\(\frac{x-2}{x+2}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{x^2-4}\left(x\ne\pm2\right)\)

\(\Leftrightarrow\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{2x-22}{\left(x-2\right)\left(x+2\right)}\)

=> x² - 4x + 4 - 3x - 6 = 2x - 22

<=> x² - 9x + 20 = 0

<=> x² - 4x - 5x + 20 = 0

<=> x( x - 4 ) - 5( x - 4 ) = 0

<=> ( x - 4 )( x - 5 ) = 0

<=> x - 4 = 0 hoặc x - 5 = 0

<=> x = 4 (tm) hoặc x = 5 (tm)

Vậy ...

\(\frac{x-2}{x+2}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{x^2-4}\)ĐKXĐ : \(x\ne\pm2\)

\(\Leftrightarrow\frac{\left(x-2\right)^2-3\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2x-22}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow x^2-4x+4-3x-6=2x-22\)

\(\Leftrightarrow x^2-7x-2=2x-22\Leftrightarrow x^2-9x+20=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\Leftrightarrow x=4;x=5\)( tmđk )

Vậy tập nghiệm phương trình là S = { 4 ; 5 } 

ĐKXĐ : \(\hept{\begin{cases}x^2+x-6\ne0\\x^2+4x+3\ne0\\2x-1\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}\left(x+3\right)\left(x-2\right)\ne0\\\left(x+1\right)\left(x+3\right)\ne0\\x\ne\frac{1}{2}\end{cases}\Rightarrow\hept{\begin{cases}x\ne2;-3\\x\ne-1;-3\\x\ne\frac{1}{2}\end{cases}}}}\)

TXĐ : \(x\ne\left\{-3;-1;\frac{1}{2};2\right\}\)

\(pt\Leftrightarrow\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{2}{\left(x+1\right)\left(x+3\right)}=\frac{-3}{2x-1}\)

\(\Leftrightarrow\frac{5\left(x+1\right)-2\left(x-2\right)}{\left(x-2\right)\left(x+1\right)\left(x+3\right)}=\frac{-3}{2x-1}\)

\(\Leftrightarrow\frac{3x+9}{\left(x-2\right)\left(x+1\right)\left(x+3\right)}=\frac{-3}{2x-1}\)

\(\Leftrightarrow\frac{3}{\left(x-2\right)\left(x+1\right)}=\frac{-3}{2x-1}\)

\(\Leftrightarrow\frac{1}{x^2-x-2}=\frac{1}{1-2x}\)

\(\Leftrightarrow x^2-x-2-1+2x=0\)

\(\Leftrightarrow x^2+x-3=0\)

\(\Leftrightarrow\left(x^2+2.\frac{1}{2}.x+\frac{1}{4}\right)-\frac{13}{4}=0\)

\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2-\left(\frac{\sqrt{13}}{2}\right)^2=0\)

\(\Leftrightarrow\left(x+\frac{1-\sqrt{13}}{2}\right)\left(x+\frac{1+\sqrt{13}}{2}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{13}-1}{2}\\x=\frac{-\sqrt{13}-1}{2}\end{cases}}\)

\(\frac{5}{x^2+x-6}-\frac{2}{x^2+4+3}=-\frac{3}{2x-1}\)

<=> \(\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{2}{\left(x+1\right)\left(x+3\right)}=-\frac{3}{2x-1}\)

<=> \(\frac{5\left(x+1\right)-2\left(x-2\right)}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)

<=> \(\frac{5x+5-2x+4}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)

<=> \(\frac{3x+9}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)

<=> \(\frac{3\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)

<=> \(\frac{1}{x-2}=-\frac{1}{2x-1}\)

<=> x-2=1-2x <=> 3x=3

=> x=1

Đáp số: x=1

a: \(\Leftrightarrow\dfrac{y+5}{y\left(y-5\right)}-\dfrac{y-5}{2y\left(y+5\right)}=\dfrac{y+25}{2\left(y-5\right)\left(y+5\right)}\)

\(\Leftrightarrow2\left(y+5\right)^2-\left(y-5\right)^2=y^2+25y\)

=>\(2y^2+20y+50-y^2+10y-25=y^2+25y\)

=>30y+25=25y

=>5y=-25

=>y=-5(loại)

b: \(\Leftrightarrow x\left(x+1\right)+x\left(x-3\right)=4x\)

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

=>2x^2-6x=0

=>2x(x-3)=0

=>x=0(nhận) hoặc x=3(loại)

c: =>x^2-9-6(2x+7)=-13(x+3)

=>x^2-9-12x-42+13x+39=0

=>x^2+x-6=0

=>(x+3)(x-2)=0

=>x=2(nhận) hoặc x=-3(loại)

\(a.\frac{x-5}{4}-2x+1=\frac{x}{3}-\frac{2-x}{6}\\\Leftrightarrow \frac{3\left(x-5\right)}{12}-\frac{24}{12}x+\frac{12}{12}=\frac{4x}{12}-\frac{2\left(2-x\right)}{12}\\\Leftrightarrow 3\left(x-5\right)-24x+12=4x-2\left(2-x\right)\\\Leftrightarrow 3x-15-24x+12=4x-4+2x\\ \Leftrightarrow3x-15-24x+12-4x+4-2x=0\\ \Leftrightarrow-27x+1=0\\ \Leftrightarrow-27x=-1\\ \Leftrightarrow x=\frac{1}{27}\)

