Giúp mình giải bài này vs : 3x(x-1)+2(1-x)+0
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(5x\left(x-3\right)=x-3\)
\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\5x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{1}{5}\end{cases}}}\)
(3x-4-x-1)(3x-4+x+1)=0
(2x-5)(4x-3)=0
2x-5 = 0 hoặc 4x-3=0
2x=5 hoặc 4x=3
x=5/2 hoặc x=3/4
a)\(2x\left(x-2016\right)-2x+4032=0\)
\(\Leftrightarrow2x\left(x-2016\right)-2\left(x-2016\right)=0\)
\(\Leftrightarrow\left(2x-2\right)\left(x-2016\right)=0\)
\(\Leftrightarrow2\left(x-1\right)\left(x-2016\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x-2016=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=2016\end{array}\right.\)
b)\(5x\left(x-3\right)=x-3\)
\(\Leftrightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-3=0\\5x-1=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\x=\frac{1}{5}\end{array}\right.\)
c)\(\left(3x-1\right)^2=\left(x+2\right)^2\)
\(\Leftrightarrow\left(3x-1\right)^2-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(3x-1+x+2\right)\left[\left(3x-1\right)-\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(4x+1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}4x+1=0\\2x-3=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{4}\\x=\frac{3}{2}\end{array}\right.\)
\(=\frac{3x^2+9x-3}{x^2+x-2}-\frac{x+1}{x+2}-\frac{x-2}{x-1}\)
\(=\frac{3x^2+9x-3}{\left(x+2\right)\left(x-1\right)}-\frac{\left(x+1\right)\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}-\frac{\left(x-2\right)\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}\)
\(=\frac{3x^2+9x-3-\left(x^2-1\right)-\left(x^2-4\right)}{\left(x-1\right)\left(x+2\right)}\)
\(=\frac{3x^2+9x-3-x^2+1-x^2+4}{\left(x-1\right)\left(x+2\right)}\)
\(=\frac{x^2+9x+2}{\left(x-1\right)\left(x+2\right)}\)
Vd1:
d) Ta có: \(\sqrt{2}\left(x-1\right)-\sqrt{50}=0\)
\(\Leftrightarrow\sqrt{2}\left(x-1-5\right)=0\)
\(\Leftrightarrow x=6\)
1/ (2x+3)(x-4)+(x+5)(x-2)=(3x-5)(x-4)
<=> 2x2 - 8x + 3x - 12 + x2 - 2x + 5x - 10 - 3x2 + 12x + 5x - 20 = 0
<=> 15x - 20 = 0
<=> 15x = 20
<=> x = 4/3
\(3x\left(x-1\right)+2\left(1-x\right)=0.\)
\(\Leftrightarrow3x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-2\right)=0\)
\(\Rightarrow x=1\) hoặc \(x=\frac{2}{3}\)