![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left\{\begin{matrix}2x^2=a^2+b^2+c^2\left(1\right)\\a+b=c\left(2\right)\end{matrix}\right.\)
(1)=>\(4x^4=\left(a^4+b^4+c^4\right)+2\left[\left(ab\right)^2+\left(ac\right)^2+\left(bc\right)^2\right]\)(3)
\(A=2\left(ac\right)^2+2\left(ab\right)^2+2\left(bc\right)^2=a^2\left(b^2+c^2\right)+c^2\left(a^2+b^2\right)+b^2\left(a^2+c^2\right)\) (*)
(2)=> \(\left\{\begin{matrix}a^2+b^2=c^2-2ab\\a^2+c^2=b^2+2ac\\b^2+c^2=a^2-2bc\\\end{matrix}\right.\)(4)
Thay (4) vào (*)
\(A=a^2\left(a^2+2bc\right)+c^2\left(c^2-2ab\right)+b^2\left(b^2+2ac\right)=a^4+2a^2bc+c^4-2abc^2+b^4+2ab^2c64\\ \)
\(A=\left(a^4+b^4+c^4\right)+2abc\left(a-c+b\right)=\left(a^4+b^4+c^4\right)+2abc.0=\left(a^4+b^4+c^4\right)\)(3)\(\Leftrightarrow4x^4=\left(a^4+b^4+c^4\right)+\left(a^4+b^4+c^4\right)=2\left(a^4+b^4+c^4\right)\)
\(\Rightarrow2x^4=\left(a^4+b^4+c^4\right)\) => dpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2=a^2+b^2+ab\)
\(\Leftrightarrow x^4=a^4+b^4+a^2b^2+2a^2b^2+2ab^3+2a^3b\)
\(\Leftrightarrow2x^4=2a^4+2b^4+6a^2b^2+4ab^3+4a^3b\)
\(\Leftrightarrow2x^4=a^4+b^4+\left(a^2\right)^2+\left(b^2\right)^2+\left(2ab\right)^2+2a^2b^2+2b^2.2ab+2.2ab.a^2\)
\(\Leftrightarrow2x^4=a^4+b^4+\left(a^2+b^2+2ab\right)^2\)
\(\Leftrightarrow2x^4=a^4+b^4+\left[\left(a+b\right)^2\right]^2\)
\(\Leftrightarrow2x^4=a^4+b^4+c^4\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(x^2=a^2+b^2+ab\)
\(\Leftrightarrow x^4=a^4+3a^2b^2+2a^3b+2ab^3+b^4\)
\(\Leftrightarrow2x^4=2a^4+2b^4+6a^2b^2+4a^3b+4ab^3\)
\(\Leftrightarrow2x^4=a^4+b^4+\left[\left(a^2+2ab+b^2\right)^2\right]\)
\(\Leftrightarrow2x^4=a^4+b^4+\left[\left(a+b\right)^2\right]^2\)
\(\Leftrightarrow2x^4=a^4+b^4+c^4\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: A = B
\(\Leftrightarrow\left(x-3\right)\left(x+4\right)-2\left(3x-2\right)=\left(x-4\right)^2\\ \Leftrightarrow x^2+4x-3x-12-6x+4=x^2-8x+16\\ \Leftrightarrow x^2-x^2+4x-3x-6x+8x=16+12-4\\ \Leftrightarrow3x=24\Leftrightarrow x=8\)
Vậy với x = 8 thì A = B
b) Ta có: A = B
\(\Leftrightarrow\left(x+2\right)\left(x-2\right)+3x^2=\left(2x+1\right)^2+2x\\ \Leftrightarrow x^2-4+3x^2=4x^2+4x+1+2x\\ \Leftrightarrow x^2+3x^2-4x^2-4x-2x=1+4\\ \Leftrightarrow-6x=5\Leftrightarrow x=-\frac{5}{6}\)
Vậy với \(x=-\frac{5}{6}\) thì A = B
c) Ta có: A = B
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-2x=x\left(x-1\right)\left(x+1\right)\\ \Leftrightarrow x^3-1-2x=x\left(x^2-1\right)\\ \Leftrightarrow x^3-1-2x=x^3-x\\ \Leftrightarrow x^3-x^3-2x+x=1\\ \Leftrightarrow-x=1\Leftrightarrow x=-1\)
Vậy với x = -1 thì A = B
d) Ta có: A = B
\(\Leftrightarrow\left(x+1\right)^3-\left(x-2\right)^3=\left(3x-1\right)\left(3x+1\right)\\ \Leftrightarrow x^3+3x^2+3x+1-x^3+6x^2-12x+8=9x^2-1\\ \Leftrightarrow x^3-x^3+3x^2+6x^2-9x^2+3x-12x=-1-1-8\\ \Leftrightarrow-9x=-10\Leftrightarrow x=\frac{10}{9}\)
Vậy với \(x=\frac{10}{9}\) thì A = B.
\(x^2=a^2+b^2+ab\Rightarrow x^4=\left(a^2+b^2+ab\right)^2\)
<=>\(x^4=a^4+b^4+a^2b^2+3a^2b^2+2a^3b+2ab^3\)
<=>\(2x^4=2a^4+2b^4+2a^2b^2+6a^2b^2+4a^3b+4ab^3\)
<=>\(2x^4=a^4+b^4+\left(a^4+4a^3b+6a^2b^2+4ab^3+b^4\right)\)
<=>\(2x^4=a^4+b^4+\left(a+b\right)^4\)
<=>\(2x^4=a^4+b^4+c^4\)(đpcm)