Tìm x , biết :
\(\frac{3}{x+5}=15\%\)
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\(\frac{-15}{12}x+\frac{3}{5}=\frac{6}{5}x-\frac{1}{2}\)
\(\Leftrightarrow\frac{-15}{12}x-\frac{6}{5}x=\frac{-1}{2}-\frac{3}{5}\)
\(\Leftrightarrow\frac{-49}{20}x=\frac{-11}{10}\)
\(\Leftrightarrow x=\frac{22}{49}\)
a) 15 + 3 ( x - 1 ) = 36
3 ( x - 1 ) = 36 - 15
3 ( x - 1 ) = 21
x - 1 = 21 : 3
x - 1 = 7
x = 7+1
x = 8
Ta có\(\frac{-15}{12}x+\frac{3}{7}=\frac{6}{5}x-\frac{1}{2}\)
=>\(\frac{-15}{12}x-\frac{6}{5}x=\frac{1}{2}-\frac{3}{7}\)
=>\(\left(\frac{-15}{12}-\frac{6}{5}\right)x=\frac{1}{14}\)
=>\(\frac{-49}{20}x=\frac{1}{14}\)
=>\(x=\frac{1}{14}:\frac{-49}{20}\)
=>\(x=\frac{-10}{343}\)
\(\frac{x+3}{97}+\frac{x+5}{95}+\frac{x+4}{96}+\frac{x+1}{99}=-4\)
\(\left(\frac{x+3}{97}+1\right)+\left(\frac{x+5}{95}+1\right)+\left(\frac{x+4}{96}+1\right)+\left(\frac{x+1}{99}+1\right)=-4+4\)
\(\frac{x+100}{97}+\frac{x+100}{95}+\frac{x+100}{96}+\frac{x+100}{99}=0\)
\(\left(x+100\right).\left(\frac{1}{97}+\frac{1}{95}+\frac{1}{96}+\frac{1}{99}\right)=0\)
=> \(\orbr{\begin{cases}x+100=0\\\frac{1}{97}+\frac{1}{95}+\frac{1}{96}+\frac{1}{99}=0\end{cases}}\)
Mà \(\frac{1}{97}+\frac{1}{95}+\frac{1}{96}+\frac{1}{99}\ne0\)
=> x + 100 = 0
=> x = -100
Vậy x = -100
Câu b trừ mỗi số đi 1 tức là trừ cả cụm đó cho 3 rùi lm tương tự câu a
\(-\frac{12}{15}=-\frac{x}{3}=-\frac{4}{y}=-\frac{z}{5}\)
\(\frac{-12}{15}=-\frac{x}{3}\)
\(\Leftrightarrow-x.15=-12.3\)
\(\Leftrightarrow-x.15=-36\)
\(\Leftrightarrow x=\frac{12}{5}\)
\(-\frac{12}{15}=-\frac{4}{y}\)
\(\Leftrightarrow-12.y=15.\left(-4\right)\)
\(\Leftrightarrow-12y=-60\)
\(\Leftrightarrow y=5\)
\(-\frac{12}{15}=\frac{z}{5}\)
\(\Leftrightarrow z.15=-12.5\)
\(\Leftrightarrow z.15=-60\)
\(\Leftrightarrow z=-4\)
\(\frac{3}{x+5}=15\%\)
\(\frac{3}{x+5}=\frac{15}{100}=\frac{3}{20}\)
=>x+5=20
=>x=15
#)Giải :
\(\frac{3}{x+5}=15\%\)
\(\frac{3}{x+5}=\frac{3}{20}\)
\(\Rightarrow\frac{3}{15+5}=\frac{3}{20}\)
\(\Rightarrow x=15\)
#~Will~be~Pens~#