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\(ĐK:y\ne0\)
\(\frac{y+5}{7-y}=\frac{2}{-5}\)
\(\Leftrightarrow\left(y+5\right).\left(-5\right)=\left(7-y\right).2\)
\(\Leftrightarrow-5y-25=14-2y\)
\(\Leftrightarrow-3y=39\)
\(\Leftrightarrow y=-13\)
Vậy: \(y=-13\)
Tíck cho mìk vs nha Bao Bui !
\(\frac{y+5}{7-y}=\frac{2}{-5}\)=> -5(y + 5) = 2(7 - y) <=> -5y - 25 = 14 - 2y => -5y - 25 + (2y + 25) = 14 - 2y + (2y + 25) <=> -3y = 39 => y = -13
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{y+5}{7-y}=\frac{2}{-5}\)
=> -5(y + 5) = 2(7 - y)
=> -5y - 25 = 14 - 2y
=> -3y = 39
=> y = -13
![](https://rs.olm.vn/images/avt/0.png?1311)
Điều kiện y khác 7
\(\frac{y+5}{7-y}=\frac{2}{-5}\Rightarrow-5\left(y+5\right)=2\left(7-y\right)\Rightarrow-5y-25=14-2y\)
\(\Rightarrow-3y=39\Rightarrow y=-13\)
![](https://rs.olm.vn/images/avt/0.png?1311)
câu 2 :
ta có ( y + 5 ) . - 5 = ( 7 - y ) . 2
-5y + -25 = 14 - 2y
-5y + 2y = 14 - ( - 25 )
- 3y = 39
=> y = 39 : ( -3 ) = -13
cau1
Vì abc chia hết cho abc
Nên 1abc = 1000 + abc
Tổng (1000 + abc) chia abc dư 3 khi 1000 – abc = 3
Suy ra abc = 1000 -3 = 997
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\(\frac{y+5}{7-y}=\frac{2}{-5}\)
<=>(y+5).(-5)=(7-y).2
<=>(-5y)+(-25)=14-2y
<=>(-5y+2y)=25+14
<=>(-3y)=39
<=>y=-13
Vậy y=-13
\(\Leftrightarrow\frac{y+5}{7-y}=\frac{-y+5}{y-7}\)
\(\Leftrightarrow\frac{-y+5}{y-7}=\frac{-2}{5}\)
\(\Rightarrow\frac{y}{7-y}+\frac{5}{7-y}=\frac{-2}{5}\)
\(\Leftrightarrow\frac{y}{7-y}+\frac{5}{7-y}+\frac{2}{5}=0\)
\(\Rightarrow\frac{-3\left(y+13\right)}{5\left(y-7\right)}=0\)
\(\Leftrightarrow\frac{1}{y-7}=0\)
=>y=13
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có :y+5.(-5)=2.(7-y)
-5y+-25=14-2y
áp dụng tìm y chuyển vế đổi dấu là giải ra
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x-2}{27}+\frac{x-3}{26}+\frac{x-4}{25}+\frac{x-5}{24}+\frac{x-44}{5}=1\)
\(\Leftrightarrow\left(\frac{x-2}{27}-1\right)+\left(\frac{x-3}{26}-1\right)+\left(\frac{x-4}{25}-1\right)+\left(\frac{x-5}{24}-1\right)\)\(+\left(\frac{x-44}{5}+3\right)=1-1\)
\(\Leftrightarrow\frac{x-29}{27}+\frac{x-29}{26}+\frac{x-29}{25}+\frac{x-29}{24}\)\(+\frac{x-29}{5}=0\)
\(\Leftrightarrow\left(x-29\right)\left(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\right)=0\)
Mà \(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\ne0\)
=> x - 29 = 0
=> x = 29.
Ta có: \(\frac{y+5}{7-y}=\frac{2}{-5}\) \(\left(y+5\right).-5=2.\left(7-y\right)\) \(-5y-25=14-2y\) \(-3y=39\) \(y=-13\) Vậy \(y=-13\)