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\(\frac{7}{3}x+x=42\)
\(\frac{7}{3}x+1x=42\)
\(\left(\frac{7}{3}+1\right)x=42\)
\(\frac{10}{3}x=42\)
\(x=42:\frac{10}{3}\)
\(x=12,6\)
\(\frac{7}{3}.x+x=42\)
\(\Rightarrow x.\left(\frac{7}{3}+1\right)=42\)
\(\Rightarrow x.\frac{10}{3}=42\)
\(\Rightarrow x=8,4\)
Vậy x = 8,4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3\frac{3}{5}x-x=4,5\)
\(\Rightarrow\frac{16}{5}x-\frac{5}{5}x=\frac{9}{2}\)
\(\Rightarrow x\left(\frac{16}{5}-\frac{5}{5}\right)=\frac{9}{2}\)
\(\Rightarrow\frac{11}{5}x=\frac{9}{2}\)
\(\Rightarrow x=\frac{9}{2}:\frac{11}{5}\)
\(\Rightarrow x=\frac{45}{22}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta co : \(\frac{\sqrt{x+1}}{\sqrt{x-3}}=\frac{\sqrt{x-3+4}}{\sqrt{x-3}}=\frac{\sqrt{x-3}}{\sqrt{x-3}}+\frac{4}{\sqrt{x-3}}\)\(=1+\frac{4}{\sqrt{x-3}}\)
De x nguyen thi \(1+\frac{4}{\sqrt{x-3}}\)nguyen
\(\Rightarrow\)\(\frac{4}{\sqrt{x-3}}\)nguyen\(\Rightarrow\)4 chia het cho \(\sqrt{x-3}\)
\(\Rightarrow\)4\(\in B\)cua \(\sqrt{x-3}\)
\(\Rightarrow\sqrt{x-3}\in\left\{4;-4;2;-2;1;-1\right\}\)
TH1 : \(\sqrt{x-3}=4\)
\(\Rightarrow x-3=16\Rightarrow x=19\)\(\Rightarrow\)chon
TH2 : \(\sqrt{x-3}=-4\) vo ly \(\Rightarrow\) loai
TH3 : \(\sqrt{x-3}=2\Rightarrow x-3=4\Rightarrow x=7\Rightarrow\)chon
TH4 : \(\sqrt{x-3}=-2\Rightarrow\)vo ly \(\Rightarrow\)chon
TH5 : \(\sqrt{x-3}=1\Rightarrow x-3=1\Rightarrow x=4\Rightarrow\)chon
TH6 : \(\sqrt{x-3}=-1\Rightarrow\)vo ly\(\Rightarrow\)loai
Vay x\(\in\){19;7;4}
. này là nhân hay phẩy vậy bạn
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tìm gì vậy x hay biểu thức