1) a) \(\left(3x-1\right)\left(9x^2+3x+1\right)-4x\left(x-5\right)\)

\(=27x^3+9x^2+3x-9x^2-3x-1-4x^2+20x\)

\(=27x^3+\left(9x^2-9x^2-4x^2\right)+\left(3x-3x+20x\right)+\left(-1\right)\)

\(=27x^3-4x^2+20x-1\)

b)\(\left(7x+2\right)\left(3-4x\right)-\left(x+3\right)\left(x^2-3x+9\right)\)

\(=21x-28x^2+6-8x-x^3+3x^2-9x-3x^2+9x-27\)

\(=\left(21x-8x-9x+9x\right)+\left(-28x^2+3x^2-3x^2\right)\)\(+\left(6-27\right)\)\(+\left(-x^3\right)\)

\(=13x-28x^2-21-x^3\)

c)\(\left(4x+3\right)\left(4x-3\right)-\left(2-x\right)\left(4+2x+x^2\right)\)

\(=16x^2-12x+12x-9-8-4x-2x^2+4x+2x^2+x^3\)

\(=\left(16x^2-2x^2+2x^2\right)+\left(-12x+12x-4x+4x\right)\)\(+\left(-9-8\right)\)\(+x^3\)

\(=16x^2-17+x^3\)

d)\(\left(3x-8\right)\left(-5x+6\right)-\left(4x+1\right)\left(3x-2\right)\)

\(=-15x^2+18x+40x-48-12x^2+8x-3x+2\)

\(=\left(-15x^2-12x^2\right)+\left(18x+40x+8x-3x\right)\)\(+\left(-48+2\right)\)

\(=-27x^2+63x-46\)

e)\(\left(3x-6\right)4x-2x\left(3x+5\right)-4x^2\)

\(=12x^2-24x-6x^2-10x-4x^2\)

\(=\left(12x^2-6x^2-4x^2\right)+\left(-24x-10x\right)\)

\(=2x^2-34x\)

f)\(\left(5x-6\right)\left(6x-5\right)-x\left(3x+10\right)\)

\(=30x^2-25x-36x+30-3x^2-10x\)

\(=\left(30x^2-3x^2\right)+\left(-25x-36x-10x\right)+30\)

\(=27x^2-71x+30\)

2) a)\(x\left(x+3\right)-x^2=6\)

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

\(\Rightarrow\left(x^2-x^2\right)+3x=6\)

\(\Rightarrow3x=6\)

\(\Rightarrow x=2\)

Vậy x=2

b) \(2x\left(x-5\right)+x\left(-2x-1\right)=6\)

\(\Rightarrow2x^2-10x-2x^2-x=6\)

\(\Rightarrow\left(2x^2-2x^2\right)+\left(-10x-x\right)=6\)

\(\Rightarrow-11x=6\)

\(\Rightarrow x=-\dfrac{6}{11}\)

\(\)Vậy \(x=-\dfrac{6}{11}\)

c) x(x+5)-(x+1)(x-2)=7

\(\Rightarrow x^2+5x-x^2+2x-x+2=7\)

\(\Rightarrow\left(x^2-x^2\right)+\left(5x+2x-x\right)=7-2\)

\(\Rightarrow6x=5\)

\(\Rightarrow x=\dfrac{5}{6}\)

Vậy x=\(\dfrac{5}{6}\)

d)\(\left(3x+4\right)\left(6x-3\right)-\left(2x+1\right)\left(9x-2\right)=10\)

\(\Rightarrow18x^2-9x+24x-12-18x^2+4x-9x+2=10\)

\(\Rightarrow\left(18x^2-18x^2\right)+\left(-9x+24x+4x-9x\right)+\left(-12+2\right)=10\)

\(\Rightarrow10x-10=10\)

\(\Rightarrow10x=20\)

\(\Rightarrow x=2\)

