K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
TU
0
18 tháng 8 2021
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, \(5x^3-20x=0\Leftrightarrow5x\left(x^2-4\right)=0\)
\(\Leftrightarrow5x\left(x-2\right)\left(x+2\right)=0\Leftrightarrow x=0;x=\pm2\)
18 tháng 8 2021
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: \(5x^3-20x=0\)
\(\Leftrightarrow5x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
TV
23 tháng 7 2018
a) \(x^3-16x=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
b) \(\left(x-1\right)\left(x+2\right)-x-2=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1-1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)
c) \(2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(2-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\2-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
tik 
NH
3
KK
1 tháng 11 2017
x3-16x=0
=> x(x2-16)=0
=> x(x-4)(x+4)=0
=> \(\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
vậy x=0 ;x=4;x=-4
3x(2-x)-2+x=0
=> 3x(2-x)-(2-x)=0
=> (2-x)(3x-1)=0
=> \(\left[{}\begin{matrix}2-x=0\\3x-1=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=2\\3x=1\Rightarrow x=\dfrac{1}{3}\end{matrix}\right.\)
vậy x=2 hoặc x=\(\dfrac{1}{3}\)
1 tháng 11 2017
c) (x+3)(x2-2x+3)=(x+3)(5-2x)
=>(x+3)(x2-2x+3) - (x+3)(5-2x)=0
=>(x+3)(x2-4x-2)=0
=>\(=>\left[{}\begin{matrix}x+3=0\\\text{x^2-4x-2}=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\\left(x-2\right)^2-6=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\\left(x-2\right)^2=6\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\x=\sqrt{6}+2\\x=-\sqrt{6}+2\end{matrix}\right.\)
x3 -16.x = 0
<=>x . ( x2 -16 ) = 0
<=> \(\orbr{\begin{cases}x=0\\x^2-16=0\end{cases}}\)
<=>\(\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)
Vậy phương trình có nghiệm { 0; 4 ; -4 }