Xét \(a+b=2\Rightarrow\left(a+b\right)^2=4\Leftrightarrow a^2+2ab+b^2=4\Leftrightarrow20+2ab=4\)

\(\Leftrightarrow2ab=-16\Leftrightarrow ab=-8\)

Vậy \(M=a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)=2.\left[20-\left(-8\right)\right]=20.28=560\)

(a+b)²=a²+2ab+b²=2²=4

=>2ab=4-a²-b²

=>2ab=4-20

=>2ab=-16

=>ab=-8

(a+b)(a²+b²)=(a+b)a²+(a+b)b²=a³+a²b+ab²+b³

=a³+b³+ab(a+b)=2.20

=>a³+b³+-16.2=40

=>a³+b³=40+32

=>a³+b³=72

Ta có:(a+b)^2=2^2

<=>a^2+2ab+b^2=4

<=>20+2ab=4

<=>ab=-8

Lại có:a^3+b^3=(a+b)(a^2-ab+b^2)

=2(20+8)=56

Vậy a^3+b^3=56

Đặt S = a + b 

P = a * b 

\(a^2+b^2=20\) 

\(a^2+2ab+b^2-2ab=20\) 

\(\left(a+b\right)^2-2ab=20\) 

\(6^2-2P=20\) 

\(36-2P=20\) 

\(2P=36-20\) 

\(2P=16\) 

\(P=8\) 

\(a^3+b^3\) 

\(=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2\) 

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

\(=S^3-3PS\) 

\(=6^3-3\cdot8\cdot6\) 

\(=216-144\) 

\(=72\)                

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

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

om=3cm,on=6cm

bao giờ vào nhà tao tao chỉ cho

biết thì trả lời cho mình chỉ cần kết quả thôi

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

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

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

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

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

1) A+B = \(-2x^2+3x^4+4x^3+1\)

A-B = \(3x^4-2x^2-4x^3+1\)

2) A+B= 0 + 0 + 5 

⇒A+B = 5

A-B = \(-4x^3+2x^2-35\)

3) A+B = \(5y^2-8xy\)

A-B = \(-2x^2-3y^2\)