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!

Yêu cầu và điều kiện đề bài là gì đây bạn?

2:

a: 5/x-y/3=1/6

=>\(\dfrac{15-xy}{3x}=\dfrac{1}{6}\)

=>\(\dfrac{30-2xy}{6x}=\dfrac{x}{6x}\)

=>30-2xy=x

=>x(2y+1)=30

=>(x;2y+1) thuộc {(30;1); (-30;-1); (10;3); (-10;-3); (6;5); (-6;-5)}

=>(x,y) thuộc {(30;0); (-30;-1); (10;1); (-10;-2); (6;2); (-6;-3)}

b: x/6-2/y=1/30

=>\(\dfrac{xy-12}{6y}=\dfrac{1}{30}\)

=>\(\dfrac{5xy-60}{30y}=\dfrac{y}{30y}\)

=>5xy-60=y

=>y(5x-1)=60

=>(5x-1;y) thuộc {(-1;-60); (4;15); (-6;-10)}(Vì x,y là số nguyên)

=>(x,y) thuộc {(0;-60); (1;15); (-1;-10)}

a) Quy đồng mẫu số ta được: 30/6x-yx2/6x=x/6x

                                         suy ra 30-yx = x           <=> x.y=30-x

      Xong bạn liệt kê ra vì có khác nhiều đáp số đó.

 Phần b) và c) tương tự vậy !!!

a)\(y=\dfrac{5}{3}-\left(\dfrac{7}{12}:\dfrac{5}{6}\right)=\dfrac{5}{3}-\dfrac{7}{10}=\dfrac{50}{30}-\dfrac{21}{30}=\dfrac{29}{30}\)

b)\(y=\dfrac{4}{15}:\left[\left(\dfrac{4}{5}+\dfrac{1}{2}\right)\times\dfrac{4}{13}\right]=\dfrac{4}{15}:\left[\left(\dfrac{8}{10}+\dfrac{5}{10}\right)\times\dfrac{4}{13}\right]\)

\(y=\dfrac{4}{15}:\left[\dfrac{13}{10}\times\dfrac{4}{13}\right]=\dfrac{4}{15}:\dfrac{2}{5}=\dfrac{2}{3}\)

a 29/30

b 20/30= 2/3

a) \(\left(3x-5\right)\left(5-3x\right)+9\left(x+1\right)^2=30\)

\(\Rightarrow15x-9x^2-25+15x+9\left(x^2+2x+1\right)-30=0\)

\(\Rightarrow30x-9x^2-25+9x^2+18x+9-30=0\)

\(\Rightarrow48x-46=0\)

\(\Rightarrow x=\frac{23}{24}\)

b) \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)

\(\Rightarrow\left(x^2+8x+16\right)-\left(x^2-1\right)=16\)

\(\Rightarrow x^2+8x+16-x^2+1=16\)

\(\Rightarrow8x+17=16\)

\(\Rightarrow8x=-1\)

\(\Rightarrow x=\frac{-1}{8}\)

c) \(\left(y-2\right)^3-\left(y-3\right)\left(y^2+3y+9\right)+6\left(y+1\right)^2=49\)

\(\Rightarrow\left(y-2\right)^3-\left(y^3-3^3\right)+6\left(y^2+2y+1\right)=49\)

\(\Rightarrow y^3-6y^2+12y-8-y^3+27+6y^2+12y+6=49\)

\(\Rightarrow\left(y^3-y^3\right)+\left(-6y^2+6y^2\right)+\left(12y+12y\right)+\left(-8+27+6\right)=49\)

\(\Rightarrow24y+25=49\)

\(\Rightarrow24y=24\)

\(\Rightarrow y=1\)

d) \(\left(y+3\right)^3-\left(y+1\right)^3=56\)

\(\Rightarrow\left(y+3-y-1\right)[\left(y+3\right)^2+\left(y+3\right)\left(y+1\right)+\left(y+1\right)^2]=56\)

\(\Rightarrow2\left(y^2+6y+9+y^2+4y+3+y^2+2y+1\right)=56\)

\(\Rightarrow3y^2+12y+13=28\)

\(\Rightarrow\left(3y^2+15y\right)-\left(3y+15\right)=0\)

\(\Rightarrow3y\left(y+5\right)-3\left(y+5\right)=0\)

\(\Rightarrow3\left(y-1\right)\left(y+5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)