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 = \(\dfrac{-34}{9}\)

Câu 3: 

a: Ta có: \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)\)

\(=6x^2-2x-6x^2-2x+18x+6\)

=14x+6

b: Ta có: \(2x\left(x+7\right)-3x\left(x+1\right)\)

\(=2x^2+14x-3x^2-3x\)

\(=-x^2+11x\)

Câu 2: 

a: Ta có: \(\left(-8x^5+12x^3-16x^2\right):4x^2\)

\(=-8x^5:4x^2+12x^3:4x^2-16x^2:4x^2\)

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

b: Ta có: \(\left(12x^3y^3-18x^2y+9xy^2\right):6xy\)

\(=12x^3y^3:6xy-18x^2y:6xy+9xy^2:6xy\)

\(=2x^2y^2-3x+\dfrac{3}{2}y\)

c: Ta có: \(\dfrac{x^3-11x^2+27x-9}{x-3}\)

\(=\dfrac{x^3-3x^2-8x^2+24x+3x-9}{x-3}\)

\(=x^2-8x+3\)

d: Ta có: \(\dfrac{6x^4-13x^3+7x^2-x-5}{3x+1}\)

\(=\dfrac{6x^4+2x^3-15x^3-5x^2+12x^2+4x-5x-\dfrac{5}{3}-\dfrac{10}{3}}{3x+1}\)

\(=2x^3-5x^2+4x-\dfrac{5}{3}-\dfrac{\dfrac{10}{3}}{3x+1}\)

A)x=10cm

B)x=11cm

1. D => bỏ

2. C

3. C

4. A => the

5. A

6. C =>  next

7. C => taught

8. C => the most

9. D => from

10. B => would

II

1. I wish I would become a singer.

2.  She used to walk to school when she was a child.

3. If I were you, I would learn Chinese

4. The man who I met yesterday was my uncle

5. The trees are watered by me everyday

Bài 34:

a: =>x+28=0

=>x=-28

b: =>27-x=0 hoặc x+9=0

=>x=27 hoặc x=-9

c: =>x(x-43)=0

=>x=0 hoặc x=43

\(4Na+O_2\rightarrow2Na_2O\)

\(Ca+\dfrac{1}{2}O_2\rightarrow CaO\)

\(S+O_2\xrightarrow[]{t^o}SO_2\)

\(Na_2O+H_2O\rightarrow2NaOH\)

\(K+H_2O\rightarrow KOH+\dfrac{1}{2}H_2\uparrow\)

\(SO_3+H_2O\rightarrow H_2SO_4\)

\(2KMnO_4\xrightarrow[]{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

\(H_2+\dfrac{1}{2}O_2\xrightarrow[]{t^o}H_2O\)

\(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)

\(H_2+PbO\xrightarrow[]{t^o}Pb+H_2O\)

\(a,4x^2-49=0\\ \Leftrightarrow\left(2x-7\right)\left(2x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\\ b,x^2+36=12x\\ \Leftrightarrow x^2-12x+36=0\\ \Leftrightarrow\left(x-6\right)^2=0\\ \Leftrightarrow x=6\\ c,\dfrac{1}{16}x^2-x+4=0\\ \Leftrightarrow\left(\dfrac{1}{4}x-2\right)^2=0\\ \Leftrightarrow\dfrac{1}{4}x-2=0\\ \Leftrightarrow x=8\\ d,x^3-3\sqrt{3}x^2+9x-3\sqrt{3}=0\\ \Leftrightarrow\left(x-\sqrt{3}\right)^3=0\\ \Leftrightarrow x=\sqrt{3}\)

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