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NT
3
19 tháng 8 2021
\(P=\left(\frac{1}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\right):\frac{\sqrt{x}}{x+\sqrt{x}}\)ĐK : x > 0
\(=\left(\frac{\sqrt{x}+1+x}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\frac{1}{\sqrt{x}+1}=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)
\(P=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\)
\(=\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{x-1}=\frac{x-2\sqrt{x}+1}{x-1}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
NT
1
22 tháng 8 2021
Với \(x\ge0;x\ne\pm16\)
\(B=\left(\frac{\sqrt{x}}{\sqrt{x}+4}+\frac{4}{\sqrt{x}-4}\right):\frac{x+16}{\sqrt{x}+2}\)
\(=\left(\frac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}\right):\frac{x+16}{\sqrt{x}-2}=\frac{\sqrt{x}-2}{x-16}\)
7 tháng 7 2017
điều kiện : \(x>0;x\ne4\)
\(H=\left(\dfrac{1}{x-4}-\dfrac{1}{x+4\sqrt{x}+4}\right).\dfrac{x+2\sqrt{x}}{\sqrt{x}}\)
\(H=\left(\dfrac{1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{1}{\left(\sqrt{x}+2\right)^2}\right)\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}}\)
\(H=\dfrac{1}{\left(\sqrt{x}-2\right)}-\dfrac{1}{\left(\sqrt{x}+2\right)}\) \(=\dfrac{\sqrt{x}+2-\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(H=\dfrac{\sqrt{x}+2-\sqrt{x}+2}{x-4}=\dfrac{4}{x-4}\)
điều kiện : \(a>0;b>0;a\ne b\)
\(K=\left(\dfrac{\sqrt{b}}{a-\sqrt{ab}}-\dfrac{\sqrt{a}}{\sqrt{ab}-b}\right)\left(a\sqrt{b}-b\sqrt{a}\right)\)
\(K=\left(\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}-\dfrac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}\right)\left(\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\right)\)
\(K=\dfrac{b-a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\)
\(K=b-a\)
rút gọn biểu thức
HT
1 tháng 5 2017
=(√x (√x -3)+2√x (√x +3)-3x-9)/(x-9)
=(x-3√x+2x+6√x-3x-9x)/(x-9)
=3(√x -3)/(√x +3)(√x -3)
=3/√x +3
12 tháng 6 2017
\(\left(\dfrac{x\sqrt{x}+1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}+\dfrac{x+1}{\sqrt{x}}\right)\) điều kiện xát định :(x > 0 ; x \(\ne\) 1 )
= \(\left(\dfrac{x\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{x\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{x+1}{\sqrt{x}}\right)\)
= \(\dfrac{\left(x\sqrt{x}+1\right)\left(\sqrt{x}+1\right)-\left(x\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(x-1\right)}+\dfrac{x+1}{\sqrt{x}}\)
= \(\dfrac{x^2+x\sqrt{x}+\sqrt{x}+1-\left(x^2-x\sqrt{x}+\sqrt{x}-1\right)}{\sqrt{x}\left(x-1\right)}+\dfrac{x+1}{\sqrt{x}}\)
= \(\dfrac{x^2+x\sqrt{x}+\sqrt{x}+1-x^2+x\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(x-1\right)}+\dfrac{x+1}{\sqrt{x}}\)
= \(\dfrac{2x\sqrt{x}+2}{\sqrt{x}\left(x-1\right)}+\dfrac{x+1}{\sqrt{x}}\) = \(\dfrac{2x\sqrt{x}+2+\left(\left(x+1\right)\left(x-1\right)\right)}{\sqrt{x}\left(x-1\right)}\)
= \(\dfrac{2x\sqrt{x}+2+x^2-1}{\sqrt{x}\left(x-1\right)}\) = \(\dfrac{2x\sqrt{x}+x^2+1}{\sqrt{x}\left(x-1\right)}\)
19 tháng 4 2017
_tớ làm chi tiết rồi nhé 
.
-biểu thức: N=\(\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right)\cdot\left(\dfrac{\sqrt{x}-2}{2}\right)\Leftrightarrow\left(\dfrac{\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{1\cdot\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{\sqrt{x}-2}{2}\Leftrightarrow\dfrac{\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}-2}{2}\Rightarrow\dfrac{2\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}-2}{2}\Rightarrow\dfrac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}-2}{2}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
AV
15 tháng 6 2018
D = \(\sqrt{7+\sqrt{33}}+\sqrt{7-\sqrt{33}}\)
=> D2 = \(7+\sqrt{33}+2\left(\sqrt{7+\sqrt{33}}\right)\left(\sqrt{7-\sqrt{33}}\right)+7-\sqrt{33}\)
D2 = \(14+2\left(49-33\right)\) = 14+32 = 48
=> D = \(\sqrt{48}\)