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a) x2 - 4x - 5 = 0
=> x2 - 5x + x - 5 = 0
=> x(x - 5) + (x - 5) = 0
=> (x + 1)(x - 5) = 0
=> \(\orbr{\begin{cases}x+1=0\\x-5=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-1\\x=5\end{cases}}\)
b) 4x2 + 7x - 11 = 0
=> 4x2 + 11x - 4x - 11 = 0
=> x(4x + 11) - (4x + 11) = 0
=> (x - 1)(4x + 11) = 0
=> \(\orbr{\begin{cases}x-1=0\\4x+11=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=1\\x=-\frac{11}{4}\end{cases}}\)
c) -7x2 + 6x + 1 = 0
=> -7x2 + 7x - x + 1 = 0
=> -7x(x - 1) - (x - 1) = 0
=> (-7x - 1)(x - 1) = 0
=> \(\orbr{\begin{cases}-7x-1=0\\x-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}-7x=1\\x=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{1}{7}\\x=1\end{cases}}\)
d) -10x2 + 7x + 3 = 0
=> -10x2 + 10x - 3x + 3 = 0
=> -10x(x - 1) - 3(x - 1) = 0
=> (-10x - 3)(x - 1) = 0
=> \(\orbr{\begin{cases}-10x-3=0\\x-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}-10x=3\\x=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{3}{10}\\x=1\end{cases}}\)
a) \(7x-10=5x-6\)
\(7x-5x=-6+10\)
\(2x=4\)
\(x=2\)
b) \(3x\left(x-2\right)+x-2=0\)
\(\left(x-2\right)\left(3x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\3x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-\frac{1}{3}\end{cases}}\)
c) \(2x^2+7x-4=0\)
\(2x^2-x+8x-4=0\)
\(x\left(2x-1\right)+2\left(2x-1\right)=0\)
\(\left(2x-1\right)\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-1=0\\x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-2\end{cases}}\)
7x-10=5x-6<=>7x-5x=-6+10<=>2x=4=>x=2
3x(x-2)+x-2=0<=>(x-2)(3x+1)=0<=>x-2=0=>x=2 HAY 3x+1=0=>x=-1/3
2x2+7x-4=0.
Câu cuối xem có lộn đề không nha bạn ơi!!!
Ta có : x2 - 7x = 0
=> x(x - 7) = 0
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-7=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=7\end{cases}}\)
Ta có : -3x2 + 5x = 0
=> \(x\left(-3x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\-3x+5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\-3x=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{5}{3}\end{cases}}\)
a) x2 - 7x = 0
x ( x - 7 ) = 0
\(\Rightarrow\orbr{\begin{cases}x=0\\x-7=0\Rightarrow x=7\end{cases}}\)
Vay x = 0 hoac x = 7
b) -3x2 + 5x = 0
x( -3x + 5 ) = 0
\(\Rightarrow\orbr{\begin{cases}x=0\\-3x+5=0\end{cases}}\)
1. x= 0
2. -3x + 5 = 0
-3x = -5
x = -5 : ( - 3 )
x = \(\frac{5}{3}\)
Vay x= 0 hoac x = \(\frac{5}{3}\)
a) \(x^2-2x=0\)
\(x\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}}\)
b) \(x^2-7x+12=0\)
\(x^2-4x-3x+12=0\)
\(\left(x^2-3x\right)-\left(4x-12\right)=0\)
\(x\left(x-3\right)-4\left(x-3\right)=0\)
\(\left(x-4\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-4=0\\x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=4\\x=3\end{cases}}}\)
a) \(x^2-2x=0\)
\(x\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
b) \(x^2-7x+12=0\)
\(x^2-3x-4x+12=0\)
\(\left(x^2-3x\right)-\left(4x-12\right)=0\)
\(x\left(x-3\right)-4\left(x-3\right)=0\)
\(\left(x-3\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-4=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=3\\x=4\end{cases}}\)
Bài này em nên đi theo hướng giải theo delta