Cho
\(A=\dfrac{1}{1.101}+\dfrac{1}{2.102}+\dfrac{1}{3.103}+...+\dfrac{1}{25.125}\)
\(B=\dfrac{1}{1.26}+\dfrac{1}{2.27}+\dfrac{1}{3.28}+...+\dfrac{1}{100.125}\)
Trong đó A có 25 số hạng,B có 100 số hạng.Tìm thương A:B
Cho A =\(\frac{1}{1.101}\) + \(\frac{1}{2.102}\) + \(\frac{1}{3.103}\) + ...... + \(\frac{1}{25.125}\)
B =\(\frac{1}{1.26}\) + \(\frac{1}{2.27}\) + \(\frac{1}{3.28}\) + ....... + \(\frac{1}{100.125}\)
A có 25 số hạng, B có 100 số hạng . Tìm thương A : B
(\(\dfrac{1}{1.101}+\dfrac{1}{2.102}+\dfrac{1}{3.103}+...+\dfrac{1}{10.110}\)).x= \(\dfrac{1}{1.11}+\dfrac{1}{2.12}+\)\(\dfrac{1}{3.13}+...+\dfrac{1}{100.110}\)
Tìm tỉ số của A và B , biết rằng :
A = \(\dfrac{1}{1.1981}+\dfrac{1}{2.1982}+.....+\dfrac{1}{n\left(1980+n\right)}.....+\dfrac{1}{25.2005}\)
B = \(\dfrac{1}{1.26}+\dfrac{1}{2.27}+......+\dfrac{1}{m\left(m+25\right)}+.......+\dfrac{1}{1980.2005}\)
Trogn đó A có 25 số hạng và B có 1980 số hạng
Ta có:
\(A=\dfrac{1}{1.1981}+\dfrac{1}{2.1982}+...+\dfrac{1}{n\left(1980+n\right)}+...+\dfrac{1}{25.2005}\)
\(=\dfrac{1}{1980}\left(\dfrac{1981-1}{1.1981}+\dfrac{1982-2}{2.1982}+...+\dfrac{1980+n-n}{n\left(1980+n\right)}+...+\dfrac{2005-25}{25.2005}\right)\)
\(=\dfrac{1}{1980}\left(1-\dfrac{1}{1981}+\dfrac{1}{2}-\dfrac{1}{1982}+...+\dfrac{1}{n}-\dfrac{1}{1980+n}+...+\dfrac{1}{25}-\dfrac{1}{2005}\right)\)
\(=\dfrac{1}{1980}\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{25}\right)-\left(\dfrac{1}{1981}+\dfrac{1}{1982}+...+\dfrac{1}{2005}\right)\right]\)
Lại có:
\(B=\dfrac{1}{1.26}+\dfrac{1}{2.27}+...+\dfrac{1}{m\left(m+25\right)}+...+\dfrac{1}{1980.2005}\)
\(=\dfrac{1}{25}\left(\dfrac{26-1}{1.26}+\dfrac{27-2}{2.27}+...+\dfrac{25+m-m}{m\left(25+m\right)}+...+\dfrac{2005-1980}{1980.2005}\right)\)
\(=\dfrac{1}{25}\left(\dfrac{1}{1}-\dfrac{1}{26}+\dfrac{1}{2}-\dfrac{1}{27}+...+\dfrac{1}{m}-\dfrac{1}{25+m}+...+\dfrac{1}{1980}-\dfrac{1}{2005}\right)\)
\(=\dfrac{1}{25}\left[\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{1980}\right)-\left(\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{2005}\right)\right]\)
\(=\dfrac{1}{25}\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{25}\right)-\left(\dfrac{1}{1981}+\dfrac{1}{1982}+...+\dfrac{1}{2005}\right)\right]\)
\(\Rightarrow\dfrac{A}{B}=\dfrac{\dfrac{1}{1980}}{\dfrac{1}{25}}=\dfrac{5}{396}\)
