a) \(x-\dfrac{3}{4}=-\dfrac{5}{8}\Rightarrow x=-\dfrac{5}{8}+\dfrac{3}{4}\Rightarrow x=\dfrac{1}{8}\)

b) \(x+\dfrac{5}{8}=-\dfrac{1}{4}\Rightarrow x=-\dfrac{1}{4}-\dfrac{5}{8}\Rightarrow x=-\dfrac{7}{8}\)

c) \(\dfrac{5}{6}+\dfrac{3}{4}x=\dfrac{5}{24}\Rightarrow x=\left(\dfrac{5}{24}-\dfrac{5}{6}\right):\dfrac{3}{4}\Rightarrow x=-\dfrac{5}{6}\)

d) \(\dfrac{3}{8}-\dfrac{2}{3}:x=-\dfrac{5}{12}\Rightarrow\dfrac{2}{3}:x=\dfrac{3}{8}+\dfrac{5}{12}\Rightarrow\dfrac{2}{3}:x=\dfrac{19}{24}\Rightarrow x=\dfrac{2}{3}:\dfrac{19}{24}=\dfrac{16}{19}\)

a) \(x-\dfrac{3}{4}=-\dfrac{5}{8}\\ \Rightarrow x=\dfrac{1}{8}\)

b) \(x+\dfrac{5}{8}=-\dfrac{1}{4}\\ \Rightarrow x=-\dfrac{7}{8}\)

c) \(\dfrac{5}{6}+\dfrac{3}{4}x=\dfrac{5}{24}\\ \Rightarrow\dfrac{3}{4}x=-\dfrac{5}{8}\\ \Rightarrow x=-\dfrac{5}{6}\)

d) \(\dfrac{3}{8}-\dfrac{2}{3}:x=-\dfrac{5}{12}\\ \Rightarrow\dfrac{2}{3}:x=\dfrac{19}{24}\\ \Rightarrow x=\dfrac{16}{19}\)

e) \(\left(6,5-2x\right):\dfrac{5}{13}=\dfrac{13}{10}\\ \Rightarrow6,5-2x=\dfrac{1}{2}\\ \Rightarrow2x=6\\ \Rightarrow x=3\)

f) \(\left|\dfrac{1}{3}x+\dfrac{1}{2}\right|-\dfrac{3}{4}=-\dfrac{1}{6}\\ \Rightarrow\left|\dfrac{1}{3}x+\dfrac{1}{2}\right|=\dfrac{7}{12}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{1}{3}x+\dfrac{1}{2}=\dfrac{7}{12}\\\dfrac{1}{3}x+\dfrac{1}{2}=-\dfrac{7}{12}\end{matrix}\right.\)

   \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{13}{4}\end{matrix}\right.\)

g) \(\dfrac{x-3}{3}=\dfrac{2x+3}{5}\\ \Rightarrow5x-15=6x+9\\ \Rightarrow-x=24\\ \Rightarrow x=-24\)

h) \(\dfrac{x-5}{6}=\dfrac{6}{x-5}\\ \Rightarrow\left(x-5\right)^2=6^2\\ \Rightarrow\left[{}\begin{matrix}x-5=-6\\x-5=6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=11\end{matrix}\right.\)

h: Ta có: \(\dfrac{5}{x+3}=\dfrac{x+3}{5}\)

\(\Leftrightarrow\left(x+3\right)^2=25\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=-5\\x+3=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=2\end{matrix}\right.\)

5: \(x:2,5=0,003:0,75\)

=>\(\frac{x}{2,5}=\frac{0.003}{0.75}=\frac{3}{750}=\frac{1}{250}\)

=>\(x=\frac{2.5}{250}=0,01\)

6: \(\frac23:0,4=x:\frac45\)

=>\(x:\frac45=\frac23:\frac25=\frac53\)

=>\(x=\frac53\cdot\frac45=\frac43\)

7: \(2,5:4x=0,5:0,2\)

=>\(\frac{2.5}{4x}=\frac{0.5}{0.2}=\frac52\)

=>\(4x=2.5\cdot\frac25=1\)

=>\(x=\frac14\)

8: \(\frac15x:3=\frac23:0,25\)

=>\(\frac{x}{15}=\frac23:\frac14=\frac23\cdot4=\frac83=\frac{40}{15}\)

=>x=40

9: \(1,25:0,8=\frac38:0,2x\)

=>\(\frac38:0,2x=\frac54:\frac45=\frac{25}{16}\)

=>\(0,2x=\frac38:\frac{25}{16}=\frac38\cdot\frac{16}{25}=\frac{3\cdot2}{25}=\frac{6}{25}\)

=>\(x=\frac{6}{25}:0,2=\frac{6}{25\cdot0,2}=\frac65\)

10: \(3x:2,7=\frac13:2\frac14\)

=>\(3x:2,7=\frac13:\frac94=\frac13\cdot\frac49=\frac{4}{27}\)

=>\(3x=\frac{4}{27}\cdot2,7=\frac{4}{10}\)

=>\(x=\frac{4}{10}:3=\frac{4}{30}=\frac{2}{15}\)

Bài 4: 

a: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{x+y}{5+6}=\dfrac{44}{11}=4\)

Do đó: x=20; y=24

bạn giúp mik bài 2 đc ko ạ=((((

Bài làm

Giả sử:  \(\frac{a}{b}>\frac{c}{d}\)

\(\Rightarrow ad>bc\)

Cộng cả hai vế với ab, ta được

ad + ab > bc + ab

=> a( b + d ) > b( a + c )

\(\Rightarrow\frac{a}{b}>\frac{a+c}{b+d}\)    (1)

Lại có: \(\frac{a}{b}>\frac{c}{d}\)

\(\Rightarrow ad>bc\)

Cộng cả hai vế với dc, ta được:

ad + dc > bc + dc

=> d( a + c ) > c( b + d )

\(\Rightarrow\frac{a+c}{b+d}>\frac{c}{d}\)            (2)

Từ (1) và (2)  \(\Rightarrow\frac{a}{b}>\frac{a+c}{b+d}>\frac{c}{d}\)( đpcm )

Cảm ơn bạn nha

1.I think playing football is boring

I find playing football boring

2 I think you are right

I find you right

3 I think doing homework is interesting

I find doing homework interesting

4 I think she is hard working

I find she is hard working

5 I think driving to school is boring

I find driving to school boring

Sorry bn mik ko có nhiều thời gian nên ko làm xong đc

Nhớ k nhé

Hok tốt

bn ơi,mỗi loại 10 câu nghĩa là 20 câu tổng cộng mak

Gọi a bằng ƯC ( m , mn + 8 )

Ta có: m chia hết cho a ( m lẻ => a lẻ )

=> mn chia hết cho a

Lại có: mn + 8 chia hết cho a

=> mn + 8 - mn chia hết cho a

=> 8 chia hết cho a

=> a ∈ Ư ( 8 ) = { 1 ; 2 ; 4 ; 8 }

Vì a lẻ 

=> a = 1

=> ƯC ( mn ; mn + 8 ) = 1

=> m và mn + 8 là hai số nguyên tố cùng nhau.