20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)

Vậy...

1) 16 - 8x = 0 ⇔ 8(2 - x) = 0⇔ 2 - x = 0 ⇔ x = 2

Vậy phương trình có nghiệm là x = 2

a) 2x² - 3x - 2 = 0.

<=> (2x + 1)(x - 2) = 0

<=> 2x + 1 = 0 hoặc x - 2 = 0

<=> x = -1/2 hoặc x = 2

b) 3x² - 7x - 10 = 0.

<=> (x + 1)(3x - 10) = 0

<=> x = -1 hoặc x = 10/3

c) 2x² - 5x + 3 = 0.

<=> (x - 1)(2x - 3) = 0

<=> x = 1 hoặc x = 3/2

\(3x+4=0\Leftrightarrow x=-\dfrac{4}{3}\\ 2x\left(x-1\right)-\left(1+2x\right)=-34\\ \Leftrightarrow2x^2-2x-1-2x=-34\\ \Leftrightarrow2x^2-4x+33=0\\ \Leftrightarrow2\left(x^2-2x+1\right)+30=0\\ \Leftrightarrow2\left(x-1\right)^2+30=0\\ \Leftrightarrow x\in\varnothing\left[2\left(x-1\right)^2+30\ge30>0\right]\\ x^2+9x-10=0\\ \Leftrightarrow x^2-x+10x-10=0\\ \Leftrightarrow\left(x-1\right)\left(x+10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-10\end{matrix}\right.\\ \left(7x-1\right)\left(2+5x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}7x-1=0\\2+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

a: (2x-10)(5x+25)=0

=>2x-10=0 hoặc 5x+25=0

=>x=5 hoặc x=-5

b: (x+15)(x-2)=0

=>x+15=0 hoặc x-2=0

=>x=-15 hoặc x=2

c: =>x(x-7)=0

=>x=0 hoặc x=7

a, (2x - 10) (5x + 25) = 0

⇒ 2x - 10 = 0 hoặc 5x + 25 = 0

⇒ x = 5 hoặc x = -5

b, (x + 15) (x - 2) = 0

⇒ x + 15 = 0 hoặc x - 2 = 0

⇒ x = -15 hoặc x = 2

c: =>x(x-7)=0

=>x=0 hoặc x=7

Mấy cái này chuyển vế đổi dấu là xong í mà :3

1,

16-8x=0

=>16=8x

=>x=16/8=2

2, 

7x+14=0

=>7x=-14

=>x=-2

3,

5-2x=0

=>5=2x

=>x=5/2

Mk làm 3 cau làm mẫu thôi

Lúc đăng đừng đăng như v :>

chi ra khỏi ngt nản

từ câu 1 đến câu 8 cs thể làm rất dễ,bn tham khảo bài của bn muwaa r làm những câu cn lại

1, 16 - 8x = 0

<=>-8x = 16

<=> x = -2

Vậy_

2, 7x + 14 = 0

<=> 7x = -14

<=> x = -2

3, 5 - 2x = 0

<=> - 2x = -5

<=> x =\(\frac{5}{2}\)

Vậy_

4, 3x - 5 = 7

<=> 3x = 7 + 5

<=> 3x = 12

<=> x = 4

Vậy...

5, 8 - 3x = 6

<=> - 3x = 6 - 8

<=> -3x = - 2

<=> x =\(\frac{2}{3}\)

Vậy......

a) \(\left(x-3\right)\left(2x+1\right)\left(4-5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\\4-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{-1}{2}\\x=\dfrac{4}{5}\end{matrix}\right.\)

Vậy ..................

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

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

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

\(\Leftrightarrow x\left[2x\left(x-1\right)-3\left(x-1\right)\right]=0\)

\(\Leftrightarrow x\left(x-1\right)\left(2x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy .................

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

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

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

\(\Leftrightarrow\left(x+4\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy .......................

d) \(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-3x-4x+12\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left[x\left(x-3\right)-4\left(x-3\right)\right]=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=3\\x=4\end{matrix}\right.\)

Vậy ...................

a,\(\left(x-3\right)\left(2x+1\right)\left(4-5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\\4-5x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\\x=\dfrac{4}{5}\end{matrix}\right.\)

Vậy...

b,\(2x^3-5x^2+3x=0\)

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

\(\Leftrightarrow x\left(2x-3\right)\left(x-1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\\x=1\end{matrix}\right.\)

Vậy...

c,Sửa đề:

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

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

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

\(\Leftrightarrow\left(3x-2\right)\left(-x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\-x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-4\end{matrix}\right.\)

Vậy...

d,\(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x+4\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x+4=0\\x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-4\\x=3\end{matrix}\right.\)

Vậy...

Sửa lại câu d, như @Gia Hân Ngô là đúng rồi đó.Tôi nhầm dấu.

1.

<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)

2.

<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

3.

<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)

4.

<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)

5. 

<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)

6,7. ko đủ điều kiện tìm

a) \(7x-10=5x-6\)

\(7x-5x=-6+10\)

\(2x=4\)

\(x=2\)

b) \(3x\left(x-2\right)+x-2=0\)

\(\left(x-2\right)\left(3x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\3x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-\frac{1}{3}\end{cases}}\)

c) \(2x^2+7x-4=0\)

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

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

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

\(\Rightarrow\orbr{\begin{cases}2x-1=0\\x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-2\end{cases}}\)

7x-10=5x-6<=>7x-5x=-6+10<=>2x=4=>x=2

3x(x-2)+x-2=0<=>(x-2)(3x+1)=0<=>x-2=0=>x=2    HAY 3x+1=0=>x=-1/3

2x²+7x-4=0.

Câu cuối xem có lộn đề không nha bạn ơi!!!

a) 7x - 10 = 5x - 6

<=> 7x - 5x = -6 + 10

<=> 2x = 4

<=> x = 2

b) 3x( x - 2 ) + x - 2 = 0

<=> 3x( x - 2 ) + 1( x - 2 ) = 0

<=> ( x - 2 )( 3x + 1 ) = 0

<=> \(\orbr{\begin{cases}x-2=0\\3x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{1}{3}\end{cases}}\)

c) 2x² + 7x - 4 = 0

<=> 2x² + 8x - x - 4 = 0

<=> 2x( x + 4 ) - 1( x + 4 ) = 0

<=> ( x + 4 )( 2x - 1 ) = 0

<=> \(\orbr{\begin{cases}x+4=0\\2x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=\frac{1}{2}\end{cases}}\)