\(\left(a^2-b^2\right)^2\) 

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

\(=\left(a^2-2ab+b^2\right)\left(a^2+2ab+b^2\right)\)

\(=\left[\left(a^2+b^2\right)-2ab\right]\left[\left(a^2+b^2\right)+2ab\right]\)

Thay \(a^2+b^2=8\) và \(ab=-2\) Ta có:

\(\left(8-2\cdot-2\right)\left(8+2\cdot-2\right)=\left(8+4\right)\left(8-4\right)=12\cdot4=48\)

N= (a² - b²)²
= - (a² + b²)²
= (-8)²
=64

 

Lời giải:

$a^2+b^2+c^2+6=2(a+2b+c)$

$\Leftrightarrow (a^2-2a+1)+(b^2-4b+4)+(c^2-2c+1)=0$

$\Leftrightarrow (a-1)^2+(b-2)^2+(c-1)^2=0$

Vì $(a-1)^2\geq 0; (b-2)^2\geq 0; (c-1)^2\geq 0$ với mọi $a,b,c\in\mathbb{R}$ nên để tổng của chúng bằng $0$ thì:

$(a-1)^2=(b-2)^2=(c-1)^2=0$

$\Rightarrow a=c=1; b=2$

$\Rightarrow K=3$

Đáp án C.

\(A=a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)=1.\left(3-ab\right)\)

ta có: \(\left(a+b\right)^2=1\Leftrightarrow a^2+2ab+b^2=1\Leftrightarrow3+2ab+1=0\Leftrightarrow ab=-1\)

=> \(A=3-\left(-1\right)=4\)

=>a^2-ab-2ab+2b^2=0

=>(a-b)(a-2b)=0

=>a=b(loại) hoặc a=2b

Khi a=2b thì G=(4b+b)/(2b+2b)=5/4

`a^2+4ab-5b^2=0`

`<=>a^2+4ab+4b^2-9b^2=0`

`<=>(a+2b)^2-9b^2=0`

`<=>(a+2b-3b)(a+2b+3b)=0`

`<=>(a-b)(a+5b)=0`

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=-5b\end{matrix}\right.\)

`Q={2a-b}/{a-b}+{3a-2b}/{a+b}`

Với `a=b` `=>` giá trị vô nghĩa

Với `a=-5b` 

`Q={-10b-b}/{-5b-b}+{-15b-2b}/{-5b+b}`

`Q={-11b}/{-6b}+{-17b}/{-4b}`

`Q=11/6+17/4`

`Q=73/12`

Dấu BĐT bị ngược, sửa đề: \(\dfrac{1}{a^4+b^4+2ab^4}+\dfrac{1}{a^2+b^4+2a^2b^2}\le\dfrac{1}{2}\).

Đặt \(b^2=x\left(x>0\right)\Rightarrow a+x=2ax\).

Khi đó ta cần chứng minh:

\(\dfrac{1}{a^4+x^2+2ax^2}+\dfrac{1}{a^2+x^4+2a^2x}\le\dfrac{1}{2}\)

Áp dụng BĐT AM-GM:

\(\dfrac{1}{a^4+x^2+2ax^2}+\dfrac{1}{a^2+x^4+2a^2x}\)

\(\le\dfrac{1}{2a^2x+2ax^2}+\dfrac{1}{2ax^2+2a^2x}\)

\(=\dfrac{2}{2ax\left(a+x\right)}\)

\(=\dfrac{1}{ax\left(a+x\right)}\)

\(=\dfrac{1}{2a^2x^2}\)

Ta thấy: \(a+x\ge2\sqrt{ax}\)

\(\Leftrightarrow2ax\ge2\sqrt{ax}\)

\(\Leftrightarrow ax-\sqrt{ax}\ge0\)

\(\Leftrightarrow\sqrt{ax}\left(\sqrt{ax}-1\right)\ge0\)

\(\Leftrightarrow\sqrt{ax}\ge1\)

\(\Rightarrow ax\ge1\)

Khi đó: \(\dfrac{1}{2a^2x^2}\le\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{a^4+x^2+2ax^2}+\dfrac{1}{a^2+x^4+2a^2x}\le\dfrac{1}{2}\)

Hay \(\dfrac{1}{a^4+b^4+2ab^4}+\dfrac{1}{a^2+b^4+2a^2b^2}\le\dfrac{1}{2}\).