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a) Ta có:
\(\frac{x}{x+4}=\frac{5}{6}\)
\(\Rightarrow6x=5\left(x+4\right)\)
\(\Rightarrow6x=5x+20\)
\(\Rightarrow5x+20-6x=0\)
\(-x=-20\)
\(x=20\)
b)Ta có: \(\frac{x-3}{x+5}=\frac{5}{7}\)
\(\Rightarrow7\left(x-3\right)=5\left(x+5\right)\)
\(\Rightarrow7x-21=5x+25\)
\(\Rightarrow7x-21-5x-25=0\)
\(2x=46\)
\(x=23\)
c)Ta có: \(\frac{x+4}{20}=\frac{5}{20}\)
\(\Rightarrow x+4=5\)
\(\Rightarrow x=1\)
a ) Ta có : \(\frac{x+11}{10}+\frac{x+21}{20}+\frac{x+31}{30}=\frac{x+41}{40}+\frac{x+101}{5}\)
\(\Leftrightarrow\left(\frac{x+11}{10}-1\right)+\left(\frac{x+21}{10}-1\right)+\left(\frac{x+31}{30}-1\right)=\left(\frac{x+41}{40}-1\right)+\left(\frac{x+101}{50}-2\right)\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}=\frac{x+1}{40}+\frac{x+1}{50}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}-\frac{x+1}{40}-\frac{x+1}{50}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)\ne0\)
Nên x + 1 = 0
=> x = -1
\(\begin{array}{l}a)x - \left( {\dfrac{5}{4} - \dfrac{7}{5}} \right) = \dfrac{9}{{20}}\\x = \dfrac{9}{{20}} + \left( {\dfrac{5}{4} - \dfrac{7}{5}} \right)\\x = \dfrac{9}{{20}} + \dfrac{{25}}{{20}} - \dfrac{{28}}{{20}}\\x = \dfrac{{6}}{{20}}\\x = \dfrac{{ 3}}{{10}}\end{array}\)
Vậy \(x = \dfrac{{ 3}}{{10}}\)
\(\begin{array}{*{20}{l}}{b)9 - x = \dfrac{8}{7} - \left( { - \dfrac{7}{8}} \right)}\\\begin{array}{l}9 - x = \dfrac{8}{7} + \dfrac{7}{8}\\9 - x = \dfrac{{64}}{{56}} + \dfrac{{49}}{{56}}\\9 - x = \dfrac{{113}}{{56}}\end{array}\\{x = 9 - \dfrac{{113}}{{56}}}\\{x = \dfrac{{504}}{{56}} - \dfrac{{113}}{{56}}}\\{x = \dfrac{{391}}{{56}}}\end{array}\)
Vậy \(x = \dfrac{{391}}{{56}}\)
Ta có: \(\frac{x-3}{x+5}=\frac{5}{7}\)
\(\Leftrightarrow\left(x-3\right).7=\left(x+5\right).5\)
\(\Leftrightarrow7x-3=5x+25\)
\(\Rightarrow7x-5x=25+3\)
\(\Rightarrow2x=28\)
\(\Rightarrow x=14\)
- c, Ta co :\(\frac{x+4}{20}=\frac{5}{x+4}\)
=> \(\left[x+4\right].\left[x+4\right]=20.5\)
=> \(\left[x+4\right]^2=100\)
=> \(\left[x+4\right]^2=10^2\)
=> \(x+4=10\)
=> \(x=10-4\)
=> \(x=6\)
k mk nha
Có lẽ bạn viết đề sai.
Câu hỏi của Vũ Mai Linh - Toán lớp 7 - Học toán với OnlineMath
\(\frac{x}{x+4}=\frac{5}{6}=>6x=5\left(x+4\right)=5x+20\)
\(=>6x-5x=20=>x=20\)
áp dụng \(\frac{a}{b}=\frac{c}{d}< =>a.d=b.c\)