\(\frac{37-x}{3}=\frac{x-13}{7}\)
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Câu a:
(4\(\frac{5}{37}\) - 3\(\frac45\) + 8\(\frac{15}{29}\)) - (3\(\frac{5}{37}\) - 6\(\frac{14}{29}\))
= 4\(\frac{5}{37}\) - 3\(\frac45\) + 8\(\frac{15}{29}\) - 3\(\frac{15}{37}\) + 6\(\frac{14}{29}\)
= (4 - 3) + (\(\frac{5}{37}-\frac{5}{37}\)) + (8 + 6) + (\(\frac{15}{29}\) + \(\frac{14}{29}\)) - 3 - \(\frac45\)
= 1 + 0 + 14 + 1 - 3 - \(\frac45\)
= 1 + 14 + 1 - 3 - \(\frac45\)
= 15 + 1 - 3 - \(\frac45\)
= 16 - 3 - \(\frac45\)
= 13 - \(\frac45\)
= \(\frac{61}{5}\)
Bài 2:
- \(\frac75\) + \(x\) = - \(\frac34\)
\(x\) = - \(\frac34+\frac75\)
\(x=-\frac{15}{20}\) + \(\frac{28}{20}\)
\(x\) = \(\frac{13}{20}\)
Vậy \(x=\frac{13}{20}\)
Bài 3:
- \(\frac74\) + \(\frac14:x\) = \(\frac32\)
\(\frac14:x\) = \(\frac32+\frac74\)
\(\frac14:x\) = \(\frac64+\frac74\)
\(\frac14:x=\frac{13}{4}\)
\(x=\frac14:\frac{13}{4}\)
\(x=\frac14\times\frac{4}{13}\)
\(x=\frac{1}{13}\)
Vậy \(x=\frac{1}{13}\)
\(\frac{37-x}{x+13}=\frac{3}{7}\)
\(\Leftrightarrow3.\left(x+13\right)=7.\left(37-x\right)\)
\(\Leftrightarrow3.x+13.3=7.37-7.x\)
\(\Leftrightarrow3x+39=-7x+259\)
\(\Leftrightarrow3x+7x=259-39\)
\(\Leftrightarrow10x=220\)
\(\Leftrightarrow x=22\)
37 −xx+13=37
7(37-x)=3(x+13)
259-7x=3x+39
259-39=3x+7x
220=10x
x=22
\(\frac{4}{7}=\frac{7}{x^2}\)
\(\Leftrightarrow4x^2=7.7\)
\(\Leftrightarrow\left(2x\right)^2=49\)
\(\Leftrightarrow\orbr{\begin{cases}2x=7\\2x=-7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{7}{2}\end{cases}}\)
a, <=> 7(37-x)=3(x+13)
<=> x =22
b, <=> (x+1)(x-1)=15
<=> x^2-1=15 <=> x^2=16 <=> x= +_4
\(\Leftrightarrow7\left(37-x\right)=3\left(x+13\right)\Leftrightarrow259-7x=3x+39\Leftrightarrow259-39=3x+7x\)
\(10x=220\Leftrightarrow x=22\)
(37-x) . 7 =3. (x+13)
=>259 -7x=3x+39
=>.259-39=3x+7x
=>220=10x
=>x=22
\(\frac{37-x}{x+13}=\frac{3}{7}\)
=> 3.(x+13)=7.(37-x)
=> 3x+39= 259-7x
=> 3x+7x=259-39
=> 10x= 220
=> x=22
\(\frac{37-x}{3}=\frac{x-13}{7}\)
\(\Rightarrow7\left(37-x\right)=3\left(x-13\right)\)
\(\Leftrightarrow259-7x=3x-39\)
\(\Leftrightarrow259+39=3x+7x\)
\(\Leftrightarrow298=10x\)
\(\Leftrightarrow x=29.8\)
\(\frac{37-x}{3}=\frac{x-13}{7}\)
\(\Rightarrow7\left(37-x\right)=3\left(x-13\right)\)
\(\Rightarrow259-7x=3x-39\)
\(\Rightarrow259+39=3x+7x\)
\(\Rightarrow298=10x\)
\(\Rightarrow x=\frac{298}{10}=\frac{149}{5}\)