a) 3/5+y=9/8

y= 21/40

b) 13/5-y= 1/8

y=99/40

cậu viết cách làm được ko?

`1/8 : y=1/5`

`=> y=1/8 : 1/5`

`=> y=1/8 xx 5`

`=>y=5/8`

\(\dfrac{1}{8}:y=\dfrac{1}{5}\)
\(y=\dfrac{1}{8}:\dfrac{1}{5}\)
\(y=\dfrac{5}{8}\)
#データネ

a: \(y\times4\frac{7}{12}=6\frac14\)

=>\(y\times\frac{55}{12}=\frac{25}{4}\)

=>\(y=\frac{25}{4}:\frac{55}{12}=\frac{25}{4}\times\frac{12}{55}=\frac{3\times5}{11}=\frac{15}{11}\)

b: \(y:3\frac78=5\frac12\)

=>\(y:\frac{31}{8}=\frac{11}{2}\)

=>\(y=\frac{11}{2}\times\frac{31}{8}=\frac{341}{16}\)

c: \(6\frac17:y=5\frac25\)

=>\(\frac{43}{7}:y=\frac{27}{5}\)

=>\(y=\frac{43}{7}:\frac{27}{5}=\frac{43}{7}\times\frac{5}{27}=\frac{215}{189}\)

`b)y:12/15=5`

`\qquad y:4/5=5`

`\qquad y=5*4/5=4`

Vậy `y=4`

`a)y-1/8=1/4`

`\qquad y=1/4+1/8`

`\qquad y=3/8`

Vậy `y=3/8`

Giải:

a) \(y-\dfrac{1}{8}=\dfrac{1}{4}\) 

            \(y=\dfrac{1}{4}+\dfrac{1}{8}\) 

            \(y=\dfrac{3}{8}\) 

b) \(y:\dfrac{12}{15}=5\) 

            \(y=5\times\dfrac{4}{5}\) 

            \(y=4\) 

Chúc em học tốt!

\(\dfrac{2}{5}\) x y : \(\dfrac{7}{4}\) = \(\dfrac{7}{8}\)

\(\dfrac{2}{5}\) x y = \(\dfrac{7}{8}\) x \(\dfrac{7}{4}\)

 \(\dfrac{2}{5}\) x y = \(\dfrac{49}{32}\)

         y = \(\dfrac{49}{32}\) : \(\dfrac{2}{5}\)

         y = \(\dfrac{245}{64}\)

2\(\dfrac{2}{5}\): y x 1\(\dfrac{1}{4}\) = 2\(\dfrac{3}{5}\)

\(\dfrac{12}{5}\): y x \(\dfrac{5}{4}\) = \(\dfrac{13}{5}\)

\(\dfrac{12}{5}\): y        = \(\dfrac{13}{5}\): \(\dfrac{5}{4}\)

 \(\dfrac{12}{5}\): y = \(\dfrac{52}{25}\)

        y = \(\dfrac{12}{5}\): \(\dfrac{52}{25}\)

        y = \(\dfrac{15}{13}\)

\(3\dfrac{1}{2}:y=\dfrac{8}{5}\\ \dfrac{7}{2}:y=\dfrac{8}{5}\\ y=\dfrac{7}{2}:\dfrac{8}{5}\\ y=\dfrac{7}{2}\times\dfrac{5}{8}\\ y=\dfrac{35}{16}\)

Vậy `y=35/16`

`3 1/2 :y=8/5`

`7/2:y=8/5`

`y=7/2:8/5`

`y=7/2xx5/8`

`y=35/16`

a: =>4y+15/16=1

=>4y=1/16

hay y=1/64

b: =>10y+1023/1024=1

=>10y=1/1024

hay  y=1/10240

a: A=-(x-7)^2-888<=-888

Dấu = xảy ra khi x=7

b: \(B=\left|2x-1\right|+\left|y-5\right|+\dfrac{8}{3}>=\dfrac{8}{3}\)

Dấu = xảy ra khi x=1/2 và y=5

c: \(C=\left(x+3\right)^2+\left|2y-5\right|-232>=-232\)

Dấu = xảy ra khi x=-3 và y=5/2

\(a,\frac{x}{8}-\frac{1}{y}=\frac{1}{4}\)

\(\Rightarrow\frac{1}{y}=\frac{x}{8}-\frac{1}{4}\)

\(\Rightarrow\frac{1}{y}=\frac{x-2}{8}\)

\(\Rightarrow\left(x-2\right)\cdot y=1\cdot8\)

\(\Rightarrow y\left(x-2\right)=8\)

xét bảng :

vậy_

b, tương tự

\(a,\frac{x}{8}-\frac{1}{y}=\frac{1}{4}\)

\(\Leftrightarrow\frac{1}{y}=\frac{x}{8}-\frac{1}{4}\)

\(\Leftrightarrow\frac{1}{y}=\frac{x}{8}-\frac{2}{8}\)

\(\Leftrightarrow\frac{1}{y}=\frac{x-2}{8}\)

\(\Leftrightarrow y(x-2)=8\)

Vì \(x,y\inℤ\)nên \(x-2\inℤ\), ta có bảng sau:

1: (x+1)(y+2)=5

mà y+2>=2(do y là số tự nhiên)

nên (x+1;y+2)∈(1;5)

=>(x;y)∈(0;3)

2: (x+1)(y+2)=6

mà x+1>=1 và y+2>=2(do x,y là các số tự nhiên)

nên (x+1;y+2)∈{(3;2);(2;3);(1;6)}

=>(x;y)∈{(2;0);(1;1);(0;4)}

3: (x+2)(y+3)=6

mà x+2>=2 và y+3>=3(do x,y là các số tự nhiên)

nên (x+2;y+3)∈{(2;3)}

=>(x;y)∈(0;0)

4: (x-1)(y+3)=6

mà y+3>=3(do y là số tự nhiên)

nên (x-1;y+3)∈{(2;3);(1;6)}

=>(x;y)∈{(3;0);(2;3)}

5: (x-1)(y-3)=5

=>(x-1;y-3)∈{(1;5);(5;1)}

=>(x;y)∈{(4;8);(6;4)}

6: (x-2)(y-1)=3

=>(x-2;y-1)∈{(1;3);(3;1)}

=>(x;y)∈{(3;4);(5;2)}

7: (x-2)(y-1)=5

=>(x-2;y-1)∈{(1;5);(5;1)}

=>(x;y)∈{(3;6);(7;2)}

8: (x-3)(y+1)=7

mà y+1>=1(do y là số tự nhiên)

nên (x-3;y+1)∈{(1;7);(7;1)}

=>(x;y)∈{(4;6);(10;0)}