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Bài 2:
1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)
=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)
=>(2x-1)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
2: \(9x^3-x=0\)
=>\(x\left(9x^2-1\right)=0\)
=>x(3x-1)(3x+1)=0
=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)
=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)
=>(2x-3)(2x-3-2)=0
=>(2x-3)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)
=>\(2x^2+10x-5x-25-10x+25=0\)
=>\(2x^2-5x=0\)
=>\(x\left(2x-5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)
Bài 1:
1: \(3x^3y^2-6xy\)
\(=3xy\cdot x^2y-3xy\cdot2\)
\(=3xy\left(x^2y-2\right)\)
2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)
\(=\left(x-2y\right)\cdot\left(x+3y\right)-2\cdot\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+3y-2\right)\)
3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)
\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)
\(=(x-2y)(3x-1+5x)\)
\(=\left(x-2y\right)\left(8x-1\right)\)
4: \(x^2-y^2-6y-9\)
\(=x^2-\left(y^2+6y+9\right)\)
\(=x^2-\left(y+3\right)^2\)
\(=\left(x-y-3\right)\left(x+y+3\right)\)
5: \(\left(3x-y\right)^2-4y^2\)
\(=\left(3x-y\right)^2-\left(2y\right)^2\)
\(=\left(3x-y-2y\right)\left(3x-y+2y\right)\)
\(=\left(3x-3y\right)\left(3x+y\right)\)
\(=3\left(x-y\right)\left(3x+y\right)\)
6: \(4x^2-9y^2-4x+1\)
\(=\left(4x^2-4x+1\right)-9y^2\)
\(=\left(2x-1\right)^2-\left(3y\right)^2\)
\(=\left(2x-1-3y\right)\left(2x-1+3y\right)\)
8: \(x^2y-xy^2-2x+2y\)
\(=xy\left(x-y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-2\right)\)
9: \(x^2-y^2-2x+2y\)
\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)
Answer:
Câu đầu bạn xem lại.
\(\left(3x+4\right)^2+\left(4x-3\right)^2+\left(2+5x\right).\left(2-5x\right)\)
\(=\left(3x\right)^2+2.2x.4+4^2+\left(4x\right)^2-2.4x.3+3^2+2^2-\left(5x\right)^2\)
\(=9x^2+24x+16+16x^2-24x+9+4-25x^2\)
\(=\left(9x^2+16x^2-25x^2\right)+\left(24x-24x\right)+\left(16+9+4\right)\)
\(=29\)
\(\left(5x+y\right).\left(25x^2-5xy+y^2\right)-\left(5x-y\right).\left(25x^2+5xy+y^2\right)\)
\(=\left(5x+y\right).[\left(5x\right)^2-5x.y+y^2]-\left(5x-y\right).[\left(5x\right)^2+5x.y+y^2]\)
\(=\left(5x\right)^3+y^3-[\left(5x\right)^3-y^3]\)
\(=\left(5x\right)^3+y^3-\left(5x\right)^3+y^3\)
\(=2y^3\)
a)
\(5x(x-2y)+2(2y-x)^2=5x(x-2y)+2(x-2y)^2\)
\(=(x-2y)[5x+2(x-2y)]=(x-2y)(7x-4y)\)
b)
\(7x(y-4)^2-(4-y)^3=7x(y-4)^2+(y-4)^3=(y-4)^2(7x+y-4)\)
c)
\((4x-8)(x^2+6)-(4x-8)(x+7)+9(8-4x)\)
\(=(4x-8)(x^2+6)-(4x-8)(x+7)-9(4x-8)\)
\(=(4x-8)[(x^2+6)-(x+7)-9]=(4x-8)(x^2-x-10)=4(x-2)(x^2-x-10)\)
d)
\(x^2-xz-9y^2+3yz=(x^2-9y^2)-(xz-3yz)\)
\(=(x-3y)(x+3y)-z(x-3y)=(x-3y)(x+3y-z)\)
e)
\(x^2(x^2-6)-x^2+9=x^4-7x^2+9=(x^4-6x^2+9)-x^2\)
\(=(x^2-3)^2-x^2=(x^2-3-x)(x^2-3+x)\)
a)
\(5x(x-2y)+2(2y-x)^2=5x(x-2y)+2(x-2y)^2\)
\(=(x-2y)[5x+2(x-2y)]=(x-2y)(7x-4y)\)
b)
\(7x(y-4)^2-(4-y)^3=7x(y-4)^2+(y-4)^3=(y-4)^2(7x+y-4)\)
c)
\((4x-8)(x^2+6)-(4x-8)(x+7)+9(8-4x)\)
\(=(4x-8)(x^2+6)-(4x-8)(x+7)-9(4x-8)\)
\(=(4x-8)[(x^2+6)-(x+7)-9]=(4x-8)(x^2-x-10)=4(x-2)(x^2-x-10)\)
d)
\(x^2-xz-9y^2+3yz=(x^2-9y^2)-(xz-3yz)\)
\(=(x-3y)(x+3y)-z(x-3y)=(x-3y)(x+3y-z)\)
e)
\(x^2(x^2-6)-x^2+9=x^4-7x^2+9=(x^4-6x^2+9)-x^2\)
\(=(x^2-3)^2-x^2=(x^2-3-x)(x^2-3+x)\)