Vậy tỉ số của \(A\) và \(B\) là \(\dfrac{5}{396}\)
Giải phương trình
a, \(x^4+2x^3-4x^2-5x-6=0\)
b, \(\left(\dfrac{1}{1.101}+\dfrac{1}{2.102}+\dfrac{1}{3.103}+...+\dfrac{1}{10.110}\right).x=\dfrac{1}{1.11}+\dfrac{1}{2.12}+\dfrac{1}{100.110}\)
a, ⇔ x4 - 2x3 + 4x3 - 8x2 + 4x2 - 8x + 3x - 6 = 0
⇔ (x - 2)(x3 + 4x2 + 4x + 3) = 0
⇔ (x - 2)(x3 + 3x2 + x2 + 3x + x + 3) = 0
⇔ (x - 2)(x + 3)(x2 + x + 1) = 0 mà x2 + x + 1 > 0 ∀ x
⇔ \(\left\{{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy tập nghiệm phương trình S = {2; -3}
Cho: \(A=\dfrac{1}{1.1981}+\dfrac{1}{2.2982}+......+\dfrac{1}{25.2005}\)
\(B=\dfrac{1}{1.26}+\dfrac{1}{2.27}+......+\dfrac{1}{1980.2005}\)
Tính A : B
Ta có: \(A=\dfrac{1}{1.1981}+\dfrac{1}{2.1982}+...+\dfrac{1}{25.2005}\)
\(\Rightarrow1980A=\dfrac{1980}{1.1981}+\dfrac{1980}{2.1982}+...+\dfrac{1980}{25.2005}\)
\(=\dfrac{1}{1}-\dfrac{1}{1981}+\dfrac{1}{2}-\dfrac{1}{1982}+...+\dfrac{1}{25}-\dfrac{1}{2005}\)
\(=\left(1+\dfrac{1}{2}+...+\dfrac{1}{25}\right)-\left(\dfrac{1}{1981}+\dfrac{1}{1982}+...+\dfrac{1}{2005}\right)\)
\(\Rightarrow A=\dfrac{1}{1980}\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{25}\right)-\left(\dfrac{1}{1981}+\dfrac{1}{1982}+...+\dfrac{1}{2005}\right)\right]\)
Mặt khác: \(B=\dfrac{1}{1.26}+\dfrac{1}{2.27}+...+\dfrac{1}{1980.2005}\)
\(\Rightarrow25B=\dfrac{25}{1.26}+\dfrac{25}{2.27}+...+\dfrac{25}{1980.2005}\)
\(=\dfrac{1}{1}-\dfrac{1}{26}+\dfrac{1}{2}-\dfrac{1}{27}+...+\dfrac{1}{1980}-\dfrac{1}{2005}\)
\(=\left(1+\dfrac{1}{2}+...+\dfrac{1}{1980}\right)-\left(\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{2005}\right)\)
\(=\left(1+\dfrac{1}{2}+...+\dfrac{1}{25}\right)-\left(\dfrac{1}{1981}+\dfrac{1}{1982}+...+\dfrac{1}{2005}\right)\)
\(\Rightarrow B=\dfrac{1}{25}\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{25}\right)-\left(\dfrac{1}{1981}+\dfrac{1}{1982}+...+\dfrac{1}{2005}\right)\right]\)
Do đó A:B=\(\dfrac{1}{1980}:\dfrac{1}{25}=\dfrac{5}{396}\)
Vậy A:B=\(\dfrac{5}{396}\)
Tìm x biết: \(\left(\dfrac{1}{1.101}+\dfrac{1}{2.102}+...+\dfrac{1}{10.110}\right)x=\dfrac{1}{1.11}+\dfrac{1}{1.12}+...+\dfrac{1}{10.110}\)
Tìm x, biết: \(\left ( \frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{10.110} \right ).x = \frac{1}{1.11}+\frac{1}{1.12}+...+\frac{1}{100.110}\)- Trường Toán Trực tuyến Pitago – Giải pháp giúp em học toán vững vàng!
