\(\frac{x^2}{\left(1+\sqrt{x+1}\right)^2}\le x-4\)
TH
Những câu hỏi liên quan
Rút gọn :
a) \(\sqrt{2x-\sqrt{4x-1}}-\sqrt{2x+\sqrt{4x-1}}\) (với \(\frac{1}{4}\le x\le\frac{1}{2}\)
b)\(\frac{\sqrt{x+\sqrt{4\left(x-1\right)}}-\sqrt{x-\sqrt{4\left(x-1\right)}}}{\sqrt{x^2-4\left(x-1\right)}}.\left(\sqrt{x-1}-\frac{1}{\sqrt{x-1}}\right)\)
B1 Cho biểu thức A=\(\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{x-3}{x+2\sqrt{x}+4}-\frac{\sqrt{x}+7}{x\sqrt{x}-8}\right):\left(\frac{\sqrt{x}+7}{x+2\sqrt{x}+4}\right)\)
1, Rút gọn A. Tìm x sao cho A<2
2, Cho 1≤a,b,c≤2. Chứng minh rằng \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le10\)
\(\sqrt{\left(x-2\right)\left(4-x\right)}+\sqrt{x-2}.\sqrt{4-x}\le\frac{x-1}{2}+\sqrt{x-1}\)
1.Chmr rằng nếu: a,b >0 thì \(\sqrt{a}+\sqrt{b}\le\sqrt{\frac{a^2}{b}}+\sqrt{\frac{b^2}{a}}\)
2. Rg biểu thức:
\(A=\frac{\sqrt{x-\sqrt{4\left(x-1\right)}}+\sqrt{x+\sqrt{4\left(x-1\right)}}}{\sqrt{x^2-4\left(x-1\right)}}.\left(1-\frac{1}{x-1}\right)\)
M^2left(sqrt{x}+sqrt{2y}right)^2left(frac{1}{_{sqrt{alpha}}}.sqrt{alpha x}+sqrt{2y}right)^2 left(frac{1}{alpha}+1right)left(alpha x+2yright)Rightarrow M^4leleft(frac{1}{alpha}+1right)^2left(alpha x+2yright)^2leleft(frac{1}{alpha}+1right)^2left(alpha^2+4right)left(x^2+y^2right)left(frac{1}{alpha}+1right)^2left(alpha^2+4right)Dấu bằng xảy ra hept{begin{cases}frac{alpha x}{frac{1}{alpha}}frac{2y}{1}frac{alpha}{x}frac{2}{y}end{cases}}Rightarrowhept{begin{cases}alpha^2x2yalphafrac{2x}{y}end{cases}R...
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\(M^2=\left(\sqrt{x}+\sqrt{2y}\right)^2=\left(\frac{1}{_{\sqrt{\alpha}}}.\sqrt{\alpha x}+\sqrt{2y}\right)^2< =\left(\frac{1}{\alpha}+1\right)\left(\alpha x+2y\right)\)
\(\Rightarrow M^4\le\left(\frac{1}{\alpha}+1\right)^2\left(\alpha x+2y\right)^2\le\left(\frac{1}{\alpha}+1\right)^2\left(\alpha^2+4\right)\left(x^2+y^2\right)=\left(\frac{1}{\alpha}+1\right)^2\left(\alpha^2+4\right)\)
Dấu bằng xảy ra => \(\hept{\begin{cases}\frac{\alpha x}{\frac{1}{\alpha}}=\frac{2y}{1}\\\frac{\alpha}{x}=\frac{2}{y}\end{cases}}\Rightarrow\hept{\begin{cases}\alpha^2x=2y\\\alpha=\frac{2x}{y}\end{cases}\Rightarrow\hept{\begin{cases}\frac{\alpha^2}{2}=\frac{y}{x}\\\frac{\alpha}{2}=\frac{x}{y}\end{cases}}}\Rightarrow\frac{\alpha^2}{2}=\frac{1}{\frac{\alpha}{2}}\Rightarrow\alpha=\sqrt[3]{4}\)
Suy ra max = \(\sqrt[4]{\left(\frac{1}{\alpha}+1\right)^2\left(\alpha^2+4\right)}\) với \(\alpha=\sqrt[3]{4}\)
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Xem thêm câu trả lời
\(A=\left(\sqrt{x}-\frac{x+2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-4}{1-x}\right)\)
b) tìm x khi \(A\le\frac{-2}{5}\)
\(A=\left(\sqrt{x}-\frac{x+2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-4}{1-x}\right)\)
\(A=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}-\frac{x+2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}\left(1-\sqrt{x}\right)}{1-x}-\frac{\sqrt{x}-4}{1-x}\right)\)
\(A=\left(\frac{x+\sqrt{x}-x-2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}-x-\sqrt{x}+4}{1-x}\right)\)
\(A=\frac{\sqrt{x}-2}{\sqrt{x}+1}:\frac{4-x}{1-x}\)
\(A=\frac{\sqrt{x}-2}{\sqrt{x}+1}.\frac{1-x}{4-x}\)
\(A=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
