find the value of x such that \(\frac{x+3}{7}=\frac{2-x}{3}\)
x=?
lamf ddusng tick cho
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\(3\cdot\left(x-3\right)=2\cdot\left(x-2\right)\)
\(3x-9=2x-4\)
\(3x-2x=4+9\)
\(x=13\)
tick nha
\(x\ne-\frac{7}{2};x\ne-3\)
\(\Rightarrow3\left(3x+9\right)=5\left(2x+7\right)\)
\(\Rightarrow9x+27=10x+35\)
\(\Rightarrow10x-9x=27-35\)
\(\Rightarrow x=-8\)
<=>3(x-3)=2(x-2)
<=>3x-9=2x-4
<=>3x-2x=-4+9
<=>x=5
tick nha
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
\(\frac{x-7}{6}+\frac{x-11}{10}=\frac{x-8}{7}+\frac{x-10}{9}\)
\(\Rightarrow\left(\frac{x-7}{6}+1\right)+\left(\frac{x-11}{10}+1\right)=\left(\frac{x-8}{7}+1\right)+\left(\frac{x-10}{9}+1\right)\)
\(\Rightarrow\frac{x-1}{6}+\frac{x-1}{10}-\frac{x-1}{7}-\frac{x-1}{9}=0\)
\(\left(x-1\right).\left(\frac{1}{6}+\frac{1}{10}-\frac{1}{7}-\frac{1}{9}\right)=0\)
\(\Rightarrow x-1=0\Rightarrow x=1\)
\(\frac{x+3}{7}=\frac{2-x}{3}=>\left(x+3\right).3=\left(2-x\right).7=>3x+9=14-7x=>3x+7x=14-9=>10x=5=>x=\frac{5}{10}=\frac{1}{2}=0,5\)
vậy x=1/2=0,5
tick tớ nhé