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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=\(\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right)\cdot\left(\dfrac{\sqrt{x}-2}{2}\right)\Leftrightarrow\left(\dfrac{\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{1\cdot\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{\sqrt{x}-2}{2}\Leftrightarrow\dfrac{\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}-2}{2}\Rightarrow\dfrac{2\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}-2}{2}\Rightarrow\dfrac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}-2}{2}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
rút gọn..help
1 tháng 8 2016
\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\) (tách \(4=\sqrt{4}+\sqrt{4}\) )
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)
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 = \(\sqrt{48}\)
s
1 tháng 5 2017
ĐKXĐ: \(x\ge0,x\ne9\)
\(M=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)
= \(\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)+2\sqrt{x}\left(\sqrt{x}+3\right)-3x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
= \(\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
= \(\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3}{\sqrt{x}+3}\)
1 tháng 5 2017
ĐKXĐ: \(x\ge0,x\ne1\)
\(M=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
= \(\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
= \(\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
= \(\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
= \(\dfrac{\left(\sqrt{x}-1\right)\left(2-5\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\)
r
1 tháng 5 2017
\(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)
ĐKXĐ: x\(\ge0;x\ne9\)
=\(\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)+2\sqrt{x}\left(\sqrt{x}+3\right)-3x-9}{x-9}=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{x-9}=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3}{\sqrt{x}+3}\)
Vậy...
G
please help me
0
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 : \(a>0;b>0;a\ne b\)
\(K=\left(\dfrac{\sqrt{b}}{a-\sqrt{ab}}-\dfrac{\sqrt{a}}{\sqrt{ab}-b}\right)\left(a\sqrt{b}-b\sqrt{a}\right)\)
\(K=\left(\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}-\dfrac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}\right)\left(\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\right)\)
\(K=\dfrac{b-a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\)
\(K=b-a\)
thank you very much