\(b.\left(2x-1\right)^2=\left(x-2\right)\left(2x-1\right)\\ \Leftrightarrow\left(2x-1\right)^2-\left(x-2\right)\left(2x-1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left[\left(2x-1\right)-\left(x-2\right)\right]=0\\ \Leftrightarrow\left(2x-1\right)\left(2x-1-x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-1\end{matrix}\right.\)

\(c.\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{-3}{25-x^2}\\\Leftrightarrow \frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{3}{x^2-25}\\\Leftrightarrow \frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{3}{\left(x-5\right)\left(x+5\right)}\\ \Leftrightarrow\frac{\left(x+5\right)\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\frac{3}{\left(x-5\right)\left(x+5\right)}\\ \Leftrightarrow\left(x+5\right)\left(x+5\right)-\left(x-5\right)\left(x-5\right)=3\\\Leftrightarrow x^2+5x+5x+25-\left(x^2-5x-5x+25\right)=3\\\Leftrightarrow x^2+5x+5x+25-x^2+5x+5x-25=3\\ \Leftrightarrow20x=3\\ \Leftrightarrow x=\frac{3}{20}\)

\(d.x^2-x-12=0\\\Leftrightarrow x^2-4x+3x-12=0\\\Leftrightarrow \left(x^2-4x\right)+\left(3x-12\right)=0\\ \Leftrightarrow x\left(x-4\right)+3\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

( 2x - 1 ) - x = 0

=> 2x - 1 = x

=> 2x - x = 1

=> x = 1 

( x - 1 )( 2x - 3) = 0

=> \(\orbr{\begin{cases}x-1=0\\2x-3=0\end{cases}}\)=> \(\orbr{\begin{cases}x=1\\x=\frac{3}{2}\end{cases}}\)

Vậy tập nghiệm của phương trình là S = { 1 ; 3/2 }

\(\frac{x}{x+1}=\frac{x+2}{x-1}\)( đkxđ : \(x\ne\pm1\))

( Chỗ này chưa học kĩ nên chưa hiểu lắm :] 

\(\left(2x-1\right)-x=0\)

\(2x-x=1\)

\(x=1\)

#hoktot

a: ĐKXĐ: \(2x-4>=0\)

=>x>=2

b: ĐKXĐ: \(\dfrac{1}{2-x}>=0\)

=>\(2-x>0\)

=>x<2

c: ĐKXĐ: \(-\dfrac{3}{2-6x}>=0\)

=>\(\dfrac{3}{6x-2}>=0\)

=>\(6x-2>0\)

=>x>1/3

d: ĐKXĐ: \(3x^2+2014>=0\)

=>\(x\in R\)

`a,5x-2=3x+1`

`<=>5x-3x=1+2`

`<=>2x=3`

`<=>x=3/2`

Vậy `x=3/2`

`b,(x+5)(2x-3)=0`

`<=>` $\left[ \begin{array}{l}x+5=0\\2x-3=0\end{array} \right.$

`<=>` $\left[ \begin{array}{l}x=\dfrac{3}{2}\\x=-5\end{array} \right.$

Vậy `S={-5,3/2}`

a: =>3(3x-7)+2(x+1)=-96

=>9x-21+2x+2=-96

=>11x-19=-96

=>11x=-96+19=-75

=>x=-75/11

b: \(x-\dfrac{x+1}{3}=\dfrac{2x+1}{5}\)

=>15x-5(x+1)=3(2x+1)

=>15x-5x-5=6x+3

=>10x-5=6x+3

=>4x=8

=>x=2

a)

\(\dfrac{3x-7}{2}+\dfrac{x+1}{3}=-16\)

\(< =>9x-21+2x+2=-96\)

\(< =>9x+2x=-96+21-2\\ < =>11x=-77\\ < =>x=-7\)

b)

\(\dfrac{x-x+1}{3}=\dfrac{2x+1}{5}\\ < =>5=6x+3\\ < =>6x=5-3\\ < =>6x=2\\ < =>x=\dfrac{1}{3}\)

\(2x^2+4x+5=2\left(x^2+2x+\frac{5}{2}\right)=2\left[\left(x^2+2.x.1+1\right)+\frac{3}{2}\right]=2\left(x+1\right)^2+3\ge3\)

Min=3 khi x=-1

Còn phần cô giáo thì zầy nè

\(\frac{1}{2x^2+4x+5}=\frac{1}{2\left(x^2+2x+\frac{5}{2}\right)}=\frac{1}{2\left[\left(x^2+2.x.1+1\right)+\frac{3}{2}\right]}=\frac{1}{2\left(x+1\right)^2+3}\)

muốn \(\frac{1}{2x^2+4x+5}\) lớn nhất thì \(2x^2+4x+5\)nhỏ nhất

\(2x^2+4x+5=2\left(x^2+2x+\frac{5}{2}\right)=2\left[\left(x^2+2.x.1+1\right)+\frac{3}{2}\right]=2\left(x+1\right)^2+3\ge3\)

Min=3 khi x=-1