\(\frac{43}{20}+\frac{17}{6}\div\left[\frac{5}{8}+\frac{7}{30}\right]\)
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\(=\frac{43}{20}+\frac{17}{65}:\frac{103}{120}\)
\(=\frac{43}{20}+\frac{408}{1339}\)
\(=2,454705004\)
tíc mình nha
\(\frac{43}{20}+\frac{17}{65}:\left(\frac{5}{8}+\frac{7}{30}\right)\)
\(=\frac{43}{20}+\frac{17}{65}:\frac{103}{120}\)
\(=\frac{43}{20}+\frac{408}{1339}\)
\(=2,4547........\)

Bài 1:
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + (\(\frac45\) - \(\frac{3}{17}\) + \(\frac13\)) - \(\frac17\) + (- \(\frac{14}{30}\))
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + \(\frac45\) - \(\frac{3}{17}\) + \(\frac13\) - \(\frac17\) - \(\frac{14}{30}\)
A = (\(\frac15\) + \(\frac45\)) + (\(\frac{3}{17}\) - \(\frac{3}{17}\)) - (\(\frac43-\frac13\)) - \(\frac{30}{210}\) - \(\frac{98}{210}\)
A = 1 + 0 - 1 - (\(\frac{30}{210}+\frac{98}{210}\))
A = 1 - 1 - \(\frac{228}{210}\)
A = 0 - \(\frac{128}{210}\)
A = - \(\frac{64}{105}\)
Bài 2:
B= (\(\frac58\) - \(\frac{4}{12}\) + \(\frac32\)) - (\(\frac58\) + \(\frac{9}{13}\)) - (\(\frac{-3}{2}\)) + \(\frac{7}{-15}\)
B = \(\frac58\) - \(\frac{4}{12}\) + \(\frac32\) - \(\frac58\) - \(\frac{9}{13}\) + \(\frac32\) - \(\frac{7}{15}\)
B = (\(\frac58\) - \(\frac58\)) + (\(\frac32\) + \(\frac32\)) - (\(\frac13\) + \(\frac{9}{13}\) + \(\frac{7}{15}\))
B = 0 + 3 - (\(\frac{65}{195}\) + \(\frac{135}{195}\) + \(\frac{91}{195}\))
B = 3 - (\(\frac{200}{195}\) + \(\frac{91}{195}\))
B = 3 - \(\frac{97}{65}\)
B = \(\frac{195}{65}\) - \(\frac{97}{65}\)
B = \(\frac{98}{65}\)

Tìm x, biết:
3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x∉−2;−5;−10;−17)
2(x−1)(x−3) +5(x−3)(x−8) +12(x−8)(x−20) −1x−20 =−34 (x∉1;3;8;20)
x+110 +2+111 x+112 =x+113 +x+114
x−1030 +x−1443 +x−595 +x−1488 =0

Bấm máy tính:
E = \(\frac{4}{3}+\frac{1}{4}+\frac{3}{5}:\frac{4}{5}\)
E = \(\frac{4}{3}+\frac{1}{4}+\frac{3}{4}\)
E = \(\frac{7}{3}\)
Vậy E = \(\frac{7}{3}\)
\(\frac{43}{20}+\frac{17}{6}:\left[\frac{5}{8}+\frac{7}{30}\right]\)
= \(\frac{43}{20}+\frac{17}{6}:\frac{103}{120}\)
= \(\frac{43}{20}+\frac{340}{103}\)
=\(\frac{11229}{2060}\).
\(\frac{43}{20}+\frac{17}{6}:\left[\frac{5}{8}+\frac{7}{30}\right]\)
\(=\frac{43}{20}+\frac{17}{6}:\left[\frac{103}{120}\right]\)
\(=\frac{43}{20}+\frac{340}{103}=\frac{11229}{2060}\)