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a/ 3x(12x-5)-6x(6x-5)=0
<=>36x2-15x-36x2+30x=0
<=>15x=0
<=>x=0
b/ x2-8x+6=0
Nghiệm lẻ xem lại câu b
a) \(3x\left(12x-5\right)-6x\left(6x-5\right)=0\)
\(\Rightarrow36x^2-15x-36x^2+30x=0\)
\(\Rightarrow15x=0\)
\(\Rightarrow x=0\)
b) \(x^2-8x+6=0\)
\(\Rightarrow x^2-8x+16-10=0\)
\(\Rightarrow x^2-8x+4^2=10\)
\(\Rightarrow\left(x-4\right)^2=10\)
\(\Rightarrow x-4=\sqrt{10}\)
\(\Rightarrow x=4+\sqrt{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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Bài 4 :
\(\left(5x-20\right)+\left(3x^2-12x\right)=0\)
\(\Leftrightarrow5\left(x-4\right)+3x\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(5+3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5+3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{5}{3}\end{matrix}\right.\)
Vậy \(x=4\) hoặc \(x=-\dfrac{5}{3}\)
Bài 5 :
\(\left(1-x\right)-3x^2+3x=0\)
\(\Leftrightarrow\left(1-x\right)-\left(3x^2-3x\right)=0\)
\(\Leftrightarrow\left(1-x\right)-3x\left(x-1\right)=0\)
\(\Leftrightarrow\left(1-x\right)+3x\left(1-x\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(1+3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}1-x=0\\1+3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(x=1\) hoặc \(x=-\dfrac{1}{3}\)
Bài 6 :
\(\left(4x+20\right)-\left(x+5\right)^2=0\)
\(\Leftrightarrow4\left(x+5\right)-\left(x+5\right)^2=0\)
\(\Leftrightarrow\left(x+5\right)\left(4-x-5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(-x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\-x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=-1\end{matrix}\right.\)
Vậy \(x=-5\) hoặc \(x=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1.6x\left(x-10\right)-2x+20=0\)
⇔\(6x\left(x-10\right)-2\left(x-10\right)=0\)
⇔ \(2\left(x-10\right)\left(3x-1\right)=0\)
⇔ x = 10 hoặc x = \(\dfrac{1}{3}\)
KL....
\(2.3x^2\left(x-3\right)+3\left(3-x\right)=0\)
⇔ \(3\left(x-3\right)\left(x^2-1\right)=0\)
⇔ \(x=+-1\) hoặc \(x=3\)
KL....
\(3.x^2-8x+16=2\left(x-4\right)\)
⇔ \(\left(x-4\right)^2-2\left(x-4\right)=0\)
⇔ \(\left(x-4\right)\left(x-6\right)=0\)
⇔ \(x=4\) hoặc \(x=6\)
KL.....
\(4.x^2-16+7x\left(x+4\right)=0\)
\(\text{⇔}4\left(x+4\right)\left(2x-1\right)=0\)
⇔ \(x=-4hoacx=\dfrac{1}{2}\)
KL.....
\(5.x^2-13x-14=0\)
⇔ \(x^2+x-14x-14=0\)
\(\text{⇔}\left(x+1\right)\left(x-14\right)=0\)
\(\text{⇔}x=14hoacx=-1\)
KL......
Còn lại tương tự ( dài quá ~ )
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![](https://rs.olm.vn/images/avt/0.png?1311)
Câu a :
\(\left(3x^2-3y^2\right)+\left(4x-4y\right)=0\)
\(\Leftrightarrow3\left(x^2-y^2\right)+4\left(x-y\right)=0\)
\(\Leftrightarrow3\left(x-y\right)\left(x+y\right)+4\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left[3\left(x+y\right)+4\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-y=0\\3\left(x+y\right)+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=y\\x+y=-\dfrac{4}{3}\Rightarrow x=-\dfrac{4}{3}-y\end{matrix}\right.\)
Vậy \(x=y\) hoặc \(x=-\dfrac{4}{3}-y\)
Câu b :
\(\left(12x^2-3xy\right)+\left(8x-2y\right)=0\)
\(\Leftrightarrow3x\left(4x-y\right)+2\left(4x-y\right)=0\)
\(\Leftrightarrow\left(4x-y\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-y=0\\3x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{y}{4}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(x=\dfrac{y}{4}\) hoặc \(x=-\dfrac{2}{3}\)
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Ta có : 12x2 + 8x = 0
<=> 4x(3x + 2) = 0
\(\Leftrightarrow\orbr{\begin{cases}4x=0\\3x+2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\3x=-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{2}{3}\end{cases}}\)
a, 4x(3x - 2) = 0
=> x=0 hoac x= 2/3
b, 2x2 + 10x - x -5 =0
<=> (x + 5)(2x-1) =0
=> x = -5 hoac x = 1/2
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( 4x - 1 )3 + ( 3 - 4x )( 9 + 12x + 16x2 ) = ( 8x - 1 )( 8x + 1 ) - ( 3x - 5 )
<=> 64x3 - 48x2 + 12x - 1 + [ 33 - ( 4x )3 ] = ( 8x )2 - 12 - 3x + 5
<=> 64x3 - 48x2 + 12x - 1 + 27 - 64x3 = 64x2 - 1 - 3x + 5
<=> 64x3 - 48x2 + 12x - 64x3 - 64x2 + 3x = -1 + 5 + 1 - 27
<=> -112x2 + 15x = -22
<=> -112x2 + 15x + 22 = 0 (*) ( lại phải xài Delta :(( )
\(\Delta=b^2-4ac=15^2-4\cdot\left(-112\right)\cdot22=225+9856=10081\)
\(\Delta>0\)nên (*) có hai nghiệm phân biệt
\(\hept{\begin{cases}x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-15+\sqrt{10081}}{-224}\\x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-15-\sqrt{10081}}{-224}\end{cases}}\)
Nghiệm xấu quá -..-
sao rối quá v bn ..