tìm x, y, z biết 

Giải

\(\dfrac{x}{y}=\dfrac{3}{4}\\ =>\dfrac{x}{3}=\dfrac{y}{4}\\ \dfrac{y}{z}=\dfrac{3}{5}\\ =>\dfrac{y}{3}=\dfrac{z}{5}\\\dfrac{x}{3}=\dfrac{y}{4}\\ =>\dfrac{x}{3.3}=\dfrac{y}{4.3}\\ =>\dfrac{x}{9}=\dfrac{y}{12}\\ \dfrac{y}{3}=\dfrac{z}{5}\\ =>\dfrac{y}{3.4}=\dfrac{z}{4.5}\\ =>\dfrac{y}{12}=\dfrac{z}{20}\\ =>\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)

Áp dụng tính chất dãy tỉ số = 

\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}=\dfrac{2x-3y+z}{2.9-3.12+20}=\dfrac{6}{2}=3\\ =>\left\{{}\begin{matrix}x=3.9=27\\y=3.12=36\\z=3.20=60\end{matrix}\right.\)

  tìm x, y, z biết 

\(2x=3y\\ =>\dfrac{x}{3}=\dfrac{y}{2}\\ 4y=5z\\ =>\dfrac{y}{5}=\dfrac{z}{4}\\ \dfrac{x}{3}=\dfrac{y}{2}\\ =>\dfrac{x}{3.5}=\dfrac{y}{2.5}\\ =>\dfrac{x}{15}=\dfrac{y}{10}\\ \dfrac{y}{5}=\dfrac{z}{4}\\ =>\dfrac{y}{5.2}=\dfrac{z}{4.2}\\ =>\dfrac{y}{10}=\dfrac{z}{8}\\ =>\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{8}\)

Áp dụng tính chất dãy tỉ số bằng nhau

\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{8}=\dfrac{x+y+z}{15+10+8}=\dfrac{11}{33}=\dfrac{1}{3}\\ =>\left\{{}\begin{matrix}x=\dfrac{1}{3}.15=5\\y=\dfrac{1}{3}.10=\dfrac{10}{3}\\z=\dfrac{1}{3}.8=\dfrac{8}{3}\end{matrix}\right.\)

  tìm x, y, z biết

\(a,\dfrac{3^{10}.11+9^5.5}{27^3.2^4}.x=-9\\ =>\dfrac{3^{10}.11+\left(3^2\right)^5.5}{\left(3^3\right)^3.2^4}.x=-9\\ =>\dfrac{3^{10}.\left(11+5\right)}{3^9.2^4}.x=-9\\ =>\dfrac{3^{10}.16}{3^9.2^4}.x=-9\\ =>\dfrac{3^{10}.2^4}{3^9.2^4}.x=-9\\ =>3^1.x=-9\\ =>x=-9:3\\ =>x=-3\)

Giải

\(a,5-\left(x-2\right)^2=-4\\ =>\left(x-2\right)^2=5-\left(-4\right)\\ =>\left(x-2\right)^2=9\\ =>\left[{}\begin{matrix}\left(x-2\right)^2=3^2\\\left(x-2\right)^2=\left(-3\right)^2\end{matrix}\right.=>\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.=>\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

Vậy \(x=5;x=-1\)

  tìm x 

Giải

\(đk:6x-2\ne0=>x\ne\dfrac{1}{3}\\\dfrac{x+2}{6x-2}=\dfrac{1}{4}\\ =>\dfrac{4\left(x+2\right)}{4\left(6x-2\right)}=\dfrac{6x-2}{4\left(6x-2\right)}\\ =>4\left(x+2\right)=6x-2\\ =>4x+8-6x+2=0\\ =>-2x+10=0\\ =>-2x=-10\\ =>x=\left(-10\right):\left(-2\right)\\ =>x=5\left(thoamanđk\right)\)

Vậy \(x=5\)

Giải

\(\dfrac{x}{3}=\dfrac{y}{4}\\ \Leftrightarrow\dfrac{x}{3.3}=\dfrac{y}{4.3}\\\Leftrightarrow\dfrac{x}{9}=\dfrac{y}{12}\\ \dfrac{y}{3}=\dfrac{z}{5}\\ \Leftrightarrow \dfrac{y}{3.4}=\dfrac{z}{5.4}\\ \Leftrightarrow\dfrac{y}{12}=\dfrac{z}{20}\\ =>\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)

Áp dụng tính chất dãy tỉ số bằng nhau 

\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}=\dfrac{2x+3x+z}{2.9+3.12+20}=\dfrac{6}{74}=\dfrac{3}{37}\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{37}\times9=\dfrac{27}{37}\\y=\dfrac{3}{37}\times12=\dfrac{36}{37}\\z=\dfrac{3}{37}\times20=\dfrac{60}{37}\end{matrix}\right.\)

Giải :

\(\dfrac{x+11}{89}+\dfrac{x+13}{87}-\dfrac{x+15}{85}-\dfrac{x+17}{83}=0\\ =>\left(\dfrac{x+11}{89}+1\right)+\left(\dfrac{x+13}{87}+1\right)-\left(\dfrac{x+15}{85}+1\right)-\left(\dfrac{x+17}{83}+1\right)=0\\ =>\left(\dfrac{x+11+89}{89}\right)+\dfrac{x+13+87}{87}-\dfrac{x+15+85}{85}-\dfrac{x+17+83}{83}=0\\ =>\dfrac{x+100}{89}+\dfrac{x+100}{87}-\dfrac{x+100}{85}-\dfrac{x+100}{83}=0\\ =>\left(x+100\right)\left(\dfrac{1}{89}+\dfrac{1}{87}-\dfrac{1}{85}-\dfrac{1}{83}\right)=0\\ =>\left[{}\begin{matrix}x+100=0\\\dfrac{1}{89}+\dfrac{1}{87}-\dfrac{1}{85}-\dfrac{1}{83}=0\left(voli\right)\end{matrix}\right.=>x=-100\)