Bài 1:
`a)4x²-y²+4x+1`
`=(4x²+4x+1)-y²`
`=[(2x)²+2.2x.1+1²]-y²`
`=(2x+1)²-y²`
`=(2x+1+y)(2x+1-y)`
`b)x³-x+y³-y`
`=(x³+y³)-(x+y)`
`=(x+y)(x²-xy+y²)-(x+y)`
`=(x+y)(x²-xy+y²-1)`
`c)x³+3x²y+x+3xy²+y+y³`
`=(x³+3x²y+3xy²+y³)+(x+y)`
`=(x+y)³+(x+y)`
`=(x+y)[(x+y)²+1]`
`=(x+y)(x²+2xy+y²+1)`
`d)x³+y(1-3x²)+x(3y²-1)-y³`
`=x³+y-3x²y+3xy²-x-y³`
`=(x³-3x²y+3xy²-y³)-(x-y)`
`=(x-y)³-(x-y)`
`=(x-y)[(x-y)²-1]`
`=(x-y)(x-y+1)(x-y-1)`
`e)x²y+xy²-x-y`
`=(x²y+xy²)-(x+y)`
`=xy(x+y)-(x+y)`
`=(x+y)(xy-1)`
`f)8xy³-5xyz-24y²+15z`
`=(8xy³-24y²)-(5xyz-15z)`
`=8y²(xy-3)-5z(xy-3)`
`=(xy-3)(8y²-5z)`
Bài 2:
`a)x(x-2)-2+x=0`
`⇔x(x-2)-(x-2)=0`
`⇔(x-2)(x-1)=0`
`⇔`\(\left[ \begin{array}{l}x-2=0\\x-1=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=2\\x=1\end{array} \right.\)
Vậy `x=2` hoặc `x=1`
`b)` Sửa đề:`5x(x-3)-(x+3)=0`
`→5x(x-3)-(x-3)=0`
`5x(x-3)-(x-3)=0`
`⇔(x-3)(5x-1)=0`
`⇔`\(\left[ \begin{array}{l}x-3=0\\5x-1=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=3\\x=\dfrac{1}{5}\end{array} \right.\)
Vậy `x=3` hoặc `x=1/5`
`c)x²+6x+8=0`
`⇔x²+2x+4x+8=0`
`⇔(x²+2x)+(4x+8)=0`
`⇔x(x+2)+4(x+2)=0`
`⇔(x+2)(x+4)=0`
`⇔`\(\left[ \begin{array}{l}x+2=0\\x+4=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-2\\x=-4\end{array} \right.\)
Vậy `x=-2` hoặc `x=-4`
`d)x³+x²+x+1=0`
`⇔(x³+x²)+(x+1)=0`
`⇔x²(x+1)+(x+1)=0`
`⇔(x+1)(x²+1)=0`
`⇔`\(\left[ \begin{array}{l}x+1=0\\x²+1=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-1\\x²=-1(vô nghiệm)\end{array} \right.\)
Vậy `x=-1`
`e)x³-x²-x+1=0`
`⇔(x³-x²)-(x-1)=0`
`⇔x²(x-1)-(x-1)=0`
`⇔(x-1)(x²-1)=0`
`⇔(x-1)(x-1)(x+1)=0`
`⇔(x-1)²(x+1)=0`
`⇔`\(\left[ \begin{array}{l}(x-1)²=0\\x+1=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x-1=0\\x=-1\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\)
Vậy `x=1` hoặc `x=-1`
Bài 3:
`a)x²-2xy-4z²+y²`
`=(x²-2xy+y²)-4z²`
`=(x-y)²-(2z)²`
`=(x-y+2z)(x-y-2z)`
Thay `x=6,y=-4,z=-45` vào biểu thức trên ta được:
`[6-(-4)+2.(-45)][6-(-4)-2.(-45)]`
`=(6+4-90)(6+4+90)`
`=(10-90)(10+90)`
`=-80.100`
`=-8000`
Vậy giá trị của biểu thức trên tại `x=6,y=-4,z=-45` là `-8000`
`b)3(x-3)(x+7)+(x-4)²+48`
`=(3x-9)(x+7)+x²-8x+16+48`
`=3x²+21x-9x-63+x²-8x+16+48`
`=(3x²+x²)+(21x-9x-8x)+(-63+16+48)`
`=4x²+4x+1`
`=(2x)²+2.2x.1+1²`
`=(2x+1)²`
Thay `x=0,5` vào biểu thức trên ta đươc:
`(2.0,5+1)²=(1+1)²=2²=4`
Vậy giá trị của biểu thức trên tại `x=0,5` là `4`
