rút gọn biểu thức:
- ( a^2 + b^2 - c^2 )^2 - ( a^2 - b^2 + c^2 )^2
- ( a + b + c )^2 + ( a + b - c )^2 - 2 *( a + b )^2
K
Khách
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TK
2
Những câu hỏi liên quan
CL
0
ND
2
9 tháng 7 2018
Ta có
(a+b+c)2+(b+c-a)2+(c+a-b)2+(a+b-c)2= [(a+b)+c]2+[(b-a)+c]2+[(a-b)+c]2+[(a+b)-c]
=(a+b)2+2c(a+b)+c2+(b-a)2+2c(b-a)+c2+(a-b)2+2c(a-b)+c2+(a+b)2-2c(a+b)+c2
=2(a+b)2+2(a-b)2+4c2( vì (a-b)2=(b-a)2)
13 tháng 7 2021
a) (a+b)3+(a-b)3=a3+3a2b+3ab2+b3+a3-3a2b+3ab2-b3
=2a3+6ab2
b) (a + b + c)2 + (a − b − c)2 + (b − c − a)2 + (c − a − b)2
=a2+b2+c2+2ab+2bc+2ca+a2+b2+c2-2ab+2bc-2ac+a2+b2+c2-2bc+2ca-2ba+a2+b2+c2-2ca+2ab-2cb
=4a2+4b2+4c2
13 tháng 7 2021
a) Ta có: \(\left(a+b\right)^3+\left(a-b\right)^3\)
\(=\left(a+b+a-b\right)\left[\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=2a\cdot\left(a^2+2ab+b^2-a^2+b^2+a^2-2ab+b^2\right)\)
\(=2a\cdot\left(a^2+3b^2\right)\)
\(=2a^3+6ab^2\)
PT
1
LT
0
MN
0
17 tháng 12 2021
\(\dfrac{a^2}{a^2-b^2-c^2}=\dfrac{a^2}{\left(a-b\right)\left(a+b\right)-c^2}=\dfrac{a^2}{\left(a-b\right)\left(-c\right)-c^2}=\dfrac{a^2}{c\left(b-a-c\right)}=\dfrac{a^2}{2bc}\\ \Leftrightarrow M=\sum\dfrac{a^2}{a^2-b^2-c^2}=\sum\dfrac{a^2}{2bc}=\dfrac{a^3+b^3+c^3}{2abc}\\ \Leftrightarrow M=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{2abc}=0\)
YN
2 tháng 1 2021
Với a + b + c = 0 , ta có :
\(A=\frac{ab}{a^2+b^2-c^2}\)\(+\frac{bc}{b^2+c^2-a^2}\)\(+\frac{ca}{c^2+a^2-b^2}\)
\(\Leftrightarrow\frac{ab}{\left(a+b\right)^2-2ab-c^2}\)\(+\frac{bc}{\left(b+c\right)^2-2ab-a^2}\)\(+\frac{ca}{\left(c+a\right)^2-2ca-b^2}\)
\(\Leftrightarrow A=\frac{ab}{\left(a+b+c\right)\left(a+b-c\right)-2ab}\)\(+\frac{bc}{\left(b+c-a\right)\left(b+c+a\right)-2ab}\)\(+\frac{ac}{\left(a+c+b\right)\left(c+a-b\right)-2ca}\)
\(\Leftrightarrow A=\frac{ab}{-2ab}\)\(+\frac{bc}{-2bc}\)\(+\frac{ac}{-2ac}\)
\(\Leftrightarrow A=\frac{-1}{2}\)\(+\frac{-1}{2}\)\(+\frac{-1}{2}\)
\(\Leftrightarrow A=\frac{-3}{2}\)
HP
0
Mình mới học lớp 6
Nên không biết nha
Chúc các bạn học giỏi
1,\(\left(a^2+b^2-c^2\right)^2\)\(-\left(a^2-b^2+c^2\right)^2\)
\(=\left(a^2+b^2-c^2\right)^2\)\(-\left(b^2-a^2-c^2\right)^2\)
\(=\left(a^2+b^2-c^2-b^2+a^2+c^2\right)\)\(\left(a^2+b^2-c^2+b^2-a^2-c^2\right)\)
\(=2a^2\left(2b^2-2c^2\right)\)
\(=4a^2b^2-4a^2c^2\)
\(=\left(2ab-2ac\right)\left(2ab+2ac\right)\)
2,\(\left(a+b+c\right)^2\)\(+\left(a+b-c\right)^2\)\(-2\left(a+b\right)^2\)
\(=\left(\left(a+b+c\right)^2-\left(a+b\right)^2\right)\)\(+\left(\left(a+b-c\right)^2-\left(a+b\right)^2\right)\)
\(=\left(a+b+c-a-b\right)\)\(\left(a+b+c+a+b\right)+\)\(\left(a+b-c-a-b\right)\)\(\left(a+b-c+a+b\right)\)
\(=c\left(2a+2b+c\right)\)\(-c\left(2a+2b-c\right)\)
\(=c\left(2a+2b+c-2a-2b+c\right)\)
\(=c.2c\)
\(=2c^2\)