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\(a^3-3ab^2=46\)\(\Rightarrow\left(a^3-3ab^2\right)=46^2\)\(\Rightarrow a^6-6a^4b^2+9a^2b^4=2116\)
\(b^3-3a^2b=9\Rightarrow\left(b^3-3a^2b\right)^2=9^2\Rightarrow b^6-6a^2b^4+9a^4b^2=81\)
\(\Rightarrow a^6-6a^4b^2+9a^2b^4+b^6-6a^2b^4+9a^4b^2=2197\)
\(\Rightarrow a^6+3a^4b^2+3a^2b^4+b^6=2197\)
\(\Rightarrow\left(a^2+b^2\right)^3=2197\)
\(\Rightarrow a^2+b^2=13\)
thôi mk tự lm đc rồi:
(a^3- 3ab^2)^2=361
=a^6- 6a^4b^2+ 9a^2 b^4
(b^3-3a^2b)^2=9604
=b^6- 6a^2b^4+9a^4 b^2
cộng 2 vế->(a^2+b^2)^3= 9604+361= 9965
mn check hộ mk nha
\(2x^2+y^2+9=6x+2xy\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(x-3\right)^2=0\Leftrightarrow\hept{\begin{cases}x-3=0\\x-y=0\end{cases}}\Leftrightarrow x=y=3\)
\(\Rightarrow A=x^{2019}.y^{2020}-x^{2020}.y^{2019}+\frac{1}{9xy}=\frac{1}{27}\)
\(3a^2+4b^2=7ab\)
\(\Rightarrow3a^2+4b^2-7ab=0\)
\(\Rightarrow3a^2-3ab-4ab+4b^2=0\)
\(\Rightarrow3a\left(a-b\right)-4b\left(a-b\right)=0\)
\(\Rightarrow\left(a-b\right)\left(3a-4b\right)=0\)
Mà \(a\ne b\Rightarrow a-b\ne0\)
Từ đó \(3a-4b=0\Rightarrow3a=4b\Rightarrow a=\frac{4}{3}b\)
\(E=\frac{a+2b}{3a-b}=\frac{\frac{4}{3}b+2b}{3.\frac{4}{3}b-b}=\frac{10}{9}\)
\(1,M=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)
\(=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\left(a+b\right)\)
\(=\left(a+b\right)\left[\left(a+b\right)^2-3ab\right]+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\left(a+b\right)\)
Thay \(a+b=1\) vào ta được:
\(1\left(1-3ab\right)+3ab\left(1-2ab\right)+6a^2b^2\)
\(=1-3ab+3ab-6a^2b^2+6a^2b^2\)
\(=1\)
Vậy ......................
`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`
Ta có:
\(a^3+b^3=3ab-1\)
\(\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)=3ab-1\)
\(\Leftrightarrow\left(a+b\right)\left(a^2+2ab+b^2-3ab\right)=3ab-1\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)=3ab-1\)
\(\Leftrightarrow\left(a+b\right)^3+1-3ab\left(a+b\right)-3ab=0\)
\(\Leftrightarrow\left(a+b+1\right)\left[a^2+2ab+b^2-a-b+1\right]-3ab\left(a+b+1\right)=0\)
\(\Leftrightarrow\left(a+b+1\right)\left(a^2+2ab+b^2-a-b+1-3ab\right)=0\)
\(\Leftrightarrow\left(a+b+1\right)\left(a^2-ab+b^2-a-b+1\right)=0\)
\(\Leftrightarrow\left(a+b+1\right)\left(2a^2+2b^2-2ab-2a-2b+2\right)=0\)
\(\Leftrightarrow\left(a+b+1\right)\left(a^2-2a+1+b^2-2b+1+a^2-2ab+b^2\right)=0\)
\(\Leftrightarrow\left(a+b+1\right)\left[\left(a-1\right)^2+\left(b-1\right)^2+\left(a-b^2\right)\right]=0\)
.......
Mình nghĩ đề a, b là 2 số dương nha, nếu a,b là 2 số dương thì mình loại được trường hợp a+b+1=0 nhé
Ta có:
\(\left(a^3+3ab^2\right)^2=a^6+6a^4b^2+9a^2b^4=196\)
\(\left(b^3+3a^2b\right)^2=b^6+6a^2b^4+9a^4b^2=169\)
Lại có:
\(\left(a^3+3ab^2\right)^2-\left(b^3+3a^2b\right)^2=27\)
\(\Leftrightarrow a^6+6a^4b^2+9ab^4-b^6-6a^2b^4-9a^4b^2=27\)
\(\Leftrightarrow a^6-3a^4b^2+3a^2b^4-b^6=27\)
\(\Leftrightarrow\left(a^2-b^2\right)^3=27\)
\(\Leftrightarrow a^2-b^2=\sqrt[3]{27}=3\)
\(a^3+3ab^2+b^3+3a^2b=27=\left(a+b\right)^3\Rightarrow a+b=3\)
\(a^3+3ab^2-b^3-3a^2b=1\Rightarrow\left(a-b\right)^3=1\Rightarrow a-b=1\)
\(\Rightarrow a^2-b^2=\left(a-b\right).\left(a+b\right)=3\)