1B

2C

3C

4C

5D

6C

7D

12B

13B

14C

15D

16A

17C

18D

19C

20D

21C

22C

23D

a: \(\Leftrightarrow\left\{{}\begin{matrix}3x+6y=4\\x+4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)

1)

ĐKXĐ: x>4

Ta có: \(\dfrac{\sqrt{x+5}}{\sqrt{x-4}}=\dfrac{\sqrt{x-2}}{\sqrt{x+3}}\)

\(\Leftrightarrow x^2+8x+15=x^2-6x+8\)

\(\Leftrightarrow8x+6x=8-15\)

\(\Leftrightarrow14x=-7\)

hay \(x=-\dfrac{1}{2}\)(loại)

2) Ta có: \(\sqrt{4x^2-9}=3\sqrt{2x-3}\)

\(\Leftrightarrow\sqrt{2x-3}\left(\sqrt{2x+3}-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)

28.A

29.C

30.B

31.B

16, Jane hasn't cooked for her family since she was at high school

17, The last time Jane cooked for her family was when she was at high school

Ta có: \(-3x^2-5x-2=0\)

Theo định lý vi-et ta có: 

\(x_1+x_2=-\dfrac{b}{a}=-\dfrac{-5}{-3}=-\dfrac{5}{3}\)

\(x_1x_2=\dfrac{c}{a}=\dfrac{-2}{-3}=\dfrac{2}{3}\) 

a) \(M=x_1+\dfrac{1}{x_1}+\dfrac{1}{x_2}+x_2\)

\(M=\left(x_1+x_2\right)+\dfrac{x_1+x_2}{x_1x_2}\)

\(M=-\dfrac{5}{3}+\dfrac{-\dfrac{5}{3}}{\dfrac{2}{3}}=-\dfrac{25}{6}\)

b) \(N=\dfrac{1}{x_1+3}+\dfrac{1}{x_2+3}\)

\(N=\dfrac{x_2+3+x_1+3}{\left(x_1+3\right)\left(x_2+3\right)}\)

\(N=\dfrac{\left(x_1+x_2\right)+6}{x_1x_2+3\left(x_1+x_2\right)+9}\)

\(N=\dfrac{-\dfrac{5}{3}+6}{\dfrac{2}{3}+3\cdot-\dfrac{5}{3}+9}=\dfrac{13}{14}\) 

c) \(P=\dfrac{x_1-3}{x^2_1}+\dfrac{x_2-3}{x^2_2}\)

\(P=\dfrac{x^2_2\left(x_1-3\right)+x^2_1\left(x_2-3\right)}{x^2_1x^2_2}\)

\(P=\dfrac{x^2_2x_1+x^2_1x_2-3x^2_2-3x^2_1}{\left(x_1x_2\right)^2}\)

\(P=\dfrac{x_1x_2\left(x_1+x_2\right)-3\left[\left(x_1+x_2\right)^2-2x_1x_2\right]}{\left(x_1x_2\right)^2}\)

\(P=\dfrac{\dfrac{2}{3}\cdot-\dfrac{5}{3}-3\cdot\left[\left(-\dfrac{5}{3}\right)^2-2\cdot\dfrac{2}{3}\right]}{\left(\dfrac{2}{3}\right)^2}=-\dfrac{49}{4}\) 

d) \(Q=\dfrac{x_1}{x_2+2}+\dfrac{x_2}{x_1+2}\)

\(Q=\dfrac{x_1\left(x_1+2\right)+x_2\left(x_2+2\right)}{\left(x_2+2\right)\left(x_1+2\right)}\)

\(Q=\dfrac{x^2_1+2x_1+x_2^2+2x_2}{x_1x_2+2x_2+2x_1+4}\)

\(Q=\dfrac{\left(x^2_1+x^2_2\right)+2\left(x_1+x_2\right)}{x_1x_2+2\left(x_1+x_2\right)+4}\)

\(Q=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2+2\left(x_1+x_2\right)}{x_1x_2+2\left(x_1+x_2\right)+4}\)

\(Q=\dfrac{\left(-\dfrac{5}{3}\right)^2-2\cdot\dfrac{2}{3}+2\cdot-\dfrac{5}{3}}{\dfrac{2}{3}+2\cdot-\dfrac{5}{3}+4}=-\dfrac{17}{12}\)

cho ảnh dễ hiểu hơn

Bài 33:

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