tính nhanh :
A = 3333/4545 + 36 / 45 + 666666 / 454545
giúp mình nha
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Câu 3 : Tính
a) 74,15 + 4,05 : 2,5
= 74,15 + 1,62
= 75,77
b) 100 - 2,6 x 3,9 + 20,97
= 100 - 10,14 + 20,97 = 110,83
Câu 4: Tính nhanh
a) \(\dfrac{1}{2}\) + \(\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)
= 2 x (1 - 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + 1/16 - 1/32 + 1/32 - 1/64)
= 2 x (1 - 1/64)
= 2 x 63/64
= 63/32
- Phần còn lại mình sẽ làm sau nhé bạn!
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#Toán-5
Câu 3:
a, 74,15 + 4,05 : 2,5
= 74,15 + 1,62
= 75,77
b, 100 - 2,6 x 3,9 + 20,97
= 100 - 10,14 + 20,97
= 89,86 + 20,97
= 110,83
\(A=\frac{3333}{4545}+\frac{1}{15}+\frac{1111}{5555}\)
\(=\frac{11}{15}+\frac{1}{15}+\frac{1}{5}\)
\(=\frac{12}{15}+\frac{3}{15}\)
\(=1\)
\(A=\frac{3333}{4545}+\frac{1}{15}+\frac{1111}{5555}\)
\(=\frac{11}{15}+\frac{1}{15}+\frac{1}{5}\)
\(=\frac{12}{15}+\frac{3}{15}\)
\(=1\)
\(A=\frac{7}{4}.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(A=\frac{231}{4}.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(A=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(A=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(A=\frac{231}{4}.\frac{4}{21}=\frac{231}{21}=11\)
k nha
\(A=\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(A=\frac{7}{4}\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}\right)\)
\(A=\frac{7}{4}\left[33\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\right]\)
\(A=\frac{7}{4}\left[33\left(\frac{1}{3}-\frac{1}{7}\right)\right]\)
\(A=\frac{7}{4}\left[33\times\frac{4}{21}\right]\)
\(A=\frac{7}{4}\times\frac{44}{7}\)
\(A=11\)
Tính bằng cách hợp lí
H=7/4x(3333/1212+3333/2020+3333/3030+3333/4242)
mình đang cần gấp, giải nhanh hộ
\(H=\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(H=\frac{7}{4}.3333.\left(\frac{1}{1212}+\frac{1}{2020}+\frac{1}{3030}+\frac{1}{4242}\right)\)
\(H=\frac{7}{4}.3333.\left(\frac{1}{12.101}+\frac{1}{20.101}+\frac{1}{30.101}+\frac{1}{42.101}\right)\)
\(H=\frac{7}{4}.3333.\frac{1}{101}.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(H=\frac{7}{4}.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(H=\frac{7}{4}.33.\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(H=\frac{7}{4}.33.\left(\frac{7}{21}-\frac{3}{21}\right)\)
\(H=33.\frac{7}{4}.\frac{4}{21}\)
\(H=11.3.\frac{1}{3}\)
\(H=11\)
Tham khảo nhé~
\(H=\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\))
\(H=\frac{7}{4}.\left(\frac{33.101}{12.101}+\frac{33.101}{20.101}+\frac{33.101}{30.101}+\frac{33.101}{42.101}\right)\)
\(H=\frac{7}{4}.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(H=\frac{7}{4}.33.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
\(H=\frac{231}{4}.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(H=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(H=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(H=\frac{231}{4}.\frac{4}{21}\)
\(H=11\)
\(A=\frac{3333}{4545}+\frac{36}{45}+\frac{666666}{454545}\)
\(=\frac{33}{45}+\frac{36}{45}+\frac{66}{45}\)
\(=\frac{69}{45}+\frac{66}{45}\)
\(=3\)
Vậy A = 3
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