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làm ra thì dài quá mk ko còn nhiều t/g
bn Áp dụng HĐT a2-b2=(a+b)(a-b) đi
Đ/a: a)x1=2;x2=6;x3,4=\(\frac{-2\pm\sqrt{452}}{14}\)
b)x1=-1;x2=1/2;x3,4=\(\frac{-2\pm\sqrt{8}}{2}\)
c)x=-5/4;x=1/2
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) \(x^2+5x+6=0\)
\(\Leftrightarrow x^2+2x+3x+6=0\)
\(\Leftrightarrow x\left(x+2\right)+3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+2=0\\x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\\x=-3\end{array}\right.\)
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) \(2x^2-3x-5=0\)
\(\Leftrightarrow2x^2+2x-5x-5=0\)
\(\Leftrightarrow2x\left(x+1\right)-5\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x-5\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+1=0\\2x-5=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-1\\x=\frac{5}{2}\end{array}\right.\)
tôi bt làm 1 câu à mấy câu kia khó quá *-*
1. 5x2+4x-2=0
\(\Leftrightarrow x\left(5x+4\right)=2\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\5x+4=2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{-2}{5}\end{cases}}}\)
\(\Rightarrow\) Nghiệm pt là :\(S=\left\{\frac{-2}{5};2\right\}\)
chúc bn sớm làm dc bài này ha
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 )
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
a) x^2 - 11x + 18 = 0
=> x^2 - 2x - 9x + 18 = 0
=> x ( x- 2 ) - 9 ( x- 2 ) = 0
=> ( x- 9 )( x- 2 )= 0
=> x- 9 = 0 hoặc x - 2 = 0
=> x= 9 hoặc x = 2
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