Giải:

a) \(\dfrac{-5}{8}=\dfrac{x}{16}\) 

\(\Rightarrow x=\dfrac{16.-5}{8}=-10\) 

\(\dfrac{3x}{9}=\dfrac{2}{6}\) 

\(\Rightarrow3x=\dfrac{2.9}{6}=3\) 

\(\Rightarrow x=1\)

b) \(\dfrac{x+3}{15}=\dfrac{1}{3}\)  

\(\Rightarrow x+3=\dfrac{1.15}{3}=5\) 

\(\Rightarrow x=2\)

\(\dfrac{6}{2x+1}=\dfrac{2}{7}\) 

\(\Rightarrow2x+1=\dfrac{6.7}{2}=21\) 

\(\Rightarrow x=10\)

c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\) 

\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\) 

\(\Rightarrow x-6=\dfrac{18.4}{-12}=-6\) 

\(\Rightarrow x=0\) 

\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\) 

\(\Rightarrow y=\dfrac{-12.24}{18}=-16\) 

 \(\dfrac{3-x}{-12}=\dfrac{16}{y+1}=\dfrac{192}{-72}\) 

\(\Rightarrow\dfrac{3-x}{-12}=\dfrac{192}{-72}\) 

\(\Rightarrow3-x=\dfrac{192.-12}{-72}=32\) 

\(\Rightarrow x=-29\) 

\(\Rightarrow\dfrac{16}{y+1}=\dfrac{192}{-72}\) 

\(\Rightarrow y+1=\dfrac{16.-72}{192}=-6\) 

d) \(\dfrac{-2}{3}< \dfrac{x}{5}< \dfrac{-1}{6}\) 

\(\Rightarrow\dfrac{-20}{30}< \dfrac{6x}{30}< \dfrac{-5}{30}\) 

\(\Rightarrow6x\in\left\{-18;-12;-6\right\}\) 

\(\Rightarrow x\in\left\{-3;-2;-1\right\}\) 

\(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\) 

\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\) 

\(\Rightarrow5x\in\left\{-5;0;5;10\right\}\) 

\(\Rightarrow x\in\left\{-1;0;1;2\right\}\) 

e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\) 

\(\Rightarrow\dfrac{x+46}{20}=x+\dfrac{2}{5}\) 

\(\Rightarrow\dfrac{x+46}{20}=\dfrac{5x+2}{5}\) 

\(\Rightarrow5.\left(x+46\right)=20.\left(5x+2\right)\) 

\(\Rightarrow5x+230=100x+40\) 

\(\Rightarrow5x-100x=40-230\) 

\(\Rightarrow-95x=-190\) 

\(\Rightarrow x=-190:-95\) 

\(\Rightarrow x=2\) 

\(y\dfrac{5}{y}=\dfrac{86}{y}\) 

\(\Rightarrow y+\dfrac{5}{y}=\dfrac{86}{y}\) 

\(\Rightarrow\dfrac{y^2+5}{y}=\dfrac{86}{y}\) 

\(\Rightarrow y^2+5=86\) 

\(\Rightarrow y^2=86-5\) 

\(\Rightarrow y^2=81\) 

\(\Rightarrow\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\) 

Chúc bạn học tốt!

\(a,\dfrac{5}{8}=\dfrac{x}{14}\)

\(\Rightarrow x=\dfrac{5.14}{8}=8,75\)

Vậy \(x=8,75\)

\(b,\dfrac{x}{6}=-\dfrac{1}{3}\)

\(\Rightarrow x=-\dfrac{1.6}{3}=-2\)

Vậy \(x=-2\)

\(c,-\dfrac{3}{5}=\dfrac{x}{10}\)

\(\Rightarrow x=-\dfrac{3.10}{5}=-6\)

Vậy \(x=-6\)

câu d đã có đáp án

mik đang cần gấp ak

\(+\text{)}\dfrac{5}{8}=14x\)⇒\(x=\dfrac{5}{112}\)

\(+\text{)}\dfrac{x}{6}=-\dfrac{1}{3}\)⇒\(x=-2\)

