\(3\left(x-1\right)^2+\left(x+5\right)\left(2-3x\right)=-25\)

\(\Leftrightarrow3x^2-6x+3-3x^2-13x+10=-25\)

\(\Leftrightarrow-19x=-38\Leftrightarrow x=2\)

\(\Rightarrow3x^2-6x+3+2x-3x^2+10-15x=-25\\ \Rightarrow-19x=-38\\ \Rightarrow x=2\)

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}

Zaza, tự làm nữa đi a~.

a) \(3x\left(x-8\right)+16=2x\)

\(\Rightarrow3x^2-24x+16=2x\)

\(\Rightarrow3x^2-26x+16=0\)

\(\Rightarrow\left(3x^2-24x\right)-\left(2x-16\right)=0\)

\(\Rightarrow3x\left(x-8\right)-2\left(x-8\right)=0\)

\(\Rightarrow\left(x-8\right)\left(3x-2\right)=0\)

Để đẳng thức xảy ra \(\Rightarrow\left[\begin{array}{nghiempt}x-8=0\\3x-2=0\end{array}\right.\)\(\Rightarrow x\in\left\{8;\frac{2}{3}\right\}\)

b) \(\left(5-x\right)^2=25=5^2=\left(-5\right)^2\)

\(\Rightarrow5-x\in\left\{\pm5\right\}\Rightarrow x\in\left\{0;10\right\}\)

2x + 3x = \(5^6\): 25

5x = \(5^4\)

5x = 625

x = 625 : 5

x = 125

\(2x+3x=5^6:25\)

\(2x+3x=5^6:5^2\)

\(2x+3x=5^4\)

\(\Leftrightarrow x.5=5^4\)

\(\Leftrightarrow x=5^3\)

2x+3x=\(5^6\):25

2x+3x =15625 : 25

2x+3x= 625

x(2+3)=625

\(x\)*5=625

\(x\)    =625 :5

 \(x\)   =125

b: \(\Leftrightarrow\left(x-5\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-6\end{matrix}\right.\)

a, 3x - 35 = 25 + 3 . (-5)

3x - 35 = 25 + (-15)

3x - 35 = 10

3x = 10 + 35

3x = 45

x = 45 : 3

x = 15

b, | x - 2 | = 0

x - 2 = 0

x = 0 + 2

x = 2

c, \(\frac{x}{5}=\frac{5}{6}+\frac{-13}{30}\)

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

\(x=\frac{2.5}{5}\)

\(x=2\)

\(\left(\dfrac{2}{5}-3x\right)^2-\dfrac{1}{5}=\dfrac{4}{25}\)

\(\Rightarrow\left(\dfrac{2}{5}-3x\right)^2=\dfrac{4}{25}+\dfrac{1}{5}=\dfrac{9}{25}\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{5}-3x=\dfrac{3}{5}\\\dfrac{2}{5}-3x=-\dfrac{3}{5}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}3x=-\dfrac{1}{5}\\3x=1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{15}\\x=\dfrac{1}{3}\end{matrix}\right.\)