y+z}\)
Đề:
Cho các số thực x, y, z thoả mãn x + y + z 1 và frac{x}{y+z}+frac{y}{x+z}+frac{z}{x+y}1
left(xne-y;yne-z;zne-xright)
Giá trị của biểu thức Pfrac{x^2}{y+z}+frac{y^2}{x+z}+frac{z^2}{x+y} là . . .
Giải:
x + y + z 1
x 1 - (y + z)
y 1 - (x + z)
z 1 - (x + y)
Thay x 1 - (y + z); y 1 - (x + z) và z 1 - (x + y) vào P, ta có:
Pfrac{xleft[1-left(y+zright)right]}{y+z}+frac{yleft[1-left(x+zright)right]}{x+z}+frac{zleft[1-left(x+yright)right]}{x+y}
frac{x-xleft(y+zright)...
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Đề:
Cho các số thực x, y, z thoả mãn x + y + z = 1 và \(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=1\)
\(\left(x\ne-y;y\ne-z;z\ne-x\right)\)
Giá trị của biểu thức \(P=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\) là . . .
Giải:
x + y + z = 1
=> x = 1 - (y + z)
y = 1 - (x + z)
z = 1 - (x + y)
Thay x = 1 - (y + z); y = 1 - (x + z) và z = 1 - (x + y) vào P, ta có:
\(P=\frac{x\left[1-\left(y+z\right)\right]}{y+z}+\frac{y\left[1-\left(x+z\right)\right]}{x+z}+\frac{z\left[1-\left(x+y\right)\right]}{x+y}\)
\(=\frac{x-x\left(y+z\right)}{y+z}+\frac{y-y\left(x+z\right)}{x+z}+\frac{z-z\left(x+y\right)}{x+y}\)
\(=\frac{x}{y+z}-\frac{x\left(y+z\right)}{y+z}+\frac{y}{x+z}-\frac{y\left(x+z\right)}{x+z}+\frac{z}{x+y}-\frac{z\left(x+y\right)}{x+y}\)
\(=\left(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\right)-\left(x+y+z\right)\)
\(=1-1\)
\(=0\)
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Trịnh Trân Trân <3
Hay quớ ak! Mơn m nhìu nha ný! <3 <3 <3 (not thả thính =))))
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CHo \(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=1\)
Tính : \(A=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\)
\(\Rightarrow\left(\frac{x}{x+y}+\frac{y}{z+x}+\frac{z}{x+y}\right)\cdot\left(x+y+z\right)=x+y+z\)
\(\Rightarrow\frac{x^2}{y+z}+\frac{xy}{y+z}+\frac{xz}{y+z}+\frac{y^2}{z+x}+\frac{xy}{z+x}+\frac{yz}{z+x}+\frac{z^2}{x+y}+\frac{xz}{x+y}+\frac{yz}{x+y}=x+y+z\)
Rồi bạn cộng 2 phân thức 2,3 5,6 8,9 lại thì được
\(\Rightarrow\frac{x^2}{y+z}+x+\frac{y^2}{z+x}+y+\frac{z^2}{x+y}+z=x+y+z\)
\(\Rightarrow\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}=0\)
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a) \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
b) \(\frac{x}{y+z+1}=\frac{y}{x+z+2}=\frac{z}{x+y-2}=x+y+z\)
Tìm x ;y;z
a) \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}=2\)
\(\Rightarrow x+y+z=\frac{1}{2}\)(do 1/(x+y+z)=2)
\(\Rightarrow y+z=\frac{1}{2}-x;z+x=\frac{1}{2}-y;x+y=\frac{1}{2}-z\)
Thay vào lần lượt ta có:
\(\frac{\frac{1}{2}-x+1}{x}=2\)\(\Rightarrow x=\frac{1}{2}\)
\(\frac{\frac{1}{2}-y+2}{y}=2\)\(\Rightarrow y=\frac{5}{6}\)
\(\frac{\frac{1}{2}-z-3}{z}=2\)\(\Rightarrow z=-\frac{5}{6}\)
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Cho x+y+z=7. Biết \frac{x}{y+z} +\frac{y}{x+z} +\frac{z}{x+y} = 3. Tính \frac{x^{2}}{y+z} +\frac{y^{2}}{x+z} +\frac{z^{2}}{x+y}