LỚP MẤY CHỨ 8 THÌ EM XIN VÁI CẢ 6 TAY

chuyên hang tư x sang 1 bên ,sô sang 1 bên va lam nhu bt /cai nay dê ma 

a) 1,2-(x-0,8)= -2(0,9+x)

VT=-(x-2)

VP=\(-\frac{10x+9}{5}\)

pt trở thành:-(x-2)=\(-\frac{10x+9}{5}\)

<=>2-x=-2x-1,8

<=>5x=-19

<=>x=-3,8

a) 1,2-(x-0,8)= -2(0,9+x)

VT=-(x-2)

VP=$-\frac{10x+9}{5}$−10x+95 

pt trở thành:-(x-2)=$-\frac{10x+9}{5}$−10x+95 

<=>2-x=-2x-1,8

<=>5x=-19

<=>x=-3,8

b) 2,3x-2(0,7+2x)=3,6-1,7x

x=-2.25179981368525*10^16

c) 3(2,2-0,3x)=2,6+(0,1x-4)

Áp dụng tc tỉ lệ thức ta có:

\(\frac{66-9x}{10}=\frac{x-14}{10}\Rightarrow\left(66-9x\right)10=10\left(x-14\right)\)

VT=-30(3x-22)

pt trờ thành -30(3x-22)=10(x-14)

<=>660-90x=10x-140

<=>-100x=-800

<=>x=8

d) 3,6-0,5(2x+1)=x-0,25(2-4x)

Áp dụng tc tỉ lệ thức ta típ tục biến đổi được

\(\frac{31-10x}{10}=\frac{4x-1}{2}\Rightarrow\left(31-10x\right)2=10\left(4x-1\right)\)

VT=-2(10x-31)

pt trở thành -2(10x-31)=40x-10

<=>-60x=-72

<=>x=1,2

Copy có khác, ko đọc đc j!!! ʌl

Câu 3:

1)

a) Ta có:

hay x=-1

Vậy: x=-1

b) Ta có:

hay y=0

Vậy: y=0

c) Ta có:

hay x=15

Vậy: x=15

d) Ta có:

Vì 3≠0

nên x-5=0

hay x=5

Vậy: x=5

a) 3x - 2 = 2x - 3

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

\(\Leftrightarrow\) x + 1 = 0

\(\Rightarrow\) x = -1

b) 3 - 4y + 24 + 6y = y + 27 + 3y

\(\Leftrightarrow\) 3 - 4y + 24 + 6y - y - 27 - 3y = 0

\(\Leftrightarrow\) -2y = 0

\(\Rightarrow\) y = 0

c)7 - 2x = 22 - 3x

\(\Leftrightarrow\) 7 - 2x - 22 + 3x = 0

\(\Leftrightarrow\) -15 + x = 0

\(\Rightarrow\) x = 15

d) 8x - 3 = 5x + 12

\(\Leftrightarrow\) 8x - 3 - 5x - 12 = 0

\(\Leftrightarrow\)3x -15 = 0

\(\Leftrightarrow\) 3x = 15

\(\Rightarrow\) x = 5

e) x - 12 + 4x = 25 + 2x - 1

\(\Leftrightarrow\) x - 12 + 4x - 25 - 2x + 1 = 0

\(\Leftrightarrow\) 3x - 36 = 0

\(\Leftrightarrow\) 3x = 36

\(\Rightarrow\) x = 12

f ) x + 2x + 3x - 19 = 3x + 5

\(\Leftrightarrow\) x + 2x + 3x - 19 - 3x - 5 = 0

\(\Leftrightarrow\)3x - 24 = 0

\(\Leftrightarrow\) 3x = 24

\(\Rightarrow\) x = 8

g) 11+ 8x - 3 = 5x - 3 +x

\(\Leftrightarrow\)8x + 8 = 6x - 3

\(\Leftrightarrow\)8x - 6x = -3 - 8

\(\Leftrightarrow\)2x = -11

\(\Rightarrow\)x = \(-\frac{11}{2}\)

h) 4 - 2x +15 = 9x + 4 -2

\(\Leftrightarrow\)19 - 2x = 7x + 4

\(\Leftrightarrow\)-2x - 7x = 4 - 19

\(\Leftrightarrow\)-9x = -15

\(\Rightarrow\)x = \(\frac{15}{9}\) = \(\frac{5}{3}\)

