Rút gọn:
A= \(\dfrac{10\sqrt{x}}{x+3\sqrt{x}-4}\) - \(\dfrac{2\sqrt{x}-3}{\sqrt{x}+4}\) + \(\dfrac{\sqrt{x}+1}{1-\sqrt{x}}\) ( với x ≥ 0; x ≠ 1)
1. Tính : \(\dfrac{12}{4-\sqrt{10}}\)-6\(\sqrt{\dfrac{5}{2}}\)+\(\dfrac{5\sqrt{2}+\sqrt{10}}{\sqrt{5}+1}\)
2,Rút gọn:A=(\(\dfrac{\sqrt{x}}{\sqrt{x}-5}\)-\(\dfrac{5}{\sqrt{x}+5}\)+\(\dfrac{10\sqrt{x}}{25-x}\)):\(\dfrac{3}{\sqrt{x}+5}\)
1: \(=8+2\sqrt{10}-3\sqrt{10}+\sqrt{10}=8\)
1,Tính \(\dfrac{12}{4-\sqrt{10}}-6\sqrt{\dfrac{5}{2}}+\dfrac{5\sqrt{2}+\sqrt{10}}{\sqrt{5}+1}\)
2,Rút gọn:A=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{5}{\sqrt{x}+5}+\dfrac{10\sqrt{x}}{25-x}\right):\dfrac{3}{\sqrt{x}+5}\)
1: \(=8+2\sqrt{10}-3\sqrt{10}+\sqrt{10}=8\)
Rút gọn:
A=\(\left(\dfrac{x-\sqrt{x}+2}{x-\sqrt{x}-2}-\dfrac{x}{x-2\sqrt{x}}\right)\div\dfrac{1-\sqrt{x}}{2-\sqrt{x}}vớix>0,x\ne1\)
B=\(\left(\dfrac{x}{3+\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right)\div\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
Lm nhanh giúp mk nhé!
a) ĐKXĐ có thêm \(x\ne4\)
\(A=\left(\dfrac{x-\sqrt{x}+2}{x-\sqrt{x}-2}-\dfrac{x}{x-2\sqrt{x}}\right):\dfrac{1-\sqrt{x}}{2-\sqrt{x}}\)
\(=\left(\dfrac{x-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{x}{\sqrt{x}\left(\sqrt{x}-2\right)}\right).\dfrac{2-\sqrt{x}}{1-\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(x-\sqrt{x}+2\right)-x\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\dfrac{2-\sqrt{x}}{1-\sqrt{x}}\)
\(=\dfrac{-2x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\dfrac{-2\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}-2}{\sqrt{x}-1}=\dfrac{-2}{\sqrt{x}+1}\)
\(B=\left(\dfrac{x}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
\(=\dfrac{x+1}{\sqrt{x}+3}:\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{x+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x+1}{\sqrt{x}+3}:\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{x+1}{\sqrt{x}+3}.\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+1}{\sqrt{x}+1}\)
Rút gọn biểu thức: A = \(\dfrac{10\sqrt{x}}{x+3\sqrt{x}-4}\) - \(\dfrac{2\sqrt{x}-3}{\sqrt{x}+4}\) + \(\dfrac{\sqrt{x}+1}{1-\sqrt{x}}\) ( Với x \(\ge\) 0, x \(\ne\) 1)
Lời giải:
$A=\frac{10\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+4)}-\frac{(2\sqrt{x}-3)(\sqrt{x}-1)}{(\sqrt{x}+4)(\sqrt{x}-1)}-\frac{(\sqrt{x}+1)(\sqrt{x}+4)}{(\sqrt{x}-1)(\sqrt{x}+4)}$
$=\frac{10\sqrt{x}-(2\sqrt{x}-3)(\sqrt{x}-1)-(\sqrt{x}+1)(\sqrt{x}+4)}{(\sqrt{x}+4)(\sqrt{x}-1)}$
$=\frac{-3x+10\sqrt{x}-7}{(\sqrt{x}+4)(\sqrt{x}-1)}$
$=\frac{-(\sqrt{x}-1)(3\sqrt{x}-7)}{(\sqrt{x}+4)(\sqrt{x}-1)}=\frac{7-3\sqrt{x}}{\sqrt{x}+4}$
Rút gọn:
A = \(\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(1-\dfrac{3-\sqrt{x}}{\sqrt{x}+1}\right)\)
B1: rút gọn:
a, \(\sqrt{4-2\sqrt{3}}-\sqrt{3}\)
b, \(\sqrt{11+6\sqrt[]{2}}-3+\sqrt{2}\)
c, \(x-4+\sqrt{16-8x+x^2}\) với x > 4
d, \(\dfrac{x^2-5}{x+\sqrt{5}}\) x khác \(-\sqrt{5}\)
