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a, Với \(x>0;x\ne1\)
\(P=\left(\frac{\sqrt{x}}{2}-\frac{1}{2\sqrt{x}}\right)^2\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)\)
\(=\left(\frac{x-1}{2\sqrt{x}}\right)^2\left(\frac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}\right)\)
\(=\frac{x^2-2x+1}{4x}.\frac{-4\sqrt{x}}{x-1}=\frac{1-x}{\sqrt{x}}\)
Thay x = 4 => \(\sqrt{x}=2\)vào P ta được :
\(\frac{1-4}{2}=-\frac{3}{2}\)
c, Ta có : \(P< 0\Rightarrow\frac{1-x}{\sqrt{x}}< 0\Rightarrow1-x< 0\)vì \(\sqrt{x}>0\)
\(\Rightarrow-x< -1\Leftrightarrow x>1\)
\(A=\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}-\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
ĐK : \(\hept{\begin{cases}x,y>0\\x\ne y\end{cases}}\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}-\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\frac{x+2\sqrt{xy}+y}{x-y}-\frac{x-2\sqrt{xy}+y}{x-y}\)
\(=\frac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y}{x-y}=\frac{4\sqrt{xy}}{x-y}\)
Với \(\hept{\begin{cases}x=7+2\sqrt{3}\\y=7-2\sqrt{3}\end{cases}}\)( tmđk )
=> \(A=\frac{4\sqrt{\left(7+2\sqrt{3}\right)\left(7-2\sqrt{3}\right)}}{7+2\sqrt{3}-\left(7-2\sqrt{3}\right)}\)
\(=\frac{4\sqrt{7^2-\left(2\sqrt{3}\right)^2}}{7+2\sqrt{3}-7+2\sqrt{3}}\)
\(=\frac{4\sqrt{49-12}}{4\sqrt{3}}\)
\(=\frac{4\sqrt{37}}{4\sqrt{3}}=\frac{\sqrt{37}}{\sqrt{3}}=\frac{\sqrt{37}\cdot\sqrt{3}}{\sqrt{3}\cdot\sqrt{3}}=\frac{\sqrt{111}}{3}\)
a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\left(\frac{1-x}{\sqrt{2}}\right)^2\)
\(=\left[\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right].\frac{\left(1-x\right)^2}{2}\)
\(=\left[\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\frac{\left(x-1\right)^2}{2}\)
\(=\left[\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{x+\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\frac{\left(\sqrt{x}-1\right)^2.\left(\sqrt{x}+1\right)^2}{2}\)
\(=\frac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2.\left(\sqrt{x}+1\right)^2}{2}\)
\(=\frac{-2\sqrt{x}.\left(\sqrt{x}-1\right)}{2}=-\sqrt{x}\left(\sqrt{x}-1\right)=-x+\sqrt{x}\)
b) Với \(0< x< 1\)\(\Rightarrow0< \sqrt{x}< 1\)
\(\Rightarrow\sqrt{x}-1< 0\)
mà \(\sqrt{x}>0\)\(\Rightarrow\sqrt{x}.\left(\sqrt{x}-1\right)< 0\)
\(\Rightarrow-\sqrt{x}.\left(\sqrt{x}-1\right)>0\)\(\Rightarrow P>0\)( đpcm )
c) \(P=-x+\sqrt{x}=-x+\sqrt{x}-\frac{1}{4}+\frac{1}{4}\)
\(=-\left(x-\sqrt{x}+\frac{1}{4}\right)+\frac{1}{4}=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\)
Vì \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\)\(\Rightarrow-\left(\sqrt{x}-\frac{1}{2}\right)^2\le0\)
\(\Rightarrow-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
Dấu " = " xảy ra \(\Leftrightarrow\sqrt{x}-\frac{1}{2}=0\)\(\Leftrightarrow\sqrt{x}=\frac{1}{2}\)\(\Leftrightarrow x=\frac{1}{4}\)( thỏa mãn ĐKXĐ )
Vậy \(maxP=\frac{1}{4}\)\(\Leftrightarrow x=\frac{1}{4}\)
ĐKXĐ \(\hept{\begin{cases}x\ne1\\x\ge0\end{cases}}\)
a, Ta có \(P=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\left(\frac{\left(1-\sqrt{x}\right).\left(1+\sqrt{x}\right)}{\sqrt{2}}\right)^2\)
\(P=\left(\frac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\left(\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}{\sqrt{2}}\right)^2\)
\(P=\left(\frac{-2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\left(\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}{\sqrt{2}}\right)^2\)
