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Câu 1:
Theo bài ra ta có:
\(a^{12}+b^{12}=a^{12}+a^{11}b-a^{11}b-ab^{11}+ab^{11}+b^{12}\)
\(=a^{11}\left(a+b\right)-ab\left(a^{10}+b^{10}\right)+b^{11}\left(a+b\right)\)
\(=\left(a+b\right)\left(a^{11}+b^{11}\right)-ab\left(a^{10}+b^{10}\right)\)
\(=\left(a+b\right)\left(a^{12}+b^{12}\right)-ab\left(a^{12}+b^{12}\right)\)(gt cho rồi nhé)
\(=\left(a^{12}+b^{12}\right)\left(a+b-ab\right)\)
\(\Rightarrow a+b-ab=1\)
\(\Leftrightarrow a+b-ab-1=0\)
\(\Leftrightarrow a\left(1-b\right)-\left(1-b\right)=0\)
\(\Leftrightarrow\left(1-b\right)\left(a-1\right)=0\)
\(\)\(\Leftrightarrow\left[{}\begin{matrix}b=1\\a=1\end{matrix}\right.\)
=> a^20 + b^20 = 2
:)) đừng ném đá nhá
2, a^3-3ab^2 = 5
<=> (a^3-3ab^2)^2 = 25
<=> a^6-6a^4b^2+9a^2b^4 = 25
b^3-3a^2b=10
<=> (b^3-3a^2b)^2 = 100
<=> b^6-6a^2b^4+9a^4b^2 = 100
=> 100+25 = a^6-6a^4b^2+9a^2b^4+b^6+6a^2b^4+9a^4b^2
<=> 125 = a^6+3a^4b^2+3a^3b^4+b^6 = (a^2+b^2)^3
<=> a^2+b^2 = 5
Khi đó : S = 2016.(a^2+b^2) = 2016.5 = 10080
Tk mk nha
1) \(x^2+6xy+5y^2-5y-x=\left(x^2+xy-x\right)+\left(5xy+5y^2-5y\right)\)
\(=x\left(x+y-1\right)+5y\left(x+y-1\right)\)
\(=\left(x+5y\right)\left(x+y-1\right)\)
2) Ta có : \(a^3-3ab^2-5\Rightarrow\left(a^3-3ab^2\right)^2=25\Rightarrow a^6-6a^4b^2+9a^2b^4=25\)
và \(b^3-3a^2b=10\Rightarrow\left(b^3-3a^2b\right)^2=100\Rightarrow b^6-6b^4a^2+9a^4b^2=100\)
\(\Rightarrow\)\(125=a^6+b^6+3a^2b^4+3a^4b^2\)
Hay \(125=\left(a^2+b^2\right)^2\Rightarrow a^2+b^2=5\)
Nên \(S=2016\left(a^2+b^2\right)=2016.5=10080\)
1a)\(\dfrac{a^2+b^2}{2}\ge\dfrac{\left(a+b\right)^2}{4}\)
\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)(luôn đúng)
b)\(\dfrac{a^2+b^2+c^2}{3}\ge\dfrac{\left(a+b+c\right)^2}{9}\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(luôn đúng)
2a)\(a^2+\dfrac{b^2}{4}\ge ab\)
\(\Leftrightarrow a^2-ab+\dfrac{b^2}{4}\ge0\)
\(\Leftrightarrow a^2-2\cdot\dfrac{1}{2}b\cdot a+\left(\dfrac{1}{2}b\right)^2\ge0\)
\(\Leftrightarrow\left(a-\dfrac{1}{2}b\right)^2\ge0\)(luôn đúng)
b)Đã cm
c)\(a^2+b^2+1\ge ab+a+b\)
\(\Leftrightarrow2a^2+2b^2+2\ge2ab+2a+2b\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\)(luôn đúng)
Dấu bằng xảy ra khi a=b=1
\(a^2\left(a+1\right)-b^2\left(b-1\right)-3ab\left(a-b+\frac{2}{3}\right)\)
\(=a^3+a^2-b^3+b^2-3a^2b+3ab^2-2ab\)
\(=\left(a^3-3a^2b+3ab^2-b^2\right)+\left(a^2-2ab+b^2\right)\)
\(=\left(a-b\right)^3+\left(a-b\right)^2\)(*)
Thay a - b = 7 vào (*), ta có:
(*) \(\Leftrightarrow7^3+7^2=392\)
Thanks bạn nhiều :3