a: =>3^x=3^4*3=3^5

=>x=5

b: =>\(2^{x+1}=2^5\)

=>x+1=5

=>x=4

c: \(\Leftrightarrow3^{x+2-3}=3\)

=>x-1=1

=>x=2

d: \(\Leftrightarrow x^2=\dfrac{32}{2}=16\)

=>x=4 hoặc x=-4

e: (2x-1)^4=81

=>2x-1=3 hoặc 2x-1=-3

=>2x=4 hoặc 2x=-2

=>x=-1 hoặc x=2

f: (2x-6)^4=0

=>2x-6=0

=>x-3=0

=>x=3

a) \(3^x=81\cdot3\)

\(\Rightarrow3^x=3^4\cdot3\)

\(\Rightarrow3^x=3^5\)

\(\Rightarrow x=5\)

b) \(2^{x+1}=32\)

\(\Rightarrow2^{x+1}=2^5\)

\(\Rightarrow x+1=5\)

\(\Rightarrow x=4\)

c) \(3^{x+2}:27=3\)

\(\Rightarrow3^{x+2}:3^3=3\)

\(\Rightarrow3^{x+2-3}=3\)

\(\Rightarrow3^{x-1}=3\)

\(\Rightarrow x-1=1\)

\(\Rightarrow x=2\)

d) \(2x^2=32\)

\(\Rightarrow x^2=16\)

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

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

e) \(\left(2x-1\right)^4=81\)

\(\Rightarrow\left(2x-1\right)^4=3^4\)

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

\(\Rightarrow\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\)

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

f)  \(\left(2x-6\right)^4=0\)

\(\Rightarrow2x-6=0\)

\(\Rightarrow2x=6\)

\(\Rightarrow x=6:2\)

\(\Rightarrow x=3\)

Bài 1:

a: \(5x\left(3x^2-4x+2\right)=5x\cdot3x^2-5x\cdot4x+5x\cdot2\)

\(=15x^3-20x^2+10x\)

b: \(\left(x^2-3x\right)\left(3x^2-x+4\right)\)

\(=x^2\cdot3x^2-x^2\cdot x+x^2\cdot4-3x\cdot3x^2+3x\cdot x-3x\cdot4\)

\(=3x^4-x^3+4x^2-9x^3+3x^2-12x=3x^4-10x^3+7x^2-12x\)

Bài 2:

a: \(x^3-6x^2+12x=0\)

=>\(x\left(x_{}^2-6x+12\right)=0\)

=>x=0

b: \(x^3+9x^2+27x+27=0\)

=>\(x^3+3\cdot x^2\cdot3+3\cdot x\cdot3^2+3^3=0\)

=>\(\left(x+3\right)^3=0\)

=>x+3=0

=>x=-3

1)

a) \(=15x^3-20x^2+10x\)

b) \(=3x^4-x^3+4x^2-9x^3+3x-12x=3x^4-10x^3+4x^2-9x\)

2) 

a) \(\Rightarrow x\left(x^2-6x+12\right)=0\)

\(\Rightarrow x=0\)(do \(x^2-6x+12=\left(x^2-6x+\dfrac{36}{4}\right)+3=\left(x-\dfrac{6}{2}\right)^2+3\ge3>0\))

b) \(\Rightarrow\left(x+3\right)^3=0\Rightarrow x=-3\)

(3x²-5x+2)+(3x²+5x)= bao nhiêu ạ

Giúp em vs ạ . Em cảm ơn

-27/5 . -4/15 + -4/15 . ( 2021 - 27/5)

=-4/15.[(-27/5)+2021-27/5]

=-4/15.2012,2

= -40244/75

 -27/5 . -4/15 + -4/15 . ( 2021 - 27/5)

=-4/15.[(-27/5)+2021-27/5]

=-4/15.2012,2

= -40244/75

a: x(2x-7)+14=4x

=>x(2x-7)-4x+14=0

=>x(2x-7)-2(2x-7)=0

=>(2x-7)(x-2)=0

=>\(\left[\begin{array}{l}2x-7=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac72\\ x=2\end{array}\right.\)

b: \(25x^3=2x\)

=>\(25x^3-2x=0\)

=>\(x\left(25x^2-2\right)=0\)

TH1: x=0

=>x=0

TH2: \(25x^2-2=0\)

=>\(x^2=\frac{2}{25}\)

=>\(\left[\begin{array}{l}x=\frac{\sqrt2}{5}\\ x=-\frac{\sqrt2}{5}\end{array}\right.\)

c: \(\left(x-5\right)^3=x^3-125\)

=>\(x^3-15x^2+75x-125=x^3-125\)

=>\(-15x^2+75x=0\)

=>-15x(x-5)=0

=>x(x-5)=0

=>\(\left[\begin{array}{l}x=0\\ x-5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=5\end{array}\right.\)

d: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)

=>\(x^2\left(x-1\right)-4\left(x^2-2x+1\right)=0\)

=>\(x^2\left(x-1\right)-4\left(x-1\right)^2=0\)

=>\(\left(x-1\right)\left(x^2-4x+4\right)=0\)

=>\(\left(x-1\right)\left(x-2\right)^2=0\)

=>\(\left[\begin{array}{l}x-1=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=2\end{array}\right.\)

a: x(2x-7)+14=4x

=>x(2x-7)-4x+14=0

=>x(2x-7)-2(2x-7)=0

=>(2x-7)(x-2)=0

=>\(\left[\begin{array}{l}2x-7=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac72\\ x=2\end{array}\right.\)

b: \(25x^3=2x\)

=>\(25x^3-2x=0\)

=>\(x\left(25x^2-2\right)=0\)

TH1: x=0

=>x=0

TH2: \(25x^2-2=0\)

=>\(x^2=\frac{2}{25}\)

=>\(\left[\begin{array}{l}x=\frac{\sqrt2}{5}\\ x=-\frac{\sqrt2}{5}\end{array}\right.\)

c: \(\left(x-5\right)^3=x^3-125\)

=>\(x^3-15x^2+75x-125=x^3-125\)

=>\(-15x^2+75x=0\)

=>-15x(x-5)=0

=>x(x-5)=0

=>\(\left[\begin{array}{l}x=0\\ x-5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=5\end{array}\right.\)

d: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)

=>\(x^2\left(x-1\right)-4\left(x^2-2x+1\right)=0\)

=>\(x^2\left(x-1\right)-4\left(x-1\right)^2=0\)

=>\(\left(x-1\right)\left(x^2-4x+4\right)=0\)

=>\(\left(x-1\right)\left(x-2\right)^2=0\)

=>\(\left[\begin{array}{l}x-1=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=2\end{array}\right.\)

A) 4¹⁰ . 8¹⁵ = (2²)¹⁰ . (2³)¹⁵ = 2²⁰ . 2⁴⁵ = 2²⁰⁺⁴⁵=2⁶⁵

B) 4¹⁵ . 5³⁰ = (2²)¹⁵ . 5³⁰ = 2³⁰ . 5³⁰ = (2 . 5)³⁰ = 10³⁰

C) 27¹⁶ : 9¹⁰ = (3³)¹⁶ : (3²)¹⁰ = 3⁴⁸ : 3²⁰ =  3^48-20=3²⁸

20²-15².3[5²+22+2022⁰