tìm x:
x/16=x/9
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=>x(x^2-9/16)=0
=>x(x-3/4)(x+3/4)=0
=>x=0; x=3/4; x=-3/4
\(x-\frac{2}{16}=-\frac{4}{2}-x\)
\(x+x=-\frac{4}{2}+\frac{2}{16}\)
\(2x=-\frac{15}{8}\)
\(x=-\frac{15}{16}\)
\(x-\frac{2}{16}=-\frac{4}{2}-x.\)
\(\Leftrightarrow x-\frac{1}{8}=-2-x\)
\(\Leftrightarrow x+x=-2+\frac{1}{8}\)(xài quy tắc chuyển vế nha)
\(\Leftrightarrow2x=\frac{-16+1}{8}\)
\(\Leftrightarrow2x=-\frac{15}{8}\Rightarrow x=-\frac{15}{8}\div2=-\frac{15}{8}\cdot\frac{1}{2}=-\frac{15}{16}\)
Mình làm hơi quá chi tiết và dài, bạn có thể lược bớt nha.
Học tốt ^3^
\(X:\dfrac{3}{9}=\dfrac{7}{8}\)
\(X=\dfrac{7}{8}\times\dfrac{3}{9}\)
\(X=\dfrac{21}{72}\)
\(\left(X-21\times13\right):11=30\)
\(X-273=330\)
\(X=603\)
`x:3/9=7/8`
`x=7/8xx3/9`
`x=7/24`
`-------------`
`(x-21xx13):11=30`
`(x-273):11=30`
`x-273=30xx11`
`x-273=330`
`x=330-273`
`x=57`
\(x=\dfrac{4}{5}\times\dfrac{4}{3}\)
\(x=\dfrac{16}{15}\)
-----------------------
\(x=\dfrac{5}{9}\times\dfrac{3}{8}\)
\(x=\dfrac{5}{24}\)
Bài 1:
a: \(\frac{x^2+x+4}{2}+\frac{x^2+x+7}{3}=\frac{x^2+x+13}{5}+\frac{x^2+x+16}{6}\)
=>\(\left(\frac{x^2+x+4}{2}-3\right)+\left(\frac{x^2+x+7}{3}-3\right)=\left(\frac{x^2+x+13}{5}-3\right)+\left(\frac{x^2+x+16}{6}-3\right)\)
=>\(\frac{x^2+x-2}{2}+\frac{x^2+x-2}{3}=\frac{x^2+x-2}{5}+\frac{x^2+x-2}{6}\)
=>\(\left(x^2+x-2\right)\left(\frac12+\frac13-\frac15-\frac16\right)=0\)
=>\(x^2+x-2=0\)
=>(x+2)(x-1)=0
=>\(\left[\begin{array}{l}x+2=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-2\\ x=1\end{array}\right.\)
b: \(\frac{2}{x^2+4x+3}+\frac{5}{x^2+11x+24}+\frac{2}{x^2+18x+80}=\frac{9}{25}\)
=>\(\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{5}{\left(x+3\right)\left(x+8\right)}+\frac{2}{\left(x+8\right)\left(x+10\right)}=\frac{9}{25}\)
=>\(\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+10}=\frac{9}{25}\)
=>\(\frac{1}{x+1}-\frac{1}{x+10}=\frac{9}{25}\)
=>\(\frac{x+10-x-1}{\left(x+1\right)\left(x+10\right)}=\frac{9}{25}\)
=>\(\frac{9}{\left(x+1\right)\left(x+10\right)}=\frac{9}{25}\)
=>(x+1)(x+10)=25
=>\(x^2+11x+10-25=0\)
=>\(x^2+11x-14=0\)
=>\(x^2+11x+\frac{121}{4}=44,25\)
=>\(\left(x+\frac{11}{2}\right)^2=\frac{177}{4}\)
=>\(x+\frac{11}{2}=\pm\sqrt{\frac{177}{2}}\)
=>\(x=-\frac{11}{2}\pm\frac{\sqrt{177}}{2}\)
x(x-3)+(x-3)=0
=>(x-3)(x+1)=0
=>\(\left[\begin{array}{l}x-3=0\\ x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=-1\end{array}\right.\)
x/16 = x/9
=> x . 9 = x.16
=> x = 0
= 0 nha bạn