\(\left(x:2,2\right)\times\dfrac{1}{6}=\dfrac{-3}{8}\times\left(0,5-1\dfrac{3}{5}\right)\)

\(\Rightarrow\left(x:2,2\right)\times\dfrac{1}{6}=\dfrac{-3}{8}\times\left(\dfrac{1}{2}-\dfrac{8}{5}\right)\)

\(\Rightarrow\left(x:2,2\right)\times\dfrac{1}{6}=\dfrac{-3}{8}\times\dfrac{11}{10}\)

\(\Rightarrow\left(x:2,2\right)\times\dfrac{1}{6}=\dfrac{33}{80}\)

\(\Rightarrow x:2,2=\dfrac{33}{80}:\dfrac{1}{6}\)

\(\Rightarrow x:2,2=\dfrac{99}{40}\)

\(\Rightarrow x=\dfrac{99}{40}\times2,2\)

\(\Rightarrow x=\dfrac{1089}{200}\)

=>(x:2,2)*1/6=-3/8(1/2-8/5)=33/80

=>x:2,2=99/40

=>x=1089/200

\(\dfrac{3}{x-2}=\dfrac{-2}{x-4}\left(dk:x\ne2;x\ne4\right)\)

\(\Rightarrow3\cdot\left(x-4\right)=-2\cdot\left(x-2\right)\)

\(\Rightarrow3x-12=-2x+4\)

\(\Rightarrow3x+2x=4+12\)

\(\Rightarrow5x=16\)

\(\Rightarrow x=\dfrac{16}{5}\left(tm\right)\)

\(ĐK:x\ne2;x\ne4\\ Có:\dfrac{3}{x-2}=\dfrac{-2}{x-4}\\ \Leftrightarrow3\left(x-4\right)=-2\left(x-2\right)\\ \Leftrightarrow3x-12=-2x+4\\ \Leftrightarrow3x+2x=4+12\\ \Leftrightarrow5x=16\\ \Leftrightarrow x=\dfrac{16}{5}\left(TM\right)\\ Vậy:x=\dfrac{16}{5}\)

ĐKXĐ: \(-2\le x\le3\)

\(\dfrac{\sqrt{-x^2+x+6}}{2x+5}-\dfrac{\sqrt{-x^2+x+6}}{x-4}\ge0\)

\(\Leftrightarrow\sqrt{-x^2+x+6}\left(\dfrac{1}{2x+5}-\dfrac{1}{x-4}\right)\ge0\)

\(\Leftrightarrow\dfrac{\left(-x-9\right)\sqrt{x^2+x+6}}{\left(2x+5\right)\left(x-4\right)}\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x^2+x+6=0\\\dfrac{-x-9}{\left(2x+5\right)\left(x-4\right)}\ge0\end{matrix}\right.\) \(\Leftrightarrow-2\le x\le3\)

Hoặc có thể biện luận như sau:

Ta có: \(\left\{{}\begin{matrix}2x+5>0;\forall x\in\left[-2;3\right]\\x-4< 0;\forall x\in\left[-2;3\right]\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{\sqrt{-x^2+x+6}}{2x+5}\ge0\\\dfrac{\sqrt{-x^2+x+6}}{x-4}\le0\end{matrix}\right.\) ; \(\forall x\in\left[-2;3\right]\)

Do đó nghiệm của BPT là \(-2\le x\le3\)

Bài 2 

b, `\sqrt{3x^2}=x+2`          ĐKXĐ : `x>=0`

`=>(\sqrt{3x^2})^2=(x+2)^2`

`=>3x^2=x^2+4x+4`

`=>3x^2-x^2-4x-4=0`

`=>2x^2-4x-4=0`

`=>x^2-2x-2=0`

`=>(x^2-2x+1)-3=0`

`=>(x-1)^2=3`

`=>(x-1)^2=(\pm \sqrt{3})^2`

`=>` $\left[\begin{matrix} x-1=\sqrt{3}\\ x-1=-\sqrt{3}\end{matrix}\right.$

`=>` $\left[\begin{matrix} x=1+\sqrt{3}\\ x=1-\sqrt{3}\end{matrix}\right.$

Vậy `S={1+\sqrt{3};1-\sqrt{3}}`

Bài 1 

a, `3x-7\sqrt{x}+4=0`            ĐKXĐ : `x>=0`

`<=>3x-3\sqrt{x}-4\sqrt{x}+4=0`

`<=>3\sqrt{x}(\sqrt{x}-1)-4(\sqrt{x}-1)=0`

`<=>(3\sqrt{x}-4)(\sqrt{x}-1)=0`

TH1 :

`3\sqrt{x}-4=0`

`<=>\sqrt{x}=4/3`

`<=>x=16/9` ( tm )

TH2

`\sqrt{x}-1=0`

`<=>\sqrt{x}=1` (tm)

Vậy `S={16/9;1}`

b, `1/2\sqrt{x-1}-9/2\sqrt{x-1}+3\sqrt{x-1}=-17`     ĐKXĐ : `x>=1`

`<=>(1/2-9/2+3)\sqrt{x-1}=-17`

`<=>-\sqrt{x-1}=-17`

`<=>\sqrt{x-1}=17`

`<=>x-1=289`

`<=>x=290` ( tm )

Vậy `S={290}`

Bài 1: 

a) Ta có: \(3x-7\sqrt{x}+4=0\)

\(\Leftrightarrow3x-3\sqrt{x}-4\sqrt{x}+4=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(3\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{9}\end{matrix}\right.\)

b) Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}\cdot\left(-1\right)=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=289\)

hay x=290

=>6/x=32/35

=>x=6:32/35=6*35/32=210/32=105/16

\(=\dfrac{\sqrt{a}+2+\sqrt{a}-2}{a-4}:\dfrac{\sqrt{a}+2-2}{\sqrt{a}+2}\)

\(=\dfrac{2\sqrt{a}}{a-4}\cdot\dfrac{\sqrt{a}+2}{\sqrt{a}}=\dfrac{2}{\sqrt{a}-2}\)

A nên đăng vào mấy h mà nhìu ng onl í 

chứ h này ko còn ai đou ạ

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

\(a,\dfrac{1}{2}x=3+2\)

\(\dfrac{1}{2}x=5\)

\(x=5\div\dfrac{1}{2}\)

\(x=10\)

\(b,\dfrac{1}{4}x^2-\sqrt{36}=10\)

\(\dfrac{1}{4}x^2-6=10\)

\(\dfrac{1}{4}x^2=10+6\)

\(\dfrac{1}{4}x^2=16\)

\(x^2=16\div\dfrac{1}{4}\)

\(x^2=64\)

\(x^2=\left(8\right)^2\)

\(\Rightarrow x=8\)