a) Ta có: \(\left|2x-1\right|=\left|2x+3\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=2x+3\left(loại\right)\\2x-1=-2x-3\end{matrix}\right.\Leftrightarrow2x+2x=-3+1\)

\(\Leftrightarrow4x=-2\)

hay \(x=-\dfrac{1}{2}\)

a) \(\left|2x-3\right|=5\)

⇒ \(\left[{}\begin{matrix}2x-3=5\\2x-3=-5\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}2x=5+3=8\\2x=\left(-5\right)+3=-2\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=8:2\\x=\left(-2\right):2\end{matrix}\right.\)

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

Vậy \(x\in\left\{4;-1\right\}.\)

b) Đề sai rồi, bạn xem lại nhé.

c) \(\left|x-1\right|+3x=11\)

⇒ \(\left|x-1\right|=11-3x\)

+) Với \(x\ge1\)

⇒ \(x-1=11-3x\)

⇒ \(x+3x=11+1\)

⇒ \(4x=12\)

⇒ \(x=12:4\)

⇒ \(x=3.\)

+) Với \(x< 1\)

⇒ \(1-x=11-3x\)

⇒ \(1-11=\left(-3x\right)+x\)

⇒ \(-10=-2x\)

⇒ \(x=\left(-10\right):\left(-2\right)\)

⇒ \(x=5.\)

Vậy \(x\in\left\{3;5\right\}.\)

Chúc bạn học tốt!

a) |2x - 3| = 5

2x - 3 = 5 Hoặc 2x - 3 = -5

* 2x - 3 = 5

2x = 5 + 3

2x = 8

x = 8 : 2

x = 4

2x - 3 = -5

2x = -5 + 3

2x = -2

x = -2 : 2

x = -1

Vậy x ϵ {-2; -1}

a)\(\left|2x-3\right|=5\\ \Rightarrow\left\{{}\begin{matrix}2x-3=5\\2x-3=-5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)

vậy ...

b) really nigga

a: =>|x-7|=3-2x

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(-2x+3\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(2x-3-x+7\right)\left(2x-3+x-7\right)=0\end{matrix}\right.\)

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

b: =>|2x-3|=4x+9

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(4x+9-2x+3\right)\left(4x+9+2x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(2x+12\right)\left(6x+6\right)=0\end{matrix}\right.\Leftrightarrow x=-1\)

c: =>3x+5=2-5x hoặc 3x+5=5x-2

=>8x=-3 hoặc -2x=-7

=>x=-3/8 hoặc x=7/2

a) \(\left|2x-3\right|=5\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3=5\\2x-3=-5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=8\\2x=-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=-1\end{cases}}\)

Vậy : \(x\in\left\{4,-1\right\}\)

b) \(\left|2x-1\right|=\left|2x-3\right|\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=2x-3\\2x-1=3-2x\end{cases}}\)

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

\(\Leftrightarrow\orbr{\begin{cases}0=-2\\4x=4\end{cases}}\)

\(\Rightarrow x=1\)

Vậy : \(x=1\)

c) \(\left|x-1\right|+3x=11\)

\(\Leftrightarrow\left|x-1\right|=11-3x\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=11-3x\\x-1=3x-11\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x+3x=11+1\\3x-x=-1+11\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}4x=12\\2x=10\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=5\end{cases}}\)

Vậy : \(x\in\left\{3,5\right\}\)

d) \(\left|5x-3\right|-x=7\)

\(\Leftrightarrow\left|5x-3\right|=7+x\)

\(\Leftrightarrow\orbr{\begin{cases}5x-3=7+x\\5x-3=-7-x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x-x=7+3\\5x+x=-7+3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}4x=10\\6x=-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{10}{4}=\frac{5}{2}\\x=-\frac{4}{6}=-\frac{2}{3}\end{cases}}\)

Vậy : \(x\in\left\{\frac{5}{2},-\frac{2}{3}\right\}\)

a,\(|2x-3|=5\)

\(\Rightarrow2x-3=5\)hoặc\(2x-3=-5\)

\(2x=5+3\)hoặc\(2x=-5+3\)

\(2x=8\)hoặc\(2x=-2\)

\(\Rightarrow x=\frac{8}{2}=4\)hoặc\(x=-\frac{2}{2}=-1\)

Vậy.......

b,Sai vế phải r bn ơi

c,\(|x-1|+3x=11\)

\(|x-1|=11-3x\)

