Câu 1:

\(\left(4x+3\right)\left(3x^2+x-2\right)\left(2x^2-3x-5\right)=0\\ \Leftrightarrow\left(4x+3\right)\left(3x-2\right)\left(x+1\right)\left(2x-5\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=-1\\x=\dfrac{2}{3}\\x=\dfrac{5}{2}\end{matrix}\right.\\ \Leftrightarrow A=\left\{-1;-\dfrac{3}{4};\dfrac{2}{3};\dfrac{5}{2}\right\}\)

Câu 2:

\(\left(x^2-4\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=3\end{matrix}\right.\Leftrightarrow A=\left\{-2;2;3\right\}\\ \left|5x\right|-11\le0\Leftrightarrow\left|5x\right|\le11\Leftrightarrow-11\le5x\le11\\ \Leftrightarrow-\dfrac{11}{5}\le x\le\dfrac{11}{5}\\ \Leftrightarrow B=\left[-\dfrac{11}{5};\dfrac{11}{5}\right]\)

\(\Leftrightarrow A\cap B=\left\{-2;2\right\}\\ A\cup B=\left[-\dfrac{11}{5};3\right]\\ A\B=\left\{3\right\}\)

save - However - of - Therefore - and - with

1 \(=\dfrac{4x^2+7+5x^2-7}{3x^2y}=\dfrac{9x^2}{3x^2y}=\dfrac{3}{y}\)

2: \(=\dfrac{6x+3x+27}{x+3}=\dfrac{9x+27}{x+3}=9\)

3: \(=\dfrac{x^2-10x+25}{x-5}=\dfrac{\left(x-5\right)^2}{x-5}=x-5\)

4: \(=\dfrac{x+y}{xy\left(x+y\right)}=\dfrac{1}{xy}\)

5: \(=\dfrac{2x-6+3x+9+1}{\left(x-3\right)\left(x+3\right)}=\dfrac{5x+4}{\left(x-3\right)\left(x+3\right)}\)

bài trên hay dưới v bn?

1.Because of the traffic jam ,I late for work this morning

2.The hare ran faster than the tortoise but the tortoise still won the race

3.You shouldn't spend more than four hours a day sitting in front of a computer or a TV

4.If Lily visit Hue in this summer ,she will enjoy royal court music performances

a: Thay x=-2 vào (1), ta được:

4+8+2m-1=0

=>2m+11=0

hay m=-11/2

b: \(\text{Δ}=\left(-4\right)^2-4\left(2m-1\right)\)

=16-8m+4

=-8m+20

Để phương trình có hai nghiệm phân biệt thì -8m+20>0

=>-8m>-20

hay m<5/2

Theo đề, ta có: \(x_1^2+x_2^2+x_1x_2=m^2-2m\)

\(\Leftrightarrow m^2-2m=\left(x_1+x_2\right)^2-x_1x_2\)

\(\Leftrightarrow m^2-2m=\left(-4\right)^2-\left(2m-1\right)\)

\(\Leftrightarrow m^2-2m=16-2m+1=17\)

hay \(m=-\sqrt{17}\)