Phân tích thành nhân tử.
a) 2 – 4a + 4bc + 2a2 – 2b2 – 2c2 ; b) 3x2 + 9cd( 2 -3cd) – 30xy – 3 + 75y2 .
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\(8a^4-2a^2-4a+2\)
\(=2\cdot\left(4a^4-a^2-2a+1\right)\)
\(=2\cdot\left(2a-1\right)\cdot\left(2a^3+a^2-1\right)\)
\(8a^4-2a^2-4a+2\)
\(=2\left(4a^4-a^2-2a+1\right)\)
\(=2\left(4a^4-2a^3+2a^3-a^2-2a+1\right)\)
\(=2\left(2a-1\right)\left(2a^3+a^2-1\right)\)
P≤√a2+2√aab+2b2+√b2+2√2bc+2c2+√c2+2√2ca+2a2P≤a2+2aab+2b2+b2+22bc+2c2+c2+22ca+2a2
P≤√(a+√2b)2+√(b+√2c)2+√(c+√2a)2P≤(a+2b)2+(b+2c)2+(c+2a)2
P≤(1+√2)(a+b+c)=1+√2P≤(1+2)(a+b+c)=1+2
Dấu "=" xảy ra khi (a;b;c)=(0;0;1)(a;b;c)=(0;0;1) và các hoán vị
Xét hiệu \(2a^2+2b^2-\left(a^3+ab^2\right)=\left(2a^2-a^3\right)+\left(2b^2-ab^2\right)\)
\(=a^2\left(2-a\right)+b^2\left(2-a\right)\)
\(=\left(a^2+b^2\right)\left(2-a\right)\)
Do \(a^2+b^2\ge0;\forall a;b\) nên:
\(2a^2+2b^2>a^3+ab^2\) khi \(\left\{{}\begin{matrix}a^2+b^2\ne0\\2-a>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2\ne0\\a< 2\end{matrix}\right.\)
\(2a^2+2b^2=a^3+ab^2\) khi \(\left[{}\begin{matrix}a^2+b^2=0\\2-a=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a=b=0\\a=2\end{matrix}\right.\)
\(2a^2+2b^2< a^3+ab^2\) khi \(\left\{{}\begin{matrix}a^2+b^2\ne0\\a>2\end{matrix}\right.\) \(\Rightarrow a>2\)
\(2a^2+2b^2\ge a^3+ab^2\) khi \(2-a\ge0\Leftrightarrow a\le2\)
a: \(\left(x-3\right)^2-\left(2x-5\right)^2\)
\(=\left(x-3-2x+5\right)\left(x-3+2x-5\right)\)
\(=\left(2-x\right)\left(3x-8\right)\)
b: \(\left(x+y\right)^2-x^2+4xy-4y^2\)
\(=\left(x+y\right)^2-\left(x-2y\right)^2\)
\(=\left(x+y+x-2y\right)\left(x+y-x+2y\right)\)
\(=3y\left(2x-y\right)\)
a) 2 - 4a + 4bc + 2a2 - 2b2 - 2c2
=> 2.( 1 - 2a + 2bc + a2 - b2 - c2)
b) 3x2 + 9cd.(2 - 3cd) - 30xy - 3 + 75y2
=> 3x2 + 18cd - 27c2d2 - 30xy - 3 + 75y2
=> 3.( x2 + 6cd - 9c2d2 - 10xy -1 + 25y2)