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câu 2 mn ơi mai mk phải hk r
29 tháng 5 2022
Câu 2:
\(B=\left(\dfrac{\sqrt{x}\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+3}{\sqrt{x}-1}\right)\cdot\dfrac{x-1}{2x+2\sqrt{x}-\sqrt{x}-1}\)
\(=\left(\dfrac{\sqrt{x}+\sqrt{x}+3}{\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}\)
\(=\dfrac{2\sqrt{x}+3}{2\sqrt{x}-1}\)
Để B<0 thì \(2\sqrt{x}-1< 0\)
=>\(2\sqrt{x}< 1\)
=>0<x<1/4
lm hộ mk vs
16 tháng 10 2017
\(A=\dfrac{2a^2+4}{1-a^3}-\dfrac{1}{1+\sqrt{a}}-\dfrac{1}{1-\sqrt{a}}\\ =\dfrac{2a^2+4}{\left(1-a\right)\left(1+a+a^2\right)}-\dfrac{1}{1+\sqrt{a}}-\dfrac{1}{1-\sqrt{a}}\\ =\dfrac{2a^2+4-\left(1-\sqrt{a}\right)\left(1+a+a^2\right)-\left(1+\sqrt{a}\right)\left(1+a+a^2\right)}{\left(1-a\right)\left(1+a+a^2\right)}\\ =\dfrac{2a^2+4-\left(1+a+a^2\right)\left(1-\sqrt{a}+1+\sqrt{a}\right)}{\left(1-a\right)\left(1+a+a^2\right)}\\ =\dfrac{2a^2+4-2\left(1+a+a^2\right)}{\left(1-a\right)\left(1+a+a^2\right)}=\dfrac{2}{1+a+a^2}\\ \)
Ta có A max <=> \(1+a+a^2min\)
Mà 1+a+a^2=\(\left(a+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\\ \)
Dấu bằng xảy ra <=> a=-1/2
=> \(A=\dfrac{2}{1+a+a^2}\le\dfrac{2}{\dfrac{3}{4}}=\dfrac{8}{3}\)
Vậy max A=8/3 <=> a=-1/2
=)) mỏi tay quá đê
NN
13 tháng 7 2017
d) \(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)
\(=\sqrt{5-2.2\sqrt{5}+4}-\sqrt{5+2.2\sqrt{5}+4}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)
\(=\left|\sqrt{5}-2\right|-\left|\sqrt{5}+2\right|\)
\(=\sqrt{5}-2-\sqrt{5}-2=-4\)
g)\(\dfrac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}\)
\(=\dfrac{\sqrt{3}+\sqrt{9+2.3.\sqrt{2}+2}-\sqrt{3+2.\sqrt{3}.\sqrt{2}+2}}{\sqrt{2}+\sqrt{5+2.\sqrt{5}.1+1}-\sqrt{5+2.\sqrt{5}.\sqrt{2}+2}}\)
\(=\dfrac{\sqrt{3}+\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}}{\sqrt{2}+\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}\)
\(=\dfrac{\sqrt{3}+3+\sqrt{2}-\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{2}+\left(\sqrt{5}+1\right)-\left(\sqrt{5}+\sqrt{2}\right)}\)
\(=\dfrac{3}{1}=3\)
13 tháng 7 2017
\(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)\(=\sqrt{9-2\cdot2\cdot\sqrt{5}}-\sqrt{9+2\cdot2\cdot\sqrt{5}}\)\(=\sqrt{2^2-2\cdot2\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}-\sqrt{2^2+2\cdot2\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}\)\(=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(2+\sqrt{5}\right)^2}\)\(=\left|2-\sqrt{5}\right|-\left|2+\sqrt{5}\right|\)\(=\left(2-\sqrt{5}\right)-\left(2+\sqrt{5}\right)\)\(=2-\sqrt{5}-2-\sqrt{5}=-2\sqrt{5}\)
\(\dfrac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}=\dfrac{\sqrt{3}+\sqrt{11+2\cdot3\cdot\sqrt{2}}-\sqrt{5+2\cdot\sqrt{2}\cdot\sqrt{3}}}{\sqrt{2}+\sqrt{6+2\cdot\sqrt{5}}-\sqrt{7+2\cdot\sqrt{2}\cdot\sqrt{5}}}=\dfrac{\sqrt{3}+\sqrt{3^2+2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}\right)^2+2\cdot\sqrt{2}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}}{\sqrt{2}+\sqrt{\left(\sqrt{5}\right)^2+2\cdot\sqrt{5}+1}-\sqrt{\left(\sqrt{2}\right)^2+2\cdot\sqrt{2}\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}}=\dfrac{\sqrt{3}+\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}}{\sqrt{2}+\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{2}+\sqrt{5}\right)^2}}=\dfrac{\sqrt{3}+\left|3+\sqrt{2}\right|-\left|\sqrt{2}+\sqrt{3}\right|}{\sqrt{2}+\left|\sqrt{5}+1\right|-\left|\sqrt{2}+\sqrt{5}\right|}=\dfrac{\sqrt{3}+3+\sqrt{2}-\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{5}+1-\sqrt{2}-\sqrt{5}}=3\)
Bạn kiểm tra lại điều kiện của đề bài nhé vì x ở dưới mẫu ko thể = 0 được
a) M = \(\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}\right)\): \(\left(\frac{2}{x}-\frac{2-x}{x\sqrt{x}+x}\right)\) ( ĐKXĐ : x > 0 , x\(\ne\)1)
M = \(\frac{\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\):\(\frac{2\left(\sqrt{x}+1\right)-2+x}{x\left(\sqrt{x}+1\right)}\)
M = \(\frac{x+2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\):\(\frac{x+2\sqrt{x}}{x\left(\sqrt{x}+1\right)}\)
M = \(\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x-1}\right)\left(\sqrt{x}+1\right)}.\frac{x\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
M = \(\frac{x}{\sqrt{x}-1}\)
b) M = \(\frac{x}{\sqrt{x}-1}\)( ĐKXĐ : x > 0, x\(\ne\)1)
Ta có : \(\frac{x}{\sqrt{x}-1}\)= \(\frac{-1}{2}\)
\(\Rightarrow\)2x = -1 ( \(\sqrt{x}-1\))
\(\Leftrightarrow\) 2x + \(\sqrt{x}\)-1 = 0
\(\Leftrightarrow\) \(\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)= 0
\(\Leftrightarrow\)\(\orbr{\begin{cases}\sqrt{x}=\frac{1}{2}\\\sqrt{x}=-1\end{cases}\Rightarrow}x=\frac{1}{4}\left(TM\text{Đ}K\right)\)
Vậy x =\(\frac{1}{4}\) để M = \(\frac{-1}{2}\)