527 + {[ 2.( 2 . 23 + 32 + 42 - 52 ) + 6780 ]3 : 332
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{4}{9\cdot11}+\dfrac{4}{13\cdot15}+...+\dfrac{4}{95\cdot97}+\dfrac{4}{97\cdot99}\\ =2\cdot\left(\dfrac{2}{9\cdot11}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{95\cdot97}+\dfrac{2}{97\cdot99}\right)\\ =2\left(\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{99}\right)\\ =2\cdot\left(\dfrac{1}{9}-\dfrac{1}{99}\right)\\ =2\cdot\dfrac{11-1}{99}\\ =2\cdot\dfrac{10}{99}\\ =\dfrac{20}{99}\)
Sửa đề: `S = 4/(9.11) + 4/(11.13) + ... + 4/(97.99)`
`S = 2 . (2/(9.11) + 2/(11.13) + ... +2/(97.99))`
`S = 2 . (1/9 - 1/11 + 1/11 - 1/13 + ... + 1/97 - 1/99)`
`S = 2 . (1/9 - 1/99)`
`S = 2 . (11/99 - 1/99)`
`S = 2 . 10/99 `
`S = 20/99`
\(\left(\dfrac{3}{2}\right)^5\cdot x=\left(\dfrac{3}{2}\right)^7\)
=>\(x=\left(\dfrac{3}{2}\right)^7:\left(\dfrac{3}{2}\right)^5=\left(\dfrac{3}{2}\right)^2=\dfrac{9}{4}\)
a: M là trung điểm của AB
=>\(MA=MB=\dfrac{AB}{2}=6\left(cm\right)\)
N là trung điểm của MA
=>\(AN=NM=\dfrac{AM}{2}=1,5\left(cm\right)\)
P là trung điểm của MB
=>\(MP=PB=\dfrac{MB}{2}=\dfrac{3}{2}=1,5\left(cm\right)\)
NP=MN+MP
=1,5+1,5=3(cm)
b: \(NP=NM+MP\)
\(=\dfrac{1}{2}\left(MA+MB\right)\)
\(=\dfrac{1}{2}\cdot AB=3\left(cm\right)\)
\(\left(\dfrac{2}{3}\right)^8:x=\left(\dfrac{2}{3}\right)^2\)
=>\(x=\left(\dfrac{2}{3}\right)^8:\left(\dfrac{2}{3}\right)^2=\left(\dfrac{2}{3}\right)^6=\dfrac{64}{729}\)
a: Để hệ có nghiệm duy nhất thì \(\dfrac{3}{m}\ne\dfrac{-1}{1}=-1\)
=>\(m\ne-3\)
b: Để hệ vô nghiệm thì \(\dfrac{3}{m}=\dfrac{-1}{1}\ne\dfrac{6}{n+3}\)
=>\(\left\{{}\begin{matrix}m=-3\\n+3\ne-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=-3\\n\ne-9\end{matrix}\right.\)
c: Để hệ có vô số nghiệm thì \(\dfrac{3}{m}=\dfrac{-1}{1}=\dfrac{6}{n+3}\)
=>\(\left\{{}\begin{matrix}m=-3\\n+3=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=-3\\n=-9\end{matrix}\right.\)
a,b,c là các số thực đôi một phân biệt
=>\(a-b;b-c;a-c\) đều khác 0
\(a^3+b^3+c^3=3bac\)
=>\(\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
=>\(\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
=>\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
=>\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
=>\(\left(a+b+c\right)\left[2a^2+2b^2+2c^2-2ab-2ac-2bc\right]=0\)
=>\(\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\right]=0\)
=>\(\left[{}\begin{matrix}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\left(loại\right)\end{matrix}\right.\)
=>a+b+c=0
=>a+b=-c; a+c=-b; b+c=-a
\(P=\dfrac{a+b}{c}\cdot\dfrac{b+c}{a}\cdot\dfrac{c+a}{b}=\dfrac{-c}{c}\cdot\dfrac{-a}{a}\cdot\dfrac{-b}{b}=-1\)
a: Vì ABCD là hình thang
nên \(\dfrac{S_{ABC}}{S_{ACD}}=\dfrac{AB}{CD}=\dfrac{2}{3}\)
b: Diện tích hình thang ABCD là:
\(S_{ABCD}=\dfrac{1}{2}\cdot3\cdot\left(2+3\right)=\dfrac{15}{2}\left(cm^2\right)\)
\(\dfrac{S_{ABC}}{S_{ADC}}=\dfrac{2}{3}\)
=>\(S_{ADC}=1,5\cdot S_{ABC}\)
\(S_{ABC}+S_{ADC}=S_{ABCD}\)
=>\(1,5\cdot S_{ABC}+S_{ABC}=7,5\)
=>\(2,5\cdot S_{ABC}=7,5\)
=>\(S_{ABC}=3\left(cm^2\right)\)
3a=5b
=>\(a=\dfrac{5b}{3}\)
a-b=-6
=>\(\dfrac{5b}{3}-b=-6\)
=>\(\dfrac{2}{3}b=-6\)
=>\(b=-6:\dfrac{2}{3}=-6\cdot\dfrac{3}{2}=-9\)
=>\(b=\dfrac{5}{3}\cdot\left(-9\right)=-15\)
\(x+\left(\dfrac{2}{5}\right)^2=\dfrac{9}{10}\)
=>\(x+\dfrac{4}{25}=\dfrac{9}{10}\)
=>\(x=\dfrac{9}{10}-\dfrac{4}{25}=\dfrac{45}{50}-\dfrac{8}{50}=\dfrac{37}{50}\)
`x + (2/5)^2 = 9/10`
`=> x + 4/25 = 9/10`
`=> x = 9/10 - 4/25`
`=> x = 45/50 - 8/50`
`=> x = 37/50`
-------------------------
`(x+2/5)^2 = 9/10`
`=> (x+2/5)^2 = (3/sqrt{10})^2`
`=> x + 2/5 = 3/sqrt{10}` hoặc `x + 2/5 = -3/sqrt{10}`
`=> x = 3/sqrt{10} - 2/5` hoặc `x = -3/sqrt{10} - 2/5`
`=> x = (-4+3sqrt{10})/10` hoặc `x = -(4+3sqrt{10})/10`
`527 + {[2 . (2 . 2^3 + 3^2 + 4^2 - 5^2) + 678^0]^3 : 33^2}`
`= 527 + {[2 . (16 + 9 + 16 - 25) + 1]^3 : 33^2}`
`= 527 + {[2 . (25 + 16 - 25) + 1]^3 : 33^2}`
`= 527 + {[2 . 16 + 1]^3 : 33^2}`
`= 527 + {[32 + 1]^3 : 33^2}`
`= 527 + {33^3 :33^2}`
`= 527 + 33^(3-2)`
`= 527 + 33`
`= 560`