a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)

\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)

b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)

\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)

c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)

\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)

\(=\dfrac{3}{\sqrt{x}-2}\)

a. \(\frac{x+1}{2}=\frac{8}{x+1}\)

\(\Leftrightarrow\left(x+1\right).\left(x+1\right)=8.2\)

\(\Leftrightarrow\left(x+1\right)^2=16\)

\(\Leftrightarrow\left(x+1\right)^2=2^4\)

\(\Leftrightarrow\left(x+1\right)=2^2\)

\(\Leftrightarrow\left(x+1\right)=4\)

\(\Leftrightarrow x=4-1=3\)

b. \(x:\left(9\frac{1}{2}-\frac{3}{2}\right)=\frac{0,4+\frac{2}{9}-\frac{2}{11}}{1,6+\frac{8}{9}-\frac{8}{11}}\)

\(\Leftrightarrow x:\left(\frac{10}{2}-\frac{3}{2}\right)=\frac{0,4+0,2-0,18}{1,6+0,8-0,72}\)

\(\Leftrightarrow x:\frac{7}{2}=\frac{\frac{21}{50}}{\frac{42}{25}}\)

\(\Leftrightarrow x=\frac{\frac{21}{50}}{\frac{42}{25}}.\frac{7}{2}\Leftrightarrow x=\frac{1}{4}.\frac{7}{2}=\frac{7}{8}\)

a )  \(\frac{x+1}{2}=\frac{8}{x+1}\)

\(\Rightarrow\left(x+1\right).\left(x+1\right)=2.8\)

\(\Rightarrow\left(x+1\right)^2=16\)

\(\Rightarrow\orbr{\begin{cases}x+1=4\\x+1=-4\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=4-1\\x=-4-1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)

Dấu " \(\orbr{\begin{cases}\\\end{cases}}\)là hoặc nha !!! 

a)\(\frac{x+1}{2}=\frac{8}{x+1}\)

\(\Rightarrow\left(x+1\right)^2=8\times2=16\)

\(\Rightarrow\left(x+1\right)^2=4^2\)

\(\Rightarrow x+1=4\)

\(\Rightarrow x=4-1=3\)

b)tớ ko biết làm . ~ sorry~

~chuk bn hok giỏi~

a) \(\left(x+1\right)^2=64\Leftrightarrow\left|x+1\right|=8\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+1=8\\x+1=-8\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=7\\x=-9\end{array}\right.\)

b) \(\frac{9}{2.4}+\frac{9}{4.6}+...+\frac{9}{96.98}+\frac{9}{98.100}=\frac{9}{4}\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{48.49}+\frac{1}{49.50}\right)\)

\(=\frac{9}{4}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{48}-\frac{1}{49}+\frac{1}{49}-\frac{1}{50}\right)\)

\(=\frac{9}{4}\left(1-\frac{1}{50}\right)=\frac{441}{200}\)

a)(x+1)²=64

x + 1      = 8

x            = 8 - 1

x            = 7

a)(x+1)²=64

=>(x+1)²=8²

=>x+1=8

=>x=7

b)\(\frac{9}{2.4}+\frac{9}{4.6}+...+\frac{9}{98.100}\)

\(=\frac{9}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\right)\)

\(=\frac{9}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)

\(=\frac{9}{2}\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(=\frac{9}{2}\cdot\frac{49}{100}\)

\(=\frac{441}{200}\)

\(\frac{x+1}{9}+\frac{x+2}{8}+\frac{x+3}{7}+...+\frac{x+9}{1}=-9\)

\(\left(\frac{x+1}{9}+1\right)+\left(\frac{x+2}{8}+1\right)+\left(\frac{x+3}{7}+1\right)+...+\left(\frac{x+9}{1}+1\right)=0\)

\(\frac{x+10}{9}+\frac{x+10}{8}+\frac{x+10}{7}+...+\frac{x+10}{1}=0\)

\(\left(x+10\right).\left(\frac{1}{9}+\frac{1}{8}+\frac{1}{7}+...+1\right)=0\)

vì \(\frac{1}{9}+\frac{1}{8}+\frac{1}{7}+...+1\ne0\)

\(\Rightarrow x+10=0\)

\(\Rightarrow x=-10\)

SKT_NTT làm đúng ùi -_- 100 %

\(\left(\frac{x+1}{9}+1\right)+\left(\frac{x+2}{8}+1\right)+...+\left(\frac{x+9}{1}+1\right)=-9+9\)

\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}+...+\frac{x+10}{1}\) =

a) \(A=\left(\frac{x+3}{x-9}+\frac{1}{\sqrt{x}+3}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)

\(=\left[\frac{x+3+\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-3}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}}\)

\(=\frac{\sqrt{x}+1}{\sqrt{x}+3}\)

c) để A>1/3 

\(\Rightarrow\frac{\sqrt{x}+3-2}{\sqrt{x}+3}>\frac{1}{3}\)

\(\Rightarrow\frac{2}{\sqrt{x}+3}>\frac{2}{3}\)

\(\Rightarrow\sqrt{x}+3>3\)

\(\Rightarrow x>0\)

\(x=0,2\)

em ko biết em mới học lớp 1

Thế mà cùng nói

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