a: \(3x\left(x-3\right)+4x-12=0\)

=>\(3x\left(x-3\right)+\left(4x-12\right)=0\)

=>\(3x\left(x-3\right)+4\left(x-3\right)=0\)

=>\(\left(x-3\right)\left(3x+4\right)=0\)

=>\(\left[{}\begin{matrix}x-3=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{4}{3}\end{matrix}\right.\)

b: Sửa đề:\(\left(x+1\right)\left(x^2-x+1\right)-x^3+2x=17\)

\(\Leftrightarrow x^3+1-x^3+2x=17\)

=>2x+1=17

=>2x=17-1=16

=>\(x=\dfrac{16}{2}=8\)

c: \(\left(x-3\right)\left(x+5\right)+\left(x-1\right)^2-6x^4y^2:3x^2y^2=15x\)

=>\(x^2+2x-15+x^2-2x+1-2x^2=15x\)

=>\(15x=-14\)

=>\(x=-\dfrac{14}{15}\)

a)   \(\left(x+3\right)^3-x.\left(3x+1\right)^2+\left(2x+1\right).\left(4x^2-2x+1\right)-3x^2=54\)

\(\Leftrightarrow x^3+9x^2+27x+27-x.\left(9x^2+6x+1\right)+8x^3+1-3x^2=54\)

\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1-3x^2=54\)

\(\Leftrightarrow26x+28=54\Leftrightarrow26x=54-28\Leftrightarrow26x=26\Leftrightarrow x=1\)

Vậy nghiệm của phương trình là x=1

b)   \(\left(x-3\right)^3-\left(x-3\right).\left(x^2+3x+9\right)+6.\left(x+1\right)^2+3x^2=-33\)

\(\Leftrightarrow x^3-9x^2+27x-27-\left(x^3-27\right)+6.\left(x^2+2x+1\right)+3x^2=-33\)

\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6x^2+12x+6+3x^2=-33\)

\(\Leftrightarrow27x+12x+6=-33\Leftrightarrow39x=-33-6\Leftrightarrow39x=-39\Leftrightarrow x=-1\)

Vậy nghiệm của phương trình là x = -1

Trần Anh: Hí hí =)) ÀI LỚP DIU CHIU CHIU CHÍU :3 CẢM ƠN PẠN NHIỀU NHÁ ;) ;) ;) 

Bài 1:

a) \(2\cdot3\cdot2\cdot3\cdot2\cdot3=2^3\cdot3^3=6^3\)

b) \(100\cdot100\cdot100=100^3=\left(10^2\right)^3=10^6\)

c) \(2x\cdot2x\cdot2x=\left(2x\right)^3=8x^3\)

d) \(2\cdot2^3\cdot2^5=2^{1+3+5}=2^9\)

e) \(3^{10}\cdot3^5\cdot3^4=3^{10+5+4}=3^{19}\)

Bài 2:

\(40-x=2^6\cdot2^2\)

\(\Rightarrow40-x=2^8\)

\(\Rightarrow40-x=256\)

\(\Rightarrow x=40-256\)

\(\Rightarrow x=-216\)

b) \(3^2\cdot3^x=81\)

\(\Rightarrow3^{2+x}=3^4\)

\(\Rightarrow2+x=4\)

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

c) \(2^x=512\)

\(\Rightarrow2^x=2^9\)

\(\Rightarrow x=9\)

d) \(x^5=243\)

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

\(\Rightarrow x=3\)

Bài 3:

a) \(3^6=3\cdot3\cdot3\cdot3\cdot3\cdot3=729\)

b) \(8^3=\left(2^3\right)^3=2^9=512\)

c) \(3^3\cdot75+3^3\cdot25=3^3\cdot\left(75+25\right)=3^3\cdot100=27\cdot100=2700\)

d) \(2^3\cdot3-\left(1^{10}+8\right):3=2^3\cdot3-9:3=2^3\cdot3-3\cdot3:3=3\cdot\left(2^3-3:3\right)=3\cdot\left(8-1\right)=21\)

e) \(32-\left[4+\left(5\cdot3^2-42\right)\right]-14=18-\left[4+\left(45-42\right)\right]\)

\(=18-\left(4+3\right)\)

\(=18-7=11\)

2:

a: =>40-x=256

=>x=40-256=-216

b: =>x+2=4

=>x=2

c: =>2^x=2^9

=>x=9

d; =>x^5=3^5

=>x=3

tìm các số nguyên x để mỗi phân số sau đây là số nguyên

Ngô Hải Nam ơi bn trả lời giúp mik ik

bài đó là bài 4^* tìm các số nguyên x để mỗi phân số sau đây là số nguyên

a) Ta có: \(\left(2x+7\right)^2=\left(x+3\right)^2\)

