Câu 9:

a) Ta có: \(9x^2-16=0\)

\(\Leftrightarrow\left(3x-4\right)\left(3x+4\right)=0\)

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

b) Ta có: \(4x^2=13\)

\(\Leftrightarrow x^2=\dfrac{13}{4}\)
\(\Leftrightarrow x\in\left\{\dfrac{\sqrt{13}}{2};-\dfrac{\sqrt{13}}{2}\right\}\)

c) Ta có: \(2x^2+9=0\)

\(\Leftrightarrow2x^2=-9\)(Vô lý)

d) Ta có: \(-x^2+324=0\)

\(\Leftrightarrow x^2=324\)

\(\Leftrightarrow\left[{}\begin{matrix}x=18\\x=-18\end{matrix}\right.\)

1.A
2.A
3.B
4.C
5.B
6.C
7.A
8.A
9.B
10.A
11.B
12.A
13.C
14.B
15.B
16.A
17.A
18.A
19.A
20.C

a: UM=AU=2,5/2=1,25cm

VA=VE=AE/2=1,25cm

b: UV=1,25+1,25=2,5cm

Câu 15: 

1: Ta có: \(\sqrt{2x+5}=\sqrt{1-x}\)

\(\Leftrightarrow2x+5=1-x\)

\(\Leftrightarrow2x+x=1-5\)

\(\Leftrightarrow3x=-4\)

hay \(x=-\dfrac{4}{3}\)

2: Ta có: \(\sqrt{2x-1}=\sqrt{x-1}\)

\(\Leftrightarrow2x-1=x-1\)

\(\Leftrightarrow x=0\)(loại

3: Ta có: \(\sqrt{x^2-x}=\sqrt{3x-5}\)

\(\Leftrightarrow x^2-x=3x-5\)

\(\Leftrightarrow x^2-4x+5=0\)

\(\Leftrightarrow\left(x-2\right)^2+1=0\)(vô lý

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

\(\Leftrightarrow2x^2=4x\)

\(\Leftrightarrow2x^2-4x=0\)

\(\Leftrightarrow2x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

5: Ta có: \(\sqrt{x^2-x}=\sqrt{3-x}\)

\(\Leftrightarrow x^2=3\)

\(\Leftrightarrow x\in\left\{\sqrt{3};-\sqrt{3}\right\}\)

\(1,=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\\ 2,=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\\ 3,=2y\left(y^2+4y+4\right)=2y\left(y+2\right)^2\\ 4,=2\left(x^2+2x+1-y^2\right)=2\left[\left(x+1\right)^2-y^2\right]\\ =2\left(x+y+1\right)\left(x-y+1\right)\\ 5,=16-\left(x-y\right)^2=\left(4-x+y\right)\left(4+x-y\right)\)

2) \(=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\)

3) \(=2y\left(y^2+4y+4\right)=2y\left(y+2\right)^2\)

4) \(=2\left[\left(x^2+2x+1\right)-y^2\right]=2\left[\left(x+1\right)^2-y^2\right]\)

\(=2\left(x+1-y\right)\left(x+1+y\right)\)

5) \(=16-\left(x^2-2xy+y^2\right)=16-\left(x-y\right)^2\)

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