a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)

\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)

\(\Leftrightarrow-9x=18\)

hay x=-2

Vậy: S={-2}

b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)

\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)

\(\Leftrightarrow14x=7\)

hay \(x=\dfrac{1}{2}\)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)

\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)

\(\Leftrightarrow5.2x=-6.5\)

hay \(x=-\dfrac{5}{4}\)

Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)

d) Ta có: \(\left(x+3\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)=6\)

\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)

\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)

\(\Leftrightarrow2x+16=6\)

\(\Leftrightarrow2x=-10\)

hay x=-5

Vậy: S={-5}

e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)

\(\Leftrightarrow14x=0\)

hay x=0

Vậy: S={0}

`a)|-2x|=18`

`<=>|2x|=18`

`**2x=18`

`<=>x=9`

`**2x=-18`

`<=>x=-9`

Vậy `x=9` hoặc `x=-9`

`b)3/5-|1x|=1/2`

`=>|x|=3/5-1/2`

`=>|x|=1/10`

`=>` \(\left[ \begin{array}{l}x=\dfrac{1}{10}\\x=-\dfrac{1}{10}\end{array} \right.\) 

Vậy `x=1/10` hoặc `x=-1/10`

a: =>2x=-18+5=-13

=>x=-13/2

b: =>3^x-1=81

=>x-1=4

=>x=5

c: =>4(5-x)=24

=>5-x=6

=>x=-1

giups mình đi ạ:((

Câu a:

Câu b:

a: Ta có: \(P=\left(a+2\right)\left(a^2-2a+4\right)-a^3+\left(a+3\right)\left(a-3\right)-a^2-18\)

\(=a^3+8-a^3+a^2-9+a^2-18\)

\(=-19\)

b: Ta có: \(Q=\left(x-5\right)\left(2x+3\right)-2x\left(x-3\right)+x+7\)

\(=2x^2+3x-10x-15-2x^2+6x+x+7\)

=-8

cac bạn cho Anh sửa đề a , kết quả = 44 chứ k phải 41 đâu nhé

@Nguyễn Huy Tú

a, \(\Leftrightarrow4x=360\\ \Leftrightarrow x=90\)

b, \(\Leftrightarrow2x-5=23\\ \Leftrightarrow2x=28\\ \Leftrightarrow x=14\)

c, \(\Leftrightarrow2^{x+2}=2^7\\ \Leftrightarrow x+2=7\\ \Leftrightarrow x=5\)

a: x=90

c: x=5

a. x + 3x = 360

<=> 4x = 360

<=> x = 90

b. 41 - (2x - 5) = 18

<=> 41 - 2x + 5 = 18

<=> 2x = -(18 - 5 - 41)

<=> 2x = 28

<=> x = 14

c. 2^x . 4 = 128

<=> 2^{x + 2} = 2⁷

<=> x + 2 = 7

<=> x = 5

Bài 1: 

a) Ta có: \(\left(2x-1\right)^{20}=\left(2x-1\right)^{18}\)

\(\Leftrightarrow\left(2x-1\right)^{20}-\left(2x-1\right)^{18}=0\)

\(\Leftrightarrow\left(2x-1\right)^{18}\left[\left(2x-1\right)^2-1\right]=0\)

\(\Leftrightarrow\left(2x-1\right)^{18}\cdot\left(2x-2\right)\cdot2x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)

b) Ta có: \(\left(2x-3\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)

c) Ta có: \(\left(x-5\right)^2=\left(1-3x\right)^2\)

\(\Leftrightarrow\left(x-5\right)^2-\left(3x-1\right)^2=0\)

\(\Leftrightarrow\left(x-5-3x+1\right)\left(x-5+3x-1\right)=0\)

\(\Leftrightarrow\left(-2x-4\right)\left(4x-6\right)=0\)

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

Bài 2: 

a) \(15^{20}-15^{19}=15^{19}\left(15-1\right)=15^{19}\cdot14⋮14\)

b) \(3^{20}+3^{21}+3^{22}=3^{20}\left(1+3+3^2\right)=3^{20}\cdot13⋮13\)

c) \(3+3^2+3^3+...+3^{2007}\)

\(=3\left(1+3+3^2\right)+...+3^{2005}\left(1+3+3^2\right)\)

\(=13\left(3+...+3^{2005}\right)⋮13\)

1e) Để \(\frac{2x-1}{x-3}\) nguyên thì \(2x-1⋮x-3\)

\(\Leftrightarrow2x-6+5⋮x-3\)

\(\Leftrightarrow2\left(x-3\right)+5⋮x-3\)

Do \(2\left(x-3\right)⋮x-3\) \(\Rightarrow5⋮x-3\)

\(\Rightarrow x-3\in\left\{-5;-1;1;5\right\}\)

\(\Leftrightarrow x\in\left\{-2;2;4;8\right\}\)

Vậy:...................