Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
bài 1
a)\(x^2+5x+6=\left(x+2\right)\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x+3=0\\x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=-2\end{cases}}}\)
\(ab\left(x-y\right)^3-8ab=ab\left[\left(x-y\right)^3-2^3\right]=ab\left(x-y-2\right)\left[\left(x-y\right)^2+2\left(x-y\right)+4\right]\)
\(36x^2-y^2+6y-9=36x^2-\left(y-3\right)^2=\left(6x-y+3\right)\left(6x+y-3\right)\)
\(8x^2+10x-3=0\)
\(8x^2-2x+12x-3=0\)
\(2x\left(4x-1\right)+3\left(4x-1\right)=0\)
\(\left(4x-1\right)\left(2x+3\right)=0\)
\(\left[\begin{array}{nghiempt}4x-1=0\\2x+3=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}4x=1\\2x=-3\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=\frac{1}{4}\\x=-\frac{3}{2}\end{array}\right.\)
\(\left(2x-5\right)^2-\left(x+4\right)^2=0\)
\(\left(2x-5+x+4\right)\left(2x-5-x-4\right)=0\)
\(\left(3x-1\right)\left(x-9\right)=0\)
\(\left[\begin{array}{nghiempt}3x-1=0\\x-9=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=\frac{1}{3}\\x=9\end{array}\right.\)
1 ) \(x\left(a-b\right)+a-b=\left(x+1\right)\left(a-b\right)\)
2 ) \(2x\left(b-a\right)+a-b=2x\left(b-a\right)-\left(b-a\right)=\left(2x-1\right)\left(b-a\right)\)
3 ) \(-2x-2y+ax+ay=-2\left(x+y\right)+a\left(x+y\right)=\left(a-2\right)\left(x+y\right)\)
4 ) \(x^2-xy-2x+2y=x\left(x-y\right)-2\left(x-y\right)=\left(x-2\right)\left(x-y\right)\)
5 ) \(5x^2y+5xy^2+a^2x+a^2y\)
\(=5xy\left(x+y\right)+a^2\left(x+y\right)\)
\(=\left(5xy+a^2\right)\left(x+y\right)\)
6 ) \(2x^2-6xy+5x-15y\)
\(=2x\left(x-3y\right)+5\left(x-3y\right)\)
\(=\left(2x+5\right)\left(x-3y\right)\)
7 ) \(ax^2-3axy+bx-3by\)
\(=\left(ax^2+bx\right)-\left(3axy+3by\right)\)
\(=x\left(ax+b\right)-3y\left(ax+b\right)\)
\(=\left(x-3y\right)\left(ax+b\right)\)
8 ) \(x^2+4x-5x-20=0\)
\(\Leftrightarrow x\left(x+4\right)-5\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-4\end{matrix}\right.\)
9 ) \(x^2+10x-2x-20=0\)
\(\Leftrightarrow x\left(x+10\right)-2\left(x+10\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-10\end{matrix}\right.\)
10 ) \(x^2-6x-4x+24=0\)
\(\Leftrightarrow x\left(x-6\right)-4\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=6\end{matrix}\right.\)
:D
a, \(x\left(x-3\right)-x^2+2=0\)
\(\Leftrightarrow x^2-3x-x^2+2=0\\ \Leftrightarrow-3x+2=0\)
\(\Leftrightarrow-3x=-2\\ \Rightarrow x=\frac{2}{3}\)
b, \(x^2-2x+1=0\\ \Leftrightarrow\left(x-1\right)^2=0^2\)
\(\Leftrightarrow x-1=0\\ \Leftrightarrow x=1\)
c, x(x-1)-(x+3)(x+4)=5x
\(\Leftrightarrow x^2-x-x^2-4x-3x-12=5x\)
\(\Leftrightarrow x^2-x-x^2-4x-3x-5x=12\\ \Leftrightarrow-13x=12\\ \Rightarrow x=\frac{-12}{13}\)
d, ko có vế phải ạ
e, \(x^2+2x=15\)
\(\Leftrightarrow\left(x^2+2x+1\right)-16=0\\ \Leftrightarrow\left(x+1\right)^2-4^2=0\)
\(\Leftrightarrow\left(x+1-4\right)\left(x+1+4\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+5\right)=0\)
\(\left[{}\begin{matrix}x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
f, \(x^4-5x^3+4x^2=0\)
\(\Leftrightarrow x^4-x^3-4x^3+4x^2=0\\ \Leftrightarrow x^3\left(x-1\right)-4x^2\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^3-4x^2\right)=0\)
\(\Leftrightarrow\left(x-1\right).x^2\left(x-4\right)=0\)
\(\left[{}\begin{matrix}x^2=0\\x-1=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=4\end{matrix}\right.\)
a) \(x^2+5x+6=x^2+2x+3x+6=x\left(x+2\right)+3\left(x+2\right)=\left(x+3\right)\left(x+2\right)\)
b) \(x^2-4x+3=x^2-x-3x+3=x\left(x-1\right)-3\left(x-1\right)=\left(x-3\right)\left(x-1\right)\)
c) \(x^2+5x+4=x^2+x+4x+4=x\left(x+1\right)+4\left(x+1\right)=\left(x+4\right)\left(x+1\right)\)
d) \(x^2-x-6=x^2+2x-3x-6=x\left(x+2\right)-3\left(x+2\right)=\left(x-3\right)\left(x+2\right)\)
\(b,x^2+4x+3=x^2+3x+x+3.\)
\(=x\left(x+3\right)+\left(x+3\right)=\left(x+1\right)\left(x+3\right)\)
\(c,16x-5x^2-3=x-5x^2+15x-3\)
\(=x\left(1-5x\right)+3\left(5x-1\right)\)
\(=\left(x+3\right)\left(1-5x\right)\)
\(d,x^4+4=x^4+4x^2+4-4x^2=\left(x+2\right)^2-4x^2\)
\(=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)
Hướng dẫn thôi :
a) x ( x + 2 ) ( x^2 - 6x + 4 )
b) ( x + 1 ) ( x + 2 ) ( x - 2 )
\(-3xy^2+x^2y^2-5x^2y\)
\(=-xy\left(3y+xy-5x\right)\)
\(x\left(y-1\right)+3\left(y^3+2y+1\right)\)
\(=3y^3+6y+3+xy-x\)
Xem lại nhé ko phân tích được
\(12xy^2-12xy+3x\)
\(=3x\left(4y^2-4y+1\right)\)
\(=3x\left(2y-1\right)^2\)
\(10x^2\left(x+y\right)-5\left(2x+2y\right)y^2\)
\(=10x^2\left(x+y\right)-10\left(x+y\right)y^2\)
\(=10\left(x+y\right)\left(x-y\right)\left(x+y\right)\)
\(=10\left(x+y\right)^2\left(x-y\right)\)
Phân tích đa thức thành nhân tử:
a) \(3a^2-3ab+9b-9a=3a\left(a-b\right)+9\left(b-a\right)=3\left(a-b\right)\left(a-3\right)\)
b) \(2xm^3-2m=2m\left(xm^2-1\right)\)
c) \(x^2-5x+6=x^2-2x-3x+6=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)
Tìm x:
a) \(8x^2+10x+3=0\)
\(\Leftrightarrow8x^2+12x-2x-3=0\Leftrightarrow4x\left(2x+3\right)-\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{3}{2}\\x=\frac{1}{4}\end{array}\right.\)
b) \(x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-2=0\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=2\end{array}\right.\)
kamasa pn nhìu lắm lun nahh^^