Vậy x=2

`a,4x-10=0   `

`<=> 4x=10`

`<=>x=10/4`

`<=>x=5/2`

`b, 7-3x=9-x     `

`<=>-3x+x=9-7`

`<=>-2x=2`

`<=>x=-1`

`c, 2x-(3-5x) = 4(x+3)`

`<=>2x-3+5x=4x+12`

`<=>2x+5x-4x=12+3`

`<=>3x=15`

`<=>x=5`

`d, 5-(6-x)=4(3-2x)     `

`<=>5-6+x=12-8x`

`<=>x+8x=12-5+6`

`<=>9x=13`

`<=>x=13/9`

`e, 4(x+3)=-7x+17   `

`<=>4x+12=-7x+17`

`<=>4x+7x=17-12`

`<=>11x=5`

`<=>x=5/11`   

`f, 5(x-3) - 4=2(x-1)+7`

`<=>5x-15-4=2x-2+7`

`<=>5x-2x=15+4-2+7`

`<=>3x=24`

`<=>x=8`

`g, 5(x-3)-4=2(x-1)+7       `

`<=>5x-15-4=2x-2+7`

`<=>5x-2x=15+4-2+7`

`<=>3x=24`

`<=>x=8`

`h,4(3x-2)-3(x-4)=7x+20`

`<=>12x-8-3x+12=7x+20`

`<=>12x-3x-7x=20+8+12`

`<=>2x=40`

`<=>x=20`

\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

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

hay x=1

a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy: \(x=\dfrac{1}{2}\)

===========

b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

\(\Leftrightarrow x=\dfrac{13}{3}\)

Vậy: \(x=\dfrac{13}{3}\)

==========

c/  \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

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

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)

\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)

\(\Leftrightarrow-2x=-2\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

\(\Leftrightarrow x=\dfrac{2}{7}\)

Vậy: \(x=\dfrac{2}{7}\)

==========

f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

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

\(\Leftrightarrow x=-2\)

Vậy: \(x=-2\)

Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu

a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14) 

=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84

=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84) 

=> 156 -  56x = 24x - 324 

=>  24x + 56x = 324 + 156 

=> 80x = 480 

=> x = 480 : 80 =  6 

Vậy x = 6 

b, 5(3x + 5) - 4(2x - 3) = 5x + 3(2x + 12) + 1 

=> 15x + 25 - 8x + 12 = 5x + 6x + 36 + 1 

=> (15x - 8x) + (25 + 12) = 11x + 37 

=> 7x + 37 = 11x + 37 

=> 11x - 7x = 0 

=>  x = 0 

tck đầu tiên chọn câu trả lời của mình đi

e, 3x(2-x) =15(x-2)

\(\Leftrightarrow3x\left(2-x\right)-15\left(x-2\right)=0\)

\(\Leftrightarrow-3x\left(x-2\right)-15\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(-3x-15\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\-3x-15=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

Vậy..

f, (x+5)(x+4)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x+5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\x=-4\end{matrix}\right.\)

Vậy..

g, x(x+4)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

,h, (2x -4)(x-2)=0

\(\Leftrightarrow2\left(x-2\right)\left(x-2\right)=0\)

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

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

i, (x+1/5)(2x-3)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{1}{5}=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-1}{5}\\x=\frac{3}{2}\end{matrix}\right.\)

k, x²-4x=0

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

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

m, 4x²-1=0

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=1\\2x=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\x=\frac{-1}{2}\end{matrix}\right.\)

n, x²-6x+9=0

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

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

<=> x=3

l, (3x-5)²-(x+4)²=0

\(\Leftrightarrow\left(3x-5-x-4\right)\left(3x-5+x+4\right)=0\)

\(\Leftrightarrow\left(2x-9\right)\left(4x-1\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-9=0\\4x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=9\\4x=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{9}{2}\\x=\frac{1}{4}\end{matrix}\right.\)

Vậy ..

o, 7x(x+2)-5(x+2)=0

\(\Leftrightarrow\left(x+2\right)\left(7x-5\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\7x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\7x=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=\frac{5}{7}\end{matrix}\right.\)

Vậy....

p, 3x(2x-5)-4x+10=0

\(\Leftrightarrow3x\left(2x-5\right)-\left(4x-10\right)=0\)

\(\Leftrightarrow3x\left(2x-5\right)-2\left(2x-5\right)=0\)

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

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

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

Vậy...

q, (2-2x)-x²+1=0

\(\Leftrightarrow2\left(1-x\right)-\left(x^2-1^2\right)=0\)

\(\Leftrightarrow2\left(1-x\right)-\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow2\left(1-x\right)+\left(1-x\right)\left(x+1\right)=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}1-x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)

Vậy ....

r, x(1-3x)=5(1-3x)

\(\Leftrightarrow x\left(1-3x\right)-5\left(1-3x\right)=0\)

\(\Leftrightarrow\left(1-3x\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}1-3x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x=-1\\x=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{3}\\x=5\end{matrix}\right.\)

s, 2x-3/4+x+1/6=3

\(\Leftrightarrow x-\frac{7}{12}=3\Leftrightarrow x=3+\frac{7}{12}=\frac{43}{12}\)

r, x(1-3x)=5(1-3x)

