a: =>4/x=y/-21=4/7

=>x=7; y=-12

b: =>xy=63

mà x>y

nên \(\left(x,y\right)\in\left\{\left(9;7\right);\left(21;3\right);\left(63;1\right);\left(-7;-9\right);\left(-3;-21\right);\left(-1;-63\right)\right\}\)

c: =>xy=45

mà x<y<0

nên \(\left(x,y\right)\in\left\{\left(-45;-1\right);\left(-15;-3\right);\left(-9;-5\right)\right\}\)

hình như thứ tự có hơi..... Mình khó hiểu để ???

biết bài nào thì giúp mình bài đó nha, 0 phải làm hết đâu

b: =>(x+1)^2=36

=>x+1=6 hoặc x+1=-6

=>x=5 hoặc x=-5

e: 5x=4y

=>x/4=y/5

mà y-x=7

nên x/4=y/5=(y-x)/(5-4)=7

=>x=28; y=35

d: =>5^2x+7=25

=>2x+7=2

=>2x=-5

=>x=-5/2

g: =>2x^2=50

=>x^2=25

=>x=5 hoặc x=-5

a) Ta có: \(\dfrac{x}{y}=\dfrac{7}{20}\Rightarrow\dfrac{x}{7}=\dfrac{y}{20}\)

\(\dfrac{y}{z}=\dfrac{5}{8}\Rightarrow\dfrac{y}{5}=\dfrac{z}{8}\Rightarrow\dfrac{y}{20}=\dfrac{z}{32}\)

\(\Rightarrow\dfrac{x}{7}=\dfrac{y}{20}=\dfrac{z}{32}\)

\(\Rightarrow\dfrac{2x}{14}=\dfrac{5y}{100}=\dfrac{2z}{64}\)

Áp dụng tc dãy tỉ số bằng nhau:

\(\dfrac{2x}{14}=\dfrac{5y}{100}=\dfrac{2z}{64}=\dfrac{2x+5y-2z}{14+100-64}=2\)

Do \(\left\{{}\begin{matrix}\dfrac{2x}{14}=2\\\dfrac{5y}{100}=2\\\dfrac{2z}{64}=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=14\\y=40\\z=64\end{matrix}\right.\).

b) \(5x=8y=20z\Rightarrow\dfrac{5x}{40}=\dfrac{8y}{40}=\dfrac{20z}{40}\)

\(\Rightarrow\dfrac{x}{8}=\dfrac{y}{5}=\dfrac{z}{2}\)

Áp dụng...

\(\dfrac{x}{8}=\dfrac{y}{5}=\dfrac{z}{2}=\dfrac{x-y-z}{8-5-2}=3\)

....

c) \(\dfrac{6}{11}x=\dfrac{9}{2}y=\dfrac{18}{5}z\Rightarrow\dfrac{x}{\dfrac{11}{6}}=\dfrac{y}{\dfrac{2}{9}}=\dfrac{z}{\dfrac{5}{18}}\)

...

\(\dfrac{9}{5}+\dfrac{5}{7}+\dfrac{7}{5}+\dfrac{3}{7}=\left(\dfrac{9}{5}+\dfrac{7}{5}\right)+\left(\dfrac{5}{7}+\dfrac{3}{7}\right)=\dfrac{16}{5}+\dfrac{8}{7}=\dfrac{112}{35}+\dfrac{40}{35}=\dfrac{152}{35}\)

\(\dfrac{1}{2}\times\dfrac{45}{33}\times\dfrac{1}{9}\times\dfrac{11}{6}=\dfrac{1}{2}\times\dfrac{15}{11}\times\dfrac{1}{9}\times\dfrac{11}{6}=\left(\dfrac{1}{2}\times\dfrac{1}{9}\right)\times\left(\dfrac{15}{11}\times\dfrac{11}{6}\right)=\dfrac{1}{18}\times\dfrac{15}{6}=\dfrac{5}{36}\)

a) \(\dfrac{-5}{6}.\dfrac{120}{25}< x< \dfrac{-7}{15}.\dfrac{9}{14}\)

\(\Rightarrow-4< x< \dfrac{-3}{10}\)

\(\Rightarrow\dfrac{-40}{10}< x< \dfrac{-3}{10}\)

\(\Rightarrow x\in\left\{\dfrac{-39}{10};\dfrac{-38}{10};\dfrac{-37}{10};...;\dfrac{-5}{10};\dfrac{-4}{10}\right\}\)

b) \(\left(\dfrac{-5}{3}\right)^2< x< \dfrac{-24}{35}.\dfrac{-5}{6}\)

\(\Rightarrow\dfrac{25}{9}< x< \dfrac{4}{7}\)

\(\Rightarrow\dfrac{175}{63}< x< \dfrac{36}{63}\)

\(\Rightarrow x=\varnothing\)

c) \(\dfrac{1}{18}< \dfrac{x}{12}< \dfrac{y}{9}< \dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{2}{36}< \dfrac{3x}{36}< \dfrac{4y}{36}< \dfrac{9}{36}\)

\(\Rightarrow x\in\left\{1;2\right\}\)

+) Với \(x=1\)

\(\Rightarrow y\in\left\{1;2\right\}\)

+) Với \(x=2\)

\(\Rightarrow y=2\)

Vậy \(x=1\) thì \(y\in\left\{1;2\right\}\); \(x=2\) thì \(y=8\).

`a)1/7xx2/7+1/7xx5/7+6/7`

`=1/7xx(2/7+5/7)+6/7`

`=1/7xx1+6/7`

`=1/7+6/7=1`

`b)6/11xx4/9+6/11xx7/9-6/11xx2/9`

`=6/11xx(4/9+7/9-2/9)`

`=6/11xx9/9`

`=6/11`

Sorry nãy ghi thiếu.

