\(\frac{7!4!}{10!}\).(\(\frac{8!}{3!5!}\)-\(\frac{9!}{2!5!}\))
Tìm A
!: VD:A!=A.1 +A.2 + A.3 ...A.A
NA
Những câu hỏi liên quan
Tìm A:B, biết:
A=\(\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+\frac{6}{4}+\frac{5}{5}+\frac{4}{6}+\frac{3}{7}+\frac{2}{8}+\frac{1}{9}\)
B=\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}+\frac{1}{10}\)
\(\frac{A}{B}=\frac{\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+\frac{6}{4}+\frac{5}{5}+\frac{4}{6}+\frac{3}{7}+\frac{2}{8}+\frac{2}{9}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{1}{9}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{\frac{10}{2}+\frac{10}{3}+\frac{10}{4}+...+\frac{10}{9}+\frac{10}{10}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{10\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}\)
\(\frac{A}{B}=10\)
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\(A=\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{2}{8}+\frac{1}{9}\)
Tách 9=1+1+...+1 ( có 9 số 1)
\(\Rightarrow A=1+\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{2}{8}+1\right)+\left(\frac{1}{9}+1\right)\)
\(A=\frac{10}{10}+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{8}+\frac{10}{9}\)
\(A=10.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)
\(\Rightarrow A:B=\frac{10.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}=10\) ( vì \(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\ne0\) )
Vậy \(A:B=10\)
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\(A=\frac{\frac{2}{5}-\frac{2}{7}+\frac{3}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
a)Tìm A
b)tìm x biết A . x + \(\frac{5}{6}\) = \(-\frac{3}{4}\)
Bạn ghi sai đề chỗ 3/11 là sai mà phải 2/11 với là chỗ 2/7 là sai mà là 2/9
\(A=\frac{\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}-\frac{7}{11}}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}=\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\)
\(=\frac{2}{7}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}.\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{10}\right)}=\frac{2}{7}:\frac{2}{7}=1\)
b,\(A.x+\frac{5}{6}=-\frac{3}{4}\)
<=>\(1.x=-\frac{3}{5}-\frac{5}{6}\)
<=>x=-43/30
Ai thấy mình làm đúng thì tích nha.Ai tích mình mình tích lại
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a,A=1863/623:2/7
A=1863/178
b,
ta có:
1863/178.x+5/6=-3/4
1863/178.x=-19/12
=>x=-1691/11178
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bài 1 So sánha)Afrac{3}{8^3}+frac{7}{8^4} ; Bfrac{7}{8^3}+frac{3}{8^4}b)Afrac{10^{1992}+1}{10^{1991}+1};Bfrac{10^{1993}+1}{10^{1992}+1}c)Afrac{10^7+5}{10^4-8};Bfrac{10^8+6}{10^8-7}d)Afrac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8};Bfrac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}e)Afrac{2011}{2012}+frac{2012}{2013};Bfrac{2011+2012}{2012+2013}
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bài 1 So sánh
a)\(A=\frac{3}{8^3}+\frac{7}{8^4}\) ; \(B=\frac{7}{8^3}+\frac{3}{8^4}\)
b)\(A=\frac{10^{1992}+1}{10^{1991}+1};B=\frac{10^{1993}+1}{10^{1992}+1}\)
