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\(N=a^3+b^3+3ab\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\)
=1
\(M=\left(a^2+b^2+2-a^2-b^2+2\right)\left[\left(a^2+b^2+2\right)^2+\left(a^2+b^2+2\right)\left(a^2+b^2-2\right)+\left(a^2+b^2-2\right)^2\right]-12\left(a^2+b^2\right)^2\\ M=4\left(a^4+b^4+4+4a^2+4b^2+2a^2b^2+\left(a^2+b^2\right)^2-4+a^4+b^4+4-4a^2-4b^2+2a^2b^2\right)-12\left(a^4+2a^2b^2+b^4\right)\\ M=4\left(3a^4+3b^4+4+6a^2b^2\right)-12\left(a^4+2a^2b^2+b^4\right)\\ M=4\left(3a^4+3b^4+4+6a^2b^2-3a^4-6a^2b^2-3b^4\right)\\ M=4\cdot4=164\)
a: Ta có: \(\frac{1}{2a-b}-\frac{a^2-1}{2a^3-b+2a-a^2b}\)
\(=\frac{1}{2a-b}-\frac{a^2-1}{a^2\left(2a-b\right)+\left(2a-b\right)}\)
\(=\frac{1}{2a-b}-\frac{a^2-1}{\left(2a-b\right)\left(a^2+1\right)}=\frac{a^2+1-a^2+1}{\left(2a-b\right)\left(a^2+1\right)}=\frac{2}{\left(2a-b\right)\left(a^2+1\right)}\)
\(\frac{4a+2b}{a^3b+ab}-\frac{2}{a}\)
\(=\frac{4a+2b}{ab\left(a^2+1\right)}-\frac{2}{a}=\frac{4a+2b-2b\left(a^2+1\right)}{ab\left(a^2+1\right)}\)
\(=\frac{4a-2a^2b}{ab\left(a^2+1\right)}=\frac{2a\left(2-ab\right)}{ab\cdot\left(a^2+1\right)}=\frac{2\left(2-ab\right)}{b\left(a^2+1\right)}\)
Ta có: \(A=\left(\frac{1}{2a-b}-\frac{a^2-1}{2a^3-b+2a-a^2b}\right):\left(\frac{4a+2b}{a^3b+ab}-\frac{2}{a}\right)\)
\(=\frac{2}{\left(2a-b\right)\left(a^2+1\right)}:\frac{2\left(2-ab\right)}{b\left(a^2+1\right)}=\frac{2b\left(a^2+1\right)}{2\left(2-ab\right)\left(2a-b\right)\left(a^2+1\right)}=\frac{b}{\left(2-ab\right)\left(2a-b\right)}\)
b:
Sửa đề: b>a>0
\(4a^2+b^2=5ab\)
=>\(4a^2-5ab+b^2=0\)
=>\(4a^2-4ab-ab+b^2=0\)
=>(a-b)(4a-b)=0
TH1: a-b=0
=>a=b
mà a>b
nên Loại
TH2: 4a-b=0
=>b=4a(nhận)
\(A=\frac{b}{\left(2-ab\right)\left(2a-b\right)}\)
\(=\frac{4a}{\left(2-a\cdot4a\right)\left(2a-4a\right)}=\frac{4a}{\left(2-4a^2\right)\left(-2a\right)}\)
\(=\frac{4a}{-2a\cdot\left(-2\right)\left(2a^2-1\right)}=\frac{1}{2a^2-1}\)
\(a,\left(2a-3\right)\left(a+1\right)+\left(a^2+6a+9\right):\left(a+3\right)\\ =2a^2-a-3+\left(a+3\right)^2:\left(a+3\right)\\ =2a^2-a-3+a+3\\ =2a^2\\ b,\left(3x-5y\right)\left(-xy\right)^2-3x^2y^2+4x^2y^3\\ =3x^3y^2-5x^2y^3-3x^2y^2+4x^2y^3\\ =3x^3y^2-3x^2y^2-x^2y^3\\ c,x\left(x-2\right)^2-\left(x+2\right)\left(x^2-2x+4\right)+4x^2\\ =x^3-4x^2+4x-x^3-8+4x^2\\ =4x-8\)
b: Ta có: \(N=a^3+b^3+3ab\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\)
\(=1-3ab+3ab\)
=1
a) M = 8ab;
b) N = [ ( 3 a + + 2 ) + ( 1 – 2 b ) ] 2 = ( 3 a – 2 b + 3 ) 2 .
a)
$A=\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}$
$\ge \dfrac{(a+b+c)^2}{a(b^2+1)+b(c^2+1)+c(a^2+1)}\qquad (\text{Cauchy Engel})$
$=\dfrac{1}{ab^2+bc^2+ca^2+1}$
$\ge \dfrac{1}{ab(a+b)+bc(b+c)+ca(c+a)+1}$
$=\dfrac{1}{(a+b+c)(ab+bc+ca)+1}
$\ge \dfrac{1}{\frac13+1}$ $=\dfrac34$
Dấu bằng khi $a=b=c=\dfrac13$.
$\boxed{A_{\min}=\dfrac34}$
b)
$B=\dfrac{a}{ab+2c}+\dfrac{b}{bc+2a}+\dfrac{c}{ca+2b}$
$\ge \dfrac{(a+b+c)^2}{a(ab+2c)+b(bc+2a)+c(ca+2b)}$
$=\dfrac4{a^2b+b^2c+c^2a+2(ab+bc+ca)}$
Lại có $a^2b+b^2c+c^2a\le (a+b+c)(ab+bc+ca)$$=2(ab+bc+ca)$
Nên $B\ge \dfrac4{4(ab+bc+ca)}$$=\dfrac1{ab+bc+ca}$
$\ge \dfrac1{\frac{(a+b+c)^2}{3}}$ $=\dfrac34$
Dấu bằng khi $a=b=c=\dfrac23$.
$B_{\min}=\dfrac34$
Ta có: \(\left(\frac{a+b}{b}-\frac{2b}{b-a}\right)\cdot\frac{b-a}{a^2+b^2}+\left(\frac{a^2+1}{2a-1}-\frac{a}{2}\right):\frac{a+2}{1-2a}\)
\(=\frac{\left(a+b\right)\left(a-b\right)+2b^2}{b\left(a-b\right)}\cdot\frac{-\left(a-b\right)}{a^2+b^2}+\frac{2\left(a^2+1\right)-a\cdot\left(2a-1\right)}{2\left(2a-1\right)}\cdot\frac{-\left(2a-1\right)}{a+2}\)
\(=\frac{a^2-b^2+2b^2}{-b}\cdot\frac{1}{a^2+b^2}+\frac{2a^2+2-2a^2+a}{2}\cdot\frac{-1}{a+2}\)
\(=\frac{-1}{b}+\frac{-1}{2}=\frac{-2-b}{2b}\)