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a. Vì 252/x=84/97 nên => 252.97=84.x hay 84.x =24444 => x= 24444: 84= 291
b 7/5 :x = 4/5-1/6
7/5 :x =19/30
x=7/5:19/30
x=42/19
c Tự làm đi dễ mà
tính ra rồi tìm
bài 1 a)
\(\frac{15}{7}\times\frac43+\frac63\)
=\(\frac{60}{21}+\frac63\)
=\(\frac{60}{21}+\frac{42}{21}\)
=\(\frac{102}{21}\)
=\(\frac{34}{7}\)
b)\(\frac{-13}{41}\times\frac{17}{11}\times\frac{11}{17}\)
=\(\frac{-13}{41}\times\left(\frac{17}{11}\times\frac{11}{17}\right)\)
=\(\frac{-13}{41}\times1\)
=\(\frac{-13}{41}\)
a) (\(6\frac{2}{7}.x+\frac{3}{7}\))=-1.\(\frac{11}{5}+\frac{3}{7}\)
(\(6\frac{2}{7}.x+\frac{3}{7}\))=\(\frac{-62}{35}\)
\(\frac{44}{7}.x\)=\(\frac{-62}{35}-\frac{3}{7}\)
\(\frac{44}{7}.x=\frac{-77}{35}\)
x=\(\frac{-77}{35}:\frac{44}{7}\)=\(\frac{539}{1540}\)
(6\(\frac27x\) + \(\frac37\)) : 2\(\frac15\) - \(\frac37\) = - 1
(\(\frac{44}{7}x\) + \(\frac37\)) : \(\frac{11}{5}\) = - 1 + \(\frac37\)
(\(\frac{44}{7}x+\) \(\frac37\)) : \(\frac{11}{5}\) = - \(\frac77+\frac37\)
\(\frac{44x+3}{7}\) : \(\frac{11}{5}\) = - \(\frac47\)
(44\(x\) + 3) : \(\frac{11}{5}\) = - 4
44\(x\) + 3 = - 4 x \(\frac{11}{5}\)
44\(x\) + 3 = - \(\frac{44}{5}\)
44\(x\) = - \(\frac{44}{5}\) - 3
44\(x\) = - \(\frac{59}{5}\)
\(x=-\frac{59}{5}:44\)
\(x\) = - \(\frac{59}{250}\)
Vậy \(x\) = - \(\frac{59}{250}\)
a: \(\Leftrightarrow-\dfrac{9}{46}+\dfrac{108}{46}-\dfrac{93}{23}:\left(\dfrac{13}{4}-\dfrac{5}{3}x\right)=1\)
\(\Leftrightarrow\dfrac{93}{23}:\left(\dfrac{13}{4}-\dfrac{5}{3}x\right)=\dfrac{53}{46}\)
\(\Leftrightarrow-\dfrac{5}{3}x+\dfrac{13}{4}=\dfrac{186}{53}\)
=>-5/3x=55/212
hay x=-33/212
c: \(\Leftrightarrow\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{18}{19}\)
\(\Leftrightarrow1-\dfrac{1}{x+3}=\dfrac{18}{19}\)
=>x+3=19
hay x=16
a) \(\frac{53}{101}\cdot-\frac{13}{97}+\frac{53}{101}\cdot-\frac{84}{97}\)
\(=\frac{53}{101}\cdot\left(-\frac{13}{97}-\frac{84}{97}\right)\)
\(=\frac{53}{101}\cdot\left(-1\right)\)
\(=-\frac{53}{101}\)
5\(\frac{8}{17}\) : x - (\(\frac{8}{17}\)) : x + 3\(\frac{1}{17}\): 17\(\frac13\) = \(\frac{4}{17}\)
(5\(\frac{8}{17}\) - \(\frac{8}{17}\)) : x + \(\frac{52}{17}\) : \(\frac{52}{3}\) = \(\frac{4}{17}\)
5:x + \(\frac{52}{17}\times\) \(\frac{3}{52}\) = \(\frac{4}{17}\)
5 : x + \(\frac{3}{17}\) = \(\frac{4}{17}\)
5 : x = \(\frac{4}{17}\) - \(\frac{3}{17}\)
5 : x = \(\frac{1}{17}\)
x = 5 : \(\frac{1}{17}\)
x = 5 x 17
x = 85
Vậy x = 85
1/1.4 + 1/4.7 + ...+1/x(x+3) = 6/19
3/1.4 + 3/4.7 +..+3/x(x+3) = 18/19
1/1 - 1/4+ 1/4 - 1/7 +...+1/x-1/(x+3) = 1 - 1/19
1- 1/(x+3) = 1 - 1/19
1/(x+3) = 1/19
x + 3 = 19
x = 19 - 3
x = 16
Vậy x = 16
5\(\frac{8}{17}\) : x - (\(\frac{8}{17}\)) : x + 3\(\frac{1}{17}\): 17\(\frac13\) = \(\frac{4}{17}\)
(5\(\frac{8}{17}\) - \(\frac{8}{17}\)) : x + \(\frac{52}{17}\) : \(\frac{52}{3}\) = \(\frac{4}{17}\)
5:x + \(\frac{52}{17}\times\) \(\frac{3}{52}\) = \(\frac{4}{17}\)
5 : x + \(\frac{3}{17}\) = \(\frac{4}{17}\)
5 : x = \(\frac{4}{17}\) - \(\frac{3}{17}\)
5 : x = \(\frac{1}{17}\)
x = 5 : \(\frac{1}{17}\)
x = 5 x 17
x = 85
Vậy x = 85
1/1.4 + 1/4.7 + ...+1/x(x+3) = 6/19
3/1.4 + 3/4.7 +..+3/x(x+3) = 18/19
1/1 - 1/4+ 1/4 - 1/7 +...+1/x-1/(x+3) = 1 - 1/19
1- 1/(x+3) = 1 - 1/19
1/(x+3) = 1/19
x + 3 = 19
x = 19 - 3
x = 16
Vậy x = 16
a) Ta có:
252/x = 84/97
==> x.84=252.97
==> x.84= 24444
==> x=24444:84
==> x=291
b) x/17 + 19/34=1/2
X/17=1/2–19/34
X/17=17/34–19/34
X/17=—2/34
==> x.34=17.(—2)
==> x.34=-34
==> x=(—34):34
==> x=—1
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