a: =>x-22=9

hay x=31

a: =>4/3x=7/9-4/9=1/3

=>x=1/4

b: =>5/2-x=9/14:(-4/7)=-9/8

=>x=5/2+9/8=29/8

c: =>3x+3/4=8/3

=>3x=23/12

hay x=23/36

d: =>-5/6-x=7/12-4/12=3/12=1/4

=>x=-5/6-1/4=-10/12-3/12=-13/12

em moi lop 4 mà

\(d,-\dfrac{5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)

\(\Leftrightarrow-\dfrac{5}{6}-x=\dfrac{7}{12}+\dfrac{-4}{12}\)

\(\Leftrightarrow-\dfrac{5}{6}-x=\dfrac{1}{4}\)

\(\Leftrightarrow-x=-\dfrac{5}{6}-\dfrac{1}{4}\)

\(\Leftrightarrow-x=-\dfrac{13}{12}\)

\(\Leftrightarrow x=\dfrac{13}{12}\)

\(c,3x+\dfrac{3}{4}=2\dfrac{2}{3}\)

\(\Leftrightarrow3x+\dfrac{3}{4}=\dfrac{8}{3}\)

\(\Leftrightarrow3x=\dfrac{8}{3}-\dfrac{3}{4}\)

\(\Leftrightarrow3x=\dfrac{23}{12}\)

\(\Leftrightarrow x=\dfrac{23}{12}:3\)

\(\Leftrightarrow x=\dfrac{23}{36}\)

Ta có :\(\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right).\left(2x-2\right)=\left(-\frac{3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)

=> \(\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right).\left(2x-2\right)=-\frac{1}{2}\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)\)

=> \(2x-2=-\frac{1}{2}\)

=> \(2x=\frac{3}{2}\)

=> \(x=\frac{3}{4}\)

1) \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12=x^4+x^3+2x^2+x^3+x^2+2x+x^2+x+2-12\)

\(=x^4+2x^3+4x^2+3x-10=\left(x^4+2x^3\right)+\left(4x^2+8x\right)+\left(-5x-10\right)\)

\(=x^3.\left(x+2\right)+4x.\left(x+2\right)-5.\left(x+2\right)=\left(x+2\right)\left(x^3+4x-5\right)\)

\(=\left(x+2\right)\left(x^3-x^2+x^2-x+5x-5\right)=\left(x+2\right)\left(x-1\right)\left(x^2+x+5\right)\)

2) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left[\left(x+2\right)\left(x+5\right)\right].\left[\left(x+3\right)\left(x+4\right)\right]-24\)

\(=\left(x^2+7x+10\right).\left(x^2+7x+12\right)-24\)

Đặt  \(a=x^2+7x+10\) thì ta có :\(a.\left(a+2\right)-24=a^2+2a-24=\left(a^2+2a+1\right)-25=\left(a+1\right)^2-5^2\)

\(=\left(a+1+5\right)\left(a+1-5\right)=\left(a+6\right)\left(a-4\right)\)

Thay a , ta có :

\(\left(x^2+7x+10+6\right)\left(x^2+7x+10-4\right)=\left(x^2+7x+16\right).\left(x^2+x+6x+6\right)\)

\(=\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)

a) => 4/3x = 7/9 - 4/9 = 1/3

=> x = 1/3 : 4/3 = 1/4

b) => 5/2 - x = 9/14 : (-4/7) = -9/8

=> x = 5/2 - (-9/8) = 5/2 + 9/8 = 29/8

c) => 3x = 2 và 2/3 - 3/4 = 8/3 - 3/4 = 23/12

=> x = 23/12 : 3 = 23/36

D) => -5/6 - x = 1/4

=> x = -5/6 - 1/4 = -13/12

a) \(\dfrac{4}{9}+\dfrac{4}{3}x=\dfrac{7}{9}\)

\(\dfrac{4}{3}x=\dfrac{7}{9}-\dfrac{4}{9}=\dfrac{1}{3}\)

\(x=\dfrac{1}{3}:\dfrac{4}{3}\)

\(x=\dfrac{1}{4}\)

b) \(\left(\dfrac{5}{2}-x\right)\left(-\dfrac{4}{7}\right)=\dfrac{9}{14}\)

\(\dfrac{5}{2}-x=\dfrac{9}{14}:\left(-\dfrac{4}{7}\right)=-\dfrac{9}{8}\)

\(x=\dfrac{5}{2}-\left(-\dfrac{9}{8}\right)\)

\(x=\dfrac{29}{8}\)

c) \(3x+\dfrac{3}{4}=2\dfrac{2}{3}\)

\(3x+\dfrac{3}{4}=\dfrac{8}{3}\)

\(3x=\dfrac{8}{3}-\dfrac{3}{4}=\dfrac{23}{12}\)

\(x=\dfrac{23}{12}:3\)

\(x=\dfrac{23}{36}\)

d) \(-\dfrac{5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)

\(-\dfrac{5}{6}-x=\dfrac{1}{4}\)

\(x=-\dfrac{5}{6}-\dfrac{1}{4}\)

\(x=-\dfrac{13}{12}\)

Ta có : 6x² - 11x + 3 

= 6x² - 2x - 9x + 3

= (6x² - 2x) - (9x - 3)

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

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

K MIK NHA BẠN !!!!!!!!!!

bÀI 1 

bÀI 2 : 

Bài 3 :

Bài 4: 

5,

6, 

7, 

8,

9, 

10,

11,

12,

13,

K MIK NHA BẠN !!!!!!!!!!

Nguyễn Quang Trung Linh Nhi : Mik cảm ơm 2 bạn nhiều nhóe ;) 

a: =(x-y)^2+2(x-y)

=(x-y)(x-y+2)

c: =(x-3)(x+3)+(x-3)^2

=(x-3)(x+3+x-3)

=2x(x-3)

d: =(x+3)(x^2-3x+9)-4x(x+3)

=(x+3)(x^2-7x+9)

e: =(x^2-8x+7)(x^2-8x+15)-20

=(x^2-8x)^2+22(x^2-8x)+85

=(x^2-8x+17)(x^2-8x+5)

a: \(x+\dfrac{3}{9}=\dfrac{7}{6}\cdot\dfrac{2}{3}\)

=>\(x+\dfrac{1}{3}=\dfrac{14}{18}=\dfrac{7}{9}\)

=>\(x=\dfrac{7}{9}-\dfrac{1}{3}=\dfrac{7}{9}-\dfrac{3}{9}=\dfrac{4}{9}\)

b: \(x-\dfrac{2}{3}=\dfrac{1}{8}:\dfrac{5}{4}\)

=>\(x-\dfrac{2}{3}=\dfrac{1}{8}\cdot\dfrac{4}{5}=\dfrac{1}{10}\)

=>\(x=\dfrac{1}{10}+\dfrac{2}{3}=\dfrac{3+20}{30}=\dfrac{23}{30}\)

TThế giới oi oi oi