Tính nhanh
a, 16 × 95 + 1,6 × 4 + 16 = ?
b, 20 × 1 + 20 × 90 + 20 × 9 = ?
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`@` `\text {Ans}`
`\downarrow`
`D={3; 4; 5; 6; 7}`
T/C đặc trưng:
`D = {x \in \text {N}` `|` `3 \le x \le 7}`
`E={0; 5; 10;...; 95}`
T/C đặc trưng:
`E = { x \in {N}` `|` `x \vdots 5, x \le` `95}`
`F = {4; 8; 12; 16; 20; 24; 28}`
T/C đặc trưng:
`F = {x \in` `\text {N*}` `|` `x \vdots 4, x \le` `28}.`
Trong tập hợp D ta thấy đây là các số tự nhiên liên tiếp lớn hơn hoặc bằng 3 và nhỏ hơn hoặc bằng 7:
\(D=\left\{x\in N|3\le x\le7\right\}\)
Trong tập hợp E ta thấy đây là tập hợp các số tự nhiên chia hết cho 5 nhưng nhỏ hơn 100
\(E=\left\{x\in N|x=5k,x< 100,k\in N\right\}\)
Trong tập hợp F ta thấy đây là tập hợp các số tự nhiên chia hết cho 4 nhưng nhỏ hơn hoặc bằng 28:
\(F=\left\{x\in N|x=4k,x\le28,k\in N\right\}\)
a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8
= 1 + 1 + 8
= 2 + 8
= 10
b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{10}{20}\)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (\(\dfrac{2}{2}\) + \(\dfrac{3}{3}\) + \(\dfrac{4}{4}\) + \(\dfrac{5}{5}\)+ \(\dfrac{6}{6}+\dfrac{7}{7}+\dfrac{8}{8}\) + \(\dfrac{10}{10}\))
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (1 + 1 +1 + 1+ 1+ 1+ 1 +1)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x 1 x 8
= \(\dfrac{1}{2}\) + \(\)\(\dfrac{1}{2}\) x 8
= \(\dfrac{1}{2}\) + 4
= \(\dfrac{9}{2}\)
a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + 8
= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8
= 1 + 1 + 8
= 2 + 8
= 10
b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{9}{18}\) + \(\dfrac{10}{20}\)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)
= \(\dfrac{1}{2}\) x 10
= 5
1/10 + 4/20 + 9/30 + 16/40 + 25/50 + 36/60 + 49/70 + 64/80 + 81/90
= 1/10 + 2/10 + 3/10 + 4/10 + 5/10 + 6/10 + 7/10 + 8/10 + 9/10
=(1+2+3+....+9) /10
=(1+9)x9:2 /10
=45/10=4,5
chắc chắn đúng đó
\(\frac{1}{5}+\frac{4}{10}+\frac{9}{15}+\frac{16}{20}+\frac{36}{30}+\frac{64}{40}+\frac{81}{45}\)
\(=\frac{33}{5}\)
\(\dfrac{1}{10}+\dfrac{4}{20}+\dfrac{9}{30}+\dfrac{16}{40}+\dfrac{25}{50}+\dfrac{36}{60}+\dfrac{49}{70}+\dfrac{64}{80}+\dfrac{81}{90}\)
\(=\dfrac{1}{10}+\dfrac{1}{5}+\dfrac{3}{10}+\dfrac{2}{5}+\dfrac{1}{2}+\dfrac{3}{5}+\dfrac{7}{10}+\dfrac{4}{5}+\dfrac{9}{10}\)
\(=\left(\dfrac{1}{10}+\dfrac{3}{10}+\dfrac{7}{10}+\dfrac{9}{10}\right)+\left(\dfrac{1}{5}+\dfrac{2}{5}+\dfrac{3}{5}+\dfrac{4}{5}\right)+\dfrac{1}{2}\)
\(=2+2+\dfrac{1}{2}\)
\(=4+\dfrac{1}{2}\)
\(=\dfrac{8}{2}+\dfrac{1}{2}=\dfrac{9}{2}\)
\(M=\frac{1}{10}+\frac{4}{20}+\frac{9}{30}+\frac{16}{40}+...+\frac{81}{90}\)
\(M=\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\frac{4}{10}+...+\frac{9}{10}\)
\(M=\frac{\left(9+1\right)\cdot\left(9-1+1\right):2}{10}\)
\(M=\frac{10\cdot9:2}{10}=4,5\)
=1/10+2/10+3/10+4/10+5/10+...+9/10
=1+2+3+4+...+9/10
=45/10
=4 1/2
=\(\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+...+\frac{9}{10}\)
\(=\frac{\left(9+1\right).\left(9-1+1\right):2}{10}=\frac{45}{10}\)
\(A=\frac{14^{16}-21^{32}.35^{68}}{10^{16}.15^2.7^{96}}=\frac{2^{16}.7^{16}-3^{32}.7^{32}.7^{68}.5^{68}}{5^{16}.2^{16}.3^2.5^2.7^{96}}\)
\(A=\frac{2^{16}.7^{16}-3^{32}.7^{100}.5^{68}}{5^{18}.2^{16}.3^2.7^{96}}=\frac{7^{16}.\left(2^{16}-3^{32}.7^{84}.5^{68}\right)}{5^{18}.2^{16}.3^2.7^{96}}=\frac{2^{16}-3^{32}.7^{84}.5^{68}}{5^{18}.2^{16}.3^2.7^{88}}\)
\(B=\frac{4^{20}-2^{20}+6^{20}}{6^{20}-3^{20}+9^{20}}=\frac{2^{40}-2^{20}+2^{20}.3^{20}}{2^{20}.3^{20}-3^{20}+3^{40}}=\frac{2^{20}.\left(2^{20}-1+3^{20}\right)}{3^{20}.\left(2^{20}-1+3^{20}\right)}\)
\(B=\frac{2^{20}}{3^{20}}\)
a, đề phải như thế này 16x95+16x4+16=16x(95+4+1)=16x100=1600
b, 20x(1+90+9)=20x100=2000
ai k mình k lại [ chỉ 3 người đầu tiên mà trên 10 điểm hỏi đáp ]