c)\(A=\frac{10^7+5}{10^4-8};B=\frac{10^8+6}{10^8-7}\)
d)\(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8};B=\frac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}\)
e)\(A=\frac{2011}{2012}+\frac{2012}{2013};B=\frac{2011+2012}{2012+2013}\)
a)\(\frac{7}{x}<\frac{x}{4}<\frac{10}{x}\)
b) Cho A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\). Chứng tỏ: \(\frac{8}{9}>A>\frac{2}{5}\)
Giải:
a) \(\dfrac{7}{x}< \dfrac{x}{4}< \dfrac{10}{x}\)
\(\Rightarrow7< \dfrac{x^2}{4}< 10\)
\(\Rightarrow\dfrac{28}{4}< \dfrac{x^2}{4}< \dfrac{40}{4}\)
\(\Rightarrow x^2=36\)
\(\Rightarrow x=6\)
b) \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\)
Ta có:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}=\dfrac{1}{4.4}< \dfrac{1}{3.4}\)
\(...\)
\(\dfrac{1}{9^2}=\dfrac{1}{9.9}< \dfrac{1}{8.9}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)
\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{9}\)
\(\Rightarrow A< \dfrac{8}{9}\left(1\right)\)
Ta có:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}>\dfrac{1}{2.3}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}>\dfrac{1}{3.4}\)
\(\dfrac{1}{4^2}=\dfrac{1}{4.4}>\dfrac{1}{4.5}\)
\(...\)
\(\dfrac{1}{9^2}=\dfrac{1}{9.9}>\dfrac{1}{9.10}\)
\(\Rightarrow A>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{10}\)
\(\Rightarrow A>\dfrac{2}{5}\left(2\right)\)
Từ (1) và (2), ta có:
\(\Rightarrow\dfrac{2}{5}< A< \dfrac{8}{9}\left(đpcm\right)\)
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So sanh A va B, biet :
a)\(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8};B=\frac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}\)
b)\(A=\frac{7^{10}}{1+7+7^2+...+7^9};B=\frac{5^{10}}{1+5+5^2+...+5^9}\)
\(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8}=\frac{1+5\left(1 +5+5^2+...+5^8\right)}{1+5+5^2+...+5^8}=5+\frac{1}{1+5+5^2+...+5^8} \)
\(B=\frac{1+3+3^2+....+3^9}{1+3+3^2+....+3^8}=\frac{1+3\left(1+3+3^2+....+3^8\right)}{1+3+3^2+....+3^8}=3+\frac{1}{1+3+3^2+....+3^8}\)
\(=5+\frac{1}{1+3+3^2+....+3^8}-2\)
Có: \(\frac{1}{1+5+5^2+...+5^8}>0\) và \(\frac{1}{1+3+3^2+....+3^8}-2< 0\)
\(\Rightarrow A>B\)
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bài 1 : tìm x biếta, frac{2}{3}timesleft(x-frac{5}{6}right)+frac{1}{4}frac{22}{9}b, frac{2}{3}:frac{x}{5}frac{10}{21}c, frac{7}{3}:frac{x}{5}frac{14}{15}d, 1-left(5frac{4}{9}times x-7frac{7}{18}right):15frac{3}{4}0bài 2 : tính gtri bta,frac{8}{7}+frac{1}{5}timesfrac{10}{9}b, frac{3}{2}+left(frac{9}{2}+frac{2}{9}right)timesleft(frac{4}{3}-frac{5}{4}right)!_ove
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bài 1 : tìm x biết
a, \(\frac{2}{3}\times\left(x-\frac{5}{6}\right)+\frac{1}{4}=\frac{22}{9}\)
b, \(\frac{2}{3}:\frac{x}{5}=\frac{10}{21}\)
c, \(\frac{7}{3}:\frac{x}{5}=\frac{14}{15}\)
d, \(1-\left(5\frac{4}{9}\times x-7\frac{7}{18}\right):15\frac{3}{4}=0\)
bài 2 : tính gtri bt
a,\(\frac{8}{7}+\frac{1}{5}\times\frac{10}{9}\)
