tính B = \(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+.......+\dfrac{1}{12}\left(1+2+3+.....+12\right)\)
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2 tháng 2 2023
a: =>1+3x-6=-x+3
=>3x-5=-x+3
=>4x=8
=>x=2(loại)
b: \(\Leftrightarrow\dfrac{3\left(x-3\right)+2\left(x-2\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\dfrac{x-1}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)
=>3x-9+2x-4=x-1
=>5x-13=x-1
=>4x=12
=>x=3(loại)
c: =>x^2-2x+4+x^3+8=12
=>x^3+x^2-2x=0
=>x(x^2+x-2)=0
=>x(x+2)(x-1)=0
=>x=0 hoặc x=1
29 tháng 5 2022
a: \(A=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}=\dfrac{25}{72}\)
b: \(B=\dfrac{8+5}{10}:\dfrac{-5}{13}=\dfrac{13}{10}\cdot\dfrac{13}{-5}=-\dfrac{169}{100}\)
c: \(C=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}+\dfrac{132}{132}-\dfrac{84}{132}\right)\)
\(=\dfrac{88-33+60}{55+132-84}=\dfrac{115}{103}\)
8 tháng 10 2017
\(a.\)
\(\left[6.\left(-\dfrac{1}{3}\right)^2-3\left(-\dfrac{1}{3}\right)+1\right]:\left(-\dfrac{1}{3}-1\right)\)
\(=\left[6.\dfrac{1}{9}+1+1\right]:\left(-\dfrac{4}{3}\right)\)
\(=\left(\dfrac{8}{3}\right):\left(-\dfrac{4}{3}\right)\)
\(=\left(\dfrac{8}{3}\right).\left(-\dfrac{3}{4}\right)\)
\(=-2\)
\(b.\)
\(\dfrac{\left(\dfrac{2}{3}\right)^3.\left(-\dfrac{3}{4}\right)^2.\left(-1\right)^{2003}}{\left(\dfrac{2}{5}\right)^2.\left(-\dfrac{5}{12}\right)^3}\)
\(=\dfrac{\dfrac{8}{27}.\dfrac{9}{16}.\left(-1\right)}{\dfrac{4}{25}.\left(-\dfrac{125}{1728}\right)}\)
\(=\dfrac{-\dfrac{1}{6}}{-\dfrac{5}{432}}\)
\(=\dfrac{72}{5}\)
TH
4 tháng 12 2018
a) 1/x(x + 1) + 1/(x + 1)(x + 2) + 1/(x + 2)(x + 3) + 1/(x + 3)(x + 4)
( 1/x - 1/x+1) + (1/x+1 - 1/x+2) + (1/x+2 - 1/ x+3) + 1/(x+3 - 1/x+4)
(1/x +1/x+4) - ( 1/x+2 - 1/x+2) - ( 1/x+3 - 1/x+3)
1/x +1/x+4
2x+4/x(x+4)
31 tháng 12 2023
a: \(\left(18\dfrac{1}{3}:\sqrt{225}+8\dfrac{2}{3}\cdot\sqrt{\dfrac{49}{4}}\right):\left[\left(12\dfrac{1}{3}+8\dfrac{6}{7}\right)-\dfrac{\left(\sqrt{7}\right)^2}{\left(3\sqrt{2}\right)^2}\right]:\dfrac{1704}{445}\)
\(=\left(\dfrac{55}{3}:15+\dfrac{26}{3}\cdot\dfrac{7}{4}\right):\left[\left(12+\dfrac{1}{3}+8+\dfrac{6}{7}\right)-\dfrac{7}{18}\right]\cdot\dfrac{445}{1704}\)
\(=\left(\dfrac{55}{45}+\dfrac{91}{6}\right):\left[20+\dfrac{101}{126}\right]\cdot\dfrac{445}{1704}\)
\(=\dfrac{295}{18}:\dfrac{2621}{126}\cdot\dfrac{445}{1704}\)
\(=\dfrac{295}{18}\cdot\dfrac{126}{2621}\cdot\dfrac{445}{1704}\simeq0,21\)
b: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
c: \(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{n+1}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{n}{n+1}\)
