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1. a) 7x2 - 5x - 2 = 7x2 - 7x + 2x - 2 = 7x(x - 1) + 2(x - 1) = (x - 1).(7x + 2)
2. 5(2x - 1)2 - 3(2x - 1) = 0
<=> (2x - 1).[5(2x - 1) - 3] = 0
<=> (2x - 1).(10x - 8) = 0
<=> (2x - 1) = 0 hoặc (10x - 8) = 0
<=> x = 1/2 hoặc x = 4/5
3. x2 - 4x + 7 = (x2 - 4x + 4) + 3 = (x - 2)2 + 3
Do: (x - 2)2 > hoặc = 0 (với mọi x)
Nên (x - 2)2 + 3 > hoặc = 3 (với mọi x)
Hay (x - 2)2 + 3 > 0 (với mọi x) => đpcm
\(d,x\left(x-3\right)-7x+21=0\)
\(\Leftrightarrow x\left(x-3\right)-7\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=7\end{cases}}}\)
\(a,2x\left(x-7\right)+5x-35=0\)
\(\Leftrightarrow2x\left(x-7\right)+5\left(x-7\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\2x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-\frac{5}{2}\end{cases}}}\)
\(c,4x^2+12x+9=0\)
\(\Leftrightarrow4x^2+6x+6x+9=0\)
\(\Leftrightarrow2x\left(2x+3\right)+3\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\)
\(\Leftrightarrow x=-\frac{3}{2}\)
a) x(x-1) - (x+1)(x+2) = 0
x\(^2\)- x -x\(^{^2}\)-2x +x+2=0
-2x+2=0
-2x=0+2
-2x=2
x=-1
Vậy x bằng -1
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 = \(\frac{5}{3}\)
1) x2 - 7x = 0
=> x(x - 7) = 0
=> \(\orbr{\begin{cases}x=0\\x=7\end{cases}}\)
2) -3x2 + 5x = 0
=> x(-3x + 5) = 0
=> \(\orbr{\begin{cases}x=0\\-3x+5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{5}{3}\end{cases}}\)
3) x2 - 19x - 20 = 0
=> x2 - 20x + x - 20 = 0
=> x(x - 20) + (x - 20) = 0
=> (x + 1)(x - 20) = 0
=> \(\orbr{\begin{cases}x+1=0\\x-20=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=20\end{cases}}\)
4) x2 - 5x - 24 = 0
=> x2 - 8x + 3x - 24 = 0
=> x(x - 8) + 3(x - 8) = 0
=> (x + 3)(x - 8) = 0
=> \(\orbr{\begin{cases}x+3=0\\x-8=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=8\end{cases}}\)
1) x2 - 7x = 0
<=> x( x - 7 ) = 0
<=> \(\orbr{\begin{cases}x=0\\x=7\end{cases}}\)
2) -3x2 + 5x = 0
<=> x( -3x + 5 ) = 0
<=> \(\orbr{\begin{cases}x=0\\x=\frac{5}{3}\end{cases}}\)
3) x2 - 19x - 20 = 0
<=> x2 + x - 20x - 20 = 0
<=> x( x + 1 ) - 20( x + 1 ) = 0
<=> ( x - 20 )( x + 1 ) = 0
<=> \(\orbr{\begin{cases}x=20\\x=-1\end{cases}}\)
4) x2 - 5x - 24 = 0
<=> x2 + 3x - 8x - 24 = 0
<=> x( x + 3 ) - 8( x + 3 ) = 0
<=> ( x - 8 )( x + 3 ) = 0
<=> \(\orbr{\begin{cases}x=8\\x=-3\end{cases}}\)
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!!!
d) \(4x^2-9-x\left(2x-3\right)=0\)
\(\Leftrightarrow4x^2-9-2x^2+3x=0\)
\(\Leftrightarrow2x^2+3x-9=0\)
\(\Delta=3^2-4.2.\left(-9\right)=9+72=81\)
Vậy pt có 2 nghiệm phân biệt
\(x_1=\frac{-3+\sqrt{81}}{4}=\frac{-3}{2}\);\(x_1=\frac{-3-\sqrt{81}}{4}=-3\)
e) \(x^3+5x^2+9x=-45\)
\(\Leftrightarrow x^3+5x^2+9x+45=0\)
\(\Leftrightarrow x^2\left(x+5\right)+9\left(x+5\right)=0\)
\(\Leftrightarrow\left(x^2+9\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+9=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm3i\\x=-5\end{cases}}\)