tim x
(x+1)^2=(x+1)^0
(3^x-4).(x-3)^3=0
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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
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
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: \(5x^3-20x=0\)
\(\Leftrightarrow5x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
\(\left(3x-4\right)\left(x-3\right)^3=0\)
\(=>\orbr{\begin{cases}3x-4=0\\x-3=0\end{cases}=>\orbr{\begin{cases}x=\frac{4}{3}\\x=3\end{cases}}}\)
Ủng hộ nha
a) \(\left(x+1\right)^2=\left(x+1\right)^0\)
\(=>\left(x+1\right)^2=1\)
\(=>\orbr{\begin{cases}x+1=-1\\x+1=1\end{cases}}\)
\(=>\orbr{\begin{cases}x=-2\\x=0\end{cases}}\)