1) Rút Gọn Các Biểu Thức Sau:
a) A = 20 + 21 + 22 + ... + 22010
b) B = 1 - 3 + 32 - 33 + ... + 3100
2) Tính Tổng Phân Số Sau:
\(\dfrac{1}{18}\) + \(\dfrac{1}{54}\) + \(\dfrac{1}{108}\) + ... + \(\dfrac{1}{990}\)
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Những câu hỏi liên quan
1 tháng 8 2023
a: \(=3\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)-3\sqrt{6}\)
=3căn 6-6-3căn 6=-6
b: \(=\dfrac{a+\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\sqrt{a}\)
\(=\dfrac{a+\sqrt{ab}-a+\sqrt{ab}}{\sqrt{a}-\sqrt{b}}=\dfrac{2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\)
22 tháng 11 2021
\(a,A=\dfrac{3x+2-3x+2+3x-6}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{3x-2}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{1}{3x+2}\\ b,B=\dfrac{1}{2}+\dfrac{x}{\dfrac{x+2-x}{x+2}}=\dfrac{1}{2}+\dfrac{x}{\dfrac{2}{x+2}}=\dfrac{1}{2}+\dfrac{x\left(x+2\right)}{2}\\ B=\dfrac{1+x^2+2x}{2}=\dfrac{\left(x+1\right)^2}{2}\)
XO
16 tháng 12 2023
a) \(A=\sqrt{18}.\sqrt{2}-\sqrt{48}:\sqrt{3}=\sqrt{18.2}-\sqrt{48:3}\)
\(=\sqrt{36}-\sqrt{16}=6-4=2\)
b) \(B=\dfrac{8}{\sqrt{5}-1}+\dfrac{8}{\sqrt{5}+1}=\dfrac{8\sqrt{5}+8+8\sqrt{5}-8}{\left(\sqrt{5}-1\right).\left(\sqrt{5}+1\right)}=\dfrac{16\sqrt{5}}{4}=4\sqrt{5}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 4 2023
b, B = \(\dfrac{1}{2}\) - \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\) - \(\dfrac{1}{2^4}\)+.....+ \(\dfrac{1}{2^{99}}\) - \(\dfrac{1}{2^{100}}\)
2 \(\times\) B = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) - \(\dfrac{1}{2^3}\) + \(\dfrac{1}{2^4}\)-.......-\(\dfrac{1}{2^{99}}\)
2 \(\times\) B + B = 1 - \(\dfrac{1}{2^{100}}\)
3B = ( 1 - \(\dfrac{1}{2^{100}}\))
B = ( 1 - \(\dfrac{1}{2^{100}}\)) : 3
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 4 2023
A = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\)+ \(\dfrac{1}{3^3}\)+......+ \(\dfrac{1}{3^{n-1}}\) + \(\dfrac{1}{3^n}\)
A\(\times\) 3 = 3 + 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^2}\)+....+ \(\dfrac{1}{3^{n-1}}\)
A \(\times\) 3 - A = 3 - \(\dfrac{1}{3^n}\)
2A = 3 - \(\dfrac{1}{3^n}\)
A = ( 3 - \(\dfrac{1}{3^n}\)) : 2
5 tháng 9 2023
21: ĐKXĐ: x>0; x<>1
\(A=\left(\dfrac{x\sqrt{x}+1-\left(x-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}}{\sqrt{x}-1}\)
\(=\dfrac{x\sqrt{x}+1-x\sqrt{x}-x+\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1}{x}\)
\(=\dfrac{-x+\sqrt{x}+2}{\sqrt{x}+1}\cdot\dfrac{1}{x}\)
\(=\dfrac{-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{x\left(\sqrt{x}+1\right)}=\dfrac{-\sqrt{x}+2}{x}\)
22:
DKXĐ: x>0; x<>1
\(A=\dfrac{x-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}:\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}+\dfrac{2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}:\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{x-1+\sqrt{x}+2-x}\)
\(=\dfrac{x}{\sqrt{x}+1}\)
23: ĐKXĐ: x>0; x<>4
\(A=\dfrac{\sqrt{x}-4+3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}:\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-x}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{-4}\)
\(=\dfrac{-4\sqrt{x}+4}{4}=-\sqrt{x}+1\)
24: ĐKXĐ: x>=0; x<>1
\(A=\dfrac{2x+1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x+\sqrt{x}+1-x+2}{x+\sqrt{x}+1}\)
\(=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{\sqrt{x}+3}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
25:
ĐKXĐ: x>=0; x<>1
\(A=1:\dfrac{x+2\sqrt{x}-2-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{2x+\sqrt{x}-1-x+1}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x+\sqrt{x}}=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)
27: ĐKXĐ: x>0; x<>4
\(P=\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4x-8\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{x-1-2\sqrt{x}+1}\)
\(=\dfrac{4\left(x-2\sqrt{x}-2\right)}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}}{-\sqrt{x}}\)
\(=\dfrac{-4\left(x-2\sqrt{x}-2\right)}{\sqrt{x}+2}\)
AH
Akai Haruma
Giáo viên
26 tháng 8 2023
Lời giải:
a.
$=2\sqrt{5}-9\sqrt{5}-2\sqrt{5}=(2-9-2)\sqrt{5}=-9\sqrt{5}$
b.
$=36\sqrt{6}-2\sqrt{6}+6\sqrt{6}=(36-2+6)\sqrt{6}=40\sqrt{6}$
15 tháng 7 2021
a) \(\dfrac{1}{2}\sqrt{20}+5=\dfrac{1}{2}\cdot2\sqrt{5}+5=5+\sqrt{5}\)
b) \(\sqrt{16}+\sqrt{64}=4+8=12\)
c) \(\sqrt{20}-\sqrt{45}+3\sqrt{18}=2\sqrt{5}-3\sqrt{5}+9\sqrt{2}=9\sqrt{2}-\sqrt{5}\)
d) \(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{2}=2-\sqrt{2}+\sqrt{2}=2\)
a) Ta có :
\(A=1+2+2^2+2^3+....................+2^{2010}\) (\(2010\) số hạng)
\(2A=2+2^2+............+2^{2010}+2^{2011}\)
\(\Rightarrow2A-A=\left(2+2^2+..........+2^{2011}\right)-\left(1+2+.............+2^{2010}\right)\)
\(A=2^{2011}-1\)
b) Ta có :
\(B=1-3+3^2-3^3+...............+3^{100}\)(\(100\) số hạng)
\(3B=3-3^2+3^3+.....+3^{99}-3^{100}+3^{101}\)
\(\Rightarrow3B+B=\left(1-3+.......+3^{100}\right)+\left(3-3^2+....-3^{100}+3^{101}\right)\)
\(4B=3^{101}+1\)
~ Chúc bn học tốt ~
2)
\(\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\)
\(=\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+...+\dfrac{1}{30.33}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{3.6}+\dfrac{3}{6.9}+\dfrac{3}{9.12}+...+\dfrac{3}{30.33}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{33}\right)\)
\(=\dfrac{1}{3}.\dfrac{10}{33}\)
\(=\dfrac{10}{99}\)