=>10x=-100

hay x=-10

⇒3x+7x=-64-36

⇒10x=-100

⇒x=-10

\(3x+36=-7x-64\)

\(10x=-100\)

\(x=-10\)

\(-5x-178=14x+145\)

\(-19x=323\)

\(x=-17\)

Bài dưới sai nha, giờ mới đúng nek

\(\frac{2^{7x}}{2^{3x}}=64\)

\(\frac{2^{3x+4x}}{2^{3x}}=64\)

\(\frac{2^{3x}.2^{4x}}{2^{3x}}=64\)

\(\frac{2^{3x}}{2^{3x}}.2^{4x}=64\)

⇒ 2^4x = 64

2^4x = 2⁶

⇒ 4x = 6

x = \(\frac{6}{4}\)

NHỚ NHA BÀI HỒI NÃY SAI NHA - CHÚC BẠN HỌC TỐT t_t

\(\frac{2^{7x}}{2^{3x}}=64\)

\(\frac{2^7.2^x}{2^3.2^x}=64\)

\(\frac{2^x}{2^x}=64:\frac{2^7}{2^3}\)

\(\frac{2^x}{2^x}=64.\frac{2^3}{2^7}\)

\(\frac{2^x}{2^x}=\frac{64.8}{2^7}=\frac{512}{128}=4\)

⇒ 1 = 4

⇒ x ∈ ∅

CHÚC BẠN HỌC TỐT ^_^

(x-1)(2x^2-8)=0

\(\Leftrightarrow\left(x-1\right)\left(2x^2-8\right)=0\\ \left(2x^3-8x-2x^2+8\right)=0\)

\(\Leftrightarrow2x\left(x-1\right)-8\left(x-1\right)=0\)

\(\Leftrightarrow x=1;x=\dfrac{8}{2}\)

3x^2-8x+5=0

áp dụng công thức bậc 2 ta có:

\(x=\dfrac{-\left(-8\right)\pm\sqrt{\left(-8\right)^2-4.3.5}}{2.3}\)

\(\Rightarrow x=\dfrac{5}{3};x=1\)

(7x-1).2x-7x+1=0

\(\Leftrightarrow\left(7x-1\right)\left(2x-1\right)=0\)

\(\Leftrightarrow x=\dfrac{1}{7};x=\dfrac{1}{2}\)

d: \(\Leftrightarrow\left(4x+2\right)\left(x-1\right)-\dfrac{1}{2}x\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(4x+2-\dfrac{1}{2}x\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-\dfrac{7}{2}x+2\right)=0\)

=>x=1 hoặc x=4/7

e: \(\Leftrightarrow2\left(5x-2\right)=3\left(5-3x\right)\)

=>10x-4=15-9x

=>19x=19

hay x=1

f: \(\Leftrightarrow\dfrac{2x-1}{x-1}+1=\dfrac{1}{x-1}\)

=>2x-1+x-1=1

=>3x-2=1

hay x=1(loại)

g: =>1+3x-6=3-x

=>3x-5-3+x=0

=>4x-8=0

=>x=2(loại)

a) \(4x-3=11-3x\)

\(\Leftrightarrow4x+3x=11+3\)

\(\Leftrightarrow7x=14\)

\(\Leftrightarrow x=2\)

Vậy .............

b) \(x^3-4x^2+3x=0\)

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

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

\(\Leftrightarrow x\left[x\left(x-1\right)-3\left(x-1\right)\right]=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=3\end{matrix}\right.\)

Vậy .................

P/s: câu c bn gõ lại dc ko

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

1, x^2 +5x+4x+20

= x(x+4)+5(x+4)

= (x+5).(x+4)

2,x^2 -x -20

= x^2 -5x+4x-20

= x(x-5)+4(x-5)

=(x+4).(x-5)

3,3x^2 +3x -2x -2 

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

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

4, 2x^2-4x-x-2

=2x(x-2)-x-2

=(2x-1)(x+2)

5,6x^2-2x+9x-3

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

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

6,3x^2 +2x+9x+6

=3x(x+3)+2(x+3)

=(3x+2)(x+3)

7,= 2(x^2-10x+3)

8,=x^4 +4^3

mk làm luôn ko chép đề nha bn 

Cảm ơn bạn kb nha

ko có j đâu bn

\(x^2+9x+20\)

\(\Leftrightarrow x^2+4x+5x+20\)

\(\Leftrightarrow\left(x^2+4x\right)+\left(5x+20\right)\)

\(\Leftrightarrow x\left(x+4\right)+5\left(x+4\right)\)

\(\Leftrightarrow\left(x+4\right)\left(x+5\right)\)

\(x^2-x-20\)

\(\Leftrightarrow x^2-5x+4x-20\)

\(\Leftrightarrow\left(x^2-5x\right)+\left(4x-20\right)\)

\(\Leftrightarrow x\left(x-5\right)+4\left(x-5\right)\)

\(\Leftrightarrow\left(x-5\right)\left(x+4\right)\)

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

\(\Leftrightarrow3x^2+3x-2x-2\)

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

\(\Leftrightarrow3x\left(x+1\right)-2x\left(x+1\right)\)

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

\(\)