a, (a+b+c) ² × (a-b-c) ² b, (a+b+c) ² +(a+b-c) ² – 2(a+b) ² c, (a+b+c) ² +(a-b+c) ² + (a+b-c) ²+(b+c-a) ² 22/07/2021 Bởi Ariana a, (a+b+c) ² × (a-b-c) ² b, (a+b+c) ² +(a+b-c) ² – 2(a+b) ² c, (a+b+c) ² +(a-b+c) ² + (a+b-c) ²+(b+c-a) ²
a, (a+b+c) ² × (a-b-c) ² =a2 + b2 + c2 + 2ab + 2bc + 2ac .a2 + b2 + c2 – 2ab + 2bc – 2ac =2a²+2b²+2c²+4ab b, (a+b+c) ² +(a+b-c) ² – 2(a+b) ² = a2 + b2 + c2 + 2ab + 2bc + 2ac+a2 + b2 + c2 + 2ab – 2bc – 2ac−2a²−4ab−2b² =2a² + 2b²+ 2c²+4ab – 2ac−2a²−4ab−2b² = 2c²-2ac c, (a+b+c) ² +(a-b+c) ² + (a+b-c) ²+(b+c-a) ² =a2 + b2 + c2 + 2ab + 2bc + 2ac+a2 + b2 + c2 + 2ab – 2bc – 2ac+a2 + b2 + c2 + 2ab – 2bc – 2ac =4a²+4b²+4c² =4.(a²+b²+c²) Chúc bạn học tốt :))) Bình luận
Đáp án: Giải thích các bước giải: a)`(a+b+c)^2.(a-b-c)^2=[(a+b+c)(a-b-c)]^2` `=(a^2-ab-ac+ab-b^2-bc+ac-bc-c^2)^2` `=(a^2-b^2-c^2)^2` b) `(a+b+c)^2+(a+b-c)^2-2(a+b)^2` `=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2+2ab-2ac-2bc-2a^2-4ab-2b^2` `=2c^2+4ab` c) `(a+b+c)^2+(a-b+c)^2+(a+b-c)^2+(b+c-a)^2` `=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2-2ab+2ac-2bc+a^2+b^2+c^2+2ab-2bc-2ac+a^2+b^2+c^2-2ab+2bc-2ac` `=4a^2+4b^2+4c^2` `=4(a^2+b^2+c^2)` Bình luận
a, (a+b+c) ² × (a-b-c) ²
=a2 + b2 + c2 + 2ab + 2bc + 2ac .a2 + b2 + c2 – 2ab + 2bc – 2ac
=2a²+2b²+2c²+4ab
b, (a+b+c) ² +(a+b-c) ² – 2(a+b) ²
= a2 + b2 + c2 + 2ab + 2bc + 2ac+a2 + b2 + c2 + 2ab – 2bc – 2ac−2a²−4ab−2b²
=2a² + 2b²+ 2c²+4ab – 2ac−2a²−4ab−2b²
= 2c²-2ac
c, (a+b+c) ² +(a-b+c) ² + (a+b-c) ²+(b+c-a) ²
=a2 + b2 + c2 + 2ab + 2bc + 2ac+a2 + b2 + c2 + 2ab – 2bc – 2ac+a2 + b2 + c2 + 2ab – 2bc – 2ac
=4a²+4b²+4c²
=4.(a²+b²+c²)
Chúc bạn học tốt :)))
Đáp án:
Giải thích các bước giải:
a)`(a+b+c)^2.(a-b-c)^2=[(a+b+c)(a-b-c)]^2`
`=(a^2-ab-ac+ab-b^2-bc+ac-bc-c^2)^2`
`=(a^2-b^2-c^2)^2`
b) `(a+b+c)^2+(a+b-c)^2-2(a+b)^2`
`=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2+2ab-2ac-2bc-2a^2-4ab-2b^2`
`=2c^2+4ab`
c) `(a+b+c)^2+(a-b+c)^2+(a+b-c)^2+(b+c-a)^2`
`=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2-2ab+2ac-2bc+a^2+b^2+c^2+2ab-2bc-2ac+a^2+b^2+c^2-2ab+2bc-2ac`
`=4a^2+4b^2+4c^2`
`=4(a^2+b^2+c^2)`