\(=\dfrac{1}{n+1}\)
d: \(-66\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{11}\right)+124\cdot\left(-37\right)+63\cdot\left(-124\right)\)
\(=-66\cdot\dfrac{33-22+6}{66}+124\left(-37-63\right)\)
\(=-17-12400=-12417\)
e: \(\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{333333}{303030}+\dfrac{33333333}{42424242}\right)\)
\(=\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)
\(=\dfrac{7}{4}\cdot33\cdot\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\right)\)
\(=33\cdot\dfrac{7}{4}\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)
\(=33\cdot\dfrac{7}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)
\(=33\cdot\dfrac{7}{4}\cdot\dfrac{4}{21}=\dfrac{33\cdot1}{3}=11\)
NT
6 tháng 8 2021
a) \(A=\dfrac{2}{3}+\dfrac{3}{4}.\left(\dfrac{-4}{9}\right)=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)
b) \(B=2\dfrac{3}{11}.1\dfrac{1}{12}.\left(-2,2\right)=\dfrac{25}{11}.\dfrac{13}{12}.\dfrac{-11}{5}=-\dfrac{65}{12}\)
c) \(C=\left(\dfrac{3}{4}-0,2\right)\left(0,4-\dfrac{4}{5}\right)=\left(\dfrac{3}{4}-\dfrac{1}{5}\right)\left(\dfrac{2}{5}-\dfrac{4}{5}\right)=\dfrac{11}{20}\left(\dfrac{-2}{5}\right)=\dfrac{-11}{50}\)
20 tháng 1 2022
Câu 1:
=>15(2x+1)-8(3x-1)=100
=>30x+15-24x+8=100
=>6x+23=100
hay x=77/6
Câu 2:
=>2(5x-3)+12-3(7x-1)=x+2
=>10x-6+12-21x+3-x-2=0
=>-12x=-7
hay x=7/12
Câu 3:
\(\Leftrightarrow2\left(x^2-1\right)+3\left(x+1\right)=2\left(x^2-4x+4\right)\)
\(\Leftrightarrow2x^2-2+3x+3-2x^2+8x-8=0\)
=>11x-7=0
hay x=-7/11
20 tháng 1 2022
Câu 4:
(x - 4)^3/6 + 1 = x(x + 1)/2 - (x - 5)(x + 5)/3
<=> (x - 4)^3 + 6/6 = x^2 + x/2 - x^2 - 25/3
<=> (x - 4)^3 + 6/6 = 3x^2 + 3x - 2x^2 + 50/6
<=> (x - 4)^3 + 6 = 3x^2 + 3x - 2x^2 + 50
<=> x^3 - 12x^2 + 48x - 58 = x^2 + 3x + 50
<=> x^3 -13x^2 + 45x - 108 = 0
Đến đây bạn bấm máy nhẩm nghiệm là ra nhé
Câu 5:
3(x + 2)^3/5 - (x - 1)^2/10 = (x - 3)(x + 3)/2
<=> 6(x + 2)^3 - (x - 1)^2/10 = 5(x^2 - 9)/10
<=> 6(x + 2)^3 - (x - 1)^2 = 5(x^2 - 9)
<=> 6x^3 + 36x^2 + 72x + 48 - x^2 + 2x - 1 - 5x^2 + 45 = 0
<=> 6x^3 + 30x^2 + 74x + 92 = 0
Đến đây bạn bấm máy nhẩm nghiệm như câu 4 nhé
Lời giải:
Ta có công thức tính tổng $n$ số hạng đầu tiên là: \(1+2+3+...+n=\frac{n(n+1)}{2}\)
\(\Rightarrow \frac{1+2+..+n}{n}=\frac{n(n+1)}{2n}=\frac{n+1}{2}\)
Áp dụng vào bài toán đã cho:
\(B=1+\frac{1}{2}(1+2)+\frac{1}{3}(1+2+3)+...+\frac{1}{12}(1+2+3+...+12)\)
\(B=1+\frac{2+1}{2}+\frac{3+1}{2}+\frac{4+1}{2}+...+\frac{12+1}{2}\)
\(B=1+\frac{1}{2}(3+4+5+...+13)\)
\(B=1+\frac{1}{2}(1+2+3+...+13)-\frac{3}{2}=1+\frac{1}{2}.\frac{13(13+1)}{2}-\frac{3}{2}\)
\(B=1+\frac{91}{2}-\frac{3}{2}=45\)