\(N=\left(a-3b\right)^2-\left(a+3b\right)^2-\left(a-1\right)\left(b-2\right)=\left(a-3b-a-3b\right)\left(a-3b+a+3b\right)-\left(ab-2a-b+2\right)=\left(-6b\right).2a-ab+2a+b-2=2a+b-13ab-2\)

Thay \(a=\dfrac{1}{2};b=-3\) vào N ta được: \(N=2a+b-13ab-2=2.\dfrac{1}{2}-3-13.\dfrac{1}{2}.\left(-3\right)-2=\dfrac{31}{2}\)

Ta có: \(N=\left(a-3b\right)^2-\left(a+3b\right)^2-\left(a-1\right)\left(b-2\right)\)

\(=a^2-6ab+9b^2-a^2-6ab-9b^2-ab+2a+b-2\)

\(=-13ab+2a+b-2\)

\(=-13\cdot\dfrac{1}{2}\cdot\left(-3\right)-1-3-2\)

\(=\dfrac{27}{2}\)

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

\(=a^2-6ab+9b^2-a^2-6ab-9b^2-ab+2a+b-2\)

\(=-13ab+2a+b-2\)

Thay \(a=\dfrac{1}{2};b=-3\) vào bt N được

\(N=\left(-13\right)\cdot\dfrac{1}{2}\cdot\left(-3\right)+2\cdot\dfrac{1}{2}+\left(-3\right)-2\)

\(=\dfrac{31}{2}\)

Vậy: Giá trị của N tại \(a=\dfrac{1}{2};b=-3\) là \(\dfrac{31}{2}\)

a) Ta có: \(\dfrac{3a^2-10a+3}{2\left(a-3\right)}\)

\(=\dfrac{3a^2-9a-a+3}{2\left(a-3\right)}\)

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

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

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

\(=\dfrac{3}{2}a-\dfrac{1}{2}\)(đpcm)

b) Ta có: \(\dfrac{b^2+3b+9}{b^3-27}\)\(=\dfrac{b^2+3b+9}{\left(b-3\right)\left(b^2+3b+9\right)}\)

\(=\dfrac{1}{b-3}\)

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

\(=\dfrac{b-2}{b^2-5b+6}\)(đpcm)

a) Ta có: \(a=\frac{3}{5}=0,6\)

\(A=\left(1^2+2^2+3^2+...+20^2\right).\left(a+b\right)\left(2a+b\right)\left(a+3b\right)\)

\(\Rightarrow A=\left(1^2+2^2+3^2+...+20^2\right)\left(a+b\right)\left(2a+b\right)\left[0,6+3.\left(-0,2\right)\right]\)

\(\Rightarrow A=\left(1^2+2^2+...+20^2\right)\left(a+b\right)\left(2a+b\right)\left(0,6-0,6\right)\)

\(\Rightarrow A=\left(1^2+2^2+...+20^2\right)\left(a+b\right)\left(2a+b\right).0\)

\(\Rightarrow A=0\)

Vậy A = 0

b) Ta có: \(\frac{a}{b}=\frac{3}{4}\Rightarrow\frac{a}{3}=\frac{b}{4}\)

Đặt \(\frac{a}{3}=\frac{b}{4}=k\Rightarrow\left\{\begin{matrix}a=3k\\b=4k\end{matrix}\right.\)

\(B=\frac{2a-3b}{a-3b}=\frac{2.3.k-3.4.k}{3k-3.4.k}=\frac{6k-12k}{3k-12k}=\frac{\left(6-12\right)k}{\left(3-12\right)k}=\frac{-6}{-9}=\frac{2}{3}\)

Vậy \(B=\frac{2}{3}\)

a) \(\left(2x+y\right)^2-\left(2x+y\right)\left(2x-y\right)+y\left(x-y\right)\)

\(=\left(2x+y\right)^2-\left(2x\right)^2+y^2+xy-y^2\)

\(=\left(2x+y+2x\right)\left(2x+y-2x\right)+xy\)

\(=\left(4x+y\right)y+xy\)

\(=\left[4\left(-2\right)+3\right].3+\left(-2\right).3\)

\(=\left(-8+3\right).3+1\)

\(=-15+1\)

\(=-14\)

thôi nha

Chắc đề ghi nhầm ngoặc sau (2 mẫu kia thực chất giống nhau, lẽ ra phải là \(\dfrac{1}{\sqrt{a+3b}}+\dfrac{1}{\sqrt{3a+b}}\)

\(VT=\sqrt{\dfrac{a}{a+3b}}+\sqrt{\dfrac{a}{3a+b}}+\sqrt{\dfrac{b}{a+3b}}+\sqrt{\dfrac{b}{3a+b}}\)

\(=\sqrt{\dfrac{a}{a+b}.\dfrac{a+b}{a+3b}}+\sqrt{\dfrac{1}{2}.\dfrac{2a}{3a+b}}+\sqrt{\dfrac{1}{2}.\dfrac{2b}{a+3b}}+\sqrt{\dfrac{b}{a+b}.\dfrac{a+b}{3a+b}}\)

\(\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a+b}{a+3b}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}+\dfrac{2a}{3a+b}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}+\dfrac{2b}{a+3b}\right)+\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{a+b}{3a+b}\right)\)

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

Dấu "=" xảy ra khi \(a=b\)