Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
anh ơi, vậy là sai đề hả anh, chứ đề kêu chứng minh phương trình vô nghiệm mà em thấy anh ghi x=2
3x.|x+1|−2x|x+2|=12
Với x < -2 ta có: 3x.(-x-1)-2x(-x-2)-12=0
<=> -3x2 - 3x + 2x2 + 4x -12 =0
<=> -x2 - x - 12=0
$\Leftrightarrow $ -(x2 +x+12)=0 ( vô lý)
Làm tương tự với 2 trường hợp còn lại:
begin{align} \begin{cases} -2 bé hơn hoặc bằng x bé hơn -1 \\ x lớn hơn hoặc bằng -1 \\ \end{cases} \end{align}
a) \(x^3-16x=0\)
\(\Leftrightarrow x\left(x^2-16\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-4=0\\x+4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=4\\x=-4\end{array}\right.\)
b) \(\left(2x-3\right)^2=\left(x-5\right)^2\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(x-5\right)^2=0\)
\(\Leftrightarrow\left(2x-3+x-5\right)\left(2x-3-x+5\right)=0\)
\(\Leftrightarrow\left(3x-8\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=\frac{8}{3}\end{array}\right.\)
c) \(x^2\left(x-1\right)-4x^2+8x-4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=2\end{array}\right.\)
a) \(4x^2-8x=0\)
\(\Rightarrow4x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=0+2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy \(x_1=0;x_2=2\)
b) \(\left(x+5\right)-3x\left(x+5\right)=0\)
\(\Rightarrow-3x^2-14x+5=0\)
\(\Leftrightarrow\left(-3x+1\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-3x+1=0\\x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-5\end{matrix}\right.\)
Vậy \(x_1=-5;x_2=\dfrac{1}{3}\)
\(a,4x^2-8x=0\Rightarrow4x\left(x-8\right)=0\Rightarrow\left[{}\begin{matrix}4x=0\\x-8=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)\(b,\left(x+5\right)-3x\left(x+5\right)=0\Leftrightarrow\left(x+5\right)\left(1-3x\right)=0\Rightarrow\left[{}\begin{matrix}x+5=0\\1-3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\3x=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{1}{3}\end{matrix}\right.\)
b: Đặt \(x^2-6x-2=a\)
Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)
=>(a+2)(a+7)=0
\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)
=>x(x-6)(x-1)(x-5)=0
hay \(x\in\left\{0;1;6;5\right\}\)
c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)
\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)
\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)
\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)
=>26x=-3
hay x=-3/26
a: \(\Leftrightarrow8x\left(x-3\right)\left(x+3\right)=0\)
hay \(x\in\left\{0;3;-3\right\}\)
b: \(\Leftrightarrow x^2-4x+4-x^2-2x+3=12\)
=>-6x=5
hay x=-5/6
\(a,x\left(8x-2\right)-8x^2+12=0\)
\(\Rightarrow8x^2-2x-8x^2+12=0\)
\(\Rightarrow-2x+12=0\)
\(\Rightarrow-2x=-12\)
\(\Rightarrow x=6\)
\(b,x\left(4x-5\right)-\left(2x+1\right)^2=0\)
\(\Rightarrow4x^2-5x-4x^2-4x-1=0\)
\(\Rightarrow-9x-1=0\)
\(\Rightarrow-9x=1\)
\(\Rightarrow x=\frac{-1}{9}\)
a) x(8 - 2) - 8x2 + 12 = 0
x(8 - 2) - 8x2 = 12 - 0
x(8 - 2) - 8x2 = 12
2x = 12
x = 6
b) x(4x - 5) - (2x + 1)2 = 0
9x - 1 = 0
9x = 0 + 1
9x = 1
x = -1/9