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) \(\frac{3}{2}-\frac{5}{6}:x=\frac{5}{15}-\frac{3}{15}\)

\(\Leftrightarrow\frac{3}{2}-\frac{5}{6}:x=\frac{2}{15}\)

\(\Leftrightarrow\frac{5}{6}:x=\frac{3}{2}-\frac{2}{15}\)

\(\Leftrightarrow\frac{5}{6}:x=\frac{41}{30}\)

\(\Leftrightarrow x=\frac{5}{6}:\frac{41}{30}\)

\(\Leftrightarrow x=\frac{25}{41}\)

b) \(x-\frac{6}{7}.\frac{14}{8}=\frac{1}{2}-\frac{2}{5}\)

\(\Leftrightarrow x-\frac{3}{2}=\frac{1}{10}\)

\(\Leftrightarrow x=\frac{1}{10}+\frac{3}{2}\)

\(\Leftrightarrow x=\frac{8}{5}\)

c) \(x:\frac{6}{5}+\frac{2}{3}=\frac{7}{3}\)

\(\Leftrightarrow x:\frac{6}{5}=\frac{7}{3}-\frac{2}{3}\)

\(\Leftrightarrow x:\frac{6}{5}=\frac{5}{3}\)

\(\Leftrightarrow x=\frac{5}{3}.\frac{6}{5}\)

\(\Leftrightarrow x=2\)

huyền ngốc

\(\trac{3}{2}

Ta có: \(\left(4,5-2x\right):\frac{3}{4}=1\frac{1}{3}\)

=> \(\left(4,5-2x\right):\frac{3}{4}=\frac{4}{3}\)

=> \(4,5-2x=1\)

=> \(2x=3,5\Rightarrow x=1,75\)

Vậy x=1,75

cảm ơn bạn

(4,5-2x):3/4=4/3

(4,5-2x)=4/3.3/4

(4,5-2x)=1

2x=4,5-1

2x=3,5

x=3,5:2

x=1,75

vậy x=1,75

|2x-5|=4

\(\Leftrightarrow\orbr{\begin{cases}2x-5=4\\2x-5=-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=9\\2x=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{2}\\x=\frac{1}{2}\end{cases}}\)

Ko hiểu đề!

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

<=>\(\left(\frac{7}{2}+2x\right).\frac{8}{3}=\frac{16}{3}\)

<=>\(\frac{28}{3}+\frac{16x}{3}=\frac{16}{3}\)

<=>\(\frac{16x}{3}=\frac{-2}{3}\)

<=>\(16x=-2\)

<=>\(x=\frac{-1}{8}\)

vậy \(x=\frac{-1}{8}\)

b,\(\left|2x+3\right|=5\)

xét x<0,ta co: \(\left|2x+3\right|=5\)<=> \(-2x+3=5\)<=>\(-2x=2\)<=>\(x=-1\)(loại)

xét x>0,ta co:\(\left|2x+3\right|=5\)<=>\(2x+3=5\)<=>\(2x=2\)<=>\(x=1\)

c,\(\frac{x-2}{4}=\frac{5+x}{3}\)

<=>\(\frac{3x-6}{12}=\frac{20+4x}{12}\)

=>\(3x-6=20+4x\)

<=>\(3x-6-20-4x=0\)

<=>\(-x-26=0\)

<=>\(-x=26\)

<=>\(x=-26\)

kl:.......

a:=>x^2-1-x=2x-1

=>x^2-x-1=2x-1

=>x^2-3x=0

=>x=0(loại) hoặc x=3(nhận)

b:=>x+2=0 hoặc 5-3x=0

=>x=-2 hoặc x=5/3

c:=>20(1-2x)+6x=9(x-5)-24

=>20-40x+6x=9x-45-24

=>-34x+20=9x-69

=>-43x=-89

=>x=89/43

d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3

=>2x^2+4x-19=-2x+7

=>2x^2+6x-26=0

=>x^2+3x-13=0

=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)

e: =>(2x-3)(2x-3-x-1)=0

=>(2x-3)(x-4)=0

=>x=4 hoặc x=3/2

ĐKXĐ ; \(x\ne\pm1\)

Ta có : \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{x^2+3}{1-x^2}=0\)

\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{x^2-1}-\dfrac{\left(x-1\right)^2}{x^2-1}+\dfrac{-x^2-3}{x^2-1}=0\)

\(\Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2-x^2-3=0\)

\(\Leftrightarrow x^2+2x+1-x^2+2x-1-x^2-3=0\)

\(\Leftrightarrow-x^2+4x-3=0\)

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

\(\Leftrightarrow-x\left(x-3\right)+\left(x-3\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(TM\right)\\x=1\left(L\right)\end{matrix}\right.\)

=> X = 3

Vậy ..

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

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

ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)

Ta có : `(x-1)/x -1/(x+1) =(2x-1)/(x(x+1))`

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

`=> x^2 +x -x-1 -x-2x+1=0`

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

`<=> x(x-3)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=3\end{matrix}\right.\)

__

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

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\5-3x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\3x=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{5}{3}\end{matrix}\right.\)

__

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

\(\Leftrightarrow\dfrac{20\left(1-2x\right)}{12}+\dfrac{6x}{12}=\dfrac{9\left(x-5\right)}{12}-\dfrac{24}{12}\)

`<=> 2x- 40x + 6x = 9x - 45 -24`

`<=>  2x- 40x + 6x-9x + 45 +24=0`

`<=>-41x+69=0`

`<=>-41x=-69`

`<=> x=69/41`

Cậu tách 2 câu 1 lượt mn trl nhanh hơn đó ạ

a:=>x^2-1-x=2x-1

=>x^2-x-1=2x-1

=>x^2-3x=0

=>x=0(loại) hoặc x=3(nhận)

b:=>x+2=0 hoặc 5-3x=0

=>x=-2 hoặc x=5/3

c:=>20(1-2x)+6x=9(x-5)-24

=>20-40x+6x=9x-45-24

=>-34x+20=9x-69

=>-43x=-89

=>x=89/43

d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3

=>2x^2+4x-19=-2x+7

=>2x^2+6x-26=0

=>x^2+3x-13=0

=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)

e: =>(2x-3)(2x-3-x-1)=0

=>(2x-3)(x-4)=0

=>x=4 hoặc x=3/2

a) \(f\left(x\right)=-x^4+3x^3-\frac{1}{3}x^2+2x+5\)

\(g\left(x\right)=x^4+3x^3-\frac{2}{3}x^2-2x-10\)

b) \(f\left(x\right)+g\left(x\right)=-x^4+3x^3-\frac{1}{3}x^2+2x+5+x^4+3x^3-\frac{2}{3}x^2-2x-10\)

                                \(=6x^3-x^2-5\)

c) +) Thay x=1 vào đa thức f(x) + g(x) ta được :

       \(6.1^3-1^2-5=0\)

Vậy x=1 là nghiệm của đa thức f(x) + g(x)

+) Thay x=-1 vào đa thức f(x) + g(x) ta được :

    \(6.\left(-1\right)^3-\left(-1\right)^2-5=-10\)

Vậy x=-1 ko là nghiệm của đa thức f(x) + g(x)