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Bài 2:
Ta có: \(f\left(a\right)=6a^5-10a^4-5a^3+23a^2-29a+2005\)
\(=\left(6a^5-10a^4-2a^3\right)-\left(3a^3-5a^2-a\right)+\left(18a^2-30a-6\right)+2011\)
\(=2a^3\left(3a^2-5a-1\right)-a\left(3a^2-5a-1\right)+6\left(3a^2-5a-1\right)+2011\)
\(=\left(2a^3-a+6\right)\left(3a^2-5a-1\right)+2011\)
Mà \(3a^2-5a-1=0\)
\(\Rightarrow f\left(a\right)=2011\)
Vậy...
e) = \(\dfrac{3}{2\left(x+3\right)}\) - \(\dfrac{x-6}{2x\left(x+3\right)}\)
= \(\dfrac{3x}{2x\left(x+3\right)}\) - \(\dfrac{x-6}{2x\left(x+3\right)}\) = \(\dfrac{3x-x+6}{2x\left(x+3\right)}\)
= \(\dfrac{2x-6}{2x\left(x+3\right)}\)
= \(\dfrac{2\left(x-3\right)}{2x\left(x+3\right)}\)
c) = \(\dfrac{2\left(a^3-b^3\right)}{3\left(a+b\right)}\) . \(\dfrac{6\left(a+b\right)}{a^2-2ab+b^2}\)
= \(\dfrac{-2\left(a+b\right)\left(a^2-2ab+b^2\right)}{3\left(a+b\right)}\) . \(\dfrac{6\left(a+b\right)}{a^2-2ab+b^2}\)
= \(\dfrac{-2\left(a+b\right)}{1}\) . \(\dfrac{2}{1}\) = -4 (a+b)
a) \(N=8a^3-27b^3\)
\(=\left(2a\right)^3-\left(3b\right)^3\)
\(=\left(2a-3b\right)^3+18ab\left(2a-3b\right)\)
\(=5^3+18\cdot12\cdot5\)
\(=125+1080=1205\)
b) \(K=a^3+b^3+6a^2b^2\left(a+b\right)+3ab\left(a^2+b^2\right)\)
\(=a^3+b^3+6a^2b^2+3a^3b+3ab^3\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\left(a^2+2ab+b^2\right)\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\left(a+b\right)^2\)
\(=\left(a+b\right)^3+3ab\left(a+b\right)\left(a+b-1\right)\)
\(=1^3+3ab\cdot1\cdot0\)
\(=1\)
a ) \(N=8a^3-27b^3\)
\(\Leftrightarrow N=\left(2a-3b\right)\left(4x^2+6ab+9b^2\right)\)
\(\Leftrightarrow N=5\left(4x^2+9b^2+72\right)\)
Ta có : \(2a-3b=5\)
\(\Leftrightarrow4a^2+9b^2=25+6ab\)
Thay vào ta được : \(N=5\left(25+6ab+72\right)=845\)
b ) \(K=a^3+b^3+6a^2b^2\left(a+b\right)+3ab\left(a^2+b^2\right)\)
\(\Leftrightarrow K=\left(a+b\right)^3-3ab\left(a+b\right)+6a^2b^2\left(a+b\right)+3ab\left(a+b\right)^2-6a^2b^2\)
\(\Leftrightarrow K=1-3ab+6a^2b^2+3ab-6a^2b^2=1\)
c ) \(P=\left(\dfrac{x}{4}\right)^3+\left(\dfrac{y}{2}\right)^3\)
\(\Leftrightarrow P=\left(\dfrac{x}{4}+\dfrac{y}{2}\right)^3-3\left[\left(\dfrac{x}{4}\right)^2\dfrac{y}{2}+\dfrac{x}{4}\left(\dfrac{y}{2}\right)^2\right]\)
\(\Leftrightarrow P=\left(\dfrac{2\left(x+2y\right)}{8}\right)^3-3\left[\dfrac{x^2y}{32}+\dfrac{xy^2}{16}\right]\)
\(\Leftrightarrow P=8-3xy\left(\dfrac{x+2y}{32}\right)\)
\(\Leftrightarrow P=8-3.4\left(\dfrac{8}{32}\right)=5\)
1: \(\Leftrightarrow\left(x+2\right)\left(x-2\right)+3\left(x+1\right)=3+x^2-x-2\)
\(\Leftrightarrow x^2-x+1=x^2-4+3x+3=x^2+3x-1\)
=>-4x=-2
hay x=1/2
2: \(\Leftrightarrow\left(x+6\right)^2+\left(x-5\right)^2=2x^2+23x+61\)
\(\Leftrightarrow x^2+12x+36+x^2-10x+25=2x^2+23x+61\)
\(\Leftrightarrow2x^2+23x+61=2x^2+2x+11\)
=>21x=-50
hay x=-50/21
3: \(\Leftrightarrow6\left(x-8\right)+\left(x+2\right)\left(x-5\right)=-18-\left(x-5\right)\left(x-8\right)\)
\(\Leftrightarrow6x-48+x^2-3x-10+18+x^2-13x+40=0\)
\(\Leftrightarrow2x^2-10x=0\)
=>2x(x-5)=0
=>x=0(nhận) hoặc x=5(loại)