Bài cúi đúng hem :))

\(\Leftrightarrow\dfrac{4\left(5x^2-3\right)+5\left(3x-1\right)}{20}< \dfrac{10x\left(2x+3\right)-100}{20}\)

\(\Leftrightarrow4\left(5x^2-3\right)+5\left(3x-1\right)< 10x\left(2x+3\right)-100\)

\(\Leftrightarrow20x^2-12+15x-5< 20x^2+30x-100\)

\(\Leftrightarrow-15x< -83\)

\(\Leftrightarrow x>\dfrac{83}{15}\)

Vậy \(S=\left\{x|x>\dfrac{83}{15}\right\}\)

\(8,\dfrac{5x^2-3}{5}+\dfrac{3x-1}{4}< \dfrac{x\left(2x+3\right)}{2}-5\)

\(\Leftrightarrow4\left(5x^2-3\right)+5\left(3x-1\right)< 10x\left(2x+3\right)-100\)

\(\Leftrightarrow20x^2-12+15x-5< 20x^2+30x-100\)

\(\Leftrightarrow20x^2-20x^2+15x-30x-12-5+100< 0\)

\(\Leftrightarrow-15x+83< 0\)

\(\Leftrightarrow-15x< -83\)

\(\Leftrightarrow x>\dfrac{83}{15}\)

Vậy \(S=\left\{x\left|x>\dfrac{83}{15}\right|\right\}\)

8d

9c

10d

11a

12d

Bài 8:

a: Ta có: \(\sqrt{4x}=\sqrt{5}\)

\(\Leftrightarrow4x=5\)

hay \(x=\dfrac{5}{4}\)

b: Ta có: \(\sqrt{4\cdot\left(1-x\right)^2}-6=0\)

\(\Leftrightarrow2\left|x-1\right|=6\)

\(\Leftrightarrow\left|x-1\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)

c: Ta có: \(\sqrt{2x-3}=\sqrt{7}\)

\(\Leftrightarrow2x-3=7\)

hay x=5

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

\(\Leftrightarrow\left|3x-2\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=4\\3x-2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=6\\3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{2}{3}\end{matrix}\right.\)

8 disappointing

12 preservation

13 seriously

1A

2B

3D

4A

5D

6B

7A

8A

6.C

7.A

8.B

\(8,\\ A=\left\{0;1;2;3\right\}\\ B=\left\{0;1;2\right\}\\ A\cap B=\left\{0;1;2\right\}\\ A\cup B=\left\{0;1;2;3\right\}\\ A\B=\left\{3\right\}\\ B\A=\varnothing\\ 9,\\ A=\left\{0;1;2;3;4\right\}\\ B=\left\{5;6\right\}\\ A\cap B=\varnothing\\ A\cup B=\left\{0;1;2;3;4;5;6\right\}\\ A\B=\left\{0;1;2;3;4\right\}\\ B\A=\left\{5;6\right\}\)

8: \(A\cap B=\left\{3\right\}\)

\(A\cup B\)=(-1;3]