CMR :
\(A=\dfrac{1}{3}+\dfrac{1}{4}+....+\dfrac{1}{32}>2\)
E đg cần gấp lắm ạ; Mong mấy ah cj ; giải dùm ạ !!!!!!!!!!
Mơn
Mn ơi giúp e bài này với ạ, e cần gấp lắm. E sắp thi cuối năm r ạ hmu-
\(\dfrac{1}{9+x}-\dfrac{1}{x}=\dfrac{1}{5}+\dfrac{1}{4}\)
E cảm ơn mn nhìu lắm!!! Mọng mn giải chi tiết cho e hiểu ạ hyhy XĐ
ĐKXĐ: \(x\notin\left\{0;-9\right\}\)
Ta có: \(\dfrac{1}{x+9}-\dfrac{1}{x}=\dfrac{1}{5}+\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{20x}{20x\left(x+9\right)}-\dfrac{20\left(x+9\right)}{20x\left(x+9\right)}=\dfrac{4x\left(x+9\right)+5x\left(x+9\right)}{20x\left(x+9\right)}\)
Suy ra: \(4x^2+36x+5x^2+45x=20x-20x-180\)
\(\Leftrightarrow9x^2+81x+180=0\)
\(\Leftrightarrow x^2+9x+20=0\)
\(\Leftrightarrow x^2+4x+5x+20=0\)
\(\Leftrightarrow x\left(x+4\right)+5\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\left(nhận\right)\\x=-5\left(nhận\right)\end{matrix}\right.\)
Vậy: S={-4;-5}
cho a,b,c>0 ,\(a^2+b^2+c^2=1\).CMR
\(\dfrac{a^3}{b+c}+\dfrac{b^3}{c+a}+\dfrac{c^3}{a+b}\ge\dfrac{1}{2}\)
Dùng bunhia nhé mn.Giúp e với e cần gấp ạ !
Đặt vế trái BĐT là P
Ta có:
\(\left(\dfrac{a^3}{b+c}+\dfrac{b^3}{c+a}+\dfrac{c^3}{a+b}\right)\left(a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)\right)\ge\left(a^2+b^2+c^2\right)^2\)
\(\Rightarrow P.\left(2ab+2bc+2ca\right)\ge1\)
\(\Rightarrow P\ge\dfrac{1}{2\left(ab+bc+ca\right)}\ge\dfrac{1}{2\left(a^2+b^2+c^2\right)}=\dfrac{1}{2}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)
(-2\(\dfrac{1}{3}\)).(-\(\dfrac{-6}{4}\))
Giup mình với ạ mình đg cần gấp!!!
`(-2 1/3 ) .( - (-6)/4 )`
`= - (2xx3+1)/3 . 6/4`
`= - 7/3 . 6/4`
`= -42/12`
`= -7/2`
\(\left(-2\dfrac{1}{3}\right).\left(-\dfrac{-6}{4}\right)\)
\(=\left(-\dfrac{7}{3}\right).\dfrac{3}{2}=-\dfrac{7}{2}\)
Tính Nhanh:
B=\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{8}\)+\(\dfrac{1}{16}\)+\(\dfrac{1}{32}\)+\(\dfrac{1}{64}\)
Giúp mình với mình cần gấp ạ
\(B=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)
=>\(B=\dfrac{32}{64}+\dfrac{16}{64}+\dfrac{6}{64}+\dfrac{2}{64}+\dfrac{1}{64}\)
=>\(B=\dfrac{32+16+6+2+1}{64}\)
=>\(B=\dfrac{63}{64}\)
\(B=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^6}\)
\(2B=1+\dfrac{1}{2}+...+\dfrac{1}{2^5}\)
\(\Rightarrow2B-B=1-\dfrac{1}{2^6}\)
\(\Rightarrow B=1-\dfrac{1}{2^6}=1-\dfrac{1}{64}=\dfrac{63}{64}\)
Tính :
a) A= \(\dfrac{1}{2010.2009}-\dfrac{1}{2009.2008}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
b) B= 63.(-124)-124.(-37)+(\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{11}\)).(-66)
Giúp mình với ạ! Mình cần gấp lắm
\(a,A=\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2009}+\dfrac{1}{2008}-...-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{2}+1\\ A=1+\dfrac{1}{2010}=\dfrac{2011}{2010}\)
