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\(\frac{7}{3}.9-\frac{9}{4}.5+\frac{11}{5}.6-\frac{13}{6}.7+\frac{15}{7}.8-\frac{17}{8}.9+\frac{19}{9}.10\)

\(=21-\frac{45}{4}+\frac{66}{5}-\frac{91}{6}+\frac{120}{7}-\frac{153}{8}+\frac{190}{9}\)

\(=26,91230159\)

39/-90 - 7/2

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1) \(\left(+15\right)+\left(+17\right)=15+17=32\)

2) \(\left(-3\right)+\left(-7\right)=-3-7=-\left(3+7\right)=-10\)

3) \(\left(-25\right)+\left(+4\right)=-25+4=-\left(25-4\right)=-21\)

4) \(\left(-6\right)+\left(-54\right)=-6-54=-\left(6+54\right)=-60\)

5) \(\left(-15\right)+20=20-15=5\)

6) \(\left(-5\right)+8+7+5\)

\(=\left(-5+5\right)+\left(8+7\right)\)

\(=15\)

7) \(\left(-8\right)+\left(-11\right)+\left(-2\right)\)

\(=\left[\left(-8\right)+\left(-2\right)\right]+\left(-11\right)\)

\(=\left(-10\right)+\left(-11\right)\)

\(=-21\)

8) \(15+\left(-5\right)+\left(-14\right)+\left(-16\right)\)

\(=\left[15+\left(-5\right)\right]+\left[\left(-14\right)+\left(-16\right)\right]\)

\(=10+\left(-30\right)\)

\(=-20\)

9) \(\left(-20\right)+\left(-14\right)+3+\left(-86\right)\)

\(=\left[\left(-20\right)+3\right]+\left[\left(-14\right)+\left(-86\right)\right]\)

\(=\left(-17\right)+\left(-100\right)\)

\(=-117\)

10) \(\left(-136\right)+123+\left(-264\right)+\left(-83\right)+240\)

\(=\left[\left(-136\right)+\left(-264\right)\right]+\left[123+\left(-83\right)\right]+240\)

\(=\left(-400\right)+40+240\)

\(=\left(-360\right)+240\)

\(=-120\)

11) \(\left(-596\right)+2001+1999+\left(-404+189\right)\)

\(=\left(-596\right)+2001+1999-404+189\)

\(=\left[\left(-596\right)-404\right]+\left(2001+189\right)+1999\)

\(=\left(-1000\right)+2190+1999\)

\(=1190+1999\)

\(=3189\)

12) \(314+\left(-153\right)+64+121+\left(-247\right)+218\)

\(=\left(314+64+121\right)+\left[\left(-153\right)+\left(-247\right)\right]+218\)

\(=\left(378+121\right)+\left(-400\right)+218\)

\(=499-400+218\)

\(=99+218\)

\(=317\)

\(\text{#}Toru\)

                                                            Bài giải

\(\frac{2}{3}+\frac{3}{4}+\frac{4}{5}=\frac{40}{60}+\frac{45}{60}+\frac{48}{60}=\frac{133}{60}\)

\(\frac{8}{5}+\frac{7}{6}+\frac{10}{9}+\frac{1}{2}=\frac{144}{90}+\frac{105}{90}+\frac{100}{90}+\frac{45}{90}=\frac{394}{90}\)

\(\frac{15}{17}-\frac{11}{13}+\frac{3}{26}=\frac{390}{442}+\frac{374}{442}+\frac{51}{442}=\frac{815}{442}\)

\(\frac{9}{12}\text{ x }\frac{4}{3}\text{ : }\frac{8}{5}=\frac{9}{12}\text{ x }\frac{4}{3}\text{ x }\frac{5}{8}=\frac{9\text{ x }4\text{ x }5}{12\text{ x }3\text{ x }8}=\frac{5}{8}\)

\(\frac{4}{5}\text{ x }\frac{15}{8}\text{ : }\frac{5}{7}=\frac{4}{5}\text{ x }\frac{15}{8}\text{ x }\frac{7}{5}=\frac{4\text{ x }15\text{ x }7}{5\text{ x }8\text{ x }5}=\frac{21}{10}\)

\(\frac{2}{3}+\frac{3}{4}+\frac{4}{5}=\frac{40}{60}+\frac{45}{60}+\frac{48}{60}=\frac{133}{60}\) 

\(\frac{8}{5}+\frac{7}{6}+\frac{10}{9}+\frac{1}{2}=\frac{144}{90}+\frac{105}{90}+\frac{100}{90}+\frac{45}{90}=\frac{197}{45}\)

\(\frac{15}{17}-\frac{11}{13}+\frac{1}{26}=\frac{390}{442}+\frac{374}{442}+\frac{51}{442}=\frac{815}{442}\)

\(\frac{9}{12}\times\frac{4}{3}:\frac{8}{5}=1:\frac{8}{5}=\frac{5}{8}\)

\(\frac{4}{5}\times\frac{15}{8}:\frac{5}{7}=\frac{3}{2}:\frac{5}{7}=\frac{21}{10}\)

Ta có: \(\dfrac{5}{9}+\dfrac{13}{7}+\dfrac{15}{3}+\dfrac{8}{7}+\dfrac{4}{9}+\dfrac{11}{13}\)

\(=\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\left(\dfrac{13}{7}+\dfrac{8}{7}\right)+5\dfrac{11}{13}\)

\(=9+\dfrac{11}{13}\)

\(=\dfrac{128}{13}\)

what

Bài 1:

a: \(\frac{5\times7}{15}=\frac{35}{15}=\frac73\)

b: \(8:\frac{16}{23}=8\times\frac{23}{16}=\frac{8}{16}\times23=23\times\frac12=\frac{23}{2}\)

c: \(\frac{8}{15}:5=\frac{8}{15\times5}=\frac{8}{75}\)

d: \(\frac{15}{19}\times3=\frac{15\times3}{19}=\frac{45}{19}\)

Bài 2:

a: \(3\times\frac49+\frac23-\frac{7}{18}\)

\(=\frac43+\frac23-\frac{7}{18}\)

\(=\frac63-\frac{7}{18}=2-\frac{7}{18}=\frac{36}{18}-\frac{7}{18}=\frac{29}{18}\)

b: \(\frac{7}{10}\times\frac23-\frac14=\frac{14}{30}-\frac14=\frac{28}{60}-\frac{15}{60}=\frac{28-15}{60}=\frac{13}{60}\)

c: \(\frac98-\frac78:\frac56=\frac98-\frac78\times\frac65\)

\(=\frac98-\frac{42}{40}=\frac{45}{40}-\frac{42}{40}=\frac{3}{40}\)

d: \(\frac23+\frac12:\frac32-\frac{11}{24}\)

\(=\frac23+\frac12\times\frac23-\frac{11}{24}\)

\(=\frac23+\frac13-\frac{11}{24}=1-\frac{11}{24}=\frac{13}{24}\)

Bài 3:

Độ dài đáy tương ứng với chiều cao là: \(\frac35\times2=\frac65\left(m\right)\)

Chu vi hình bình hành là:

\(\left(\frac65+\frac45\right)\times2=\frac{10}{5}\times2=2\times2=4\left(m\right)\)

Diện tích bình hành là;

\(\frac65\times\frac35=\frac{18}{25}\left(m^2\right)\)