Tính:
\(4.\left(\dfrac{-1}{2}\right)^3-2.\left(\dfrac{-1}{2}\right)^2+3.\left(\dfrac{-1}{2}\right)+1\)
Giải chi tiết giúp mik nha!
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
30 tháng 10 2023
\(E=\dfrac{\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{2002}-1\right)\left(\dfrac{1}{2003}-1\right)}{\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot...\cdot\dfrac{9999}{10000}}\)
\(=\dfrac{\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2002}\right)\left(1-\dfrac{1}{2003}\right)}{\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{100^2}\right)}\)
\(=\dfrac{\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2002}\right)\left(1-\dfrac{1}{2003}\right)}{\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{100}\right)\left(1+\dfrac{1}{100}\right)}\)
\(=\dfrac{\dfrac{100}{101}\cdot\dfrac{101}{102}\cdot...\cdot\dfrac{2002}{2003}}{\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1+\dfrac{1}{100}\right)}\)
\(=\dfrac{100}{2003}:\left(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\right)\)
\(=\dfrac{100}{2003}:\left(\dfrac{101}{2}\right)=\dfrac{100}{2003}\cdot\dfrac{2}{101}=\dfrac{200}{202303}\)
NM
22 tháng 8 2023
\(\left(-\dfrac{2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{3}+\dfrac{4}{7}\right)+\dfrac{4}{5}\\ =-\dfrac{5}{21}:\dfrac{4}{5}+\dfrac{5}{21}\\ =\left(-\dfrac{5}{21}+\dfrac{5}{21}\right):\dfrac{4}{5}\\ =0:\dfrac{4}{5}\\ =0.\)
NM
22 tháng 8 2023
Sửa cho mk dòng đầu là :4/5 và dòng tiếp theo mk thiếu :4/5
ND
Nguyễn Đức Trí
VIP
29 tháng 8 2023
\(...A=\left(-\dfrac{1}{2}\right).\left(-\dfrac{2}{3}\right).\left(-\dfrac{3}{4}\right)....\left(-\dfrac{1998}{1999}\right).\)
Số dấu trừ là : \(\left(1998-1\right):1+1=1998\) là số chẵn
\(\Rightarrow A=\dfrac{1.2.3...1998}{2.3.4...1999}\)
\(\Rightarrow A=\dfrac{1}{1999}\)
DT
29 tháng 8 2023
gợi ý nè
tính hết mấy cái hiệu trong ngoặc rồi nhân lại
vì kết thúc ở số 1999
nên sẽ có 1999 dấu -
nên kq là âm
nhân ra rồi triệt tiêu đi
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
16 tháng 7 2023
\(\dfrac{1}{15}\) + \(\dfrac{1}{21}\) + \(\dfrac{1}{28}\) + \(\dfrac{1}{36}\) +...+ \(\dfrac{2}{x\left(x+1\right)}\) = \(\dfrac{11}{40}\) (\(x\in\) N*)
\(\dfrac{1}{2}\).(\(\dfrac{1}{15}\)+\(\dfrac{1}{21}\)+\(\dfrac{1}{28}\)+\(\dfrac{1}{36}\)+.....+ \(\dfrac{2}{x\left(x+1\right)}\)) = \(\dfrac{11}{40}\) \(\times\) \(\dfrac{1}{2}\)
\(\dfrac{1}{30}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\) + \(\dfrac{1}{72}\)+...+ \(\dfrac{1}{x\left(x+1\right)}\) = \(\dfrac{11}{80}\)
\(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + \(\dfrac{1}{7.8}\)+...+ \(\dfrac{1}{x\left(x+1\right)}\) = \(\dfrac{11}{80}\)
\(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)+...+ \(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\) = \(\dfrac{11}{80}\)
\(\dfrac{1}{5}\) - \(\dfrac{1}{x+1}\) = \(\dfrac{11}{80}\)
\(\dfrac{1}{x+1}\) = \(\dfrac{1}{5}\) - \(\dfrac{11}{80}\)
\(\dfrac{1}{x+1}\) = \(\dfrac{1}{16}\)
\(x\) + 1 = 16
\(x\) = 16 - 1
\(x\) = 15
XO
16 tháng 7 2023
