Đáp án:

$a)A=\dfrac{311}{1197}\\ b)B=\dfrac{1}{8}\\ c)C=\dfrac{99}{50}\\ d)D=\dfrac{44}{13}$

Giải thích các bước giải:

$a)A=\dfrac{7}{35}.\dfrac{10}{9}+\dfrac{7}{19}.\dfrac{9}{35}-\dfrac{2}{35}\\ =\dfrac{7}{35}.\dfrac{10}{9}+\dfrac{7}{35}.\dfrac{9}{19}-\dfrac{2}{35}\\ =\dfrac{7}{35}\left(\dfrac{10}{9}+\dfrac{9}{19}\right)-\dfrac{2}{35}\\ =\dfrac{7}{35}\left(\dfrac{10.19}{9.19}+\dfrac{9.9}{19.9}\right)-\dfrac{2}{35}\\ =\dfrac{7}{35}.\dfrac{10.19+9.9}{19.9}-\dfrac{2}{35}\\ =\dfrac{7}{35}.\dfrac{271}{171}-\dfrac{2}{35}\\ =\dfrac{7.271}{35.171}-\dfrac{2.171}{35.171}\\ =\dfrac{1555}{35.171}\\ =\dfrac{311}{7.171}\\ =\dfrac{311}{1197}\\ b)B=\dfrac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^{10}.6^{19}-7.2^{29}.27^6}\\ =\dfrac{5.2^{2.15}.3^{2.9}-2^2.3^{20}.2^{3.9}}{5.2^{10}.2^{19}.3^{19}-7.2^{29}.3^{3.6}}\\ =\dfrac{5.2^{30}.3^{18}-3^{20}.2^{29}}{5.2^{29}.3^{19}-7.2^{29}.3^{18}}\\ =\dfrac{2^{29}.3^{18}\left(5.2-3^{2}\right)}{2^{27}.3^{18}\left(5.3-7\right)}\\ =\dfrac{5.2-3^{2}}{5.3-7}\\ =\dfrac{10-9}{15-7}\\ =\dfrac{1}{8}\\ c)C=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)\cdots \left(1+\dfrac{1}{98.100}\right)\\ =\dfrac{4}{1.3}.\dfrac{9}{2.4}.\dfrac{16}{3.5}\cdots \dfrac{9801}{98.100}\\ =\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{16}{3.5}\cdots \dfrac{99^2}{98.100}\\ =\dfrac{2.3.4 \cdots 99}{1.2.3.4 \cdots 98}.\dfrac{2.3.4 \cdots 99}{3.4 \cdots 100}\\ =99.\dfrac{2}{100}\\ =\dfrac{99}{50}\\ d)D=\dfrac{155-\dfrac{10}{7}-\dfrac{5}{11}+\dfrac{5}{23}}{403-\dfrac{26}{7}-\dfrac{13}{11}+\dfrac{13}{23}}+\dfrac{\dfrac{3}{5}+\dfrac{3}{13}-0,9}{\dfrac{7}{91}+0,2-\dfrac{3}{10}}\\ =\dfrac{5\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}{13\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}+\dfrac{\dfrac{3}{5}+\dfrac{3}{13}-\dfrac{9}{10}}{\dfrac{1}{13}+\dfrac{1}{5}-\dfrac{3}{10}}\\ =\dfrac{5}{13}+\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{13}-\dfrac{3}{10}\right)}{\dfrac{1}{13}+\dfrac{1}{5}-\dfrac{3}{10}}\\ =\dfrac{5}{13}+3\\ =\dfrac{44}{13}$

Đáp án:

`a,`

`A = 7/35 × 10/9 + 7/19 × 9/35 - 2/35`

`⇔ A = 7/35 × 10/9 + 7/35 × 9/19 - 2/35`

`⇔ A = 7/35 × (10/9 + 9/19) - 2/35`

`⇔ A = 7/35 × 271/171 - 2/35`

`⇔ A = 271/855 - 2/35`

`⇔ A = 311/1197`

Vậy `A = 311/1197`

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$b,$

`B = (5 × 4^{15} × 9^9 - 4 × 3^{20} × 8^9)/(5 × 2^{10} × 6^{19} - 7 × 2^{29} × 27^6)`

`⇔ B = (5 × (2^2)^{15} × (3^2)^9 - 2^2 × 3^{20} × (2^3)^9)/(5 × 2^{10} × 2^{19} × 3^{19} - 7 × 2^{29} × (3^3)^6)`

`⇔ B = (5 × 2^{30} × 3^{18} -  3^{20} × 2^{29})/(5  × 2^{29} × 3^{19} - 7 × 2^{29} × 3^{18})`

`⇔ B = (5 × 2^{29} × 2 × 3^{18} - 3^2 × 3^{18} × 2^{29})/(5 × 2^{29}  × 3^{18}×3 - 7 × 2^{29} × 3^{18})`

`⇔ B = (2^{29} × 3^{18} × (5 × 2 - 3^2) )/(2^{29} × 3^{18} × (5×3-7) )`

`⇔ B = (5 × 2 - 9)/(5 × 3 - 7)`

`⇔ B = (10 - 9)/(15 - 7)`

`⇔ B = 1/8`

Vậy `B = 1/8`

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$c,$

`C = (1 + 1/(1×3) ) (1 + 1/(2×4) ) (1 + 1/(3×5) ) (1 + 1/(4×6) ) ... (1 + 1/(98×100) )`

`⇔ C = ( (1×3)/(1×3) + 1/(1×3) ) ( (2×4)/(2×4) + 1/(2×4) ) ( (3×5)/(3×5) + 1/(3×5) ) ( (4×6)/(4×6) + 1/(4×6) ) ... ( (98×100)/(98×100) + 1/(98×100) )`

`⇔ C = (3/(1×3) + 1/(1×3) ) (8/(2×4) + 1/(2×4) ) (15/(3×5) + 1/(3×5) ) (24/(4×6) + 1/(4×6) ) ... ( 9800/(98×100) + 1/(98 × 100) )`

`⇔ C = 4/(1×3) × 9/(2×4) × 16/(3×5) × 25/(4×6) ... 9801/(98×100)`

`⇔ C = (2×2)/(1×3) × (3×3)/(2×4) × (4×4)/(3×5) × (5×5)/(4×6) ... (99×99)/(98×100)`

`⇔ C = (2×2×3×3×4×4×5×5...99×99)/(1×3×2×4×3×5×4×6...98×100)`

`⇔  C = (2×3×4...98×99)/(1×2×3×4...97×98) × (2×3×4...98×99)/(3×4×5...99×100)`

`⇔ C = 99 × 2/100`

`⇔ C = 99/50`

Vậy `C = 99/50`

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$d,$

`D = (155 - 10/7 - 5/11 + 5/23)/(403 - 26/7 - 13/11 + 13/23) + (3/5 + 3/13 - 0,9)/(7/91 + 0,2 - 3/10)`

`⇔ D = (5 × 31 - 5×2/7 - 5×1/11 + 5 × 1/23)/(13 × 31 - 13 × 2/7 - 13× 1/11 + 13 × 1/23) + (3/5 + 3/13 - 9/10)/(1/13 + 1/5 - 3/10)`

`⇔ D = (5 × (31 - 2/7 - 1/11 + 1/23) )/(13 × (31 - 2/7 - 1/11 + 1/23) ) + (3 × (1/5 + 1/13 - 3/10) )/(1/5 + 1/13 - 3/10)`

`⇔ D = 5/13 + 3`

`⇔ D = 44/13`

Vậy `D = 44/13`