Bài 1:
\(16x^2-\left(4x-5\right)^2=15\)
\(\Leftrightarrow\left(4x-4x+5\right)\left(4x+4x-5\right)=15\)
\(\Leftrightarrow5\left(8x-5\right)=15\)
\(\Leftrightarrow8x=8\Leftrightarrow x=1\)
Vậy x = 1
Bài 2:
\(VT=\left(7x+1\right)^2-\left(x+7\right)^2\)
\(=\left(7x+1-x-7\right)\left(7x+1+x+7\right)\)
\(=\left(6x-6\right)\left(8x+8\right)\)
\(=48\left(x-1\right)\left(x+1\right)\)
\(=48\left(x^2-1\right)=VP\)
\(\Rightarrowđpcm\)
Đúng 1
Bình luận (1)