a: \(4x^3+12=120\)

=>\(4x^3=108\)

=>\(x^3=27=3^3\)

=>x=3

b: \(\left(x-4\right)^2=64\)

=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)

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

=>\(\left(x+1\right)^3=25+2=27\)

=>x+1=3

=>x=2

d: 136-(x+5)^2=100

=>(x+5)^2=36

=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)

e: \(4^x=16\)

=>\(4^x=4^2\)

=>x=2

f: \(7^x\cdot3-147=0\)

=>\(3\cdot7^x=147\)

=>\(7^x=49\)

=>x=2

g: \(2^{x+3}-15=17\)

=>\(2^{x+3}=32\)

=>x+3=5

=>x=2

h: \(5^{2x-4}\cdot4=10^2\)

=>\(5^{2x-4}=\dfrac{100}{4}=25\)

=>2x-4=2

=>2x=6

=>x=3

i: (32-4x)(7-x)=0

=>(4x-32)(x-7)=0

=>4(x-8)*(x-7)=0

=>(x-8)(x-7)=0

=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)

k: (8-x)(10-2x)=0

=>(x-8)(x-5)=0

=>\(\left[{}\begin{matrix}x-8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)

m: \(3^x+3^{x+1}=108\)

=>\(3^x+3^x\cdot3=108\)

=>\(4\cdot3^x=108\)

=>\(3^x=27\)

=>x=3

n: \(5^{x+2}+5^{x+1}=750\)

=>\(5^x\cdot25+5^x\cdot5=750\)

=>\(5^x\cdot30=750\)

=>\(5^x=25\)

=>x=2

\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

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

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

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

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

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

\(\Leftrightarrow-8x=-8\)

hay x=1

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

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy: \(x=\dfrac{1}{2}\)

===========

b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

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

\(\Leftrightarrow3x=13\)

\(\Leftrightarrow x=\dfrac{13}{3}\)

Vậy: \(x=\dfrac{13}{3}\)

==========

c/  \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

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

\(\Leftrightarrow-8x=-8\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)

\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)

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

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

\(\Leftrightarrow x=\dfrac{2}{7}\)

Vậy: \(x=\dfrac{2}{7}\)

==========

f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

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

\(\Leftrightarrow x=-2\)

Vậy: \(x=-2\)

a.

\(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)

\(6x^2+21x-2x-7-6x^2+5x-6x+5=16\)

\(\left(6x^2-6x^2\right)+\left(21x-2x+5x-6x\right)-\left(7-5\right)=16\)

\(18x-2=16\)

\(18x=16+2\)

\(18x=18\)

\(x=\frac{18}{18}\)

\(x=1\)

b.

\(\left(10x+9\right)x-\left(5x-1\right)\left(2x+3\right)=8\)

\(10x^2+9x-10x^2-15x+2x+3=8\)

\(\left(10x^2-10x^2\right)-\left(15x-9x-2x\right)+3=8\)

\(-4x=8-3\)

\(-4x=5\)

\(x=-\frac{5}{4}\)

c.

\(\left(3x-5\right)\left(7-5x\right)+\left(5x+2\right)\left(3x-2\right)-2=0\)

\(21x-15x^2-35+25x+15x^2-10x+6x-4-2=0\)

\(\left(15x^2-15x^2\right)+\left(25x+21x-10x+6x\right)-\left(35+4+2\right)=0\)

\(42x=41\)

\(x=\frac{41}{42}\)

- mọi người ơi giúp mình với ạ. ai mình cx cho đúng

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

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

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

a) \(5x\left(\frac{1}{5}x-2\right)+3\left(6-\frac{1}{3}x^2\right)=12\)

=> \(x^2-10x+18-x^2=12\)

