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Bài 2:
a: \(x^2+5x-6=\left(x+6\right)\left(x-1\right)\)
b: \(5x^2+5xy-x-y\)
\(=5x\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(5x-1\right)\)
c:\(-6x^2+7x-2\)
\(=-6x^2+3x+4x-2\)
\(=-3x\left(2x-1\right)+2\left(2x-1\right)\)
\(=\left(2x-1\right)\left(-3x+2\right)\)
1.
a) \(=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)
b) \(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
c) \(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]=5\left[\left(x-y\right)^2-4z^2\right]\)
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
2.
a) \(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)
b) \(=5x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(5x-1\right)\)
c) \(=-\left[3x\left(2x-1\right)-2\left(2x-1\right)\right]=-\left(2x-1\right)\left(3x-2\right)\)
3.
b) \(=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)\)
c) \(=-\left[5x\left(x-3\right)-1\left(x-3\right)\right]=-\left(x-3\right)\left(5x-1\right)\)
4.
a) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
b) \(16x-5x^2-3=5x\left(3-x\right)-\left(3-x\right)=\left(3-x\right)\left(5x-1\right)\)
c) \(2x^2+3x-5=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)\)
d) \(2x^2+3x-5=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)\)
a) \(2x^2+5x+2\)
\(=2x^2+4x+x+2\)
\(=2x\left(x+2\right)+\left(x+2\right)\)
\(=\left(x+2\right)\left(2x+1\right)\)
b) \(4x^2-4x-9y^2+12y-3\)
\(=\left(4x^2-4x+1\right)-\left(9y^2-12y+4\right)\)
\(=\left(2x-1\right)^2-\left(3y-2\right)^2\)
\(=\left(2x-1+3y-2\right)\left(2x-1-3y+2\right)\)
\(=\left(2x+3y-3\right)\left(2x-3y+1\right)\)
c) \(x^4-2x^3-4x^2+4x-3\)
\(=x^4+x^3-x^2+x-3x^2-3x+3x-3\)
\(=\left(x^4+x^3-x^2+x\right)-\left(3x^2+3x-3x+3\right)\)
\(=x\left(x^3+x^2-x+1\right)-3\left(x^3+x^2-x+1\right)\)
\(=\left(x^3+x^2-x+1\right)\left(x-3\right)\)
d) \(x^3-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
a: \(15x^2-5x^3=5x^2\left(3-x\right)\)
b: \(8x^3-y^3+4x^2y-2xy^2\)
\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)+2xy\left(2x-y\right)\)
\(=\left(2x-y\right)\left(4x^2+4xy+y^2\right)\)
\(=\left(2x-y\right)\left(2x+y\right)^2\)
c: Ta có: \(x^8+64y^4\)
\(=x^8+16x^4y^2+64y^4-16x^4y^2\)
\(=\left(x^4+8y^2\right)^2-\left(4x^2y\right)^2\)
\(=\left(x^2-4x^2y+8y^2\right)\left(x^2+4x^2y+8y^2\right)\)
\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11
e: Ta có: \(x^2-6x+y^2+4y+2=0\)
\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Dấu '=' xảy ra khi x=3 và y=-2
a/ 2x^3 -5x^2 + 8x -3
= 2x^3 -x^2 -4x^2 +2x +6x -3
= x^2 .[2x-1] - 2x[2x-1] +3. [2x-1]
= [x^2-2x+3] [2x-1]
b/ 3x^3 - 14x^2 +4x +3
= 3x^3 +x^2 -15 x^2 -5x +9x +3
= x^2 [3x+1] -5.x [3x+1] +3. [3x+1]
= [x^2 -5x+3] [3x+1]
c/ Đặt C = 12x^2 + 5 x -12 y^2 +12y -10xy -3
= -[12y^2+10xy+3-12x^2-5x-12y]
12y^2 + 10xy +3-12x^2-5x-12y = 18xy +12y^2 -6y - 12x^2 -8xy +4x -9x -6y +3
= 6y [3x+2y-1] - 4.x[3x+2y-1] -3.[3x+2y-1]
= [6y-4x-3] [3x+2y-1]
=> C = -[6y-4x-3]. [ 3x+2y-1]
tom lai minh ra
12x2+5x-12y2+12y-10xy-3=12(x+(2y-1)/3)(x-(6y-3)/4)) co dung ko nha.