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

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

\(\Leftrightarrow\left(x+4\right)\cdot\left(3x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\3x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\3x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-\dfrac{10}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{-4;-\dfrac{10}{3}\right\}\)

b) Ta có: \(\left(4x+14\right)^2=\left(7x+2\right)^2\)

\(\Leftrightarrow\left(4x+14\right)^2-\left(7x+2\right)^2=0\)

\(\Leftrightarrow\left(4x+14-7x-2\right)\left(4x+14+7x+2\right)=0\)

\(\Leftrightarrow\left(-3x+12\right)\left(11x+16\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-3x+12=0\\11x+16=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x=-12\\11x=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{16}{11}\end{matrix}\right.\)Vậy: \(S=\left\{4;-\dfrac{16}{11}\right\}\)

(2x+7)²=(x+3)²

=>(2x+7)²-(x+3)²=0

=>(2x+7-x-3)(2x+7+x+3)=0

=>(x-4)(3x+10)=0

=>x-4=0 hoặc 3x+10=0

TH1:x-4=0=>x=4

TH2:3x+10=0=>x=-10/3

(4x+14)²=(7x+2)²

(4x+14)²-(7x+2)²=0

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

(-3x+12)(11x+16)=0

TH1:-3x+12=0=>x=4

TH2:11x+16=0=>x=-16/11

Đề bài là gì bạn nhỉ ?

\(\left(\dfrac{1}{3}+\dfrac{1}{6}\right)\cdot2^{x+4}-2^x=2^{13}-2^{10}\)

\(\Rightarrow\dfrac{1}{2}\cdot2^{x+4}-2^x=2^{13}-2^{10}\)

\(\Rightarrow2^{x+3}-2^x=2^{13}-2^{10}\)

\(\Rightarrow x+3=13;x+0=10\)

\(\Rightarrow x=10\)

(\(\dfrac{1}{3}\) +\(\dfrac{1}{6}\) ) . 2x+4 - 2x = 213 - 210

(\(\dfrac{2}{6}\)  + \(\dfrac{1}{6}\)) .   \(2^{x+4}\) -   \(2^x\) = 8192 - 1024

\(\dfrac{3}{6}\) . 2x . \(2^4\) -\(2^x\) = 7168

8 . 2x  - 2x . 1   = 7168

 2x . ( 8 - 1 ) = 7168

 2x . 7 = 7168

 2x = 7168 : 7

 2x = 1024

 2x = \(2^{10}\)

⇒ x = 10

\(a,\Leftrightarrow x^3=\dfrac{20}{3}\Leftrightarrow x=\sqrt[3]{\dfrac{20}{3}}\\ b,\Leftrightarrow x-1=9\Leftrightarrow x=10\\ c,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ d,\Leftrightarrow2x+1=5\Leftrightarrow x=2\\ e,\Leftrightarrow2x-4=4\Leftrightarrow x=4\)

Câu a) xem lại đề giùm nhé em

b) \(\left(x-1\right)^3=9^3\)

\(x-1=9\)

\(x=10\)

Vậy \(x=10\)

c) \(\left(x-1\right)^2=25\)

\(x-1=5\) hoặc \(x-1=-5\)

* \(x-1=5\)

\(x=6\)

* \(x-1=-5\)

\(x=-4\)

Vậy \(x=-4\); \(x=6\)

d) \(\left(2x+1\right)^3=125\)

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

\(2x+1=5\)

\(2x=4\)

\(x=2\)

Vậy \(x=2\)

e) Sửa đề: \(\left(2x+4\right)^3=64\)

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

\(2x+4=4\)

\(2x=0\)

\(x=0\)

Vậy \(x=0\)

3 . ( 2x - 1 ) - 2 = 13 

3 . ( 2x - 1 ) = 12 + 3 

3 . ( 2x - 1 ) = 15 

      2x - 1 = 15 : 3 

      2x - 1 = 5 

     2x = 5 + 1 = 6 

x = 6 : 2 = 3 

Vậy x = 3

\(3\left(2x-1\right)-2=13\)

\(3\left(2x-1\right)=15\)

\(2x-1=5\)

\(2x=6\)

\(x=3\)

3.(2x - 1 ) - 2 = 13

\(\Leftrightarrow\) 3.( 2x - 1) = 15

\(\Leftrightarrow\) 2x- 1 = 15 : 3

\(\Leftrightarrow\) 2x-1 = 5

\(\Leftrightarrow\)2x = 6

\(\Leftrightarrow\) x = 3

Vậy x = 3