Câu hỏi của Nguyễn Thị Trà

Question

Chứng minh các biểu thức sau

a)   $\left(a^2-1\right)^2+4a^2=\left(a^2+1\right)^2$

b)    $\left(x-y\right)^2+\left(x+y\right)^2+2\left(x^2-y^2\right)=4x^2$

Khôi Bùi · Accepted Answer

a ) $VT=\left(a^2-1\right)^2+4a^2$

$=a^4-2a^2+1+4a^2$

$=a^4+2a^2+1$

$=\left(a^2+1\right)^2=VP\left(đpcm\right)$

b ) $VT=\left(x-y\right)^2+\left(x+y\right)^2+2\left(x^2-y^2\right)$

$=\left(x-y\right)^2+\left(x+y\right)^2+2\left(x-y\right)\left(x+y\right)$

$=\left(x-y+x+y\right)^2$

$=\left(2x\right)^2=4x^2=VP\left(đpcm\right)$Câu trả lời: a ) $VT=\left(a^2-1\right)^2+4a^2$

$=a^4-2a^2+1+4a^2$

$=a^4+2a^2+1$

$=\left(a^2+1\right)^2=VP\left(đpcm\right)$

b ) $VT=\left(x-y\right)^2+\left(x+y\right)^2+2\left(x^2-y^2\right)$

$=\left(x-y\right)^2+\left(x+y\right)^2+2\left(x-y\right)\left(x+y\right)$

$=\left(x-y+x+y\right)^2$

$=\left(2x\right)^2=4x^2=VP\left(đpcm\right)$

Sáng · Answer

a, Ta có:

$VT=\left(a^2-1\right)^2+4a^2=a^4-2a^2+1+4a^2=a^4+2a^2+1=\left(a^2+1\right)^2=VP$

$\Rightarrow dpcm$

b, Ta có:

$VT=\left(x-y\right)^2+\left(x+y\right)^2+2\left(x^2-y^2\right)$

$=x^2-2xy+y^2+x^2+2xy+y^2+2x^2-2y^2=4x^2=VP$

$\Rightarrow dpcm$Câu trả lời: a, Ta có:

$VT=\left(a^2-1\right)^2+4a^2=a^4-2a^2+1+4a^2=a^4+2a^2+1=\left(a^2+1\right)^2=VP$

$\Rightarrow dpcm$

b, Ta có:

$VT=\left(x-y\right)^2+\left(x+y\right)^2+2\left(x^2-y^2\right)$

$=x^2-2xy+y^2+x^2+2xy+y^2+2x^2-2y^2=4x^2=VP$

$\Rightarrow dpcm$

Bài 3: Những hằng đẳng thức đáng nhớ