Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
cái bài 2 câu 1 câu 2 và câu 3 sửa cái vế phải lại thành 3/2-1-2x/4 và -15/5 và 2.(x-1)/5
\(\Leftrightarrow-2x+1-x-2=8\cdot\left(-4x^2+6x-2x\right)+4\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow-3x-1+32x^2-48x+16x-4x^2+8x-4=0\)
\(\Leftrightarrow28x^2-27x-5=0\)
\(\text{Δ}=\left(-27\right)^2-4\cdot28\cdot\left(-5\right)=1289>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{27-\sqrt{1289}}{56}\\x_2=\dfrac{27+\sqrt{1289}}{56}\end{matrix}\right.\)
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) \(2x^2-3x-5=0\)
\(\Leftrightarrow2x^2+2x-5x-5=0\)
\(\Leftrightarrow2x\left(x+1\right)-5\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x-5\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+1=0\\2x-5=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-1\\x=\frac{5}{2}\end{array}\right.\)
\(1,\)\(3x-3y-x^2+2xy-y^2\)
\(=-\left(x^2-2xy+y^2\right)+\left(3x-3y\right)\)
\(=-\left(x-y\right)^2+3\left(x-y\right)\)
\(=\left(x-y\right)\left[-\left(x-y\right)+3\right]\)
\(=\left(x-y\right)\left(3-x+y\right)\)
\(2,\)\(49\left(x-4\right)^2-9\left(x+2\right)^2\)
\(=\left[7\left(x-4\right)\right]^2-\left[3\left(x-2\right)\right]^2\)
\(=\left(7x-28-3x+6\right)\left(7x-28+3x-6\right)\)
\(=\left(4x-22\right)\left(10x+34\right)\)
\(3,\)\(x^4+4x^2-5\)
\(=x^4-x^2+5x^2-5\)
\(=x^2\left(x^2-1\right)+5\left(x^2-1\right)\)
\(=\left(x^2+5\right)\left(x-1\right)\left(x+1\right)\)
a, \(x^2-4x+3=0\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)
TH1 : x = 3 ; TH2 : x = 1
b, \(2x^2-3x-2=0\Leftrightarrow\left(x-2\right)\left(x+\frac{1}{2}\right)=0\)
TH1 : x = 2 ; TH2 : x = -1/2
c, Đặt \(x^2=t\left(t\ge0\right)\)
\(t^2+2t-8=0\Leftrightarrow\left(t-2\right)\left(t+4\right)=0\)
TH1 : t = 2 ; TH2 : t = -4
Tương tự ...
1a)
x2 - 4x + 3 = x2 - x - 3x + 3
= x( x - 1 ) - 3( x - 1 )
= ( x - 1 )( x - 3 )
2c)
2x2 - 3x - 2 = 2x2 + x - 4x - 2
= x( 2x +1 ) - 2( 2x + 1 )
= ( 2x + 1 )( x - 2 )
3e)
x4 + 2x2 - 8 (*)
Đặt t = x2
(*) <=> t2 + 2t - 8
= t2 - 2t + 4t - 8
= t( t - 2 ) + 4( t - 2 )
= ( t - 2 )( t + 4 )
= ( x2 - 2 )( x2 + 4 )
4b) x2 + 4x - 12 = x2 - 2x + 6x - 12
= x( x - 2 ) + 6( x - 2 )
= ( x - 2 )( x + 6 )
d) 2x3 + x - 2x2 - 1 = 2x2( x - 1 ) + 1( x - 1 )
= ( x - 1 )( 2x2 + 1 )
f) x2 - 2xy - 3y2 = ( x2 - 2xy + y2 ) - 4y2
= ( x - y )2 - ( 2y )2
= ( x - y - 2y )( x - y + 2y )
= ( x - 3y )( x + y )