CMR:
\(\sqrt{X-2\sqrt{X-1}}+\sqrt{X+2\sqrt{X-1}}=2\sqrt{X-1}\)
(x>0,x # 1)
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MC
30 tháng 10 2018
VẬy bạn giải ra cho mọi người xem được ko?
Lớn hơn hoặc bằng kí hiệu trong Latex là \geq nha!
25 tháng 8 2022
a: Sửa đề: \(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+x+1\)
\(=x-\sqrt{x}-x-\sqrt{x}+x+1=x-2\sqrt{x}+1\)
b: \(=\left(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{\left(\sqrt{x}+1\right)\left(x-1\right)}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\sqrt{x}}=2\)
S
5 tháng 7 2019
\(\frac{\sqrt{2}-1}{\sqrt{2}+2}-\frac{1}{1+\sqrt{2}}+\frac{\sqrt{2}+1}{\sqrt{2}}=\frac{\sqrt{2}-1}{\sqrt{2}+2}-\frac{\sqrt{2}}{\left(1+\sqrt{2}\right)\sqrt{2}}+\frac{\left(\sqrt{2}+1\right)^2}{\sqrt{2}\left(\sqrt{2}+1\right)}=\frac{\sqrt{2}-1}{2+\sqrt{2}}-\frac{\sqrt{2}}{2+\sqrt{2}}+\frac{3+2\sqrt{2}}{2+\sqrt{2}}=\frac{\sqrt{2}-1-\sqrt{2}+3+2\sqrt{2}}{2+\sqrt{2}}=\frac{2+2\sqrt{2}}{2+\sqrt{2}}\) \(b,\sqrt{x}-2+\frac{10-x}{\sqrt{x}+2}=\left(\sqrt{x}-2\right)+\frac{10-x}{\sqrt{x}+2}=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+10-x}{\sqrt{x}+2}=\frac{x-4+10-x}{\sqrt{x}+2}=\frac{6}{\sqrt{x}+2}\)
\(c,\frac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x^3}-\sqrt{y^3}}{\sqrt{x}-\sqrt{y}}=\frac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\sqrt{x}-\sqrt{y}}=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y\)
14 tháng 8 2016
Áp dụng BĐT Cô si ta có:
\(x+y\ge2\sqrt{xy}=2\cdot\frac{1}{\sqrt{z}};y+z\ge2\sqrt{yz}=2\cdot\frac{1}{\sqrt{x}};z+x\ge2\sqrt{xz}=2\cdot\frac{1}{\sqrt{y}}.\)( vì xyz=1)
=> P\(\ge\)\(\frac{2x\sqrt{x}}{y\sqrt{y}+2z\sqrt{z}}\)+ \(\frac{2y\sqrt{y}}{z\sqrt{z}+2x\sqrt{x}}+\frac{2z\sqrt{z}}{x\sqrt{x}+2y\sqrt{y}}\)
Đặt \(\hept{\begin{cases}a=y\sqrt{y}+2z\sqrt{z}\\b=z\sqrt{z}+2x\sqrt{x}\\c=x\sqrt{x}+2y\sqrt{y}\end{cases}\left(a;b;c\ge0\right)}\)<=> \(\hept{\begin{cases}4a+b=2c+9z\sqrt{z}\\4b+c=2a+9x\sqrt{x}\\4c+a=2b+9y\sqrt{y}\end{cases}}\)
<=> \(\hept{\begin{cases}z\sqrt{z}=\frac{4a+b-2c}{9}\\x\sqrt{x}=\frac{4b+c-2a}{9}\\y\sqrt{y}=\frac{4c+a-2b}{9}\end{cases}}\)
Do đó:
P \(\ge\)\(\frac{2}{9}\cdot\left(\frac{4a+b-2c}{c}+\frac{4b+c-2a}{a}+\frac{4c+a-2b}{b}\right)\)
<=> P \(\ge\)\(\frac{2}{9}\left(4\left(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\right)+\left(\frac{b}{c}+\frac{c}{a}+\frac{a}{b}\right)-6\right)\)
<=> P \(\ge\frac{2}{9}\cdot\left(4\cdot3\cdot\sqrt[3]{\frac{a}{c}\cdot\frac{b}{a}\cdot\frac{c}{b}}+3\cdot\sqrt[3]{\frac{b}{c}\cdot\frac{c}{a}\cdot\frac{a}{b}}-6\right)\)( Áp dụng BĐT Cô si cho 3 số ko âm)
<=> P \(\ge\frac{2}{9}\left(12+3-6\right)=2\)( đpcm)
Dấu = khi x=y=z=1.
N
1 tháng 12 2018
a)ĐK: \(x\ge0;x\ne1\)
\(A\Leftrightarrow\left(\dfrac{x+2+\sqrt{x}\left(\sqrt{x}+1\right)-1\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\left(\dfrac{\sqrt{x}-1}{2}\right)\)
\(\Leftrightarrow\left(\dfrac{x-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}-1}{2}\)
\(\Leftrightarrow\dfrac{x-1}{2\left(x+\sqrt{x}+1\right)}\)
3 tháng 9 2016
a) \(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-\left(x-2\sqrt{xy}+y\right)\)
\(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y=\sqrt{xy}\)
b) \(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\frac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
c) \(4x-\sqrt{8}+\frac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}=4x-\sqrt{8}+\frac{\sqrt{x^2\left(x+2\right)}}{x+2}=4x-\sqrt{8}+x=5x-\sqrt{8}\)
ĐK : \(x\inℕ^∗\left|x\ne1\right|min=2|x=2\)
\(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}=2\sqrt{x-1}\)
\(\sqrt{0\sqrt{1}}+\sqrt{4\sqrt{1}}=2\sqrt{1}\)
\(\sqrt{0.-1}+\sqrt{4.-1}=2.-1\)
\(0+-2=-2\)
\(\Rightarrow\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}=2\sqrt{x-1}\)