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Chứng minh giúp mình mấy câu bất đẳng thức này nhaa) frac{2sqrt{ab}}{sqrt{a}+sqrt{b}}lesqrt[4]{ab}left(a,b0right)b) left(sqrt{a}+sqrt{b}right)^8ge64ableft(a+bright)^2left(a,b0right)c) yleft(frac{1}{x}+frac{1}{x}right)+frac{1}{y}left(x+zright)leleft(frac{1}{x}+frac{1}{z}right)left(x+zright)left(0 xle yle zright)d) a+b+cge3left(frac{1}{a}+frac{1}{b}+frac{1}{c}right)left(a,b,c0;a+b+cabcright)
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Chứng minh giúp mình mấy câu bất đẳng thức này nha
a) \(\frac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\le\sqrt[4]{ab}\left(a,b>0\right)\)
b) \(\left(\sqrt{a}+\sqrt{b}\right)^8\ge64ab\left(a+b\right)^2\left(a,b>0\right)\)
c) \(y\left(\frac{1}{x}+\frac{1}{x}\right)+\frac{1}{y}\left(x+z\right)\le\left(\frac{1}{x}+\frac{1}{z}\right)\left(x+z\right)\left(0< x\le y\le z\right)\)
d) \(a+b+c\ge3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(a,b,c>0;a+b+c=abc\right)\)
a, Đặt \(\sqrt[4]{a}=x;\sqrt[4]{b}=y.\)Bất đẳng thức ban đầu trở thành: \(\frac{2x^2y^2}{x^2+y^2}\le xy.\)
ta có : \(x^2+y^2\ge2xy\Rightarrow\frac{2x^2y^2}{x^2+y^2}\le\frac{2x^2y^2}{2xy}=xy.\)(đpcm )
dấu " = " xẩy ra khi x = y > 0
vậy bất đăng thức ban đầu đúng. dấu " = " xẩy ra khi a = b >0
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Rút gọn biểu thức frac{left(sqrt{x-4sqrt{x-4}}+sqrt{x+4sqrt{x-4}}right)left(sqrt{x-1}-1right)}{sqrt{x-2sqrt{x-1}}}b,sqrt{x-2sqrt{x-1}+sqrt{x+2sqrt{x-1}}}1le xle2c, sqrt{x+6sqrt{x-9}}+sqrt{x-6sqrt{x-9}}x18d, frac{1}{2left(1+sqrt{x+2}right)}+frac{1}{2left(1-2sqrt{x+2}right)}e,frac{1}{sqrt{x+2sqrt{x-1}}}-frac{1}{sqrt{x-2sqrt{x-1}}}
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Rút gọn biểu thức
\(\frac{\left(\sqrt{x-4\sqrt{x-4}}+\sqrt{x+4\sqrt{x-4}}\right)\left(\sqrt{x-1}-1\right)}{\sqrt{x-2\sqrt{x-1}}}\)
b,\(\sqrt{x-2\sqrt{x-1}+\sqrt{x+2\sqrt{x-1}}}1\le x\le2\)
c, \(\sqrt{x+6\sqrt{x-9}}+\sqrt{x-6\sqrt{x-9}}x>18\)
d, \(\frac{1}{2\left(1+\sqrt{x+2}\right)}+\frac{1}{2\left(1-2\sqrt{x+2}\right)}\)
e,\(\frac{1}{\sqrt{x+2\sqrt{x-1}}}-\frac{1}{\sqrt{x-2\sqrt{x-1}}}\)
Giải PT:
\(\sqrt{x^4+x^2+1}+\sqrt{x\left(x^2-x+1\right)}\le\sqrt{\frac{\left(x^2+1\right)^3}{x}}\)
Tìm max
\(A=3\sqrt{2x-1}+x\sqrt{5-4x^2}\left(\frac{1}{2}\le x\le\frac{\sqrt{5}}{2}\right)\)
\(B=\frac{xyz\left(x+y+z+\sqrt{x^2+y^2+z^2}\right)}{\left(x^2+y^2+z^2\right)\left(xy+yz+zx\right)}\left(x,y,z>0\right)\)
A
Áp dụng BĐT cosi ta có
\(\sqrt{\left(2x-1\right).1}\le\frac{2x-1+1}{2}=x\)
\(x\sqrt{5-4x^2}\le\frac{x^2+5-4x^2}{2}=\frac{-3x^2+5}{2}\)
Khi đó
\(A\le3x+\frac{-3x^2+5}{2}=\frac{-3x^2+6x+5}{2}=\frac{-3\left(x-1\right)^2}{2}+4\le4\)
MaxA=4 khi \(\hept{\begin{cases}2x-1=1\\x^2=5-4x^2\\x=1\end{cases}\Rightarrow}x=1\)
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B
Áp dụng BĐT cosi ta có :
\(x^2+y^2+z^2\ge\frac{1}{3}\left(x+y+z\right)^2\)
=> \(x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}\)
=> \(B\le\frac{xyz.\left(\sqrt{3\left(x^2+y^2+z^2\right)}+\sqrt{x^2+y^2+z^2}\right)}{\left(x^2+y^2+z^2\right)\left(xy+yz+xz\right)}=\frac{xyz.\left(\sqrt{3}+1\right)}{\left(xy+yz+xz\right)\sqrt{x^2+y^2+z^2}}\)
Lại có \(x^2+y^2+z^2\ge3\sqrt[3]{x^2y^2z^2}\); \(xy+yz+xz\ge3\sqrt[3]{x^2y^2z^2}\)
=> \(\sqrt{x^2+y^2+z^2}\left(xy+yz+xz\right)\ge3\sqrt[3]{x^2y^2z^2}.\sqrt{3\sqrt[3]{x^2y^2z^2}}=3\sqrt{3}.xyz\)
=> \(B\le\frac{\sqrt{3}+1}{3\sqrt{3}}=\frac{3+\sqrt{3}}{9}\)
\(MaxB=\frac{3+\sqrt{3}}{9}\)khi x=y=z
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