x(3-x)=0

Th1: x=0

Th2: 3-x=0     =>       x=3

Vậy x=0 và x=3

(x+2)(4x-8)=0

Th1: x+2= 0        =>         x=-2

Th2: 4x-8 =0           =>             4x =8

                                                x= 2

Vậy x= +- 2 (cộng trừ 2 nhé)

(x+1)+(x+2)+(x+3)+....+(x+199)=5750

199x +(1+2+3+...199) =5750

199x+ {(199+1)* [(199-1)+1] : 2} =5750

199x + 19900= 5750

199x = -14150

x= -14150/199

 20 . 2^x + 1 = 10.4^2 + 1 

 20 . 2^x + 1 = 10 . 16 + 1

20 . 2^x + 1 = 161

20 . 2^x = 161 - 1

20 . 2^x = 160

2^x = 8

2^x = 2^3

=> x = 3

Ban co giai dc phan 2 ko ?

a. \(\left(2x-3\right)\left(x+1\right)+\left(2x-3\right)\left(3x-7\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(x+1+3x-7\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(4x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\4x-6=0\end{matrix}\right.\)\(\Leftrightarrow x=\dfrac{3}{2}\)

b. \(\left(x-4\right)\left(3x-2\right)+x^2-16=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\4x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{2}\end{matrix}\right.\)

(2x-3)(x+1)+(2x+3)(3x-7)=0

<=> (2x-3)(x+1)-(2x-3)(3x-7)=0

<=> (2x-3)(x+1-3x+7)=0

<=> (2x-3)(-2x+8)=0

<=> 2x-3=0 => x=3/2

Hoặc -2x+8=0 => x= 4

Vậy x thuộc{3/2;4}

a: \(\dfrac{x-5}{x-3}>0\)

=>x-5>0 hoặc x-3<0

=>x>5 hoặc x<3

b: \(\dfrac{x+8}{x-9}< 0\)

=>x+8>0 và x-9<0

=>-8<x<9

c: \(\dfrac{x+1}{2017}+\dfrac{x+2}{2016}+\dfrac{x+3}{2015}+\dfrac{x+4}{2014}+4=0\)

\(\Leftrightarrow\left(\dfrac{x+1}{2017}+1\right)+\left(\dfrac{x+2}{2016}+1\right)+\left(\dfrac{x+3}{2015}+1\right)+\left(\dfrac{x+4}{2014}+1\right)=0\)

=>x+2018=0

hay x=-2018

1: \(16x^2-9\left(x+1\right)^2=0\)

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

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

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

=>\(\left[\begin{array}{l}x-3=0\\ 7x+3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=-\frac37\end{array}\right.\)

2: \(\left(5x-4\right)^2-49x^2=0\)

=>\(\left(5x-4\right)^2-\left(7x\right)^2=0\)

=>(5x-4-7x)(5x-4+7x)=0

=>(12x-4)(-2x-4)=0

=>4(3x-1)*(-2)(x+2)=0

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

=>\(\left[\begin{array}{l}3x-1=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac13\\ x=-2\end{array}\right.\)

3: \(5x^3-20x=0\)

=>\(5x\left(x^2-4\right)=0\)

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

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

a, \(16x^2-9\left(x+1\right)^2=0\)

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

\(\Leftrightarrow\left(x-3\right)\left(6x+3\right)=0\Leftrightarrow x=-\frac{1}{2};x=3\)

b, \(\left(5x-4\right)^2-49x^2=0\Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\)

\(\Leftrightarrow\left(-2x-4\right)\left(12x-4\right)=0\Leftrightarrow x=-2;x=\frac{1}{3}\)

c, \(5x^3-20x=0\Leftrightarrow5x\left(x^2-4\right)=0\)

\(\Leftrightarrow5x\left(x-2\right)\left(x+2\right)=0\Leftrightarrow x=0;x=\pm2\)

1: Ta có: \(16x^2-9\left(x+1\right)^2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{7}{3}\end{matrix}\right.\)

2: Ta có: \(\left(5x-4\right)^2-49x^2=0\)

\(\Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\)

\(\Leftrightarrow\left(2x+4\right)\left(12x-4\right)=0\)

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

3: Ta có: \(5x^3-20x=0\)

\(\Leftrightarrow5x\left(x-2\right)\left(x+2\right)=0\)

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