Question

1.cho a^2-b^2=4c^2.CM: (5a-3b+8c)(5a-3b-8c)=(3a-5b)^2

2.cho a^2+b^2+c^2=2017. Tính M=(2a+2b-c)^2+(2b+2c-a)^2+(2c+2a-b)^2

Answer 1

a, Vì \(a^2-b^2=4c^2\Rightarrow16a^2-16b^2=64c^2\) (1)

Ta có:\(\left(5a-3b+8c\right)\left(5a-3b-8c\right)=\left(5a-3b\right)^2-\left(8c\right)^2\)

\(=25a^2-30ab+9b^2-64c^2\) (2)

Thay (1) vào (2) ta được

\(\left(5a-3b+8c\right)\left(5a-3b-8c\right)=25a^2-30ab+9b^2-16a^2+16b^2\)

\(=9a^2-30ab+25b^2=\left(3a-5b\right)^2\)

=> đpcm

b, \(M=\left(2a+2b-c\right)^2+\left(2b+2c-a\right)^2+\left(2c+2b-b\right)^2\)

\(=4a^2+4b^2+c^2+4b^2+4c^2+a^2+4c^2+4a^2+b^2\)

\(+8ab-4ac-4bc+8bc-4ab-4ac+8ac-4bc-4ab\)

\(=9.\left(a^2+b^2+c^2\right)=9.2017=18153\)

Vậy M=18153