a)\(25,5+y-12,5=4.7\)

⇔\(13+y=28\)

⇔\(y=15\)

b)\(76,22-y-25,7=30+5,52\)

⇔\(50,52-y=35,52\)

⇔\(y=15\)

c)\(4,5-y+1,2=3,5\)

⇔\(5,7-y=3,5\)

⇔\(y=2,2\)

\(a,\Rightarrow10+y=28\\ \Rightarrow y=18\\ b,\Rightarrow50,52-y=35,52\\ \Rightarrow y=15\\ c,\Rightarrow5,7-y=3,5\\ \Rightarrow y=2,2\)

a,Ta có:

\(\dfrac{x}{y}=\dfrac{7}{4}=\dfrac{x}{7}=\dfrac{y}{4}\)

ÁP dụng tcdtsbn , ta có:

\(\dfrac{x}{7}=\dfrac{y}{4}=\dfrac{x+y}{7+4}=\dfrac{33}{11}=3\)

\(\Rightarrow\left\{{}\begin{matrix}x=21\\y=12\end{matrix}\right.\)

b,

\(\Rightarrow3.\left(x-1\right)=-24\)

\(\Rightarrow x-1=-8\)

\(\Rightarrow x=-7\)

A)\(\dfrac{x}{y}=\dfrac{7}{4}\Rightarrow\dfrac{x}{7}=\dfrac{y}{4}\)

Áp dụng t/c dtsbn ta có:

\(\dfrac{x}{7}=\dfrac{y}{4}=\dfrac{x+y}{7+4}=\dfrac{33}{11}=3\)

\(\dfrac{x}{7}=3\Rightarrow x=21\\ \dfrac{y}{4}=3\Rightarrow y=12\)

B) \(3\left(x-1\right)+5=-19\\ \Rightarrow3\left(x-1\right)=-24\\ \Rightarrow x-1=-8\\ \Rightarrow x=-7\)

\(a.\)

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

Áp dụng tính chất dãy tỉ số bằng nhau : 

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{35}{7}=5\)

\(\Rightarrow x=5\cdot2=10\\ y=5\cdot5=25\)

\(b.\)

\(\dfrac{x+2}{y+10}=\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{x+2}{1}=\dfrac{y+10}{5}\)

\(\Leftrightarrow\dfrac{3x+6}{3}=\dfrac{y+10}{5}\)

Áp dụng tính chất dãy tỉ số bằng nhau : 

\(\Leftrightarrow\dfrac{3x+6}{3}=\dfrac{y+10}{5}=\dfrac{y+10-3x-6}{5-3}=\dfrac{2-4}{2}=-1\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x+6=-3\\y+10=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-15\end{matrix}\right.\)

\(c.\)

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

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

Áp dụng tính chất dãy tỉ số bằng nhau : 

\(\dfrac{2x}{8}=\dfrac{y}{5}=\dfrac{2x-y}{8-5}=\dfrac{15}{3}=5\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=5\cdot8\\y=5\cdot5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=20\\y=25\end{matrix}\right.\)

a) Ta có: \(\dfrac{x}{2}=\dfrac{y}{5}\)

mà x+y=35

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

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{35}{7}=5\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{x}{2}=5\\\dfrac{y}{5}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=25\end{matrix}\right.\)

Vậy: (x,y)=(10;25)

b) Ta có: \(\dfrac{x+2}{y+10}=\dfrac{1}{5}\)

nên \(\dfrac{x+2}{1}=\dfrac{y+10}{5}\)

hay \(\dfrac{3x+6}{3}=\dfrac{y+10}{5}\)

mà y-3x=2 

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

\(\dfrac{3x+6}{3}=\dfrac{y+10}{5}=\dfrac{y-3x+10-6}{5-3}=\dfrac{2+4}{2}=3\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{3x+6}{3}=3\\\dfrac{y+10}{5}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+6=9\\y+10=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=3\\y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)

Vậy: (x,y)=(1;5)

c) Ta có: \(\dfrac{x}{4}=\dfrac{y}{5}\)

nên \(\dfrac{2x}{8}=\dfrac{y}{5}\)

mà 2x-y=15

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

\(\dfrac{2x}{8}=\dfrac{y}{5}=\dfrac{2x-y}{8-5}=\dfrac{15}{3}=5\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{x}{4}=5\\\dfrac{y}{5}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=20\\y=25\end{matrix}\right.\)

Vậy: (x,y)=(20;25)

Giải:

a) \(y^2=3-\left|2x-3\right|\) 

