c. \(\left|\dfrac{8}{4}-\left|x-\dfrac{1}{4}\right|\right|-\dfrac{1}{2}=\dfrac{3}{4}\)

\(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{8}{4}-x+\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{8}{4}+x-\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{9}{4}-x\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{7}{4}+x\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}\dfrac{9}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\\x=\dfrac{9}{4}-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\\\left[{}\begin{matrix}\dfrac{7}{4}+x-\dfrac{1}{2}=\dfrac{3}{4}\\-\dfrac{7}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\end{matrix}\right.\\\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-3\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\\x=-3\end{matrix}\right.\) 

Ở nơi x=9/4-1/2 là x-9/4-1/2 nha

a. -1,5 + 2x = 2,5

<=> 2x = 2,5 + 1,5

<=> 2x = 4

<=> x = 2

b. \(\dfrac{3}{2}\left(x+5\right)-\dfrac{1}{2}=\dfrac{4}{3}\)

<=> \(\dfrac{3}{2}x+\dfrac{15}{2}-\dfrac{1}{2}=\dfrac{4}{3}\)

<=> \(\dfrac{9x}{6}+\dfrac{45}{6}-\dfrac{3}{6}=\dfrac{8}{6}\)

<=> 9x + 45 - 3 = 8

<=> 9x = 8 + 3 - 45

<=> 9x = -34

<=> x = \(\dfrac{-34}{9}\)

Bài 5:

N=(x-y)(x-2y)(x-3y)(x-4y)\(+y^4\)

\(=\left(x^2-5xy+4y^2\right)\left(x^2-5xy+6y^2\right)+y^4\)

\(=\left(x^2-5xy\right)^2+10y^2\left(x^2-5xy\right)+24y^4+y^4\)

\(=\left(x^2-5xy\right)^2+2\left(x^2-5xy\right)\cdot5y^2+\left(5y^2\right)^2\)

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

=>N là số chính phương

BÀi 3:

a: \(6x^2-\left(2x-3\right)\left(3x+2\right)=1\)

=>\(6x^2-\left(6x^2+4x-9x-6\right)=1\)

=>\(6x^2-\left(6x^2-5x-6\right)=1\)

=>5x+6=1

=>5x=1-6=-5

=>x=-1

b: \(\left(x+1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-2=0\)

=>\(x^3+3x^2+3x+1-\left(x^3-1\right)-2=0\)

=>\(x^3+3x^2+3x-1-x^3+1=0\)

=>\(3x^2+3x=0\)

=>3x(x+1)=0

=>x(x+1)=0

=>x=0 hoặc x=-1

9. Has the work been done by him?

10. The boxes were opened and cigarettes were taken out by us

11. She was given a new one

12. He is proved wrong

13. We were promised higher wages

14. This is the third time we have been written to about this by them

15. We were asked to be there at 8 o'clock

16. She is being shown how to do it

\(3\sqrt{6^2.2}-\dfrac{2}{5}\sqrt{10^2.2}+\sqrt{8^2.2}=18\sqrt{2}-4\sqrt{2}+8\sqrt{2}=22\sqrt{2}\)

2x+5=x-1

x=-6

vậy...

\(2x+5=x-1\)

\(2x-x=-1-5\)

\(x=-6\)

Bài 1:

a: Ta có: \(\hat{A_1}=\hat{B_1}\left(=50^0\right)\)

mà hai góc này là hai góc ở vị trí so le trong

nên a//b

b: Không có hai đường thẳng nào song song

BÀi 2:

Vẽ lại hình:

Cách 1: Ta có: \(\hat{A_1}+\hat{B_2}=130^0+50^0=180^0\)

mà hai góc này là hai góc ở vị trí trong cùng phía

nên a//b

Cách 2: Ta có; \(\hat{B_3}+\hat{B_2}=180^0\) (hai góc kề bù)

=>\(\hat{B_3}=180^0-130^0=50^0\)

Ta có: \(\hat{B_3}=\hat{A_1}\left(=50^0\right)\)

mà hai góc này là hai góc ở vị trí so le trong

nên a//b

Cách 3: Ta có; \(\hat{B_1}+\hat{B_2}=180^0\) (hai góc kề bù)

=>\(\hat{B_1}=180^0-130^0=50^0\)

Ta có: \(\hat{B_1}=\hat{A_1}\left(=50^0\right)\)

mà hai góc này là hai góc ở vị trí đồng vị

nên a//b

Câu 1 : 

\(1) C_2H_4 + H_2O \xrightarrow{t^o,xt} C_2H_5OH\\ 2) C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O\\ 3) CH_3COOH + NaOH \to CH_3COONa + H_2O\\ 4) CH_3COOH + C_2H_5OH \buildrel{{H_2SO_4,t^o}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O\)

Câu 2 : 

Trích mẫu thử

Cho qùy tím vào các mẫu thử

- mẫu thử chuyển màu đỏ là axit axetic

Cho Natri vào hai mẫu thử còn : 

- mẫu thử nào tan, xuất hiện khí là rượu etylic

\(2C_2H_5OH + 2Na \to 2C_2H_5ONa + H_2\)

- mẫu thử không hiện tượng gì là chất béo