chọn C vì advise sb to do sth ( công thức: advise somebody to do sth: Khuyên ai đó làm j)

Advise + O + (not) + to V : khuyên ai đó (không) làm gì

-> chọn C

Bài 2: 

\(a^2+b^2=\left(a+b\right)^2-2ab=5^2-2\cdot\left(-2\right)=9\)

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

\(=\dfrac{5^3-3\cdot5\cdot\left(-2\right)}{\left(-2\right)^3}=\dfrac{125+30}{8}=\dfrac{155}{8}\)

\(a-b=-\sqrt{\left(a+b\right)^2-4ab}=-\sqrt{5^2-4\cdot\left(-2\right)}=-\sqrt{33}\)

a, a = 40; b = 226

    a + b = 40 + 226 = 266

b, a = 160; b = 347

    a + b = 160 + 347 = 507 

a/ Nếu a=40, b=226 thì a+b=40+226=266

b/ Nếu a=160,b=347 thì a+b=160+347=507

dễ nhưng ko chỉ

Có: \(a+b=a^2+b^2=a^3+b^3\)

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

\(\Rightarrow\left(a-2a^2+a^3\right)+\left(b-2b^2+b^3\right)=0\)

\(\Rightarrow a\left(1-2a+a^2\right)+b\left(1-2b+b^2\right)=0\)

\(\Rightarrow a\left(1-a\right)^2+b\left(1-b\right)^2=0\) (1)

Vì: \(a>0;\left(1-a\right)^2\ge0\)

=> \(a\left(1-a\right)^2\ge0\)

Vì: \(b>0;\left(1-b\right)^2\ge0\)

=> \(b\left(1-b\right)^2\ge0\)

Do đó:

\(\left(1\right)\Leftrightarrow\begin{cases}a\left(1-a\right)^2=0\\b\left(1-b\right)^2=0\end{cases}\)\(\Leftrightarrow\begin{cases}1-a=0\\1-b=0\end{cases}\)\(\Leftrightarrow a=b=1\)

Khi đó; \(a^{2015}+b^{2015}=1^{2015}+1^{2015}=2\)

\(3a^2+3b^2=10ab\)

\(\Leftrightarrow3a^2-10ab+3b^2=0\)

\(\Rightarrow3a^2-9ab-ab+3b^2=0\)

\(\Leftrightarrow3a\left(a-3b\right)-b\left(a-3b\right)=0\)

\(\Leftrightarrow\left(a-3b\right)\left(3a-b\right)=0\)

Trường hợp 1: a=3b

\(A=\dfrac{a-b}{a+b}=\dfrac{3b-b}{3b+b}=\dfrac{2}{4}=\dfrac{1}{2}\)

Trường hợp 2: b=3a

\(A=\dfrac{a-b}{a+b}=\dfrac{a-3a}{a+3a}=\dfrac{-2}{4}=-\dfrac{1}{2}\)

\(=>A+B-C+D=a+b-5-b-c+1-b+c+4+b-a\)

\(=-5+4=-1\)

\(\left(-a\right)\times b=-\left(a\times b\right)=-15\)

\(\left(-a\right)\times\left(-b\right)=a\times b=15\)

\(a\times\left(-b\right)=-\left(a\times b\right)=-15\)