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x.\left(2x+1\right)}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{2x.\left(2x+1\right)}=\dfrac{1}{20}\)
\(\Leftrightarrow\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2x.\left(2x+1\right)}=\dfrac{1}{20}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2x}-\dfrac{1}{2x+1}=\dfrac{1}{20}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x+1}=\dfrac{1}{20}\)
\(\Leftrightarrow\dfrac{1}{2x+1}=\dfrac{9}{20}\)
\(\Leftrightarrow2x+1=\dfrac{20}{9}\Leftrightarrow x=\dfrac{11}{18}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
16 tháng 7 2023
Em giải như XYZ olm em nhé
Sau đó em thêm vào lập luận sau:
\(x\) = \(\dfrac{11}{18}\)
Vì \(\in\) N*
Vậy \(x\in\) \(\varnothing\)
18 tháng 10 2021
\(\dfrac{\dfrac{4}{5}:\left(\dfrac{4}{5}\cdot\dfrac{5}{4}\right)}{\dfrac{16}{25}-\dfrac{1}{25}}+\dfrac{\left(\dfrac{27}{25}-\dfrac{2}{25}\right):\dfrac{4}{7}}{\left(\dfrac{59}{9}-\dfrac{13}{4}\right)\cdot\dfrac{36}{17}}+\left(\dfrac{6}{5}\cdot\dfrac{1}{2}\right):\dfrac{4}{5}\)
\(=\dfrac{4}{5}:\dfrac{3}{5}+\dfrac{7}{4}:7+\dfrac{3}{5}:\dfrac{4}{5}\)
\(=\dfrac{4}{3}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\dfrac{7}{3}\)
ND
Nguyễn Đức Trí
VIP
20 tháng 8 2023
\(A=\left(\dfrac{1}{4}-1\right).\left(\dfrac{1}{9}-1\right)....\left(\dfrac{1}{100}-1\right).\)
\(\Rightarrow A=\left(-\dfrac{3}{4}\right).\left(-\dfrac{8}{9}\right)....\left(-\dfrac{99}{100}\right)\)
mà A có 9 dấu - \(\left(4;9;16;25;36;49;64;81;100\right)\)
\(\Rightarrow0>A=\left(-\dfrac{3}{4}\right).\left(-\dfrac{8}{9}\right)....\left(-\dfrac{99}{100}\right)=-\dfrac{1}{2}\)
Ta lại có \(\left\{{}\begin{matrix}\dfrac{1}{2}=\dfrac{21}{42}\\\dfrac{11}{21}=\dfrac{22}{42}\end{matrix}\right.\) \(\Rightarrow\dfrac{1}{2}< \dfrac{11}{21}\Rightarrow-\dfrac{1}{2}>-\dfrac{11}{21}\)
\(\Rightarrow A>-\dfrac{11}{21}\)
20 tháng 8 2023
\(A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)...\left(\dfrac{1}{100}-1\right)\)
\(A=\left(-\dfrac{2^2-1}{2^2}\right)\left(-\dfrac{3^2-1}{3^2}\right)...\left(-\dfrac{10^2-1}{10^2}\right)\)
\(A=\left[-\dfrac{1\cdot3}{2\cdot2}\right]\left[-\dfrac{2\cdot4}{3\cdot3}\right]...\left[-\dfrac{9\cdot11}{10\cdot10}\right]\)
Dễ thấy A có 9 thừa số, suy ra
\(A=-\dfrac{1\cdot3\cdot2\cdot4\cdot...\cdot9\cdot11}{2\cdot2\cdot3\cdot3\cdot...\cdot10.10}=-\dfrac{1\cdot11}{2\cdot10}=\dfrac{-11}{20}\)
Vì 20 < 21 nên \(\dfrac{11}{20}>\dfrac{11}{21}\), suy ra \(\dfrac{-11}{20}< \dfrac{-11}{21}\)
Vậy \(A< \dfrac{-11}{21}\)
8 tháng 8 2023
a) Ta có: x2\(\ge0,\forall x\)
=> x2 +3/4 \(\ge\dfrac{3}{4}\) , mọi x
Vậy min A = 3/4
Dấu "=" xảy ra <=> x =0
b) ( x- 3/2)2 -0,4
Ta có ( x-3/2)2 lớn hơn hoặc bằng 0, mọi x
=> ( x-3/2)2 - 0,4 lớn hơn hoặc bằng 0 - 0;4 = -0,4
Vậy min B =-0,4
Dấu "=" xảy ra <=> x = 3/2
Chúc bạn học tốt !
\(=4.\left(-\dfrac{1}{8}\right)-2.\dfrac{1}{4}-\dfrac{3}{2}+1=\)
\(=-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{3}{2}+1=-\dfrac{3}{2}\)
= 4 . -1/8 - 2 . -1/4 + 3 . -1/2 + 1
= -1/2 - -1/2 + -3/2 + 1
= -1/2