\(1,\\ a,2^x=16=2^4\Rightarrow x=4\\ b,3^{x+1}=9^x=3^{2x}\\ \Rightarrow x+1=2x\Rightarrow x=1\\ c,2^{3x+2}=4^{x+5}=2^{2\left(x+5\right)}\\ \Rightarrow3x+2=2x+10\Rightarrow x=8\\ d,3^{2x-1}=243=3^5\\ \Rightarrow2x-1=5\Rightarrow x=3\\ 2,\\ a,2^{225}=8^{75}< 9^{75}=3^{150}\\ b,2^{91}=\left(2^{13}\right)^7=8192^7>3125^7=\left(5^5\right)^7=5^{35}\\ c,99^{20}=\left(99^2\right)^{10}< \left(99\cdot101\right)^{10}=9999^{10}\\ 3,\\ a,12^8\cdot9^{12}=2^{16}\cdot3^8\cdot3^{24}=2^{16}\cdot3^{32}=\left(2\cdot3^2\right)^{16}=18^{16}\\ b,75^{20}=\left(3\cdot5^2\right)^{20}=3^{20}\cdot5^{40}=\left(3^{20}\cdot5^{10}\right)\cdot5^{30}=\left(3^2\cdot5\right)^{10}\cdot5^{30}=45^{10}\cdot5^{30}\)

Bài 1:

a) \(\Rightarrow2^x=2^4\Rightarrow x=4\)

b) \(\Rightarrow3^{x+1}=3^{2x}\Rightarrow x+1=2x\Rightarrow x=1\)

c) \(\Rightarrow2^{3x+2}=2^{2x+10}\Rightarrow3x+2=2x+10\Rightarrow x=8\)

d) \(\Rightarrow3^{2x-1}=3^5\Rightarrow2x-1=5\Rightarrow x=3\)

Bài 2:

a) \(2^{225}=\left(2^3\right)^{75}=8^{75}< 9^{75}=\left(3^2\right)^{75}=3^{150}\)

b) \(2^{91}=\left(2^{13}\right)^7=8192^7>3125^7=\left(5^5\right)^7=5^{35}\)

c) \(99^{20}=\left(99^2\right)^{10}=9801^{10}< 9999^{10}\)

Bài 3:

a) \(12^8.9^{12}=\left(4.3\right)^8.9^{12}=4^8.3^8.9^{12}=2^{16}.9^4.9^{12}=2^{16}.9^{16}=\left(2.9\right)^{16}=18^{16}\)

b) \(75^{20}=\left(75^2\right)^{10}=5625^{10}=\left(45.125\right)^{10}=45^{10}.125^{10}=45^{10}.5^{30}\)

a) \(x^3-9x^2+15x+25\)

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

\(=x^2\left(x+1\right)-10x\left(x+1\right)+25\left(x+1\right)\)

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

b) \(x^8-4x^2-11x+30:\text{đề sai thì phải bạn ạ!}\)

c) \(x^4+x^3-22x^2+15x-36\)

\(=\left(2x^4-6x^3\right)+\left(7x^3-21x^2\right)-\left(x^2-3x\right)+12x-36\)

\(=2x^3\left(x-3\right)+7x^2\left(x-3\right)-x\left(x-3\right)+12\left(x-3\right)\)

\(=\left(x-3\right)\left(2x^3+8x^2-x^2-x+12\right)\)

\(=\left(x-3\right)\left[\left(2x^3+8x^2\right)-\left(x^2+4x\right)+3\left(x+4\right)\right]\)

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

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

1) 120 + ( 999 - 9x) : 6) = 20

                999 - 9x : 6    = 140

                  999- 9x        = 840

                           9x       = 999 - 840

                           x          = 17 dư 7

giúp mình với nha chiều nộp rồi

câu 1:

\(\left(9x+2\right)\cdot3=50\)

\(\Leftrightarrow9x+2=\frac{50}{3}\)

\(\Leftrightarrow9x=\frac{44}{3}\)

\(\Leftrightarrow x=\frac{44}{27}\)

Câu2

\(\left(9x+5\right)\cdot4=200\)

\(\Leftrightarrow9x+5=50\)

\(\Leftrightarrow9x=45\)

\(\Leftrightarrow x=5\)

Câu 1 :\(\left(9x+2\right)3=50\)

\(\Rightarrow9x+2=\frac{50}{3}\)

\(\Rightarrow9x=\frac{50}{3}-2\)

\(\Rightarrow9x=\frac{44}{3}\)

\(\Rightarrow x=\frac{44}{3}:9\)

\(\Rightarrow x=\frac{44}{27}\)

Câu 2

\(\left(9x+5\right)4=200\)

\(\Rightarrow\left(9x+5\right)=\frac{200}{4}\)

\(\Rightarrow9x+5=50\)

\(\Rightarrow9x=50-5\)

\(\Rightarrow9x=45\)

\(\Rightarrow x=\frac{45}{9}\)

\(\Rightarrow x=5\)

\(x^3=9x\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)

a: =>(3x+1)(3x-1)-(3x+1)(2x-3)=0

=>(3x+1)(3x-1-2x+3)=0

=>(3x+1)(x+2)=0

=>x=-1/3 hoặc x=-2

b: =>(3x+1)(6x+2)-(3x+1)(x-2)=0

=>(3x+1)(6x+2-x+2)=0

=>(3x+1)(5x+4)=0

=>x=-1/3 hoặc x=-4/5

Lời giải:
c. 

$(x-3)(x^2+3x+9)-x^3=x^3-3^3-x^3=-27$ không phụ thuộc vào giá trị của biến

Ta có đpcm

d. 

$(3x+2)(9x^2-6x+4)-9x(3x^2+1)+9x$

$=(3x)^3+2^3-27x^3-9x+9x$

$=27x^3+8-27x^3=8$ không phụ thuộc vào giá trị của biến 

Ta có đpcm

c) Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-x^3\)

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

=-27

d) Ta có: \(\left(3x+2\right)\left(9x^2-6x+4\right)-9x\left(3x^2+1\right)+9x\)

\(=27x^3+8-27x^3-9x+9x\)

=8