`c)4/25xx5/8xx25/4xx24`

`=(4xx5xx25xx24)/(25xx8xx4)`

`=(4xx5xx24)/(4xx8)`

`=(5xx24)/8`

`=5xx3=15`

a, \(\dfrac{1}{7}.\dfrac{2}{7}+\dfrac{1}{7}.\dfrac{5}{7}+\dfrac{6}{7}\)

\(=\dfrac{1}{7}.\left(\dfrac{2}{7}+\dfrac{5}{7}\right)+\dfrac{6}{7}\)
\(=\dfrac{1}{7}.1+\dfrac{6}{7}\)

\(=\dfrac{1}{7}+\dfrac{6}{7}=1\)

b, \(\dfrac{6}{11}.\dfrac{4}{9}+\dfrac{6}{11}.\dfrac{7}{9}-\dfrac{6}{11}.\dfrac{2}{9}\)

\(=\dfrac{6}{11}.\left(\dfrac{4}{9}+\dfrac{7}{9}-\dfrac{2}{9}\right)\)

\(=\dfrac{6}{11}.1=\dfrac{6}{11}\)

c, \(\dfrac{4}{25}.\dfrac{5}{8}.\dfrac{25}{4}.24\)

\(=\left(\dfrac{4}{25}.\dfrac{25}{4}\right).\left(\dfrac{5}{8}.24\right)\)

\(=1.15=15\)

\(405:27\times516=15\times516=7740\)

\(\dfrac{6}{7}+\dfrac{2}{3}\times\dfrac{5}{4}=\dfrac{6}{7}+\dfrac{5}{6}=\dfrac{71}{42}\)

\(\dfrac{9}{5}-\dfrac{6}{7}:\dfrac{11}{7}=\dfrac{9}{5}-\dfrac{6}{11}=\dfrac{69}{55}\)

mt đou bn

7740

71/42

69/55

M~~1/4

M=1/4

a) ĐK xác định : x≠0;y≠0

ta có : \(\left\{{}\begin{matrix}\dfrac{5}{x}+\dfrac{6}{y}=9\\\dfrac{2}{x}-\dfrac{6}{y}=7\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\dfrac{7}{x}=16\\\dfrac{2}{x}-\dfrac{6}{y}=7\end{matrix}\right.< =>\left\{{}\begin{matrix}x=\dfrac{7}{16}\\y=-\dfrac{42}{17}\end{matrix}\right.\)

Vậy S = {(\(\dfrac{7}{16};-\dfrac{42}{17}\))}

b) Đk xác định : x≠0;y≠0

ta có : \(\left\{{}\begin{matrix}\dfrac{5}{x}+\dfrac{1}{y}=14\\\dfrac{8}{x}-\dfrac{1}{y}=-8\end{matrix}\right.< =>\left\{{}\begin{matrix}\dfrac{13}{x}=6\\\dfrac{5}{x}+\dfrac{1}{y}=14\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=\dfrac{13}{6}\\y=\dfrac{13}{152}\end{matrix}\right.\)

Vậy S={(\(\dfrac{13}{6};\dfrac{13}{152}\))}

c) ĐK xác định : x≠0;y≠0

ta có : \(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{7}{y}=21\\-\dfrac{2}{x}-\dfrac{5}{y}=-11\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\dfrac{2}{y}=10\\\dfrac{2}{x}+\dfrac{7}{y}=21\end{matrix}\right.< =>\left\{{}\begin{matrix}y=\dfrac{1}{5}\\x=-\dfrac{1}{7}\end{matrix}\right.\)

Vậy S={(\(-\dfrac{1}{7};\dfrac{1}{5}\))}

d) ĐK xác định : x≠0;y≠0

ta có : \(\left\{{}\begin{matrix}\dfrac{9}{x}+\dfrac{2}{y}=22\\\dfrac{5}{x}-\dfrac{2}{y}=13\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\dfrac{14}{x}=35\\\dfrac{5}{x}-\dfrac{2}{y}=13\end{matrix}\right.< =>\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-4\end{matrix}\right.\)

Vậy S={(0,4;-4)}

e) ĐKXĐ : x≠0;y≠0

ta có : \(\left\{{}\begin{matrix}\dfrac{3}{x}+\dfrac{5}{y}=10\\-\dfrac{3}{x}-\dfrac{7}{y}=8\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}-\dfrac{2}{y}=18\\\dfrac{3}{x}+\dfrac{5}{y}=10\end{matrix}\right.< =>\left\{{}\begin{matrix}y=-\dfrac{1}{9}\\x=\dfrac{3}{55}\end{matrix}\right.\) 'Vậy....

a) _{^{\(\dfrac{x}{y}\times\dfrac{3}{4}=\dfrac{5}{6}+\dfrac{1}{3}\)}}

    _{^{\(\dfrac{x}{y}\times\dfrac{3}{4}=\dfrac{7}{6}\)}}

     _{^{\(\dfrac{x}{y}=\dfrac{7}{6}:\dfrac{3}{4}\)}}

     _{^{\(\dfrac{x}{y}=\dfrac{14}{9}\)}}

b) ^{_{\(\dfrac{7}{9}:\dfrac{x}{y}=\dfrac{10}{7}-\dfrac{13}{14}\)}}

    ^{_{\(\dfrac{7}{9}:\dfrac{x}{y}=\dfrac{1}{2}\)}}

          _{^{\(\dfrac{x}{y}=\dfrac{7}{9}:\dfrac{1}{2}\)}}

            _{^{\(\dfrac{x}{y}=\dfrac{14}{9}\)}}