a)\(\text{ 4x - 15 = -75 - x}\)

\(4x-15+75+x=0\)

\(5x+60=0\)

\(5x=-60\)

\(x=-14\)

Vậy....

Thêm dấu suy ra trc mỗi dòng nha

Học tốt

b)\(|2x-7|+2=13\)

\(|2x-7|=11\)

\(\Leftrightarrow\hept{\begin{cases}2x-7=11\\2x-7=-11\end{cases}\Leftrightarrow\hept{\begin{cases}2x=18\\2x=4\end{cases}\Leftrightarrow}\hept{\begin{cases}x=9\\x=2\end{cases}}}\)

vậy x=9 hoặc x=2

a) 4x - 15 = -75 - x

4x + x = -75 + 15

5x = -60

x= -60:5

x=-12

b) |2x - 7| + 2 = 13

| 2x - 7 | = 11

th1: 2x-7= 11

2x = 18

x = 9

th2: 2x - 7 = -11

2x = -4

x = -2

c)(2x - 1)^2 = 9

(2x-1)^2= 3^2

2x-1 = 3

2x = 4

x = 2

d)2x (x - 3) = 0

Th1: 2x=0

x= 0

Th2: x- 3 =0

x= 3

Hok tốt!!!

\(1,\dfrac{3x+2}{6}-\dfrac{3x-2}{4}=\dfrac{15}{8}\\ \Leftrightarrow\dfrac{4\left(3x+2\right)}{24}-\dfrac{6\left(3x-2\right)}{24}-\dfrac{45}{24}=0\\ \Leftrightarrow12x+24-18x+12-45=0\\ \Leftrightarrow-6x-9=0\\ \Leftrightarrow x=-\dfrac{3}{2}\)

2, ĐKXĐ:\(x\ne\pm3\)

\(\dfrac{x+2}{3+x}-\dfrac{x}{3-x}=\dfrac{8x-6}{9-x^2}\\ \Leftrightarrow\dfrac{\left(x+2\right)\left(3-x\right)}{\left(3+x\right)\left(3-x\right)}-\dfrac{x\left(3+x\right)}{\left(3+x\right)\left(3-x\right)}-\dfrac{8x-6}{\left(3+x\right)\left(3-x\right)}=0\\ \Leftrightarrow\dfrac{-x^2+x+6-3x-x^2-8x+6}{\left(3+x\right)\left(3-x\right)}=0\\ \Leftrightarrow-2x^2-10x+12=0\\ \Leftrightarrow x^2+5x-6=0\\ \Leftrightarrow\left(x-1\right)\left(x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-6\left(tm\right)\end{matrix}\right.\)

\(a,\dfrac{3x+2}{6}-\dfrac{3x-2}{4}=\dfrac{15}{8}\)

\(\Leftrightarrow4\left(3x+2\right)-6\left(3x-2\right)=45\)

\(\Leftrightarrow12x+8-18x+12=45\)

\(\Leftrightarrow12x-18x=45-12-8\)

\(\Leftrightarrow-6x=25\)

\(\Leftrightarrow x=\dfrac{-25}{6}\)

Vậy \(S=\left\{\dfrac{-25}{6}\right\}\)

\(b,\dfrac{x+2}{3+x}-\dfrac{x}{3-x}=\dfrac{8x-6}{9-x^2}\left(ĐKXĐ:x\ne3;x\ne-3\right)\)

\(\Leftrightarrow\left(x+2\right)\left(3-x\right)-x\left(3+x\right)=8x-6\)

\(\Leftrightarrow3x-x^2+6-2x-3x-x^2=8x-6\)

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

\(\Leftrightarrow-2x^2-10x=0\)

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

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

Vậy \(S=\left\{0;5\right\}\)

a) (2x - 1)(x^2 - 1 + 1) = 2x^3 - 3x^2 + 2

(2x - 1).x^2 = 2x^3 - 3x^2 + 2

2x^3 - x^2 = 2x^3 - 3x^2 + 2

-x^2 = -3x^2 + 2

2x^2 = 2

x^2 = 1

=> x = 1; -1

b) (x + 2)(x + 2) - (x - 2)(x - 2) = 8x

(x + 2)^2 - (x - 2)^2 = 8x

x^2 + 4x + 4 - x^2 + 4x - 4 = 8x

8x = 8x

=> x thuộc N*

c) (x + 1)(x + 2)(x + 5) - x^3 - 8x^2 = 27

x^3 + 5x^2 + 2x^3 + 10x + x^2 + 5x + 2x + 10x - x^3 - x^2 = 27

17x + 10 = 27

17x = 27 - 10

17x = 17

=> x = 1

d) (x + 1)(x^2 + 2x + 4) - x^3 - 3x^2 + 16 = 0

x^3 + 2x^2 + 4x + x^2 + 2x + 4 - x^3 - 3x^2 + 16 = 0

6x + 20 = 0

6x = -20

x = -20/6

=> x = -10/3

rút gọn à banj

đúng rồi á

Mình ko ghi lại đề , bạn ghi ra xong rồi suy ra như mình nha .

1) \(=>A=\left(6x^2+3x-10x-5\right)-\left(6x^2+14x-9x-21\right)\)

\(=>A=-12x+16\)

2) \(=>B=8x^3+27-8x^3+2=29\)

3)\(=>C=[\left(x-1\right)-\left(x+1\right)]^3=\left(-2\right)^3=-8\)

4)\(=>D=[\left(2x+5\right)-\left(2x\right)]^3=5^3=125\)

5)\(=>E=\left(3x+1\right)^2-\left(3x+5\right)^2+12x+2\left(6x+3\right)\)

\(=>E=\left(3x+1+3x+5\right)\left(3x+1-3x-5\right)+12x+12x+6\)

\(=>E=\left(6x+6\right)\left(-4\right)+24x+6=-24x-24+24x+6=-18\)

6)\(=>F=\left(2x^2+3x-10x-15\right)-\left(2x^2-6x\right)+x+7=-8\)

k cho mik nha ,