Câu 1:

a) 1 + x = 0

c) 1 - 2t = 0

d) 3y = 0

câu 1 bạn tự làm nha

câu 2:

a) 4x-20=0

<=> 4x =20

<=>x =\(\dfrac{20}{4}\)

<=> x=5

b) 2x+x+12=0

<=>3x+12=0

<=>3x =-12

<=>x =\(\dfrac{-12}{3}\)

<=> x=-4

c) x-5=3-x

<=>x+x=3+5

<=>2x=8

<=>x =\(\dfrac{8}{2}\)

<=>x=4

d) 7-3x=9-x

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

<=>-2x=2

<=>x=\(\dfrac{2}{-2}\)

<=>x=-1

Mọi người giúp mình nhanh nhé ! Mình đang cần gấp

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 \(S=\left\{5\right\}\)

`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`

b,2x.(x-5)-x.(3+2x)=26

2x² - 10x - 3x - 2x² = 26

-13x = 26

x = -2

c, (x+7)²-x.(x-3)=12

x² +14x +49 - x² + 3x = 12

17x + 49 = 12

17x = - 37

x = \(\dfrac{-37}{17}\)

d, 9( x -2018) - x+ 2018 =0

9( x -2018) - (x -2018) = 0

( 9-1)(x -2018) = 0

8( x -2018) = 0

x -2018 = 0

x = 2018

a: =>2x+10-x^2-5=0

=>-x^2+2x+5=0

=>\(x\in\left\{1+\sqrt{6};1-\sqrt{6}\right\}\)

e: =>4x^2+4x+9x^2-4=15

=>13x^2+4x-19=0

=>\(x\in\left\{\dfrac{-2+\sqrt{251}}{13};\dfrac{-2-\sqrt{251}}{13}\right\}\)

g) \(\left(2x-1\right)^2-\left(2x+4\right)^2=0\)

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

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

\(\Leftrightarrow4x+3=0\)

\(\Leftrightarrow4x=-3\)

\(\Leftrightarrow x=\frac{-3}{4}\)

Vậy tập nghiệm của pt là \(S=\left\{\frac{-3}{4}\right\}\)

h) \(\left(2x-3\right)\left(3x+1\right)-x\left(6x+10\right)=30\)

\(\Leftrightarrow3x\left(2x-3\right)+\left(2x-3\right)-6x^2-10x=30\)

\(\Leftrightarrow6x^2-9x+2x-3-6x^2-10x=30\)

\(\Leftrightarrow-9x+2x-3-10x=30\)

\(\Leftrightarrow-17x-3=30\)

\(\Leftrightarrow-17x=33\)

\(\Leftrightarrow x=\frac{-33}{17}\)

Vậy tập nghiệm của pt là \(S=\left\{\frac{-33}{17}\right\}\)

k) \(\left(2x-1\right)\left(8x+5\right)-\left(4x+3\right)^2=20\)

\(\Leftrightarrow8x\left(2x-1\right)+5\left(2x-1\right)-\left(16x^2+24x+9\right)=20\)

\(\Leftrightarrow16x^2-8x+10x-5-16x^2-24x-9=20\)

\(\Leftrightarrow-8x+10x-5-24x-9=20\)

\(\Leftrightarrow-22x-14=20\)

\(\Leftrightarrow-11x-7=10\)

\(\Leftrightarrow x=\frac{-11}{17}\)

Vậy tập nghiệm của pt là \(S=\left\{\frac{-11}{17}\right\}\)