a) \(12+\frac{x}{42}=\frac{5}{6}\)

\(\frac{x}{42}=\frac{5}{6}-12\)

\(\frac{x}{42}=\frac{-67}{6}\)

\(\frac{x}{42}=\frac{-469}{42}\)

\(\Rightarrow x=-469\)

Ta có nếu x=0 hoặc y=0 hoặc z=0 thì hpt vô nghiệm. Vậy x,y,z khác 0

\(\hept{\begin{cases}\frac{xy}{x+y}=\frac{6}{5}\\\frac{yz}{y+z}=\frac{4}{3}\\\frac{zx}{z+x}=\frac{12}{7}\end{cases}}\)nghịch đảo ta có (nghịch đảo đc vì x,y,z khác 0)\(\hept{\begin{cases}\frac{x+y}{xy}=\frac{5}{6}\\\frac{y+z}{yz}=\frac{3}{4}\\\frac{z+x}{xz}=\frac{7}{12}\end{cases}}\)<=>\(\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}=\frac{5}{6}\\\frac{1}{y}+\frac{1}{z}=\frac{3}{4}\\\frac{1}{z}+\frac{1}{x}=\frac{7}{12}\end{cases}}\)

Đặt a=\(\frac{1}{x}\),b=\(\frac{1}{y}\),c=\(\frac{1}{z}\)ta có \(\hept{\begin{cases}a+b=\frac{5}{6}\\b+c=\frac{3}{4}\\c+a=\frac{7}{12}\end{cases}}\) <=>\(\hept{\begin{cases}a+b+c=\left(\frac{5}{6}+\frac{3}{4}+\frac{7}{12}\right):2\\b=\frac{5}{6}-a\\c=\frac{7}{12}-a\end{cases}}\)

Thay vào giải ta có \(a+b+c=\frac{13}{12}\)

\(a+\frac{5}{6}-a+\frac{7}{12}-a=\frac{13}{12}\) => \(a=\frac{1}{3}\)=>\(x=3\)

tiếp tục tìm đc \(b=\frac{1}{2}\)=>\(y=2\)

                      \(c=\frac{1}{4}\)=>\(z=4\)

Vậy nghiệm hpt là \(\hept{\begin{cases}x=3\\y=2\\z=4\end{cases}}\)

Đặt \(M=\hept{\begin{cases}\frac{xy}{x+y}=\frac{6}{5}\\\frac{yz}{y+z}=\frac{4}{3}\\\frac{zx}{z+x}=\frac{12}{7}\end{cases}}\)

Ta có: \(\frac{xy}{x+y}=\frac{6}{5}\Leftrightarrow xy=6\&x+y=5\)

\(\Rightarrow x=5-6=\left(-1\right)\)

     \(\frac{yz}{y+z}=\frac{4}{3}\Leftrightarrow yz=4\&y+z=3\)

\(\Rightarrow y=3-4=\left(-1\right)\)

\(\frac{zx}{z+x}=\frac{12}{7}\Leftrightarrow zx=12;z+x=7\Rightarrow z=7-12=-5\)

\(\Rightarrow\hept{\begin{cases}x=-1\\y=-1\\z=-5\end{cases}}\)

\(\hept{\begin{cases}\frac{xy}{x+y}=\frac{6}{5}\\\frac{yz}{y+z}=\frac{4}{3}\\\frac{zx}{z+x}=\frac{12}{7}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}=\frac{5}{6}\\\frac{1}{y}+\frac{1}{z}=\frac{3}{4}\\\frac{1}{z}+\frac{1}{x}=\frac{7}{12}\end{cases}}\)

Đặt \(\hept{\begin{cases}\frac{1}{x}=a\\\frac{1}{y}=b\\\frac{1}{z}=c\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a+b=\frac{5}{6}\\b+c=\frac{3}{4}\\c+a=\frac{7}{12}\end{cases}}\)

Tự làm nốt

\(\frac{|x-2|}{12}\)\(+\)\(\frac{|x-2|}{20}+\)\(\frac{|x-2|}{30}+\)\(\frac{|x-2|}{42}\)\(=\frac{70^5}{2^3.21^6}\)

\(\Rightarrow|x-2|.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=\frac{2^5.5^5.7^5}{2^3.7^6.3^6}\)

\(\Rightarrow|x-2|.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)=\frac{2^2.5^5}{7.3^6}\)

\(\Rightarrow|x-2|.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)=\frac{4.5^5}{21.3^5}\)

\(\Rightarrow|x-2|\left(\frac{1}{3}-\frac{1}{7}\right)=\frac{4.5^5}{21.3^5}\)\(\Rightarrow|x-2|=\frac{5^5}{3^5}\)

ĐẾN ĐÂY DỄ RÙI TỰ GIẢI TIẾP

\(\frac{5}{14}\frac{5}{14}\) LÀ NHÂN À

Mình lỡ đánh nhầm 2 lần \(\frac{5}{14}\)nha :)) chỉ 1 lần thôi

\(\frac{x-199}{6}+\frac{x-199}{12}+\frac{x-199}{20}+\frac{x-199}{30}+\frac{x-199}{42}=\frac{5}{14}\)

\(\left(x-199\right)\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=\frac{5}{14}\)

\(\left(x-199\right)\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)=\frac{5}{14}\)

\(\left(x-199\right)\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{5}{14}\right)\)

\(\left(x-199\right)\left(\frac{1}{2}-\frac{1}{7}\right)=\frac{5}{14}\)

\(\left(x-199\right).\frac{5}{14}=\frac{5}{14}\)

\(\Leftrightarrow x-199=1\)

\(\Leftrightarrow x=200\)

\(\Leftrightarrow\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\times\frac{x}{3}=\frac{5}{21}\)

\(\Leftrightarrow\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\times\frac{x}{3}=\frac{5}{21}\)

\(\Leftrightarrow\left(\frac{1}{2}-\frac{1}{7}\right)\times\frac{x}{3}=\frac{5}{21}\)

\(\Leftrightarrow\frac{5}{14}\times\frac{x}{3}=\frac{5}{21}\)

\(\Leftrightarrow\frac{x}{3}=\frac{2}{3}\)

\(\Leftrightarrow x=2\)

\(\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right).\frac{x}{3}=\frac{5}{21}\)

\(\Rightarrow\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right).\frac{x}{3}=\frac{5}{21}\)

\(\Rightarrow\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}\right).\frac{x}{3}=\frac{5}{21}\)

\(\Rightarrow\left(\frac{1}{2}-\frac{1}{7}\right).\frac{x}{3}=\frac{5}{21}\)

\(\Rightarrow\left(\frac{7}{14}-\frac{2}{14}\right).\frac{x}{3}=\frac{5}{21}\)

\(\Rightarrow\frac{5}{14}.\frac{x}{3}=\frac{5}{21}\)

\(\Rightarrow\frac{x}{3}=\frac{5}{21}:\frac{5}{14}\)

\(\Rightarrow\frac{x}{3}=\frac{2}{3}\)

\(\Rightarrow x=2\)

Vậy \(x=2\)

Đặt \(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)

\(\Rightarrow A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)

\(\Rightarrow A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}\)

\(\Rightarrow A=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)

\(\Leftrightarrow\frac{5}{14}.\frac{4}{3}x=\frac{5}{14}\)

\(\Rightarrow\frac{4}{3}.x=\frac{5}{21}:\frac{5}{14}\)

\(\Rightarrow x=2\)

a) 42 x 5 : 6 - 12 = 23

b) 25- 40 x 3 : 8 = 10

a) x = 23

b) x =10

a=23

b=10