➜x(1-3x)-5(1-3x)=0

➜(x-5)(1-3x)=0

➜\(\left[{}\begin{matrix}x-5=0\\1-3x=0\end{matrix}\right.\text{➜}\left[{}\begin{matrix}x=5\\x=\frac{1}{3}\end{matrix}\right.\)

Mk lười lắm mai nha!!!~~~~~~~~~~~~

Làm dần:

a, 3x+6=0

➜3x=-6

➜x=2

b, 2x-10=0

➜2x=10

➜x=5

c, 3x-7=11

➜3x=11+7

➜3x=18

➜x=6

d, 3x-9=0

➜3x=9

➜x=3

a, \(-\left(x+3\right)\left(x-4\right)+\left(x+1\right)\left(x-1\right)=10\)

\(\Rightarrow-\left(x^2-4x+3x-12\right)+x^2-1=10\)

\(\Rightarrow-x^2+x+12+x^2-1=10\)

\(\Rightarrow x=10+1-12\Rightarrow x=-1\)

b, \(\left(2x-1\right)\left(x-2\right)-\left(x+3\right)\left(2x-7\right)=3\)

\(\Rightarrow2x^2-4x-x+2-\left(2x^2-7x+6x-21\right)=3\)

\(\Rightarrow2x^2-5x+2-2x^2+x+21=3\)

\(\Rightarrow-4x=3-21-2\Rightarrow-4x=-20\)

\(\Rightarrow x=5\)

Các câu còn lại làm tương tự! Phá ngoặc ra!

Chúc bạn học tốt!!!

a) 5 - 4x = 3x - 9

\(\Leftrightarrow5-4x-3x+9=0\)

\(\Leftrightarrow14-7x=0\)

\(\Leftrightarrow7x=14\Leftrightarrow x=2\)

Vậy \(S=\left\{2\right\}\)

b) \(\left(x-4\right)\left(3x+9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

Vậy \(S=\left\{-3;4\right\}\)

c) \(\dfrac{x}{x+4}+\dfrac{12}{x-4}=\dfrac{4x+48}{x\cdot x-16}\)(1)

ĐKXĐ: \(x\ne\pm4\)

\(\left(1\right)\Leftrightarrow\dfrac{x\left(x-4\right)+12\left(x+4\right)-4x-48}{\left(x+4\right)\left(x-4\right)}=0\)

\(\Leftrightarrow x^2-4x+12x+48-4x-48=0\)

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

\(\Leftrightarrow x\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-4\left(KTM\right)\end{matrix}\right.\)

Vậy \(S=\left\{0\right\}\)

d) \(4-2x=7-x\)

\(\Leftrightarrow4-2x-7+x=0\)

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

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

Vậy \(S=\left\{-3\right\}\)

e) \(\left(x+4\right) \left(8-4x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\8-4x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)

Vậy \(S=\left\{-4;2\right\}\)

f) \(\dfrac{x}{x+5}+\dfrac{11}{x-5}=\dfrac{x+55}{x\cdot x-25}\left(2\right)\)

ĐKXĐ: \(x\ne\pm5\)

\(\left(2\right)\Leftrightarrow\dfrac{x\left(x-5\right)+11\left(x+5\right)-x-55}{\left(x+5\right)\left(x-5\right)}=0\)

\(\Leftrightarrow x^2-5x+11x+55-x-55=0\)

\(\Leftrightarrow x^2+5x=0\)

\(\Leftrightarrow x\left(x+5\right)=0\)

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

Vậy \(S=\left\{0\right\}\)

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

\(\Leftrightarrow\dfrac{3\left(3x+2\right)-3x-1-10-12x}{6}=0\)

\(\Leftrightarrow9x+6-3x-1-10-12x=0\)

\(\Leftrightarrow-6x-5=0\)

\(\Leftrightarrow-6x=5\)

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

Vậy \(S=\left\{-\dfrac{5}{6}\right\}\)

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

\(\Leftrightarrow2x-3+5x-4x-12=0\)

\(\Leftrightarrow3x-15=0\)

\(\Leftrightarrow x=5\)

Vậy \(S=\left\{5\right\}\)

i) \(3x-6+x=9-x\)

\(\Leftrightarrow3x-6+x-9+x=0\)

\(\Leftrightarrow5x-15=0\)

\(\Leftrightarrow x=3\)

Vậy \(S=\left\{3\right\}\)

k)\(2t-3+5t=4t+12\)

\(\Leftrightarrow2t-3+5t-4t-12=0\)

\(\Leftrightarrow3t-15=0\)

\(\Leftrightarrow t=5\)

Vậy \(S=\left\{5\right\}\)