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

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

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

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

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

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

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

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

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

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

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

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

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

\(6\)

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

=>\(\left[\begin{array}{l}a^2-3a+1=1\\ a^2-3a+1=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}a^2-3a=0\\ a^2-3a+2=0\end{array}\right.\)

=>\(\left[\begin{array}{l}a\left(a-3\right)=0\\ \left(a-1\right)\left(a-2\right)=0\end{array}\right.\Rightarrow a\in\left\lbrace0;1;2;3\right\rbrace\)

ĐKXĐ: a<>2

=>a∈{0;1;3}

\(A=\frac{2a^3-12a^2+17a-2}{a-2}\)

\(=\frac{2a^3-4a^2-8a^2+16a+a-2}{a-2}=\frac{2a^2\left(a-2\right)-8a\left(a-2\right)+\left(a-2\right)}{a-2}\)

\(=2a^2-8a+1\)

Khi a=0 thì \(A=2a^2-8a+1=2\cdot0^2-8\cdot0+1=1\)

Khi a=1 thì \(A=2a^2-8a+1=2\cdot1^2-8\cdot1+1=2-8+1=3-8=-5\)

Khi a=3 thì \(A=2a^2-8a+1=2\cdot3^2-8\cdot3+1=18-24+1=19-24=-5\)

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

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

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

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

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

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

\(=-a^2\)

a: ĐKXĐ: a∉{-1/3;-3}

\(\frac{3a-1}{3a+1}+\frac{a-3}{a+3}=2\)

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

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

=>\(2\left(3a^2+9a+a+3\right)=3a^2+9a-a-3+3a^2-9a+a-3\)

=>\(6a^2+20a+6=6a^2-6\)

=>20a=-12

=>a=-3/5(nhận)

b: ĐKXĐ: a∉{5/2;2/3}

\(\frac{2a-9}{2a-5}+\frac{3a}{3a-2}=2\)

=>\(\frac{2a-5-4}{2a-5}+\frac{3a-2+2}{3a-2}=2\)

=>\(1-\frac{4}{2a-5}+1+\frac{2}{3a-2}=2\)

=>\(\frac{2}{3a-2}=\frac{4}{2a-5}\)

=>\(\frac{4}{6a-4}=\frac{4}{2a-5}\)

=>6a-4=2a-5

=>4a=-1

=>a=-1/4(nhận)

c: ĐKXĐ: a<>-3

\(\frac{10}{3}-\frac{3a-1}{4a+12}-\frac{7a+2}{6a+18}=2\)

=>\(\frac{3a-1}{4a+12}+\frac{7a+2}{6a+18}=\frac{10}{3}-2=\frac43\)

=>\(\frac{3\left(3a-1\right)}{12\left(a+3\right)}+\frac{2\left(7a+2\right)}{12\left(a+3\right)}=\frac43\)

=>\(\frac{9a-3+14a+4}{12\left(a+3\right)}=\frac{4\cdot4\cdot\left(a+3\right)}{12\left(a+3\right)}\)

=>23a+1=16(a+3)=16a+48

=>7a=47

=>a=47/7(nhận)