giúp với
Tìm X biết
\(\frac{X}{5}\)=\(\frac{-14}{35}\)
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a)\(\frac{-5}{6}\).\(\frac{120}{25}\)<x<\(\frac{-7}{15}\).\(\frac{9}{14}\)
-4 <x<\(\frac{-3}{10}\)
\(\frac{-40}{10}\)< x <\(\frac{-3}{10}\)=>x E {-39:-38:-37:.....:-4}
b)\(\left(\frac{-5}{3}\right)^3\)<x<\(\frac{-24}{35}.\frac{-5}{6}\)
\(\frac{-875}{189}< x< \frac{108}{189}\)
=> x E {\(\frac{-874}{189},\frac{-873}{189},......,\frac{107}{189}\)}
\(\frac{x+32}{11}+\frac{x+33}{12}=\frac{x+34}{13}+\frac{x+35}{14}\)
\(\Leftrightarrow\left(\frac{x+32}{11}-1\right)+\left(\frac{x+33}{12}-1\right)=\left(\frac{x+34}{13}-1\right)+\left(\frac{x+35}{14}-1\right)\)
\(\Leftrightarrow\frac{x-21}{11}+\frac{x-21}{12}=\frac{x-21}{13}+\frac{x-21}{14}\)
\(\Leftrightarrow\left(x-21\right)\left(\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Vì \(\left(\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\ne0\)
\(\Rightarrow x-21=0\Rightarrow x=21\)
\(\frac{x+32}{11}+\frac{x+33}{12}=\frac{x+34}{13}+\frac{x+35}{14}\)
\(\Leftrightarrow\left(\frac{x+32}{11}-1\right)+\left(\frac{x+33}{12}-1\right)=\left(\frac{x+34}{13}-1\right)+\left(\frac{x+35}{14}-1\right)\)( trừ cả hai vế cho 2 )
\(\Leftrightarrow\frac{x-21}{11}+\frac{x-21}{12}=\frac{x-21}{13}+\frac{x-21}{14}\)
\(\Leftrightarrow\frac{x-21}{11}+\frac{x-21}{12}-\frac{x-21}{13}-\frac{x-21}{14}=0\)
\(\Leftrightarrow\left(x-21\right)\left(\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Mà \(\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\)
\(\Rightarrow x-21=0\)
\(\Leftrightarrow x=21\)
Vậy \(x=21\)
1)
a)
\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)
\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)
\(\frac{-20}{5}< x< \frac{-3}{10}\)
\(\frac{-40}{10}< x< \frac{-3}{10}\)
\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)
\(\frac{x-3}{53}+\frac{x+14}{36}=\frac{x+15}{35}-\frac{x}{50}\)
\(\left(1+\frac{x-3}{53}\right)+\left(1+\frac{x+14}{36}\right)=\left(\frac{x+15}{35}+1\right)-\left(1+\frac{x}{50}\right)\)
\(\frac{x+50}{53}+\frac{x+50}{36}-\frac{x+15}{35}+\frac{x+50}{50}=0\)
\(\left(x+50\right).\left(\frac{1}{53}+\frac{1}{36}-\frac{1}{35}+\frac{1}{50}\right)=0\)
\(\left(\frac{1}{53}+\frac{1}{36}-\frac{1}{35}+\frac{1}{50}\right)\ne0\)
\(\Rightarrow\left(x+50\right)=0\Rightarrow x=-50\)
Vậy x=-50
p/s: bn viết sai đề nên mk sửa lại
\(\frac{3}{2}+\frac{3}{14}+\frac{3}{15}+...+\frac{6}{\left(x-3\right).x}=\frac{96}{49}\)
\(\frac{6}{\left(1.2\right).2}+\frac{6}{\left(2.7\right).2}+...+\frac{6}{\left(x-3\right).x}=\frac{96}{49}\)
\(\frac{6}{1.4}+\frac{6}{4.7}+...+\frac{6}{\left(x-3\right).x}=\frac{96}{49}\)
\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{\left(x-3\right).x}=\frac{96}{49.2}\)
\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{\left(x-3\right)}-\frac{1}{x}=\frac{96}{98}\)
=> \(1-\frac{1}{x}=\frac{48}{49}\)
=> \(\frac{1}{x}=\frac{1}{49}\)
=> \(x=49\)
a)\(\frac{5}{21}\)+\(\frac{-3}{7}\)<\(\frac{x}{21}\)<\(\frac{-2}{7}\)+\(\frac{8}{21}\)
\(\Rightarrow\)\(\frac{-4}{21}\)<\(\frac{x}{21}\)<\(\frac{2}{21}\)
\(\Rightarrow\)\(\frac{x}{21}\)\(\in\)\(\left\{\frac{-3}{21};\frac{-2}{21};\frac{-1}{21};\frac{0}{21};\frac{1}{21}\right\}\)
vậy x\(\in\)\(\left\{-3;-2;-1;0;1\right\}\)
\(\frac{x}{15}=\frac{3}{5}+\frac{-2}{3}\)
\(\frac{x}{15}=\frac{-1}{15}\)
=> \(x=-1\)
\(\frac{x}{182}=\frac{-6}{14}\cdot\frac{35}{91}\)
\(\frac{x}{182}=\frac{-15}{91}\)
=> \(91x=182\cdot\left(-15\right)\)
=> \(91x=-2730\)
=> \(x=-30\)
g, \(\frac{x}{15}=\frac{3}{5}+\frac{-2}{3}\Leftrightarrow\frac{x}{15}=\frac{3}{5}-\frac{2}{3}\Leftrightarrow\frac{x}{15}=-\frac{1}{15}\)
\(\Leftrightarrow x=-1\)
h, \(\frac{x}{182}=\frac{-6}{14}.\frac{35}{91}\Leftrightarrow\frac{x}{182}=-\frac{15}{91}\Leftrightarrow\frac{x}{182}=\frac{-30}{182}\)
\(\Leftrightarrow x=-30\)