Vì \(-\left|2x-3\right|\le0\forall x\) nên \(3-\left|2x-3\right|\le3\forall x\) nên \(y^2\le3\rightarrow y^2\in\left\{0;1\right\}\) (vì \(y\in Z\) )

TH1:

\(y^2=0\) 

\(\Rightarrow y=0\) 

\(\Rightarrow\left|2x-3\right|=3\) 

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\) 

TH2:

\(y^2=1\) 

\(\Rightarrow y=\pm1\)

ơ xin lỗi lỡ nhấn gửi

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

\(\dfrac{x}{4}=\dfrac{y}{-5}=\dfrac{-3x+2y}{-12-10}=\dfrac{55}{-22}=\dfrac{-5}{2}\)

Do đó: \(\left\{{}\begin{matrix}x=\dfrac{-20}{2}=-10\\y=\dfrac{25}{2}\end{matrix}\right.\)

b: Ta có: \(\dfrac{x}{y}=\dfrac{-7}{4}\)

nên \(\dfrac{x}{-7}=\dfrac{y}{4}\)

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

\(\dfrac{x}{-7}=\dfrac{y}{4}=\dfrac{4x-5y}{-28-20}=\dfrac{72}{-48}=\dfrac{-3}{2}\)

Do đó: \(\left\{{}\begin{matrix}x=\dfrac{21}{2}\\y=\dfrac{-12}{2}=-6\end{matrix}\right.\)

c) \(\dfrac{x}{-3}=\dfrac{y}{8}\)   

⇒\(\dfrac{x^2}{-9}=\dfrac{y^2}{64}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x^2}{-9}=\dfrac{y^2}{64}=-\dfrac{44}{\dfrac{5}{-9+64}}=-\dfrac{44}{\dfrac{5}{55}}=-484\)

` 242/363 + 1616/2121 = 2/7 xxy`

`2/7 xxy= 2/3 + 16/21`

`2/7 xxy= 14/21 +16/21`

`2/7 xxy= 30/21`

`y=10/7 : 2/7`

`y=10/7 xx 7/2`

`y=70/14`

`y=5`

__

` (y + 1/4) + (y + 1/16) + (y + 1/16) =2`

`(y+y+y)+(1/4 + 1/16+1/16)=2`

`3y + (4/16 +1/16 +1/16)=2`

`3y + 6/16=2`

`3y=2-6/16`

`3y= 32/16-6/16`

`3y= 26/16`

`y=26/16 : 3`

`y=26/48`

`y=13/24`

\(a,\dfrac{242}{363}+\dfrac{1616}{2121}=\dfrac{2}{7}\times y\)

\(\dfrac{2}{7}\times y=\dfrac{2\times121}{3\times121}+\dfrac{16\times101}{21\times101}\)

\(\dfrac{2}{7}\times y=\dfrac{2}{3}+\dfrac{16}{21}\)

\(\dfrac{2}{7}\times y=\dfrac{14}{21}+\dfrac{16}{21}\)

\(\dfrac{2}{7}\times y=\dfrac{30}{21}\)

\(\dfrac{2}{7}\times y=\dfrac{10}{7}\)

\(y=\dfrac{10}{7}:\dfrac{2}{7}\)

\(y=\dfrac{10}{7}\times\dfrac{7}{2}\)

\(y=5\)

\(---\)

\(b,\left(y+\dfrac{1}{4}\right)+\left(y+\dfrac{1}{16}\right)+\left(y+\dfrac{1}{16}\right)=2\)

\(\left(y+y+y\right)+\left(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{16}\right)=2\)

\(3\times y+\left(\dfrac{4}{16}+\dfrac{2}{16}\right)=2\)

\(3\times y+\dfrac{6}{16}=2\)

\(3\times y+\dfrac{3}{8}=2\)

\(3\times y=2-\dfrac{3}{8}\)

\(3\times y=\dfrac{16}{8}-\dfrac{3}{8}\)

\(3\times y=\dfrac{13}{8}\)

\(y=\dfrac{13}{8}:3\)

\(y=\dfrac{13}{8}\times\dfrac{1}{3}\)

\(y=\dfrac{13}{24}\)

#\(Toru\)

=2

y=2

a) y x 42 + y x 57 + y = 25 400

y x ( 42 + 57 + 1 ) = 25 400

y x 100 = 25 400

y = 25 400 : 100

y = 254

b) 142 x y - 41 x y - y = 408 000

y x ( 142 - 41 - 1 ) = 408 000

y x 100 = 408 000

y = 408 000 : 100

y = 4 080

a.y*(42+57+1)=25400

y*100=25400

y=25400:100=254

b.(142-41-1)*y=408000

100*y=408000

y=408000:100=4080