A=16-4/√16 +1
Những câu hỏi liên quan
A=1/4+1/9+1/16+...+1/2024.2024
A=1/4+1/9+1/16+...+1/2024
rút gọn biểu thức
a) A=16^8 -1/(2+1)(2^2+1)(2^4+1)(2^8+1(3^16+1)
b) B=(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)/9^16-1
giúp mk vs ah mk đang cần gấp ah
a) Ta có: \(A=\dfrac{16^8-1}{\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^{16}-1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{2^{32}-1}=1\)
b) Ta có: \(B=\dfrac{\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{9^{16}-1}\)
\(=\dfrac{\left(3^2-1\right)\cdot\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}=\dfrac{1}{2}\)
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A=1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+...+1/16(1+2+3+4+...+16)
A= \(A=1*2*3+2*4*6+4*8*12+8*16*24 \over 2*3*4+4*6*8+8*12*16+16*24*32\)
mình xin nhầm đề là: A = \(1*2*3+2*4*6+4*8*12+8*16*24 \over2*3*4+4*6*8+8*12*16+16*24*32\)
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Tính :a)
3
+
−
5
+
−
1
+
−
7
b)
16
+
−
5
−
−
16
+
4
Đọc tiếp
Tính :
a) 3 + − 5 + − 1 + − 7
b) 16 + − 5 − − 16 + 4
a) 3 + − 5 + − 1 + − 7 = 3 + 5 + 1 + 7 = 16
b) 16 + − 5 − − 16 + 4 = 16 + 5 − 16 + 4 = 9
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Câu 1 : A=1+3+3^2+3^3+3^4...+3^300+3^301+3^302 có chia hết cho 13 ko
Câu 2: A=6+16+16^2+16^3+...+16^8+16^9 chứng tỏ rằng A vừa chia hết cho 2 vừa chia hết cho 5
a) 5√1,96 - 4√2,25 + 3√1,69
b) 1/3√49 + 3/4√324 - 2/3√400
c) 5/6√289 - 1/2√256 + 2/5√225
d) 15√16/25 + 12√9/16 - 18√16/81
a: =5*1,4-4*1,5+3*1,3
=7-6+3,9=4,9
b: =1/3*7+3/4*18-2/3*20
=7/3+54/4-40/3
=-11+54/4
=2,5
c: =5/6*17-1/2*16+2/5*15
=85/6-8+6
=85/6-2
=73/6
d: =15*4/5+12*3/4-18*4/9
=12+9-8
=12+1=13
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tính nhanh
a) 1/1*2 + 1/2*3 + 1/3*4 + 1/4*5 + ......... + 1/98*99 + 1/99*100
b) 5/11*16 + 5/16*21 + 5/21*16 + .......... + 5/61*66
a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
b)\(\frac{5}{11.16}+\frac{5}{16.21}+...+\frac{5}{61.66}\)
\(=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+....+\frac{1}{61}-\frac{1}{66}\)
\(=\frac{1}{11}-\frac{1}{66}\)
\(=\frac{5}{66}\)
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a,\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
ta có:
\(\frac{1}{1.2}=\frac{2-1}{1.2}=\frac{2}{1.2}-\frac{1}{1.2}=1-\frac{1}{2}\)
\(\frac{1}{2.3}=\frac{3-2}{2.3}=\frac{3}{2.3}-\frac{2}{2.3}=\frac{1}{2}-\frac{1}{3}\)
...
\(\frac{1}{99.100}=\frac{1}{99}-\frac{1}{100}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
=\(1-\frac{1}{100}=\frac{99}{100}\)
b,
\(\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.16}+...+\frac{5}{61.66}\)
ta có:
\(\frac{5}{11.16}=\frac{16-11}{11.16}=\frac{16}{11.16}-\frac{11}{11.16}=\frac{1}{11}-\frac{1}{16}\)
\(\frac{5}{16.21}=\frac{21-16}{16.21}=\frac{21}{16.21}-\frac{16}{16.21}=\frac{1}{16}-\frac{1}{21}\)
...
\(\frac{5}{61.66}=\frac{66-61}{61.66}=\frac{66}{61.66}-\frac{61}{61.66}=\frac{1}{61}-\frac{1}{66}\)
= \(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}\)
=\(\frac{1}{11}-\frac{1}{66}\)=\(\frac{5}{66}\)
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Tính nhanh
A= 1+ 1/2 (1+2) +1/3 (1+2+3) +1/4 (1+2+3+4) +...+ 1/16 (1+2+3+4+...+16)
A=1+1/2x3+1/3X6+1/4X10+...+1/16X136
A=1+3/2+2+5/2+3+...+17/2
A=2/2+3/2+4/2+5/2+6/2+...+17/2
A=2+3+4+5+...+16+17/2
A=(2+17)x16:2/2
A=19x16:2/2
A=304:2/2
A=152/2
A=76
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