a: Thay x=-3 vào A, ta được:

\(A=\frac{-3+2}{-3}=\frac{-1}{-3}=\frac13\)

\(x=\sqrt{\left(-3\right)^2}=\sqrt9=3\)

Thay x=3 vào A, ta được:

\(A=\frac{3+2}{3}=\frac53\)

b: \(B=\frac{3}{x+5}+\frac{20-2x}{x^2-25}\)

\(=\frac{3}{x+5}+\frac{20-2x}{\left(x+5\right)\left(x-5\right)}\)

\(=\frac{3\left(x-5\right)+20-2x}{\left(x+5\right)\left(x-5\right)}=\frac{3x-15+20-2x}{\left(x+5\right)\left(x-5\right)}=\frac{x+5}{\left(x+5\right)\left(x-5\right)}\)

\(=\frac{1}{x-5}\)

c: \(A=B\cdot\left|x-4\right|\)

=>\(\frac{x+2}{x}:\frac{1}{x-5}=\left|x-4\right|\)

=>\(\frac{\left(x+2\right)\left(x-5\right)}{x}=\left|x-4\right|\)

=>\(\begin{cases}\frac{\left(x+2\right)\left(x-5\right)}{x}\ge0\\ \left(x+2\right)^2\cdot\frac{\left(x-5\right)^2}{x^2}=\left(x-4\right)^2\end{cases}\Rightarrow\begin{cases}\left[\begin{array}{l}-2\le x<0\\ x\ge5\end{array}\right.\\ \left(x+2\right)^2\cdot\left(x-5\right)^2=x^2\cdot\left(x-4\right)^2\end{cases}\)

Ta có: \(\left(x+2\right)^2\cdot\left(x-5\right)^2=x^2\cdot\left(x-4\right)^2\)

=>\(\left(x^2-3x-10\right)^2=\left(x^2-4x\right)^2\)

=>\(\left(x^2-4x-x^2+3x+10\right)\left(x^2-4x+x^2-3x-10\right)=0\)

=>(-x+10)\(\left(2x^2-7x-10\right)=0\)

TH1: -x+10=0

=>-x=-10

=>x=10(nhận)

TH2: \(2x^2-7x-10=0\)

=>\(x^2-\frac72x-5=0\)

=>\(x^2-2\cdot x\cdot\frac74+\frac{49}{16}-\frac{129}{16}=0\)

=>\(\left(x-\frac74\right)^2=\frac{129}{16}\)

=>\(\left[\begin{array}{l}x-\frac74=\frac{\sqrt{129}}{4}\\ x-\frac74=-\frac{\sqrt{129}}{4}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\sqrt{129}+7}{4}\left(loại\right)\\ x=\frac{-\sqrt{129}+7}{4}\left(nhận\right)\end{array}\right.\)

\(a^2+b^2=20\Leftrightarrow\left(a+b\right)^2-2ab=20\Leftrightarrow2^2-2ab=20\Rightarrow ab=-8\)

\(M=a^3+b^3=\left(a+b\right)^3-3a^2b-3ab^2=\left(a+b\right)^3-3ab\left(a+b\right)=2^3-3.\left(-8\right).2=56\)

om=3cm,on=6cm

a: \(x^2-8x+5\)

\(=x^2-8x+16-11\)

\(=\left(x-4\right)^2-11\ge-11\forall x\)

Dấu '=' xảy ra khi x-4=0

=>x=4

b: \(a^3+b^3+c^3=3bac\)

=>\(\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)

=>\(\left(a+b+c\right)\left\lbrack\left(a+b\right)^2-c\left(a+b\right)+c^2\right\rbrack-3ab\left(a+b+c\right)=0\)

=>\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)

=>\(a^2+b^2+c^2-ab-ac-bc=0\)

=>\(2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)

=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

=>a=b=c

\(N=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}\)

\(=\frac{a^2+a^2+a^2}{\left(a+a+a\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3}{3^2}=\frac13\)

a, a x 6 = 3 x 6 = 18

b, a + b = 4 + 2 = 6

c, b + a = 2 + 4 = 6

d, a - b = 8 - 5 = 3

e, m x n = 5 x 9 = 45

a. 18

b. 6

c. 6

d. 3

e. 45

Ta có: \(M=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\cdot\left(a+b\right)\)

\(\Leftrightarrow M=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left(a^2+b^2\right)+6a^2b^2\)

\(\Leftrightarrow M=a^2-ab+b^2+3ab\left(a^2+2ab+b^2\right)\)

\(\Leftrightarrow M=a^2-ab+b^2+3ab\cdot\left(a+b\right)^2\)

\(\Leftrightarrow M=a^2-ab+3ab+b^2\)

\(\Leftrightarrow M=\left(a+b\right)^2=1^2=1\)

Vậy: Khi a+b=1 thì M=1

M=(a+b)^3-3ab(a+b)+3ab[(a+b)^2-2ab]+6a^2b^2

=1-3ab+3ab(1-2ab)+6a^2b^2

=1