\(\Rightarrow x-1=+\left(11-3x\right)\)hoặc\(\Rightarrow x-1=-\left(11-3x\right)\)

\(x-1=11-3x\)hoặc\(x-1=-11+3x\)

\(x+3x=11+1\)hoặc\(x-3x=-11+1\)

\(4x=12\)hoặc\(-2x=-10\)

\(\Rightarrow x=\frac{12}{4}=3\)hoặc\(x=\frac{-10}{-2}=5\)

Vậy.....

d,\(|5x-3|-x=7\)

\(|5x-3|=7+x\)

\(\Rightarrow5x-3=+\left(7+x\right)\)hoặc\(5x-3=-\left(7+x\right)\)

\(5x-3=7+x\)hoặc\(5x-3=-7-x\)

\(5x-x=7+3\)hoặc\(5x+x=-7+3\)

\(4x=10\)hoặc\(6x=-4\)

\(\Rightarrow x=\frac{10}{4}=2,5\)hoặc\(x=-\frac{4}{6}=-\frac{2}{3}\)

Vậy......

1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)

=>-13x=0

=>x=0

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

=>3x=13

=>x=13/3

3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)

=>-2x^2=0

=>x=0

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

=>-8x=6-14=-8

=>x=1

`1)2x(x-5)-(3x+2x^2)=0`

`<=>2x^2-10x-3x-2x^2=0`

`<=>-13x=0`

`<=>x=0`

___________________________________________________

`2)x(5-2x)+2x(x-1)=13`

`<=>5x-2x^2+2x^2-2x=13`

`<=>3x=13<=>x=13/3`

___________________________________________________

`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`

`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`

`<=>x=0`

___________________________________________________

`4)5x(x-1)-(x+2)(5x-7)=0`

`<=>5x^2-5x-5x^2+7x-10x+14=0`

`<=>-8x=-14`

`<=>x=7/4`

___________________________________________________

`5)6x^2-(2x-3)(3x+2)=1`

`<=>6x^2-6x^2-4x+9x+6=1`

`<=>5x=-5<=>x=-1`

___________________________________________________

`6)2x(1-x)+5=9-2x^2`

`<=>2x-2x^2+5=9-2x^2`

`<=>2x=4<=>x=2`

1,

a,\(2x\left(3x^2-5x+3\right)\)

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

b,\(-2x\left(x^2+5x-3\right)\)

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

c,\(-\dfrac{1}{2}x\left(2x^3-4x+3\right)\)

\(=-x^4+2x^2-\dfrac{3}{2}x\)

Bài 2:

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

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

\(=2x^3-18x-x^2+9\)

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

\(=-\left(10x^2+15x-8x-12\right)\)

\(=-10x^2-7x+12\)

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

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

Bài 3: (chỉ cần kết quả ko chứa biến là ta có đpcm, nói chung bài này yêu cầu ta rút gọn)

a) \(x\left(3x+12\right)-\left(7x-20\right)+x^2\left(2x-3\right)-x\left(2x^2+5\right)\)

\(=3x^2+12x-7x+20+2x^3-3x^2-2x^3-5x\)

\(=20\)

b) \(3\left(2x-1\right)-5\left(x-3\right)+6\left(3x-4\right)-19x\)

\(=6x-3-5x+15+18x-24-19x\)

\(=-12\)

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

\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)

\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x+10=3x^2-12x+20\)

\(\Leftrightarrow3x^2-7x-2=3x^2-12x+20\)

\(\Leftrightarrow-7x+12x=20+2\)

\(\Leftrightarrow5x=22\)

\(\Rightarrow x=\dfrac{22}{5}\)

tick cho mk nha

b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)

\(\Leftrightarrow24x^2+16x-9x-6-4x^2-23x-28=10x^2+3x-1\)

\(\Leftrightarrow20x^2-16x-34-10x^2-3x+1=0\)

\(\Leftrightarrow10x^2-19x-33=0\)

\(\Delta=\left(-19\right)^2-4.10.\left(-33\right)=1320\)

\(x_1=3;x_2=\dfrac{-11}{10}\)

Tick cho mk nha

c) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)

\(\Leftrightarrow21x-15x^2-35+25x-4x+15x^2-4=4\)

\(\Leftrightarrow42x-39=4\)

\(\Leftrightarrow42x=4+39\)

\(\Leftrightarrow42x=43\)

\(\Rightarrow x=\dfrac{43}{42}\)

Tick cho mk nha