a: Ta có: \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{8}\\x=-\dfrac{3}{8}\end{matrix}\right.\)

b: Ta có: \(\left(\dfrac{1}{2}\right)^x=\dfrac{1}{32}\)

nên x=5

Part 1

1.girls

2.are

3.reading

4.boy

5.she's

Part 2

2. No, they aren't. Some girls are making puppets

3. They are red, blue and yellow

4. The girl is reading a text

5. The teacher is in the office

6. Yes, she is reading a book

6.No, they aren't

7.They are painting red, blue and yellow.

8.The girl is reading a text.

9.Our teacher is in the office.

10.Yes, she is.

`#3107.101107`

a)

\(\dfrac{-22}{25}+\left(\dfrac{3}{7}-0,12\right)\\ =\dfrac{-22}{25}+\dfrac{3}{7}-0,12\\ =-\dfrac{79}{175}-0,12\\ =-\dfrac{4}{7}\)

b)

\(\dfrac{3}{8}-\left(1,2-\dfrac{5}{8}\right)\\ =\dfrac{3}{8}-1,2+\dfrac{5}{8}\\ =\dfrac{8}{8}-1,2\\ =1-1,2\\ =-0,2\)

c)

\(\dfrac{4}{15}-\left(2,9-\dfrac{11}{15}\right)\\ =\dfrac{4}{15}-2,9+\dfrac{11}{15}\\=\dfrac{15}{15}-2,9\\ =1-2,9\\ =-1,9\)

d)

\(\dfrac{17}{11}-\left(\dfrac{6}{5}-\dfrac{16}{11}\right)+\dfrac{26}{5}\\ =\dfrac{17}{11}-\dfrac{6}{5}+\dfrac{16}{11}+\dfrac{26}{5}\\ =\left(\dfrac{17}{11}+\dfrac{16}{11}\right)+\left(-\dfrac{6}{5}+\dfrac{26}{5}\right)\\ =\dfrac{33}{11}+\dfrac{20}{5}\\ =3+4\\ =7\)

e)

\(\dfrac{39}{5}+\left(\dfrac{9}{4}-\dfrac{9}{5}\right)-\left(\dfrac{5}{4}+\dfrac{6}{7}\right)\\ =\dfrac{39}{5}+\dfrac{9}{4}-\dfrac{9}{5}-\dfrac{5}{4}-\dfrac{6}{7}\\ =\left(\dfrac{39}{5}-\dfrac{9}{5}\right)+\left(\dfrac{9}{4}-\dfrac{5}{4}\right)-\dfrac{6}{7}\\ =\dfrac{30}{5}+\dfrac{4}{4}-\dfrac{6}{7}\\ =6+1-\dfrac{6}{7}\\=\dfrac{43}{7}\)

1.C                                                                                             5.C

2.A                                                                                             6.B

3.C                                                                                             7.A

4.A                                                                                               8.A

c. \(\left(x+2\right)^4-6\left(x+2\right)^2+5=0\)

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

\(\Leftrightarrow\left(x+2\right)^2\left[\left(x+2\right)^2-1\right]-5\left[\left(x+2\right)^2-1\right]=0\)

\(\Leftrightarrow\left[\left(x+2\right)^2-1\right]\left[\left(x+2\right)^2-5\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left(x+1\right)\left(x+2+\sqrt{5}\right)\left(x+2-\sqrt{5}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+1=0\\x+2+\sqrt{5}=0\\x+2-\sqrt{5}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\\x=-\sqrt{5}-2\\x=\sqrt{5}-2\end{matrix}\right.\)

Vậy: Phương trình có tập nghiệm \(S=\left\{-3;-1;-\sqrt{5}-2;\sqrt{5}-2\right\}\)

lm cho em câu 4 nữa đc ko

Bài 1: 

b: Ta có: \(18^n:2^n=\left(\sqrt{81}\right)^2\)

\(\Leftrightarrow9^n=81\)

hay n=2

Ko vì -5/2 khác -3

a: f(0)=1/2

f(1/2)=3/2

a: góc ACB=180-50-65=65 độ

b: góc yAC=180-50=130 độ

góc yAx=góc ABC=65 độ=1/2*góc yAC

=>Ax là phân giác của góc yAC