a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)

b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)

\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)

\(\Leftrightarrow24x=-13\)

hay \(x=-\dfrac{13}{24}\)

a)  3x – 15 = 25 – 5x 

=> 3x + 5x = 25 + 15

=> 8x = 40

=> x = 5

 b) 3x - 17 = 2x – 7     

=> 3x - 2x = -7 + 17

=> x = 10

 c) 2x – 17 =  – (3x – 18)

=> 2x - 17 = -3x + 18

=> 2x + 3x = 18 + 17

=> 5x = 35

=> x = 7

d) 3x – 14 = 2(x – 9) + 1

=> 3x - 14 = 2x - 18 + 1

=> 3x - 2x = -18 + 1 + 14

=> x = -3

f) (x – 5)² = 9          

\(\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)

a) Ta có: \(3x-15=25-5x\)

\(\Leftrightarrow3x-15-25+5x=0\)

\(\Leftrightarrow8x-40=0\)

\(\Leftrightarrow8x=40\)

hay x=5

Vậy: x=5

b) Ta có: \(3x-17=2x-7\)

\(\Leftrightarrow3x-17-2x+7=0\)

\(\Leftrightarrow x-10=0\)

hay x=10

Vậy: x=10

c) Ta có: \(2x-17=-\left(3x-18\right)\)

\(\Leftrightarrow2x-17=-3x+18\)

\(\Leftrightarrow2x-17+3x-18=0\)

\(\Leftrightarrow5x-35=0\)

\(\Leftrightarrow5x=35\)

hay x=7

Vậy: x=7

d) Ta có: \(3x-14=2\left(x-9\right)+1\)

\(\Leftrightarrow3x-14=2x-18+1\)

\(\Leftrightarrow3x-14-2x+18-1=0\)

\(\Leftrightarrow x+3=0\)

\(\Leftrightarrow x=-3\)

Vậy: x=-3

f) Ta có: \(\left(x-5\right)^2=9\)

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

Vậy: \(x\in\left\{2;8\right\}\)

Bài 2. Tìm x, biết :                                

a) \(3x-15=25-5x\)

\(\Leftrightarrow8x=40\)

\(\Leftrightarrow x=5\)

Vậy x = 5

b) \(3x-17=2x-7\)

\(\Leftrightarrow x=10\)

Vậy x = 10

c) \(2x-17=-\left(3x-18\right)\)

\(\Leftrightarrow2x-17=18-3x\)

\(\Leftrightarrow5x=35\)

\(\Leftrightarrow x=7\)

Vậy x = 7

d) \(3x-14=2\left(x-9\right)+1\)

\(\Leftrightarrow3x-14=2x-18+1\)

\(\Leftrightarrow3x-14=2x-17\)

\(\Leftrightarrow x=-3\)

Vậy x = -3

e) \(\left(x-5\right)^2=9\)

\(\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)

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

Vậy x = {8; 2}

a,2x-(-17)=15

2x+17=15

2x=15-17

2x=-2

=>x=-1

a, 2x -(-17) = 15

2x + 17 = 15 

2x        = 15 - 17 

2x        =(-2)

 x         = (-2) : 2 

x          = (-1)

\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2-2\right)=15\)

\(\Leftrightarrow x^3+27-x^3+2x=15\)

\(\Leftrightarrow2x=-12\)

\(\Leftrightarrow x=-6\)

a) Rút gọn VT = 45x + 8. Từ đó tìm được x = 2 15 .  

b) Rút gọn VT = -25x – 8. Từ đó tìm được x = − 11 25 .

\(\left(x^3+3^3\right)-x\left(x^2-4\right)=15\)

\(x^3+27-x^3+4x=15\)

\(4x=15-27\)

\(4x=-12\)

\(x=-3\)

(x+3)(x²-3x+9)-x(x-2)(x+2)=15

=>x³+27-4x-x³=15

=>(x³-x³)+4x+27=15

=>4x+27=15

=>4x=-12

=>x=-3.Vậy x=-3

câu này sai đề  (x -3)³ -(x-3)(x² +3x+9) +9(x+1)² = 15

g: \(\Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)-4=0\)

\(\Leftrightarrow\left(x^2+6x\right)^2+13\left(x^2+6x\right)+36=0\)

\(\Leftrightarrow\left(x+3\right)^2\left(x^2+6x+4\right)=0\)

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

\(=>x^3+6x^2+12x+8-x^3+27+6x^2+12x+6=15\)

\(=>12x^2+24x+41-15=0\)

\(=>12x^2+24x+26=0\)

\(=>12\left(x^2+2x+1\right)+14=0\)

\(=>12\left(x+1\right)^2+14=0\)

\(=>2[6\left(x+1\right)^2+7]=0\)

\(=>6\left(x+1\right)^2+7=0\)

Mà \(\left(x+1\right)^2\ge0\)nên \(6\left(x+1\right)^2+7>0\)

Vậy ko có giá trị x nào thỏa mãn đề bài

Trả lời:

a, \(\left(3x+1\right)\left(x-3\right)-x\left(3x-14\right)=15\)

\(\Leftrightarrow3x^2-9x+x-3-3x^2+14x=15\)

\(\Leftrightarrow6x-3=15\)

\(\Leftrightarrow6x=18\)

\(\Leftrightarrow x=3\)

Vậy x = 3 là nghiệm của pt.

b, \(\left(x-3\right)^2=9-x^2\)

\(\Leftrightarrow\left(x-3\right)^2-9+x^2=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(x^2-9\right)=0\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=0\end{cases}}}\)

Vậy x = 3; x = 0 là nghiệm của pt.

c, \(\left(2x-\frac{1}{2}\right)^2-\left(1-2x\right)^2=2\)

\(\Leftrightarrow4x^2-2x+\frac{1}{4}-\left(1-4x+4x^2\right)=2\)

\(\Leftrightarrow4x^2-2x+\frac{1}{4}-1+4x-4x^2=2\)

\(\Leftrightarrow2x-\frac{3}{4}=2\)

\(\Leftrightarrow2x=\frac{11}{4}\)

\(\Leftrightarrow x=\frac{11}{8}\)

Vậy x = 11/8 là nghiệm của pt.

d, \(4x^2+4x-3=0\)

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

\(\Leftrightarrow2x\left(2x-1\right)+3\left(2x-1\right)=0\)

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

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

Vậy x = 1/2; x = - 3/2 là nghiệm của pt.