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a) \(7x^2+34x-5=7x\left(x+5\right)-1\left(x+5\right)\)
\(=\left(x+5\right)\left(7x-1\right)\)
b) \(12a^2-3ab+8ac-2bc=3a\left(4a-b\right)+2c\left(4a-b\right)\)
\(=\left(4a-b\right)\left(3a+2c\right)\)
\(a,=7x^2-x+35x-5=x\left(7x-1\right)+5\left(7x-1\right)=\left(x+5\right)\left(7x-1\right)\\ b,=3a\left(4a-b\right)+2c\left(4a-b\right)=\left(3a+2c\right)\left(4a-b\right)\)
\(a,Sửa:x^2-xy-13x+13y=x\left(x-y\right)-13\left(x-y\right)=\left(x-13\right)\left(x-y\right)\\ b,=\left(x+y\right)^2-\left(2z\right)^2=\left(x+y-2z\right)\left(x+y+2z\right)\\ c,=\left(x^2-2x\right)-\left(3x-6\right)=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)
\(x^2-2xy+x-2y=x\left(x-2y\right)+x-2y=\left(x-2y\right)\left(x+1\right)\)
\(3x^3+6x+3-3y^2=3\left[\left(x^2+2x+1\right)-y^2\right]=3\left[\left(x+1\right)^2-y^2\right]=3\left(x-y+1\right)\left(x+y+1\right)\)
a) \(25a^2-b^2=\left(5a-b\right)\left(5a+b\right)\)
b) \(z^2+z-mz-m=z\left(z+1\right)-m\left(z+1\right)=\left(z+1\right)\left(z-m\right)\)
a , 25a2 - b2
= (5a -b ) ( 5a+ b )
b , z2 + z - m z -m
= z ( z + 1) - m( z + 1 )
= ( z + 1 ) ( z - m )
\(5x^4y+10x^3y+10x^2y^3+5xy^4\)
\(=5xy.x^3+5xy.2x^3+5xy.2xy^3+5xy.y^3\)
\(=5xy\left(x^3+2x^3+2xy^3+y^3\right)\)
Ht pt
a) \(7\left(3x-2\right)+y\left(3x-2\right)=\left(3x-2\right)\left(7+y\right)\)
b) \(x\left(y-x\right)-3\left(x-y\right)=x\left(y-x\right)+3\left(y-x\right)=\left(y-x\right)\left(x+3\right)\)
c) \(x^2-6xy+9y^2=\left(x-3y\right)^2\)
a) \(=\left(3ax+3bx\right)-\left(4by+4ay\right)=3x\left(a+b\right)-4y\left(a+b\right)=\left(a+b\right)\left(3x-4y\right)\)
b) \(=3\left[\left(x^2-2xy+y^2\right)-4t^2\right]=3\left[\left(x-y\right)^2-4t^2\right]=3\left(x-y-2t\right)\left(x-y+2t\right)\)
c) Không phân tích được
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^^