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d=43/37d=43/37 × 23/15−43/3723/15−43/37 × 8/15+31/378/15+31/37
=43/37=43/37 x 23/15+43/3723/15+43/37 x (−8/15)+31/37(−8/15)+31/37
=43/37(23/15−8/15)+31/37=43/37(23/15−8/15)+31/37
=43/37.1+31/37=43/37.1+31/37
=2=2
E=−5/24E=−5/24 × 9/11−5/249/11−5/24 × 2/11+7/122/11+7/12
=−5/24(9/11+2/11)+7/12=−5/24(9/11+2/11)+7/12
=−5/24.1+7/12=−5/24.1+7/12
=3/8=3/8
a)3/7–−11/14−7/3×6/7a)3/7–−11/14−7/3×6/7
=3/7−−11/14−2=3/7−−11/14−2
=3/7+11/14−2=3/7+11/14−2
=−11/14=−11/14
b)6/5:3/2+−2/3×9/14b)6/5:3/2+−2/3×9/14
=4/5−3/7=4/5−3/7
=13/35
d=43/37d=43/37 × 23/15−43/3723/15−43/37 × 8/15+31/378/15+31/37
=43/37=43/37 x 23/15+43/3723/15+43/37 x (−8/15)+31/37(−8/15)+31/37
=43/37(23/15−8/15)+31/37=43/37(23/15−8/15)+31/37
=43/37.1+31/37=43/37.1+31/37
=2=2
E=−5/24E=−5/24 × 9/11−5/249/11−5/24 × 2/11+7/122/11+7/12
=−5/24(9/11+2/11)+7/12=−5/24(9/11+2/11)+7/12
=−5/24.1+7/12=−5/24.1+7/12
=3/8=3/8
a)3/7–−11/14−7/3×6/7a)3/7–−11/14−7/3×6/7
=3/7−−11/14−2=3/7−−11/14−2
=3/7+11/14−2=3/7+11/14−2
=−11/14=−11/14
b)6/5:3/2+−2/3×9/14b)6/5:3/2+−2/3×9/14
=4/5−3/7=4/5−3/7
=13/35

a. A = 2 / 3 . 5 + 2 / 5 . 7 + 2 / 7 . 9 + ... + 2 / 37 . 39
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}=\frac{1}{3}-\frac{1}{39}\)
\(=\frac{13}{39}-\frac{1}{39}=\frac{12}{39}\)
b. B = 4 / 5 . 8 + 4 / 8 . 11 + ... + 4 / 305 . 208
\(=\frac{4}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{305.308}\right)=\frac{4}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{305}-\frac{1}{308}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{5}-\frac{1}{308}\right)=\frac{4}{3}.\frac{303}{1540}=\frac{101}{385}\)
2. a. \(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\)
\(=\frac{1}{3}-\frac{1}{39}=\frac{4}{13}\)
b. \(B=\frac{4}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{305.308}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{305}-\frac{1}{308}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{5}-\frac{1}{308}\right)\)
\(=\frac{4}{3}.\frac{303}{1540}=\frac{101}{385}\)

\(A=\frac{1}{1\cdot2}+\frac{2}{2\cdot4}+\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+...+\frac{10}{46\cdot56}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{56}\)
\(A=1-\frac{1}{56}\)
\(A=\frac{55}{56}\)
\(B=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}\)
\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\)
\(B=\frac{1}{3}-\frac{1}{27}\)
\(B=\frac{8}{27}\)
\(C=\frac{4}{3\cdot6}+\frac{4}{6\cdot9}+\frac{4}{9\cdot12}+...+\frac{4}{99\cdot102}\)
\(C=\frac{4}{3}\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{99\cdot102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\cdot\frac{33}{102}\)
\(C=\frac{22}{51}\)

\(c=2^3.5^3-\left\{7^2.2^3-5^2.\left[4^3::8+11^2:121-2\left(37-5.7\right)\right]\right\}\)
\(c=8.5-\left\{49.8-25.\left[8:8+121:121-2\left(37-5.7\right)\right]\right\}\)
\(c=8.5-\left\{49.8-25.\left[8:8+121:121-2\left(37-35\right)\right]\right\}\)
\(c=8.5-\left\{49.8-25.\left[8:8+121:121+2.2\right]\right\}\)
\(c=8.5-\left\{1+1+2.2\right\}\)
\(c=8.5-\left\{1+1+4\right\}\)
\(c=8.5-6\)
\(c=40-6\)
\(c=34\)
nhầm
\(c=2^3.5^3-\left\{7^2.2^3-5^2.\left[4^3:8+11^2:121-2\left(37-5.7\right)\right]\right\}\)
\(c=8.125-\left\{49.8-25.\left[8:8+121:121-2\left(37-35\right)\right]\right\}\)
\(c=8.125-\left\{49.8-25.\left[8:8+121:121-2.2\right]\right\}\)
\(c=8.125-\left\{49.8-25.\left[1+1-2.2\right]\right\}\)
\(c=8.125-\left\{49.8-25.\left[1+1-4\right]\right\}\)
\(c=8.125-\left\{49.8-25.-2\right\}\)
\(c=8.125-\left\{392+50\right\}\)
\(c=8.125-442\)
\(c=1000-442\)
\(c=558\)

10) 37.(27+25)-27.(37+25)=37.27+37.25-27.37-27.25
=37.25-27.25=25(37-27)=25.10=250
11)3.(-5+3)+7(2+5)=3.(-2)+7.7=-6+49=43
12)-3.(2-7)-6(8-5)=-3.(-5)-6.3=15-18=-3

1)\(1\frac{3}{4}-1\frac{3}{8}:\left(\frac{3}{8}+\frac{9}{20}\right)=\frac{7}{4}-\frac{11}{8}:\frac{33}{40}=\frac{7}{4}-\frac{5}{3}=\frac{1}{12}\)

a; -2\(x\) - 3.(\(x-17\)) = 34 - 2.( - \(x\) + 25)
- 2\(x\) - 3\(x\) + 51 = 34 + 2\(x\) - 50
2\(x\) + 2\(x\) + 3\(x\) = - 34 + 50 + 51
7\(x\) = 67
\(x\) = 67 : 7
\(x\) = \(\dfrac{67}{7}\)
Vậy \(x\) = \(\dfrac{67}{7}\)
b; 17\(x\) + 3.(- 16\(x\) - 37) = 2\(x\) + 43 - 4\(x\)
17\(x\) - 48\(x\) - 111 = 2\(x\) - 4\(x\) + 43
- 31\(x\) - 2\(x\) + 4\(x\) = 111 + 43
- \(x\) x (31 + 2 - 4) = 154
- \(x\) x (33 - 4) = 154
- \(x\) x 29 = 154
- \(x\) = 154 : (-29)
\(x\) = - \(\dfrac{154}{29}\)
Vậy \(x=-\dfrac{154}{29}\)
`2/3+4/37+5/11-2/7`
`=5698/8547+924/8547+3885/8547-2442/8547`
`=8065/8547`
\(\dfrac{2}{3}+\dfrac{4}{37}+\dfrac{5}{11}-\dfrac{2}{7}\)
\(=\dfrac{86}{111}+\dfrac{5}{11}-\dfrac{2}{7}\)
\(=\dfrac{1501}{1221}-\dfrac{2}{7}\)
\(=\dfrac{8065}{8547}\)
Chúc e học tốt nha!☘