Tìm x, biết:
a).\(\left(\dfrac{1}{1.101}+\dfrac{1}{2.102}+...+\dfrac{1}{10.110}\right).x=\dfrac{1}{1.11}+\dfrac{1}{2.12}+...+\dfrac{1}{100.110}\)
b).\(x-\dfrac{20}{11.13}-\dfrac{20}{13.15}-\dfrac{20}{15.17}-...-\dfrac{20}{53.55}=\dfrac{3}{11}\)
c).\(\dfrac{x-1}{99}+\dfrac{x-2}{98}+\dfrac{x-5}{95}=3+\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{95}\)
Mấy bạn tính nhanh, hợp lí, giải ra từng bước dùm mik nha
Thanks m.n
b: \(\Leftrightarrow x-10\left(\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{53\cdot55}\right)=\dfrac{3}{11}\)
\(\Leftrightarrow x-10\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)=\dfrac{3}{11}\)
\(\Leftrightarrow x-10\cdot\dfrac{4}{55}=\dfrac{3}{11}\)
=>x=3/11+20/55=3/11+4/11=7/11
c: \(\Leftrightarrow\left(\dfrac{x-1}{99}-1\right)+\left(\dfrac{x-2}{98}-1\right)+\left(\dfrac{x-5}{95}-1\right)=\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{95}\)
\(\Leftrightarrow x-100=1\)
hay x=101
Cho: B = \(\frac{1}{101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{25.125}\)
C =\(\frac{1}{26}+\frac{1}{2.27}+\frac{1}{3.28}+...+\frac{1}{100.125}\)
Tìm thương B : C
Giải phương trình:
b) \(\dfrac{7}{2}-\left(\dfrac{x}{5}-\dfrac{1}{4}\right)=\dfrac{9}{2}\)
c) (x+2) . (x-5). (x-6) (x+3) = 180
d) \(x-\dfrac{\dfrac{x}{2}-\dfrac{3+x}{4}}{2}=\dfrac{2x-\dfrac{10-7x}{3}}{2}-x-1\)
e) \(\left(\dfrac{1}{1.101}+\dfrac{1}{2.102}+........+\dfrac{1}{10.110}\right).\left(x-3\right)=\dfrac{1}{1.11}+\dfrac{1}{2.12}+.......+\dfrac{1}{100.110}\)
b) \(\dfrac{7}{2}-\left(\dfrac{x}{5}-\dfrac{1}{4}\right)=\dfrac{9}{2}\)
<=> \(\dfrac{7}{2}-\dfrac{x}{5}+\dfrac{1}{4}=\dfrac{9}{2}\)
<=> \(\dfrac{15}{4}-\dfrac{x}{5}-\dfrac{9}{2}=0\)
<=> \(\dfrac{x}{5}=\dfrac{5}{4}\)
<=> x = 6,25
Vậy,...
c) ( x + 2)( x + 3)( x - 5)( x - 6) = 180
<=> ( x + 2)( x - 5)( x + 3)( x - 6) = 180
<=> ( x2 - 3x - 10 )( x2 - 3x - 18 ) = 180
Đặt : x2 - 3x - 14 = a , ta có :
( a + 4)( a - 4) = 180
<=> a2 - 16 - 180 = 0
<=> a2 - 196 = 0
<=> ( a - 14)( a + 14 ) = 0
<=> a = 14 hoặc a = -14
* Với , a = 14 , ta có :
x2 - 3x - 14 = 14
<=> x2 - 3x - 28 = 0
<=> x2 - 7x + 4x - 28 = 0
<=> x( x - 7) + 4( x - 7) = 0
<=> ( x + 4)( x - 7) = 0
<=> x = -4 hoặc : x = 7
* Với : a = -14 , ta có :
x2 - 3x - 14 = -14
<=> x( x - 3) = 0
<=> x = 0 hoặc : x = 3
Vậy,...