\(P=\frac{2\sqrt{x}-2x}{\sqrt{2}}\)
\(P=\sqrt{2x}-\sqrt{2}x\)
\(P=\sqrt{2x}\left(1-\sqrt{x}\right)\)
b, Vì \(0< x< 1\Rightarrow\sqrt{x}< 1\Rightarrow1-\sqrt{x}< 1\)
\(\Rightarrow\sqrt{2x}\left(1-\sqrt{x}\right)>0\)
c, Ta có \(P=-\sqrt{2}\left(x-\sqrt{x}\right)\)
\(P=-\sqrt{2}\left(x-\frac{1}{2}.2.\sqrt{x}+\frac{1}{4}-\frac{1}{4}\right)\)
\(P=-\sqrt{2x}\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{\sqrt{8}}\le\frac{1}{\sqrt{8}}\)
Dấu = xảy ra \(\Leftrightarrow\)\(\sqrt{x}-\frac{1}{2}=0\)
\(\Rightarrow x=\frac{1}{4}\)
vậy GTLN của P là \(\frac{1}{\sqrt{8}}\)với x=\(\frac{1}{4}\)
Đặt Q = \(\frac{x^3}{4\left(y+2\right)}+\frac{y^3}{4\left(x+2\right)}\) = \(\frac{x^3\left(x+2\right)}{4\left(x+2\right)\left(y+2\right)}+\frac{y^3\left(y+2\right)}{4\left(x+2\right)\left(y+2\right)}\)
Q = \(\frac{x^4+y^4+2x^3+2y^3}{4\left(x+2\right)\left(y+2\right)}\) = \(\frac{x^4+y^4+2\left(x+y\right)\left(x^2-xy+y^2\right)}{4\left(xy+2x+2y+4\right)}\)
Q = \(\frac{x^4+y^4+2\left(x+y\right)\left(x^2-xy+y^2\right)}{4\left(2x+2y+8\right)}\) = \(\frac{x^4+y^4+2\left(x+y\right)\left(x^2-xy+y^2\right)}{8\left(x+y+4\right)}\)
Áp dụng bất đẳng thức AM-GM ta có:
\(x^4+y^4\ge2\sqrt{x^4y^4}=2x^2y^2\)
\(x^2+y^2\ge2\sqrt{x^2y^2=}2xy\)
\(\Leftrightarrow\)Q = \(\frac{x^4+y^4+2\left(x+y\right)\left(x^2-xy+y^2\right)}{8\left(x+y+4\right)}\ge\frac{2x^2y^2+2xy\left(x+y\right)}{8\left(x+y+4\right)}=\frac{2xy\left(xy+x+y\right)}{8\left(x+y+4\right)}\)
\(\Leftrightarrow\)Q = \(\frac{8\left(x+y+4\right)}{8\left(x+y+4\right)}\)= \(1\)
Đẳng thức xảy ra : \(\Leftrightarrow\hept{\begin{cases}x,y>0\\x=y\Rightarrow\\xy=4\end{cases}x=y=2}\)
Vậy giá trị nhỏ nhất của Q là 1 \(\Leftrightarrow x=y=2\)
CMR: \(\left(2+\sqrt{3}\right)^{2021}+\left(2-\sqrt{3}\right)^{2021}⋮4\)
đặt \(a=2+\sqrt{3}\); \(b=2-\sqrt{3}\)
suy ra: \(a+b=2+\sqrt{3}+2-\sqrt{3}=4\)
và : \(ab=\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)=1\)
Ta có: \(a^{2021}+b^{2021}=\left(a+b\right)\left(a^{2020}-a^{2019}b+a^{2018}b^2-...+a^{1010}b^{1010}-...-ab^{2019}+b^{2020}\right)\)
\(=\left(a+b\right)\left(a^{2020}-a^{2018}ab+a^{2016}a^2b^2-...+a^{1010}b^{1010}-...-abb^{2018}+b^{2020}\right)\)
Vì \(a+b=4\);\(ab=1\)nên:
\(a^{2021}+b^{2021}=4\left(a^{2020}-a^{2018}+a^{2016}-...+1-...-b^{2018}+b^{2020}\right)\)
\(=4\left(a^{2020}+b^{2020}-\left(a^{2018}+b^{2018}\right)+a^{2016}+b^{2016}-...+1\right)\)
\(=4\left(\left(a+b\right)^{2020}-2\left(ab\right)^{1010}-\left(a+b\right)^{2018}+2\left(ab\right)^{1009}+\left(a+b\right)^{2016}-2\left(ab\right)^{1008}-...+1\right)\)\(=4\left(4^{2020}-2-4^{2018}+2+4^{2016}-2-...+1\right)\)
\(=4S\)(Với \(S\inℕ^∗\))
suy ra \(a^{2021}+b^{2021}=4S⋮4\)
Vậy \(\left(2+\sqrt{3}\right)^{2021}+\left(2-\sqrt{3}\right)^{2021}⋮4\left(đpcm\right)\)
\(P=\frac{1}{x^2-\sqrt{x}}:\frac{\sqrt{x}+1}{x\sqrt{x}+x+\sqrt{x}}\)\(=\frac{1}{\sqrt{x}\left(\sqrt{x^3}-1^3\right)}:\frac{\sqrt{x}+1}{\sqrt{x}\left(x+\sqrt{x}+1\right)}\)
\(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\frac{\sqrt{x}\left(x+\sqrt{x}+1\right)}{\sqrt{x}+1}\)\(\frac{1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{1}{x-1}\)
2) De 3P=1+x
=> \(3\left(\frac{1}{x-1}\right)=x+1\)<=>\(\frac{3}{x-1}=x+1\)<=>\(\left(x+1\right)\left(x-1\right)=3\)
<=> \(x^2-1=3\)<=> \(x^2=4\)<=> \(x=2\)hoac \(x=-2\)
Vay voi x =2 va x=-2 ta co 3P=x+1
Đáp án là A