\(+\text{)}-\dfrac{3}{5}=\dfrac{x}{10}\)⇒\(x=-6\)

\(+\text{)}\dfrac{3}{5}=-\dfrac{9}{11}\)⇒\(\dfrac{33}{55}=\dfrac{45}{55}\)

\(+\text{)}\dfrac{x}{2}=\dfrac{2}{x}\)⇒\(x=+-2\)

\(+\text{)}\dfrac{x}{-5}=\dfrac{-5}{x}\)⇒\(x=+-5\)

a: -2x(x+3)+x(2x-1)=10

=>-2x^2-6x+2x^2-x=10

=>-7x=10

=>x=-10/7

b: Sửa đề: 2/3x(9/2x+1/4)-(3x^2+2)=3

=>3x^2+1/6x-3x^2-2=3

=>1/6x-2=3

=>x=30

Xin lỗi m.n nhé gửi nhầm tí

\(a,PT\Leftrightarrow x^3-6x^2+12x-8-x^3+x+6x^2-18x-10=0\)

\(\Leftrightarrow-5x-18=0\)

\(\Leftrightarrow x=-\dfrac{18}{5}\)

Vậy ...

\(b,PT\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+10=0\)

\(\Leftrightarrow12x+6=0\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy ...

\(c,PT\Leftrightarrow\left(x+1\right)^3+3^3=0\)

\(\Leftrightarrow\left(x+1+3\right)\left(x^2+2x+1-3x-3+9\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x^2-x+7\right)=0\)

Thấy : \(x^2-\dfrac{2.x.1}{2}+\dfrac{1}{4}+\dfrac{27}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}>0\)

\(\Rightarrow x+4=0\)

\(\Leftrightarrow x=-4\)

Vậy ...

\(d,PT\Leftrightarrow\left(x-2\right)^3+1^3=0\)

\(\Leftrightarrow\left(x-2+1\right)\left(x^2-4x+4-x+2+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-5x+7\right)=0\)

Thấy : \(x^2-5x+7=x^2-\dfrac{5.x.2}{2}+\dfrac{25}{4}+\dfrac{3}{4}=\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)

\(\Rightarrow x-1=0\)

\(\Leftrightarrow x=1\)

Vậy ...

a x=4

\(a,\dfrac{x}{2}=\dfrac{8}{x}\\ \Rightarrow x^2=16\\ \Rightarrow x=\pm4\\ b,\dfrac{x+1}{5}=\dfrac{x+1}{5}\left(luôn.đúng\right)\\ c,\dfrac{x+1}{5}=\dfrac{x+3}{10}\\ \Rightarrow\dfrac{2x+2}{10}=\dfrac{x+3}{10}\\ \Rightarrow2x+2=x+3\\ \Rightarrow2x-x=3-2\\ \Rightarrow x=1\\ d,\dfrac{x}{4}=\dfrac{18}{x+1}\\ \Rightarrow x\left(x+1\right)=4.18\\ \Rightarrow x^2+x=72\\ \Rightarrow x^2+x-72=0\\ \Rightarrow\left(x^2+9x\right)-\left(8x+72\right)=0\\ \Rightarrow x\left(x+9\right)-8\left(x+9\right)=0\\ \Rightarrow\left(x-8\right)\left(x+9\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-9\end{matrix}\right.\)

Bài 2: 

a: =>x-1=1 hoặc x-1=-1

=>x=2 hoặc x=0

b: =>x+1=-1

hay x=-2

c: =>(135-7x):9=8

=>135-7x=72

=>7x=63

hay x=9

d: =>(x+7)(x-3)<0

=>-7<x<3

e: \(\Leftrightarrow3^{x-3}=18+9=27\)

=>x-3=3

hay x=6

f: =>4-2x=0

hay x=2

b: =>x(8-7)=-33

=>x=-33

c: =>-12x+60+21-7x=5

=>-19x=-76

hay x=4

d: =>-2x-2-x+5+2x=0

=>3-x=0

hay x=3