e, \(\dfrac{x^2+2\sqrt{2}x+2}{x+\sqrt{2}}\) x khác \(-\sqrt{2}\)
g, \(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\)
giúp em với ạ , em cảm ơn
a) \(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\sqrt{3}-1-\sqrt{3}=-1\)
b) \(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}=\sqrt{\left(3+\sqrt{2}\right)^2}-3+\sqrt{2}=3+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}\)
c) \(x-4+\sqrt{16-8x+x^2}=x-4+\sqrt{\left(x-4\right)^2}=x-4+\left|x-4\right|\)
\(=x-4+x-4\left(x>4\right)=2x-8\)
d) \(\dfrac{x^2-5}{x+\sqrt{5}}=\dfrac{\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)}{x+\sqrt{5}}=x-\sqrt{5}\)
e) \(\dfrac{x^2+2\sqrt{2}x+2}{x+\sqrt{2}}=\dfrac{\left(x+\sqrt{2}\right)^2}{x+\sqrt{2}}=x+\sqrt{2}\)
g) \(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\dfrac{1}{\sqrt{2}}\)
a) Ta có: \(\sqrt{4-2\sqrt{3}}-\sqrt{3}\)
\(=\sqrt{3}-1-\sqrt{3}\)
=-1
b) Ta có: \(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\)
\(=3+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}\)
c) Ta có: \(x-4+\sqrt{x^2-8x+16}\)
\(=x-4+x-4=2x-8\)
d) Ta có: \(\dfrac{x^2-5}{x+\sqrt{5}}\)
\(=\dfrac{\left(x+\sqrt{5}\right)\left(x-\sqrt{5}\right)}{x+\sqrt{5}}\)
\(=x-\sqrt{5}\)
( \(\dfrac{\sqrt{x}}{\sqrt{x}-2}\) + \(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\) - \(\dfrac{3\sqrt{x}+2}{x-4}\) ) : \(\dfrac{\sqrt{x}-2}{x-4}\) ( với x ≥ 0; x ≠ 4)
RÚT GỌN Ạ
Với \(x\ge0;x\ne4\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)+\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{x-4}{\sqrt{x}-2}\)
\(=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}-\sqrt{x}-2-3\sqrt{x}+2}{x-4}.\dfrac{x-4}{\sqrt{x}-2}\)
\(=\dfrac{2x-4\sqrt{x}}{x-4}.\dfrac{x-4}{\sqrt{x}-2}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-2}=2\sqrt{x}\)
(\(\dfrac{\sqrt{x}}{\sqrt{x}-2}\) + \(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)-\(\dfrac{3\sqrt{x}+2}{x-4}\) ) : \(\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\) ( với x ≥ 0; x ≠ 4)
RÚT GỌN Ạ
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)+\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-3\sqrt{x}-2}{x-4}\right):\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\\ =\dfrac{x+2\sqrt{x}+x-\sqrt{x}-2\sqrt{x}+2-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\times\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\\ =\dfrac{2x-4\sqrt{x}}{\sqrt{x}-2}\times\dfrac{1}{\sqrt{x}-2}\\ =\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-2}\times\dfrac{1}{\sqrt{x}-2}=\dfrac{2\sqrt{x}}{\sqrt{x}-2}\)
rút gọn:
A=\(x-4-\sqrt{16-8x^2+x^4}\left(x>4\right)\)
B=\(\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\dfrac{1}{\sqrt{a}+\sqrt{b}}\left(a,b>0,a\ne b\right)\)
\(A=x-4-\sqrt{x^4-8x^2+16}=x-4-\sqrt{[\left(x-2\right)\left(x+2\right)]^2}\)
\(A=x-4-\left(x-2\right)\left(x+2\right)=x-4-\left(x^2-4\right)=-x^2+x\)
\(B=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}.\left(\sqrt{a}+\sqrt{b}\right)=a-b\)