find the value of x such that \(\frac{x+3}{7}=\frac{2-x}{3}\)
x=?
lamf ddusng tick cho
Find the value of x such that: \(\frac{5}{7-2x}=\frac{3}{x-3}\)
Find the value of x such that\(\frac{3}{x-2}=\frac{2}{x-3}\)
\(3\cdot\left(x-3\right)=2\cdot\left(x-2\right)\)
\(3x-9=2x-4\)
\(3x-2x=4+9\)
\(x=13\)
tick nha
Find the value of x such that .\(\frac{x+3}{7}=\frac{2-x}{3}\)
Answer: .x=.....
(write your answer by decimal in simplest form)
=>3(x+3)=7.(2-x)
=>3x+9=14-2x
=>3x+2x=14-9
=>5x=5=>x=1
Find the value of x such that: \(\frac{3\left(x+2\right)}{2x+3}=\frac{7}{8},\left(x\ne-\frac{3}{2}\right)\) . Answer: x = ...
( write your answer by decimal in simplest form )
Find the value of x such that:\(\frac{3}{2x+7}=\frac{5}{3x+9}\)
\(x\ne-\frac{7}{2};x\ne-3\)
\(\Rightarrow3\left(3x+9\right)=5\left(2x+7\right)\)
\(\Rightarrow9x+27=10x+35\)
\(\Rightarrow10x-9x=27-35\)
\(\Rightarrow x=-8\)
ôg giúp t cái nài nữa :
Find the value of x such that \(\frac{3}{x-2}=\frac{2}{x-3}\)
Answer x = ......
<=>3(x-3)=2(x-2)
<=>3x-9=2x-4
<=>3x-2x=-4+9
<=>x=5
tick nha
Find the value of x such that \(\frac{x-7}{6}+\frac{x-11}{10}=\frac{x-8}{7}+\frac{x-10}{9}\)
.
Answer: .x=..
Ta có
\(\frac{x-7}{6}+1+\frac{x-11}{10}+1=\frac{x-8}{7}+1+\frac{x-10}{9}+1\)
\(\frac{x-1}{6}+\frac{x-1}{10}-\frac{x-1}{7}-\frac{x-1}{9}=0\)
<=>\(\left(x-1\right)\left(\frac{1}{6}+\frac{1}{10}-\frac{1}{7}-\frac{1}{11}\right)=0\)
=>x-1=0
<=>x=1
Find the value of x such that\(\frac{x^3}{3}=\frac{3^2}{2^3}\) .
Answer x=: .
(write your answer by decimal in simplest form)
Find the value of x, such that:
\(\frac{x-7}{6}+\frac{x-11}{10}=\frac{x-8}{7}+\frac{x-10}{9}\)
Help cái koi, Nhọ Nồi ôg onl thì lm giúp cái !
\(\frac{x-7}{6}+\frac{x-11}{10}=\frac{x-8}{7}+\frac{x-10}{9}\)
\(\Rightarrow\left(\frac{x-7}{6}+1\right)+\left(\frac{x-11}{10}+1\right)=\left(\frac{x-8}{7}+1\right)+\left(\frac{x-10}{9}+1\right)\)
\(\Rightarrow\frac{x-1}{6}+\frac{x-1}{10}-\frac{x-1}{7}-\frac{x-1}{9}=0\)
\(\left(x-1\right).\left(\frac{1}{6}+\frac{1}{10}-\frac{1}{7}-\frac{1}{9}\right)=0\)
\(\Rightarrow x-1=0\Rightarrow x=1\)