Câu 5:
`VT=(x+y)²+(x-y)²-2(x+y)(x-y)`
`=(x+y)²-2(x+y)(x-y)+(x-y)²`
`=[(x+y)-(x-y)]²`
`=(x+y-x+y)`
`=(2y)²`
`=4y²`
`=VP`
`⇒đpcm`
Câu 6:
`a)4-(x-2)²=0`
`⇔2²-(x-2)²=0`
`⇔(2+x-2)(2-x+2)=0`
`⇔x(4-x)=0`
`⇔`\(\left[ \begin{array}{l}x=0\\4-x=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=0\\x=4\end{array} \right.\)
Vậy `x=0` hoặc `x=4`
`b)3/4x²-16-1/3x²-1/6x²=0`
`⇔(3/4x²-1/3x²-1/6x²)-16=0`
`⇔[(3/4-1/3-1/6)x²]-4²=0`
`⇔[(9/12-4/12-2/12)x²]-4²=0`
`⇔1/4x²-4²=0`
`⇔(1/2x)^2-4²=0`
`⇔(1/2x+4)(1/2x-4)=0`
`⇔`\(\left[ \begin{array}{l}\dfrac{1}{2}x+4=0\\\dfrac{1}{2}x-4=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}\dfrac{1}{2}x=-4\\\dfrac{1}{2}x=4\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-8\\x=8\end{array} \right.\)
Vậy `x=-8` hoặc `x=8`
`c)(2x-1)²+(x+1)²=7(x²+1)-2(x²+1/2x)`
`⇔4x²-4x+1+x²+2x+1=7x²+7-2x²-x`
`⇔5x²-2x+2=5x²-x+7`
`⇔5x²-2x-5x²+x=7-2`
`⇔-x=5`
`⇔x=-5`
Vậy `x=-5`
