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A
5 tháng 9 2020
Bài 1 :
a, PT <=> \(-16x^2+52x-12=0\)
\(\left(x-\frac{1}{4}\right)\left(x-3\right)=0\)
TH1 : x = 1/4 ; TH2 : x =3
b, \(x^2+x+90=0\)( vô nghiệm )
c, \(x^2-x+2=0\)( vô nghiệm )
5 tháng 9 2020
1.
\(\left(3x+2\right)^2-\left(5x-4\right)^2=0\)
\(\left[3x+2-\left(5x-4\right)\right]\left(3x+2+5x-4\right)=0\)
\(\left(-2x+6\right)\left(8x-2\right)=0\)
\(\orbr{\begin{cases}-2x+6=0\\8x-2=0\end{cases}}\)
\(\orbr{\begin{cases}x=3\\x=\frac{1}{4}\end{cases}}\)
2.
\(x^2+x+90=0\)
\(x^2+2\cdot x\cdot\frac{1}{2}+\left(\frac{1}{2}\right)^2+90-\left(\frac{1}{2}\right)^2=0\)
\(\left(x+\frac{1}{2}\right)^2+\frac{359}{4}=0\)
\(\left(x+\frac{1}{2}\right)^2=\frac{-359}{4}\) ( sai vì \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\) )
Suy ra phương trình vô nghiệm
3.
\(x^2-x+2=0\)
\(x^2-2\cdot x\cdot\frac{1}{2}+\left(\frac{1}{2}\right)^2+2-\left(\frac{1}{2}\right)^2=0\)
\(\left(x-\frac{1}{2}\right)^2+\frac{7}{4}=0\)
\(\left(x-\frac{1}{2}\right)^2=\frac{-7}{4}\) ( sai vì \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\) )
Suy ra phương trình vô nghiệm
ES
3
Q
1 tháng 10 2016
1/ x² - 5x + 6 = 0
⇔ x² - 2x - 3x + 6 = 0
⇔ x(x - 2) - 3(x - 2) = 0
⇔ (x - 2)(x - 3) = 0
⇒S = {2 ; 3}.
1 tháng 10 2016
1) \(x^2+5x+6=0\)
\(\Leftrightarrow x^2+2x+3x+6=0\)
\(\Leftrightarrow x\left(x+2\right)+3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+2=0\\x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\\x=-3\end{array}\right.\)
2) \(2\left(x+3\right)-x^2-3x=0\)
\(\Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(2-x\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+3=0\\2-x=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-3\\x=2\end{array}\right.\)
3) \(x^2+4x+3=0\)
\(\Leftrightarrow x^2+x+3x+3=0\)
\(\Leftrightarrow x\left(x+1\right)+3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+1=0\\x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-1\\x=-3\end{array}\right.\)
4) \(2x^2-3x-5=0\)
\(\Leftrightarrow2x^2+2x-5x-5=0\)
\(\Leftrightarrow2x\left(x+1\right)-5\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x-5\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+1=0\\2x-5=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-1\\x=\frac{5}{2}\end{array}\right.\)
ES
1
IM
1 tháng 10 2016
a)
\(10x^2+10xy+5x+5y\)
\(=10x\left(x+y\right)+5\left(x+y\right)\)
\(=5\left(x+y\right)\left(2x+1\right)\)
b)
\(x^3+x^2-x-1\)
\(=x^2\left(x+1\right)-\left(x+1\right)\)
\(=\left(x-1\right)\left(x^2-1\right)\)
\(=\left(x-1\right)^2\left(x+1\right)\)
c)
\(x+2a\left(x-y\right)-y\)
\(=\left(x-y\right)+2a\left(x-y\right)\)
\(=\left(x-y\right)\left(2a+1\right)\)
d)
\(x^2-y^2+7x-7y\)
\(=\left(x+y\right)\left(x-y\right)+\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y+1\right)\)
ES
1
IM
1 tháng 10 2016
a)
\(x\left(x-2\right)+x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-1\end{array}\right.\)
Vậy x = 2 ; x = - 1
b)
\(x^3+x^2+x+1=0\)
\(\Leftrightarrow x\left(x^2+1\right)+\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\)
Vì x2+1 > 0
=> x + 1 = 0
=> x = - 1
Vậy x = - 1
c)
\(\left(x+3\right)-x^2-3x=0\)
\(\Leftrightarrow\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-3\end{array}\right.\)
Vậy x = 1 ; x = - 3
d)
\(2x\left(3x-5\right)=10-6x\)
\(\Leftrightarrow2x\left(3x-5\right)+2\left(3x-5\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{5}{3}\\x=-\frac{1}{2}\end{array}\right.\)
Vậy x = 5 / 3 ; x = - 1 / 2
L
0
9 tháng 2 2017
Bài 3a)
\(a+b+c=0\Leftrightarrow a+b=-c\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)\)
mà \(a+b=-c\Rightarrow a^3+b^3+c^3=3abc\)
Phân tích ? -.-
1. x2 + x - 90 = x2 - 9x + 10x - 90 = x( x - 9 ) + 10( x - 9 ) = ( x - 9 )( x + 10 ) < -90 mới ra nhé :v >
2. x2 - x + 2 = x2 + x - 2x + 2 = x( x + 1 ) - 2( x + 1 ) = ( x + 1 )( x - 2 )
3. x2 + 19x + 90 = x2 + 9x + 10x + 90 = x( x + 9 ) + 10( x + 9 ) = ( x + 9 )( x + 10 )
4. x2 - 23x + 132 = x2 - 12x - 11x + 132 = x( x - 12 ) - 11( x - 12 ) = ( x - 12 )( x - 11 )