a) 2 + 1/3 - x = 1 + 1/4

7/3 -x = 5/4

x = 7/3 - 5/4

x = 13/12

b) (2/7 x 2) : x = 1 :7/2 

4/7 : x = 2/7

x = 4/7 : 2/7

x = 2

a) 2 + 1/3 - x = 1 + 1/4

           7/3 -x = 5/4

                  x = 7/3 - 5/4

                  x = 13/12

b) (2/7 x 2) : x = 1 :7/2 

          4/7 : x = 2/7

                  x = 4/7 : 2/7

                  x = 2

a) 2 + 1/3 - x = 1 + 1/4

2 + 1/3 - x =  5/4

7/3 - x =  5/4 

x = 7/3 - 5/4

x = 13/12
b) (2/7 x 2) : x = 1 :7/2  

(2/7 x 2) : x = 2/7

4/7 : x = 2/7

x = 4/7 : 2/7

x = 2

`c)1/4x+2/5=7/5`

`=>1/4x=7/5-1/5=1`

`=>x=1:1/4=4`

Vậy `x=4` 

`a)2x-2/3=-3/4`

`=>2x=-3/4+2/3=-1/12`

`=>x=-1/24`

Vậy `x=-1/24`

`b)x:3/4+1/4=-2/3`

`=>x:3/4=-2/3-1/4=-11/4`

`=>x=-11/4 xx 3/4=-33/16`

Vậy `x=-33/16`

Bài 2: 

a: =>-5x=-200

hay x=40

b: =>x*2/3=1244

hay x=1866

a, Thay x=-3 vào A ta có:

\(A=2x^2-6x=2.\left(-3\right)^2-6.\left(-3\right)=2.9+6.3=18+18=39\)

Thay x=4 vào A ta có:

\(A=2x^2-6x=2.4^2-6.4=2.16-24=32-24=8\)

b, \(A=0\)

\(\Leftrightarrow2x^2-6x=0\\ \Leftrightarrow2x\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

c, \(A=4x\)

\(\Leftrightarrow2x^2-6x=4x\\ \Leftrightarrow2x^2-10x=0\\ \Leftrightarrow2x\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

a, Thay x = -3 vào A ta được 

\(A=2.9-6.4=18-24=-6\)

Thay x = 4 vào A ta được 

\(A=2.16-6.4=32-24=6\)

b, Ta có \(A=2x\left(x-3\right)=0\Leftrightarrow x=0;x=3\)

c, Ta có \(A=2x^2-10x=0\Leftrightarrow2x\left(x-5\right)=0\Leftrightarrow x=0;x=5\)

a) \(x^2-x+x=4\)

\(x^2=4\)

\(x=\pm2\)

b) \(3x\left(x-5\right)-2\left(x-5\right)=0\)

\(\left(x-5\right)\left(3x-2\right)=0\)

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

c) Ta có: \(a+b+c=5-3-2=0\)

\(\left[{}\begin{matrix}x=1\\x=\dfrac{c}{a}=\dfrac{-2}{5}\end{matrix}\right.\)

d) Đặt \(x^2=t\left(t\ge0\right)\) . Lúc đó phương trình trở thành :

\(t^2-11t+18=0\)

\(\left[{}\begin{matrix}t=9\left(tmđk\right)\\t=2\left(tmđk\right)\end{matrix}\right.\)

\(t=9\rightarrow x^2=9\rightarrow x=\pm3\)

\(t=2\rightarrow x^2=2\rightarrow x=\pm\sqrt{2}\)

b) Ta có: \(\dfrac{x-2}{4}=\dfrac{2x+1}{3}\)

\(\Leftrightarrow3\left(x-2\right)=4\left(2x+1\right)\)

\(\Leftrightarrow3x-6=8x+4\)

\(\Leftrightarrow3x-8x=4+6\)

\(\Leftrightarrow-5x=10\)

hay x=-2

Vậy: x=-2

a) x³-9x²-4x-36=0

⇔ x²(x-9)-4(x-9)=0

⇔ (x-9)(x²-4)=0

⇒ Xảy ra 2 trường hợp:

- TH1: x-9=0 ⇔ x=9

- TH2: x²-4=0 ⇔ x=2 hoặc x=-2

Vậy x=9 hoặc x=2 hoặc x=-2.

a) x + \(\dfrac{3}{4}\) = \(\dfrac{5}{3}\)

    x = \(\dfrac{5}{3}\) - \(\dfrac{3}{4}\)

    x = \(\dfrac{20}{12}\) - \(\dfrac{9}{12}\)  

    x =  \(\dfrac{11}{12}\) 

b) x - \(\dfrac{2}{3}\) = \(\dfrac{7}{2}\)

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

    x = \(\dfrac{21}{6}\) + \(\dfrac{4}{6}\) 

    x = \(\dfrac{25}{6}\)

nhanh lên nhé 

sos

a: =>-12<x<2y<-9

=>x=-11; y=-5

b: =>-7<3(x-1)<8

\(\Leftrightarrow3\left(x-1\right)\in\left\{-6;-3;0;3;6\right\}\)

\(\Leftrightarrow x-1\in\left\{2;1;0;-1;-2\right\}\)

hay \(x\in\left\{3;2;1;0;-1\right\}\)

a: \(x\left(x+7\right)-\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow x^2+7x-x^2-x+6=0\)

hay x=-1

b: Ta có: \(\left(x+2\right)^2-\left(x^2-4\right)=0\)

\(\Leftrightarrow x+2=0\)

hay x=-2

b. (x + 2)² - x² + 4 = 0

<=> (x + 2 - x)(x + 2 + x) + 4 = 0

<=> 2(2 + 2x) + 4 = 0

<=> 4(1 + x) + 4 = 0

<=> 4(1 + x) = -4

<=> 1 + x = -1

<=> x = -1 - 1

<=> x = -2

\(a,\) \(x\left(x+7\right)-\left(x-2\right)\left(x+3\right)\)

\(\Leftrightarrow x^2+7x-x^2-3x+2x+6\\ \Leftrightarrow6x=0\\ \Leftrightarrow x=0\)

    \(Vậy...\)

\(b,\) \(\left(x+2\right)^2-\left(x^2-4\right)=0\)

\(\Leftrightarrow x^2+4x+4-x^2+4=0\\ \Leftrightarrow4x+8=0\\ \Leftrightarrow x=-2\)

Bài 2:

x^3+6x^2+12x+m chia hết cho x+2

=>x^3+2x^2+4x^2+8x+4x+8+m-8 chia hết cho x+2

=>m-8=0

=>m=8