=> -10x + 18 = 12

=> -10x = -6

=> -5x = -3

=> x = 3/5

b) 7x(x - 2) - 5(x - 1) = 7x² + 3

=> 7x² - 14x - 5x + 5 = 7x² + 3

=> 7x² - 14x - 5x + 5 - 7x² - 3 = 0

=> -19x + 2 = 0

=> -19x = -2

=> x = \(\frac{2}{19}\)

c) 2(5x - 8) - 3(4x - 5) = 4(3x - 4) + 11

=> 10x - 16 - 12x + 15 = 12x - 16 + 11

=> 10x - 16 - 12x + 15 - 12x + 16 - 11 = 0

=> (10x - 12x - 12x) + (-16 + 15 + 16 - 11) = 0

=> -14x + 4 = 0

=> -14x = -4

=> -7x = -2

=> x = 2/7

Đăng ít một thôi bạn :v

a) 3x - (3 - 2x) = 0

3x - 3 + 2x = 0

5x - 3 = 0

5x = 0 + 3

5x = 3

x = 3/5

b) (x + 2).3 - 4x.3 = 0

3.(x + 2) - 12.x = 0

3[x + 2 - (4x)] = 0

x + 2 - 4 = 0

-3x + 2 = 0

-3x = 0 - 2

-3x = -2

x = 2/3

c) (x - 2)(x - 4)(1 - 7x) = 0

x - 2 = 0 hoặc x - 4 = 0 hoặc 1 - 7x = 0

x = 0 + 2         x = 0 + 4          -7x = 0 - 1

x = 2               x = 4                 -7x = -1

                                                 x = 1/7

d) 4x² - 1/4 = 0

4x² = 0 + 1/4

4x² = 1/4

x² = 1/4 : 4

x² = 1/16

x² = (1/4)²

x = 1/4 hoặc x = -1/4

e) -3x² + 48 = 0

3x² - 48 = 0

3x² = 0 + 48 

3x² = 48

x² = 48 : 3

x² = 16

x² = 4²

x = 4 hoặc x = -4

g) 3(1/2 - 1/3x)³ - 1/9 = 0

3(1/2 - x/3)³ - 1/9 = 0

3(1/2 - x/3)³ = 0 + 1/9

3(1/2 - x/3)³ = 1/9

(1/2 - x/3)³ = 1/9 : 3

(1/2 - x/3)³ = 1/27

(1/2 - x/3)³ = (1/3)³

1/2 - x/3 = 1/3

-x/3 = 1/3 - 1/2

-x/3 = -1/6

-x = -1/6.3

-x = -3/6 = -1/2

x = -1/2

m) 4x³ + 5x⁴ = 0

x³(4 + 5x) = 0

x = 0 hoặc 4 + 5x = 0

x = 0          5x = 0 - 4

                  5x = -4

                  x = -4/5

h) -x³ + 1/64x = 0

-x³ + x/64 = 0

x/64 - x³ = 0

x(1/64 - x³) = 0

x = 0 hoặc 1/64 - x² = 0

x = 0           -x² = 0 - 1/64

                   -x² = -1/64

                    x² = 1/64 = -+1/8

k) (x² + 1)² + 3x(x² + 1) + 2 = 0

x⁴ + 2x² + 1 + 3x³ + 3x + 2 = 0

x⁴ + 2x² + 3 + 3x³ + 3x = 0

(x³ + 2x² + 3)(x + 1) = 0

Mà x³ + 2x² + 3 # 0 nên

x + 1 = 0

x = -1

a: \(P=-3x^3+5x\)

\(=x\cdot\left(-3x^2\right)+x\cdot5\)

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

b: \(Q=\left(2x-1\right)+\left(x-2\right)\left(2x-1\right)\)

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

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

c: \(R=4-16x^2\)

\(=4\cdot1-4\cdot4x^2\)

\(=4\left(1-4x^2\right)\)

\(=4\left(1-2x\right)\left(1+2x\right)\)

d: \(S=36-4x^2\)

\(=4\cdot9-4\cdot x^2\)

\(=4\left(9-x^2\right)\)

\(=4\left(3-x\right)\left(3+x\right)\)

e: \(T=8x^3-1\)

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

\(=\left(2x-1\right)\left(4x^2+2x+1\right)\)

f: \(Q=8-x^3\)

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

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

g: \(N=64-x^3\)

\(=4^3-x^3\)

\(=\left(4-x\right)\left(16+4x+x^2\right)\)