Tìm x \(\in\)Q biết
a,\(\dfrac{-7}{12}-\left(\dfrac{3}{5}+x\right)=\dfrac{3}{4}\)
b,\(2007.x\left(x-\dfrac{2006}{7}\right)=0\)
c,\(5\left(x-2\right)+3x\left(2-x\right)=0\)
help me, bạn nào làm xong nhanh và đúng sẽ đc tick
Thực hiện phép tính :
A = \(\left(1-\dfrac{1}{1+2}\right)\cdot\left(1-\dfrac{1}{1+2+3}\right)\cdot...\cdot\left(1-\dfrac{1}{1+2+3+...+2006}\right)\)
So sánh :
A = \(\dfrac{2006}{2007}\) - \(\dfrac{2007}{2008}\) + \(\dfrac{2008}{2009}\) - \(\dfrac{2009}{2010}\)
B = \(-\dfrac{1}{2006.2007}\) - \(\dfrac{1}{2008.2009}\)
help me ~ mai mik nộp rồi
Câu 5.Thực hiện phép tính:
a) \(\left(\dfrac{-1}{2}\right)^4+|-\dfrac{2}{3}|-2007^0\)
b)4\(\left(\dfrac{-1}{2}\right)^3+|\dfrac{1}{2}|:5\)
1) Không dùng máy tính hãy so sánh:
A=\(\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2006}\) với 4
2) So sánh A và B biết:
A=\(\left(100^{99}+99^{99}\right)^{100}\)
B=\(\left(100^{100}+99^{100}\right)^{99}\)
3) Chứng tỏ rằng:
\(\left(2003^n+1\right)\left(2003^n+2\right)⋮6\forall n\in N\)
Cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}\)và \(a+b+c\ne0\). Tính giá trị của \(A=\dfrac{a^4.\left(2b^2\right)^3.c^{2007}}{b^{2017}}\).
tính
a) \(\left[\dfrac{0.8\div\left(\dfrac{4}{5}\cdot1025\right)}{0.64-1}+\dfrac{\left(1.08-\dfrac{2}{25}\right)\div\dfrac{4}{7}}{\left(6\dfrac{5}{7}-3\dfrac{1}{4}\right)\cdot2\dfrac{2}{17}}+\left(1.2\cdot0.5\right)\div\dfrac{4}{5}\right]\)
b) \(\left(0.2\right)^{-3}\left[\left(-\dfrac{1}{5}\right)^{-2}\right]^{-1}+\left[\left(\dfrac{1}{2}\right)^{-3}\right]^{-2}\div\left(2^{-3}\right)^{-1}-\left(0.175\right)^{-2}\)
c) \(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)
d) \(\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{3}\)
e) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2\div2\)
f) \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
g) \(\dfrac{1}{-\left(2017\right)\left(-2015\right)}+\dfrac{1}{\left(-2015\right)\left(-2013\right)}+...+\dfrac{1}{\left(-3\right)\cdot\left(-1\right)}\)
h) \(\left(1-\dfrac{1}{1\cdot2}\right)+\left(1-\dfrac{1}{2\cdot3}+...+\left(1-\dfrac{1}{2017\cdot2018}\right)\right)\)
Thu gọn các biểu thức sau
A = \(\left(-2\right).\left(-1\dfrac{1}{2}\right).\left(-1\dfrac{1}{3}\right).\left(-1\dfrac{1}{4}\right)...\left(-1\dfrac{1}{214}\right)\)
B = \(\left(-1\dfrac{1}{2}\right).\left(-1\dfrac{1}{3}\right).\left(-1\dfrac{1}{4}\right)...\left(-1\dfrac{1}{299}\right)\)
C = \(-\dfrac{7}{4}.\left(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{3333}{3030}+\dfrac{333333}{424242}\right)\)
A= \(\left(\dfrac{1}{2}-1\right)\)\(\left(\dfrac{1}{3}-1\right)\).........\(\left(\dfrac{1}{10}-1\right)\). So sánh A với \(\dfrac{-1}{9}\)
B= \(\left(\dfrac{1}{4}-1\right)\)\(\left(\dfrac{1}{9}-1\right)\)...........\(\left(\dfrac{1}{100}-1\right)\). So sánh B với \(\dfrac{-11}{21}\)