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

Bài 2:

a) \(x-\dfrac{3}{4}=-\dfrac{5}{8}\)

    \(x=-\dfrac{5}{8}-\dfrac{3}{4}\)

    \(x=-\dfrac{11}{8}\)

b) \(x+\dfrac{5}{8}=\dfrac{-1}{4}\)

     \(x=\dfrac{-1}{4}-\dfrac{5}{8}\)

     \(x=\dfrac{-7}{8}\)

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

Bài 1.

a) Áp dụng dãy tỉ số bằng nhau, ta được

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{x+y}{3+4}=\dfrac{35}{7}=5\) \(\Rightarrow\left\{{}\begin{matrix}x=15\\y=20\end{matrix}\right.\)

Vậy \(x=15\) và \(y=20\)

b) Áp dụng dãy tỉ số bằng nhau, ta được

\(\dfrac{x}{3}=\dfrac{3x}{9}=\dfrac{y}{4}=\dfrac{3x+y}{9+4}=\dfrac{39}{13}=3\) \(\Rightarrow\left\{{}\begin{matrix}x=9\\y=12\end{matrix}\right.\)

Vậy \(x=9\) và \(y=12\)

c) Ta có \(\left\{{}\begin{matrix}xy=48\\\dfrac{x}{3}=\dfrac{y}{4}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{48}{y}\\3x=4y\end{matrix}\right.\) 

\(\Rightarrow4x=3.\dfrac{48}{y}\) \(\Rightarrow x^2=\dfrac{3.48}{4}=36\) \(\Rightarrow x=\pm6\)

Với \(x=6\), ta có \(\dfrac{6}{3}=\dfrac{y}{4}\Rightarrow y=\dfrac{6.4}{3}=8\)

Với \(x=-6\), ta có \(\dfrac{-6}{3}=\dfrac{y}{4}\Rightarrow y=\dfrac{-6.4}{3}=-8\)

Vậy \(\left(x,y\right)\in\left\{\left(6;8\right),\left(-6;-8\right)\right\}\)

Bài 1: 

c: Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=3k\\y=4k\end{matrix}\right.\)

Ta có: xy=48

\(\Leftrightarrow12k^2=48\)

\(\Leftrightarrow k^2=4\)

Trường hợp 1: k=2

\(\Leftrightarrow\left\{{}\begin{matrix}y=3k=3\cdot2=6\\y=4k=4\cdot2=8\end{matrix}\right.\)

Trường hợp 2: k=-2

\(\Leftrightarrow\left\{{}\begin{matrix}y=3k=3\cdot\left(-2\right)=-6\\y=4k=4\cdot\left(-2\right)=-8\end{matrix}\right.\)

Bài 4:

a) Ta có: \(\widehat{yOz}+\widehat{xOy}=180^0\)(2 góc kề bù)

\(\Rightarrow\widehat{yOz}=180^0-\widehat{xOy}=180^0-50^0=130^0\)

b) Ta có: \(\widehat{zOt}=\widehat{yOt}=\dfrac{1}{2}\widehat{yOz}=\dfrac{1}{2}.130^0=65^0\)(do Ot là tia phân giác \(\widehat{yOz}\))

c) Ta có: \(\widehat{xOt}=\widehat{yOt}+\widehat{xOy}=65^0+50^0=115^0\)

Bài 5: 

 a) Ta có: \(\widehat{xOz}+\widehat{xOy}=180^0\)(2 góc kề bù)

\(\Rightarrow\widehat{xOz}=180^0-\widehat{xOy}=180^0-110^0=70^0\)

b) Ta có: \(\widehat{zOt}=\dfrac{1}{2}\widehat{xOz}=\dfrac{1}{2}.70^0=35^0\)( Ot là tia phân giác \(\widehat{xOz}\))

c) Ta có: \(\widehat{xOt}=\widehat{zOt}=35^0\)( Ot là tia phân giác \(\widehat{xOz}\))

Bài 4: 

a: Ta có: \(\widehat{xOy}+\widehat{yOz}=180^0\)

\(\Leftrightarrow\widehat{yOz}=180^0-50^0\)

\(\Leftrightarrow\widehat{yOz}=130^0\)

b: \(\widehat{zOt}=\dfrac{\widehat{yOz}}{2}=65^0\)

=23 x (58-30) + 28 x 77

=23 x 28 +28 x 77

=28 x (23+77)

=28 x 100

=2800

học tốt bạn nhé

bài này là dạng nâng cao về toán tính nhanh, mik nghĩ là ẽ ít bạn trả lời đc 

bằng 2800

đề bài là gì ạ ? 

8/49 x 8/5 = 64/245

15/9 - 10/27 + 1/6 = 79/54

7/15 + ( 5/6 - 2/6) = 7/15 + 3/6 = 7/15+ 1/2 = 14/30 + 15/30 = 29/30

2 câu kia tôi làm rồi nhé !

\(\dfrac{7}{15}+\dfrac{5-2}{6}=\dfrac{7}{15}+\dfrac{1}{2}=\dfrac{14+15}{30}=\dfrac{29}{30}\)

đăng từng câu thì đầy đủ

Bài 3: 

a: =>1<x<=4

hay \(x\in\left\{2;3;4\right\}\)

b: =>-104<x<-99

hay \(x\in\left\{-103;-102;-101;-100\right\}\)