\(\frac{\frac{4}{115}-\frac{4}{5}-\frac{4}{6115}}{\frac{7}{115}-\frac{7}{5}-\frac{7}{6115}}+\frac{3}{7}\)

\(=\frac{4.\left(\frac{1}{115}-\frac{1}{5}-\frac{1}{6115}\right)}{7.\left(\frac{1}{115}-\frac{1}{5}-\frac{1}{6115}\right)}+\frac{3}{7}\)

\(=\frac{4}{7}+\frac{3}{7}=1\)

\(a)\)\(\frac{\frac{4}{115}-\frac{4}{5}-\frac{4}{6115}}{\frac{7}{115}-\frac{7}{5}-\frac{7}{6115}}+\frac{3}{7}\)

\(=\)\(\frac{4.\left(\frac{1}{115}-\frac{1}{5}-\frac{1}{6115}\right)}{7.\left(\frac{1}{115}-\frac{1}{5}-\frac{1}{6115}\right)}+\frac{3}{7}\)

\(=\)\(\frac{4}{7}+\frac{3}{7}\)

\(=\)\(1\)

Theo đề ra, ta có: 

\(\hept{\begin{cases}x\left(x+y+z\right)=-5\\y\left(x+y+z\right)=9\\z\left(x+y+z\right)=5\end{cases}}\)

\(\Rightarrow x\left(x+y+z\right)+y\left(x+y+z\right)+z\left(x+y+z\right)=\left(-5\right)+9+5\)

\(\Rightarrow\left(x+y+z\right)^2=9\)

\(\Rightarrow\orbr{\begin{cases}x+y+z=3\\x+y+z=-3\end{cases}}\)

Trường hợp 1: \(x+y+z=3\Rightarrow\hept{\begin{cases}3x=\left(-5\right)\\3y=9\\3z=5\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{-5}{3}\\y=3\\z=\frac{5}{3}\end{cases}}\)

Trường hợp 2: \(x+y+z=-3\Rightarrow\hept{\begin{cases}\left(-3x=-5\right)\\\left(-3y=9\right)\\\left(-3z=5\right)\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\y=-3\\z=\frac{-5}{3}\end{cases}}\)

pls fast

{k} FATS PLSSSSSSSSSSSSSSSS

Ta có \(\frac{a}{b}=\frac{c}{d}\)

a) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}+1=\frac{c}{d}+1\Rightarrow\frac{a+b}{b}=\frac{c+d}{d}\)

b) \(\frac{a}{c}=\frac{a+b}{c+d}\Rightarrow\frac{a}{a+b}=\frac{c}{c+d}\)

c) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}-1=\frac{c}{d}-1\Rightarrow\frac{a-b}{b}=\frac{c-d}{d}\)

d) \(\frac{a}{b}=\frac{c}{d}\Rightarrow1:\frac{a}{b}=1:\frac{c}{d}\Rightarrow\frac{b}{a}=\frac{d}{c}\Rightarrow1-\frac{b}{a}=1-\frac{d}{c}\Rightarrow\frac{a-b}{a}=\frac{c-d}{c}\Rightarrow1:\frac{a-b}{a}=1:\frac{c-d}{c}\)

\(\Rightarrow\frac{a}{a-b}=\frac{c}{c-d}\)

Đặt `a/b=c/d =k ->a=bk, c=dk`

`a,`

`(a+b)/b=(bk +b)/b=(b (k+1) )/b=k+1`

`(c+d)/d=(dk +d)/d=(d (k+1) )/d=k+1`

`-> (a+b)/b=(c+d)/d`

`b,`

`a/(a+b)=(bk)/(bk+b)=(bk)/(b(k+1) )=k/(k+1)`

`c/(c+d)=(dk)/(dk+d)=(dk)/(d(k+1) ) = k/(k+1)`

`-> a/(a+b)=c/(c+d)`

`c,`

`(a-b)/b=(bk-b)/b=(b(k-1) )/b=k-1`

`(c-d)/d=(dk-d)/d=(d(k-1) )/d=k-1`

`-> (a-b)/b=(c-d)/d`

`d,`

`a/(a-b) =(bk)/(bk-b)=(bk)/(b(k-1) )=k/(k-1)`

`c/(c-d)=(dk)/(dk-d)=(dk)/(d(k-1) )=k/(k-1)`

`-> a/(a-b)=c/(c-d)`

21/10 nha bạn

\(\frac{3}{5}-\left(-\frac{6}{4}\right)=\frac{3}{5}+\frac{3}{2}=\frac{6+15}{10}=\frac{21}{10}\)

a) \(-\left(\frac{13}{25}-\frac{4}{191}+\frac{2}{51}\right)+\left(-\frac{4}{191}+\frac{2}{51}+\frac{3}{5}\right)\)

\(=-\frac{13}{25}+\frac{4}{191}-\frac{2}{51}-\frac{4}{191}+\frac{2}{51}+\frac{3}{5}\)

\(=\left(-\frac{13}{25}+\frac{3}{5}\right)+\left(\frac{4}{191}-\frac{4}{191}\right)-\left(\frac{2}{51}+\frac{2}{51}\right)\)

\(=\frac{2}{25}+0-0=\frac{2}{25}\)

b) \(12\frac{3}{5}:\left(-\frac{5}{7}\right)+2\frac{2}{5}:\left(-\frac{5}{7}\right)\)

\(=\frac{63}{5}.\frac{-7}{5}+\frac{12}{5}.\frac{-7}{5}\)

\(=\left(\frac{63}{5}+\frac{12}{5}\right).\frac{-7}{5}\)

\(=15.\frac{-7}{5}=-21\)

\(\left|5x-3\right|=\left|7-x\right|\)

\(\Rightarrow\orbr{\begin{cases}5x-3=7-x\\5x-3=x-7\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}5x+x=7+3\\5x-x=\left(-7\right)+3\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}6x=10\\4x=-4\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}\)

`|5x-3|=|7-x|`

TH1 : `5x-3=7-x -> 5x+x=7+3-> 6x=10 ->x=5/3`

TH2 : `5x-3=-7 +x -> 5x-x=-7+3 -> 4x=-4 ->x=-1`

Vậy `x=5/3,x=-1`

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