b, \(\frac{3}{2}+\left(\frac{9}{2}+\frac{2}{9}\right)\times\left(\frac{4}{3}-\frac{5}{4}\right)\)
!_ove
a) x = 99/20
b) x = 7
c) x = 2
( chỉ lm đc đến đó thui nk )
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Bài 1:a) Afrac{4^5.9-2.6^9}{2^{10}.3^8+6^8.20}b) Bfrac{2^{12}.3^5-4^6.9^2}{left(2^2.3right)+8^4.3^5}-frac{5^{10}.7^3-25^5.49^2}{left(125.7right)^3+5^9.14^3}Bài 2: Tìm x biết:left|x-frac{1}{3}right|+frac{4}{5}left|left(-3,2right)+frac{2}{5}right|Bài 3: Tìm các số: a1, a2, a3,..., a9 biết:frac{a_1-1}{9}frac{a_2-2}{8}...frac{a_9-9}{1}và a_1+a_2+a_3+...+a_990Cứu với!!! ╥_╥
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Bài 1:
a) \(A=\frac{4^5.9-2.6^9}{2^{10}.3^8+6^8.20}\)
b) \(B=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
Bài 2: Tìm x biết:
\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\left(-3,2\right)+\frac{2}{5}\right|\)
Bài 3: Tìm các số: a1, a2, a3,..., a9 biết:
\(\frac{a_1-1}{9}=\frac{a_2-2}{8}=...=\frac{a_9-9}{1}\)và \(a_1+a_2+a_3+...+a_9=90\)
Cứu với!!! ╥_╥
Bài 2
| x - \(\frac{1}{3}\)| + \(\frac{4}{5}\)= | ( -3,2) + \(\frac{2}{5}\)|
=> | x - \(\frac{1}{3}\)| + \(\frac{4}{5}\)= | -2,8|
=> | x - \(\frac{1}{3}\)| + \(\frac{4}{5}\)= -2,8
=> | x - \(\frac{1}{3}\)| = -2,8 - \(\frac{4}{5}\)
=> | x - \(\frac{1}{3}\)| = - 3,6
=> x - \(\frac{1}{3}\)= -3,6
=> x = -3,6 + \(\frac{1}{3}\)
=> x = \(\frac{-49}{15}\)
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Bài 3 :
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{a_1-1}{9}=\frac{a_2-2}{8}=...=\frac{a_9-9}{1}=\frac{a_1-1+a_2-2+...+a_9-9}{9+8+...+1}\)
\(=\frac{\left[a_1+a_2+...+a_9\right]-\left[1+2+...+9\right]}{9+8+...+1}=\frac{90-45}{45}=1\)
Ta có : \(\frac{a_1-1}{9}=1\Rightarrow a_1=10\)
Tương tự : \(a_1=a_2=....=a_9=10\)
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Bài 1:Tìm x biết Bài 2:So sánh
a, x+frac{1}{2}frac{3}{8}.frac{4}{5} a, Afrac{10^{10}-1}{10^{11}-1}vaBfrac{10^9-1}{10^{10}-1}
b, frac{5}{16}:x-frac{1}{4}frac{5}{8} b, B frac{10^{10}}{10^{10}+1}vaBfrac{10^{10}+1}{10^{10}+2}
c, frac{-1}{4}.x+frac{3}{7}.x2
d, frac{22}{9}-left(x+frac{1}{2}right)^2frac{7}{3}
e, left|frac{1}{4}-xright|+5frac{1}{8}6frac{1}{8}
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Bài 1:Tìm x biết Bài 2:So sánh
a, \(x+\frac{1}{2}=\frac{3}{8}.\frac{4}{5}\) a, \(A=\frac{10^{10}-1}{10^{11}-1}vaB=\frac{10^9-1}{10^{10}-1}\)
b, \(\frac{5}{16}:x-\frac{1}{4}=\frac{5}{8}\) b, B =\(\frac{10^{10}}{10^{10}+1}vaB=\frac{10^{10}+1}{10^{10}+2}\)
c, \(\frac{-1}{4}.x+\frac{3}{7}.x=2\)
d, \(\frac{22}{9}-\left(x+\frac{1}{2}\right)^2=\frac{7}{3}\)
e, \(\left|\frac{1}{4}-x\right|+5\frac{1}{8}=6\frac{1}{8}\)
Tính hợp lí: \(A=\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
Đáp số là 1 vì cả hai phân số đều bằng 2/7 và 2/7:2/7=14/14=1.Đúng 100000000000000000000% luôn bạn ơi!
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đưa tử về 1 thử coi bn
trước đây cô giáo cũng bày mk thế
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