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a/ \(\left(2x-3\right)^2-\left(3x+2\right)^2=5x\left(2-x\right)\)
<=> \(\left(2x-3-3x-2\right)\left(2x-3+3x+2\right)=5x\left(2-x\right)\)
<=> \(\left(-x-5\right)\left(5x-1\right)=5x\left(2-x\right)\)
<=> \(-5x^2-25x+x+5=10x-5x^2\)
<=> \(10x+25x-x=5\)
<=> \(34x=5\)
<=> \(x=\frac{5}{34}\)
b/ pt <=> \(2^3x^3-3.2^2.x^2.1+3.2.x.1^2-1^3=0\)
<=> \(\left(2x-1\right)^3=0\)
<=> 2 x - 1 = 0
<=> x = 1/2.
\(a,x\left(8x-2\right)-8x^2+12=0\)
\(\Rightarrow8x^2-2x-8x^2+12=0\)
\(\Rightarrow-2x+12=0\)
\(\Rightarrow-2x=-12\)
\(\Rightarrow x=6\)
\(b,x\left(4x-5\right)-\left(2x+1\right)^2=0\)
\(\Rightarrow4x^2-5x-4x^2-4x-1=0\)
\(\Rightarrow-9x-1=0\)
\(\Rightarrow-9x=1\)
\(\Rightarrow x=\frac{-1}{9}\)
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
\(x^3+9x=0\)
<=> \(x\left(x^2+9\right)=0\)
<=> \(\orbr{\begin{cases}x=0\\x^2+9=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=0\\x\in\varnothing\end{cases}}\)
<=> \(x=0\)
\(9x^2-4-2\left(3x-2\right)^2=0\)
<=> \(\left(9x^2-4\right)-2\left(3x-2\right)^2=0\)
<=> \(\left[\left(3x\right)^2-2^2\right]-2\left(3x-2\right)^2=0\)
<=> \(\left(3x-2\right)\left(3x+2\right)-2\left(3x-2\right)^2=0\)
<=> \(\left(3x-2\right)\left[\left(3x+2\right)-2\left(3x-2\right)\right]=0\)
<=> \(\left(3x-2\right)\left(3x+2-6x+4\right)=0\)
<=> \(\left(3x-2\right)\left(-3x+6\right)=0\)
<=> \(\left(3x-2\right)3\left(-x+2\right)=0\)
<=> \(3\left(3x-2\right)\left(2-x\right)=0\)
<=> \(\orbr{\begin{cases}3x-2=0\\2-x=0\end{cases}}\)
<=> \(\orbr{\begin{cases}3x=2\\x=2\end{cases}}\)
<=> \(\orbr{\begin{cases}x=\frac{2}{3}\\x=2\end{cases}}\)
\(\left(x^3-x^2\right)-4x+8x-4=0\)
<=> \(\left(x^3-x^2\right)+\left(4x-4\right)=0\)
<=> \(x^2\left(x-1\right)+4\left(x-1\right)=0\)
<=> \(\left(x-1\right)\left(x^2+4\right)=0\)
<=> \(\orbr{\begin{cases}x-1=0\\x^2+4=0\end{cases}}\)
<=> \(x=1\)
\(\left(25x^2-10x\right):\left(-5x\right)-3\left(x-2\right)=4\)
<=> \(5x\left(5x-2\right)\left(-\frac{1}{5x}\right)-3\left(x-2\right)=4\)
<=> \(-\left(5x-2\right)-3\left(x-2\right)=4\)
<=> \(\left(5x-2\right)+3\left(x-2\right)=-4\)
<=> \(5x-2+3x-6=-4\)
<=> \(8x-8=-4\)
<=> \(8\left(x-1\right)=-4\)
<=> \(x-1=-\frac{1}{2}\)
<=> \(x=-\frac{3}{2}\)
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
\(\frac{3x^4-8x^3-10x^2+8x-5}{3x^2-2x+1}\)
\(=\frac{x^2\left(3x^2-2x+1\right)-2x\left(3x^2-2x+1\right)-5\left(3x^2-2x+1\right)}{3x^2-2x+1}\)
\(=\frac{\left(3x^2-2x+1\right)\cdot\left(x^2-2x-5\right)}{3x^2-2x+1}\)
\(=x^2-2x-5\)
\(\frac{2x^3-9x^2+19x-15}{x^2-3x+5}\)
\(=\frac{2x\left(x^2-3x+5\right)-3\left(x^2-3x+5\right)}{x^2-3x+5}\)
\(=\frac{\left(x^2-3x+5\right)\left(2x-3\right)}{x^2-3x+5}\)
\(=2x-3\)
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) \(x^2\left(x-1\right)-4x^2+8x-4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=2\end{array}\right.\)
Đặt 8x^2=t => t>=0
\(t^2-9-\left(t-1\right)=54\Leftrightarrow t^2-t+1-9=54\)
\(t^2-t+\frac{1}{4}=54+8+\frac{1}{4}=\frac{249}{4}\) lẻ thế nhỉ
\(\left(t-\frac{1}{2}\right)^2=\frac{249}{4}\Rightarrow\left[\begin{matrix}t=\frac{1-\sqrt{249}}{2}< 0\left(loai\right)\\t=\frac{1+\sqrt{249}}{2}\end{matrix}\right.\)
\(\left\{\begin{matrix}x< 0\\8x^2=\frac{1+\sqrt{249}}{2}\end{matrix}\right.\Rightarrow x=\frac{-\sqrt{1+\sqrt{249}}}{16}\)
thanks