Rút gon biểu thức sau: \(\sqrt{18\left(\sqrt{2}-\sqrt{3}\right)^2}\)
Rút gọn biểu thức
\(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}\)
\(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}=\sqrt{3}-\sqrt{2}+3\sqrt{2}=\sqrt{3}+2\sqrt{2}\)
\(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}\)
\(=\sqrt{3}-\sqrt{2}+3\sqrt{2}\)
\(=\sqrt{3}+2\sqrt{2}\)
(3)
a) rút gon biểu thức: A= \(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\) vs \(x>0;x\ne1\)
giúp mk vs
\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(\Rightarrow A=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(\Rightarrow A=\dfrac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{x+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(\Rightarrow A=\dfrac{x+1}{\sqrt{x}}\)
Rút gọn các biểu thức sau :
a) \(\sqrt{18\left(\sqrt{2}-\sqrt{3}\right)^2}\) - \(\sqrt{54}\)
b) \(\dfrac{a+\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\) - a\(\sqrt{\dfrac{1}{a}}\)
c) ( \(\sqrt{28}\) - 2\(\sqrt{3}\) +7) \(\sqrt{7}\) +\(\sqrt{84}\)
a: \(=3\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)-3\sqrt{6}\)
=3căn 6-6-3căn 6=-6
b: \(=\dfrac{a+\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\sqrt{a}\)
\(=\dfrac{a+\sqrt{ab}-a+\sqrt{ab}}{\sqrt{a}-\sqrt{b}}=\dfrac{2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\)
rút gon biểu thức sau A= \(\frac{8}{\sqrt{5}-1}-\left(2\sqrt{5}-1\right)\)
cho biểu thức B=\(\frac{\left(1-\sqrt{x}\right)^2+4\sqrt{x}}{1+\sqrt{x}}\)với \(x\ge0\). rút gọn biểu thức A
Ai bt làm ơn giải giúp mk, mk đag rất cần
A = \(\frac{8}{\sqrt{5}-1}\) - (\(2\sqrt{5}-1\) ) ( chúng ta cần trục căn thức lên để khử mẫu )
= \(\frac{8\left(\sqrt{5}+1\right)}{5-1}\)- \(\left(2\sqrt{5}-1\right)\)
= \(2\sqrt{5}\)+ 2 - \(2\sqrt{5}\)+1
= 3
B = \(\frac{\left(1-\sqrt{x}\right)^2+4\sqrt{x}}{1+\sqrt{x}}\)( x \(\ge\)0 )
= \(\frac{1-2\sqrt{x}+x+4\sqrt{x}}{1+\sqrt{x}}\)
= \(\frac{1+2\sqrt{x}+x}{1+\sqrt{x}}\)
= \(\frac{\left(1+\sqrt{x}\right)^2}{1+\sqrt{x}}\)
= 1 +\(\sqrt{x}\)
#mã mã#
rút gọn biểu thức : \(\left(2\sqrt{2-\sqrt{5}+3\sqrt{2}}\right)\left(\sqrt{18}-\sqrt{20+2\sqrt{2}}\right)\)
cho biểu thức M = \(\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right).\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)\) . Khi x > 0 ; x≠1.
Rút gon biểu thức M
\(M=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}}\cdot\dfrac{x-1+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ M=\dfrac{\left(\sqrt{x}-1\right)\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ M=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}+1}\)
\(M=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\cdot\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}-\dfrac{\sqrt{x}-1}{x+\sqrt{x}}\right)\)
\(=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{x-1-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}\cdot\dfrac{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}\right)}{\sqrt{x}}\)
\(=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}}\)
Rút gon biểu thức:
\(D=\left(\frac{2+\sqrt{x}}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right).\left(\frac{x\sqrt{x}+x-\sqrt{x}-1}{\sqrt{x}}\right)\)
rút gon biểu thức \(Q=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\left(\frac{1-x}{\sqrt{2}}\right)^2\)
rút gọn và tính biểu thức sau
\(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{\left(3-\sqrt{15}\right)^2}\)
\(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{\left(3-\sqrt{15}\right)^2}\)
\(=\left|4-\sqrt{15}\right|+\left|3-\sqrt{15}\right|\)
\(=4-\sqrt{15}+\sqrt{15}-3=1\)