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28 tháng 6 2017

Phép nhân các phân thức đại số

19 tháng 7 2017

Bài 2:
Ta có: \(f\left(a\right)=6a^5-10a^4-5a^3+23a^2-29a+2005\)

\(=\left(6a^5-10a^4-2a^3\right)-\left(3a^3-5a^2-a\right)+\left(18a^2-30a-6\right)+2011\)

\(=2a^3\left(3a^2-5a-1\right)-a\left(3a^2-5a-1\right)+6\left(3a^2-5a-1\right)+2011\)

\(=\left(2a^3-a+6\right)\left(3a^2-5a-1\right)+2011\)

\(3a^2-5a-1=0\)

\(\Rightarrow f\left(a\right)=2011\)

Vậy...

30 tháng 7 2018

e) = \(\dfrac{3}{2\left(x+3\right)}\) - \(\dfrac{x-6}{2x\left(x+3\right)}\)

= \(\dfrac{3x}{2x\left(x+3\right)}\) - \(\dfrac{x-6}{2x\left(x+3\right)}\) = \(\dfrac{3x-x+6}{2x\left(x+3\right)}\)

= \(\dfrac{2x-6}{2x\left(x+3\right)}\)

= \(\dfrac{2\left(x-3\right)}{2x\left(x+3\right)}\)

30 tháng 7 2018

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)

28 tháng 6 2018

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) \(K=a^3+b^3+6a^2b^2\left(a+b\right)+3ab\left(a^2+b^2\right)\)

\(=a^3+b^3+6a^2b^2+3a^3b+3ab^3\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\left(a^2+2ab+b^2\right)\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\left(a+b\right)^2\)

\(=\left(a+b\right)^3+3ab\left(a+b\right)\left(a+b-1\right)\)

\(=1^3+3ab\cdot1\cdot0\)

\(=1\)

28 tháng 6 2018

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 ) \(P=\left(\dfrac{x}{4}\right)^3+\left(\dfrac{y}{2}\right)^3\)

\(\Leftrightarrow P=\left(\dfrac{x}{4}+\dfrac{y}{2}\right)^3-3\left[\left(\dfrac{x}{4}\right)^2\dfrac{y}{2}+\dfrac{x}{4}\left(\dfrac{y}{2}\right)^2\right]\)

\(\Leftrightarrow P=\left(\dfrac{2\left(x+2y\right)}{8}\right)^3-3\left[\dfrac{x^2y}{32}+\dfrac{xy^2}{16}\right]\)

\(\Leftrightarrow P=8-3xy\left(\dfrac{x+2y}{32}\right)\)

\(\Leftrightarrow P=8-3.4\left(\dfrac{8}{32}\right)=5\)

29 tháng 12 2018

\(\dfrac{5a-b}{3a+7}\)-\(\dfrac{3b-2a}{2b-7}\)

=\(\dfrac{5a-b}{3a+2a-b}\)-\(\dfrac{3b-2a}{2b-\left(2a-b\right)}\)

=\(\dfrac{5a-b}{5a-b}\)-\(\dfrac{3b-2a}{2b-2a+b}\) (vì 2a-b=7)

=\(\dfrac{5a-b}{5a-b}\)-\(\dfrac{3b-2a}{3b-2a}\)

=1-1

=0

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)