2)

a) \(5-\left(x-6\right)=4\cdot\left(3-2\right)\)

\(\Leftrightarrow5-x+6=12-8\)

\(\Leftrightarrow11-x=4\)

\(\Rightarrow x=7\)

b) \(2x\cdot\left(x+2\right)^2-8x^2=2\cdot\left(x-2\right)\cdot\left(x^2+2x+4\right)\)

\(\Leftrightarrow2x\cdot\left(x^2+4x+4\right)-8x^2=2\cdot\left(x^3-8\right)\)

\(\Leftrightarrow2x^3+8x^2+8x-8x^2-2x^3+16=0\)

\(\Leftrightarrow8x+16=0\)

\(\Rightarrow x=-2\)

c) \(7-\left(2x+4\right)=-\left(x+4\right)\)

\(\Leftrightarrow7-2x-4=-x-4\)

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

\(\Leftrightarrow-x=-7\)

\(\Rightarrow x=7\)

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

\(\Leftrightarrow x^3-6x^2+12x-8+9x^2-1-x^3-3x^2-3x-1=0\)

\(\Leftrightarrow9x-10=0\)

\(\Rightarrow x=\frac{10}{9}\)

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

\(\Leftrightarrow2x^3-3x+2x-3-2x^2-10x+x+5=0\)

\(\Leftrightarrow2-10x=0\)

\(\Rightarrow x=\frac{2}{10}=\frac{1}{5}\)

f)\(\left(x-1\right)^3-x\cdot\left(x+1\right)^2=5x\cdot\left(2-x\right)-11\cdot\left(x+2\right)\)

\(\Leftrightarrow x^3-3x^2+3x-1-x^3-2x^2-x-10x+5x^2+11x+22=0\)

\(\Leftrightarrow3x+21=0\)

\(\Rightarrow x=-7\)

g)\(\left(x-1\right)-\left(2x-1\right)=9-x\)

\(\Leftrightarrow x-1-2x+1-9+x=0\)

\(\Leftrightarrow-9=0\)

\(\Rightarrow\) Phương trình vô nghiệm

h)\(\left(x-3\right)\cdot\left(x+4\right)-2\cdot\left(3x-2\right)=\left(x-4\right)^2\)

\(\Leftrightarrow x^2+4x-3x-12-6x+4=x^2-8x+16\)

\(\Leftrightarrow x^2-5x-8=x^2-8x+16\)

\(\Leftrightarrow x^2-5x-8-x^2+8x-16=0\)

\(\Leftrightarrow3x-24=0\)

\(\Rightarrow x=8\)

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

\(\Leftrightarrow x^3+6x^2+9x-3x=x^3+6x^2+12x+8+1\)

\(\Leftrightarrow x^3+6x^2+6x=x^3+6x^2+12x+9\)

\(\Leftrightarrow x^3+6x^2+6x-x^3-6x^2-12x-9=0\)

\(\Leftrightarrow-6x-9=0\)

\(\Rightarrow x=-\frac{3}{2}\)

j)\(\left(x+1\right)\cdot\left(x^2-x+1\right)-2x=x\cdot\left(x+1\right)\cdot\left(x-1\right)\)

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

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

\(\Leftrightarrow1-x=0\)

\(\Rightarrow x=1\)

a)   1,2 - ( x - 0,8 ) = -2( 0,9+ x ) 

<=> 1,2 -  x + 0,8  = -1.8 - 2x  

<=> x = -3,8

Vậy x = -3,8

b)   2,3x - 2(0,7 + 2x ) = 3,6 - 1,7x 

<=>  2,3x - 1,4 - 4x  = 3,6 - 1,7x 

<=>  -3,4x = 5

<=>  x = \(\dfrac{-25}{17}\)

Vậy x = \(\dfrac{-25}{17}\)

c)    3(2,2 - 0,3x ) = 2,6 + (0,1x - 4 ) 

<=> 6,6 - 0,9x = 2,6 + 0,1x - 4

<=> -x = -8

<=> x = 8

Vậy x = 8

d)    3,6 - 0,5(2x + 1) = x- 0,25(2-4x)

<=> 3,6 - x - 0.5 = x - 0,5 + x

<=> -3x = -3,6

<=>  x = 1.2

Vậy x = 1.2