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17 tháng 1 2018

Giải tắt quá bạn ơi!

\(\dfrac{a^3-3a+2}{2a^3-7a^2+8a-3}\)

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

\(=\dfrac{a\left(a-1\right)\left(a+1\right)-2\left(a-1\right)}{2a^2\left(a-1\right)-5a\left(a-1\right)+3\left(a-1\right)}\)

\(=\dfrac{\left(a-1\right)\left(a^2+a-2\right)}{\left(a-1\right)\left(2a^2-5a+3\right)}\)

\(=\dfrac{\left(a+2\right)\left(a-1\right)}{\left(a-1\right)\left(2a-3\right)}\)

\(=\dfrac{a+2}{2a-3}\)

6 tháng 7 2016

Tử = \(a^3-3a+2=a^3-1-3a+3\)

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

                                \(=\left(a-1\right)\left(a^2+a-2\right)\)

                                \(=\left(a-1\right)\left(a-1\right)\left(a+2\right)=\left(a-1\right)^2\left(a+2\right)\)

Mẫu =\(2a^3-7a^2+8a-3=2a\left(a^2-2a+1\right)-3\left(a^2-2a+1\right)\)

                                                 \(=\left(a-1\right)^2\left(2a-3\right)\)

=>\(\frac{a^3-3a+2}{2a^3-7a^2+8a-3}=\frac{\left(a-1\right)^2\left(a+2\right)}{\left(a-1\right)^2\left(2a-3\right)}=\frac{a+2}{2a-3}\)

Nhớ h cho mik nhé

6 tháng 9 2016

a) \(\sqrt{3a^3}\cdot\sqrt{12a}=\sqrt{3a^3\cdot12a}=\sqrt{36a^4}=6a^2\)

b) \(\sqrt{2a\cdot32ab^2}=\sqrt{64a^2b^2}=8ab\)

25 tháng 11 2021

\(\sqrt{4a^2+12a+9}+\sqrt{4a^2-12a+9}\) với \(-\dfrac{3}{2}\le a\le\dfrac{3}{2}\)

\(\sqrt{\left(2a+3\right)^3}+\sqrt{\left(2a-3\right)^3}\)

\(\left|2a+3\right|+\left|2a-3\right|\)

\(2a+3-2a+3\)

\(6\)

\(7a\left(3a-5\right)+\left(2a-3\right)\left(4a+1\right)-\left(6a-2\right)^2\)

\(=21a^2-35a+8a^2+2a-12a-3-36a^2+24a-4\)

\(=-7a^2+4a-7\)

a) Ta có: \(A=\dfrac{a^2-1}{3}\cdot\sqrt{\dfrac{9}{\left(1-a\right)^2}}\)

\(=\dfrac{\left(a+1\right)\cdot\left(a-1\right)}{3}\cdot\dfrac{3}{\left|1-a\right|}\)

\(=\dfrac{\left(a+1\right)\left(a-1\right)}{1-a}\)

=-a-1

b) Ta có: \(B=\sqrt{\left(3a-5\right)^2}-2a+4\)

\(=\left|3a-5\right|-2a+4\)

\(=5-3a-2a+4\)

=9-5a

c) Ta có: \(C=4a-3-\sqrt{\left(2a-1\right)^2}\)

\(=4a-3-\left|2a-1\right|\)

\(=4a-3-2a+1\)

\(=2a-2\)

d) Ta có: \(D=\dfrac{a-2}{4}\cdot\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\)

\(=\dfrac{a-2}{4}\cdot\dfrac{4a^2}{\left|a-2\right|}\)

\(=\dfrac{a^2\left(a-2\right)}{-\left(a-2\right)}\)

\(=-a^2\)