18/5 + 1< x < 1+ 30/7
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\(\frac{5}{7}\)x18x\(\frac{1}{3}\)-\(\frac{5}{7}\)x 30 x\(\frac{1}{3}\)
=\(\frac{90}{7}\)x\(\frac{1}{3}\)-\(\frac{5}{7}\)x 30 x\(\frac{1}{3}\)
=\(\frac{30}{7}\)-\(\frac{5}{7}\)x 30 x\(\frac{1}{3}\)
=\(\frac{30}{7}\)-\(\frac{150}{7}\)x\(\frac{1}{3}\)
=\(\frac{30}{7}\)-\(\frac{50}{7}\)
=\(\frac{-20}{7}\)
Hok tốt!
@Kaito Kid
![](https://rs.olm.vn/images/avt/0.png?1311)
18/5 + 1 = 18/5 + 5/5 = 23/5 = 161/7
1 + 30/7 = 7/7 + 30/7 = 37/7 = 185/7
vậy ta có: \(\frac{161}{7}\)< x < \(\frac{185}{7}\)
x = \(\frac{162}{7};\frac{163}{7};....;\frac{184}{7}\)
18/5 + 1 = 18/5 + 5/5 = 23/5 = 161/7
1 + 30/7 = 7/7 + 30/7 = 37/7 = 185/7
vậy ta có: $\frac{161}{7}$1617 < x < $\frac{185}{7}$1857
x = $\frac{162}{7};\frac{163}{7};....;\frac{184}{7}$1627 ;1637 ;....;1847
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{18}{5}+1=\frac{23}{5}=\frac{23\cdot7}{5\cdot7}=\frac{161}{35}\)
\(\frac{30}{7}+1=\frac{37}{7}=\frac{37\cdot5}{7\cdot5}=\frac{185}{35}\)
=>161/35<x<185/35
=>x=162/35 ;163/35 ;..............184/35
18/5 + 1=23/5
1+30/7=37/7
Quy đòng tử 23/5 và 37/7 được 851/185 và 851/161
Vậy x = 851/184;851/183;.....;851/162
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) Ta có: \(\left(-5+x\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-5+x=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=7\end{matrix}\right.\)
Vậy: \(x\in\left\{5;7\right\}\)
2) Ta có: \(\left(30-x\right)\left(2x-16\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}30-x=0\\2x-16=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=-30\\2x=16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=30\\x=8\end{matrix}\right.\)
Vậy: \(x\in\left\{30;8\right\}\)
3) Ta có: \(\left(-5-x\right)\left(17+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-5-x=0\\17+x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=5\\x=0-17\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-17\end{matrix}\right.\)
Vậy: \(x\in\left\{-5;-17\right\}\)
4) Ta có: \(\left(-3x+18\right)\left(-5x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-3x+18=0\\-5x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x=-18\\-5x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{6;-2\right\}\)
Bài nay ta có hai vế bạn hãy đặt giả sử một trong hai vế bằng 0 rồi giải phương trình cho mỗi vế bằng o
![](https://rs.olm.vn/images/avt/0.png?1311)
B)
( x + 4/3 ) * 7/4 = 5 - 7/6
( x + 4/3) * 7/4 = 5/1 - 7/6
( x + 4/3) * 7/4 = 30/6 - 7/6
( x + 4/3) * 7/4 = 23/6
( x + 4/3) = 23/6 : 7/4
( x+ 4/3) = 23/6 . 4/7
( x+ 4/3) = 46/21
x = 46/21 - 4/3
x = 46/21 - 28/21
x = 18/21 = 6/7
Tk vs kb hộ mk nha! thank u
may cau khac ko kho lam dau chi co cau d thoi.
(100%-50%+40%):x=18%+30%:x
90%:x=18%+30%:x
18%=90%:x-30%:x
18%=60%:x
X=60%:18%
x=..................................................................................................
xong roi do a cho minh nhe