\(b,B=\left(-124\right)\left(63-37\right)+\dfrac{17}{66}\left(-66\right)=-124\cdot26+17=-3224+17=-3207\)
Cho: \(\dfrac{a}{(b)^{2}} = \dfrac{b^{2}}{(c)^{3}} = \dfrac{c^{3}}{(a)^{4}}\)
Tính P =\((1 + \dfrac{a}{b}).(1+\dfrac{b}{c}).(1+\dfrac{c}{a})\)
Giúp mk với mk đg cần gấp
Đặt \(\dfrac{a}{b^2}=\dfrac{b^2}{c^3}=\dfrac{c^3}{a^4}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=k.b^2\\b^2=k.c^3\\c^3=k.a^4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=k.k.c^3=k^2c^3\\c^3=k.a^4\end{matrix}\right.\)
\(\Rightarrow a=k^2.k.a^4\)
\(\Rightarrow a=k^3a^4\)
\(\Rightarrow\left(ka\right)^3=1\)
\(\Rightarrow ka=1\)
\(\Rightarrow a=\dfrac{1}{k}\) (1)
Thế vào \(c^3=k.a^4\Rightarrow c^3=k.\dfrac{1}{k^4}=\dfrac{1}{k^3}\)
\(\Rightarrow c=\dfrac{1}{k}\) (2)
Thế vào \(b^2=kc^3\Rightarrow b^2=k.\dfrac{1}{k^3}=\dfrac{1}{k^2}\)
\(\Rightarrow b=\dfrac{1}{k}\) hoặc \(b=-\dfrac{1}{k}\) (3)
(1);(2);(3) \(\Rightarrow\left[{}\begin{matrix}a=b=c\\a=c=-b\end{matrix}\right.\)
TH1: \(a=b=c\)
\(\Rightarrow P=\left(1+\dfrac{a}{a}\right)\left(1+\dfrac{a}{a}\right)\left(1+\dfrac{a}{a}\right)=2.2.2=8\)
Th2: \(a=c=-b\)
\(\Rightarrow P=\left(1+\dfrac{-b}{b}\right)\left(1+\dfrac{b}{-b}\right)\left(1+\dfrac{-b}{-b}\right)=0.0.2=0\)
Cho 2 phân thức A=\(\dfrac{2x^2+20}{x^2-4}+\dfrac{3}{x+2}-\dfrac{7}{x-2}\) và B=\(\dfrac{2x^2}{3x+6}\) (x≠2, x≠-2, x≠0)
a, Tính B khi x=-3
b, Rút gọn phân thức A
"giúp e với ạ e đg cần gấp, cảm ơn ạ"
a: Thay x=-3 vào B, ta được:
\(B=\dfrac{2\cdot\left(-3\right)^2}{3\cdot\left(-3\right)+6}=\dfrac{2\cdot9}{-9+6}=\dfrac{18}{-3}=-6\)
b: \(A=\dfrac{2x^2+20+3x-6-7x-14}{\left(x+2\right)\left(x-2\right)}=\dfrac{2x^2-4x}{\left(x+2\right)\left(x-2\right)}=\dfrac{2x}{x+2}\)
a) \(\dfrac{1}{2}\) x - 2 = 3 b) \(\dfrac{1}{4}\) \(x^2\) - \(\sqrt{36}\) = 10
Mong cao nhân nào giúp em hai câu này với ạ hiện tại em đang cần gấp lắm ạ
\(a,\dfrac{1}{2}x=3+2\)
\(\dfrac{1}{2}x=5\)
\(x=5\div\dfrac{1}{2}\)
\(x=10\)
\(b,\dfrac{1}{4}x^2-\sqrt{36}=10\)
\(\dfrac{1}{4}x^2-6=10\)
\(\dfrac{1}{4}x^2=10+6\)
\(\dfrac{1}{4}x^2=16\)
\(x^2=16\div\dfrac{1}{4}\)
\(x^2=64\)
\(x^2=\left(8\right)^2\)
\(\Rightarrow x=8\)
1. Tính
a, \(\dfrac{2}{3}\) + \(\dfrac{5}{2}\) - \(\dfrac{3}{4}\)
b, \(\dfrac{2}{5}\) x \(\dfrac{1}{2}\) : \(\dfrac{1}{3}\)
c, \(\dfrac{2}{9}\) : \(\dfrac{2}{9}\) x \(\dfrac{1}{3}\)
Giúp tui với mình cần gấp lắm.
a)
`2/3+5/2-3/4`
`=10/4-3/4+2/3`
`=7/4+2/3`
`=21/12+8/12`
`=29/12`
b)
`2/5xx1/2:1/3`
`=2/10xx3/1`
`=6/10=3/5`
c)
`2/9:2/9xx1/3`
`=2/9xx9/2xx1/3`
`=1xx1/3`
`=1/3`
a, \(\dfrac{2}{3}\) + \(\dfrac{5}{2}\) - \(\dfrac{3}{4}\)
= \(\dfrac{8}{12}\) + \(\dfrac{30}{12}\) - \(\dfrac{9}{12}\)
= \(\dfrac{38-9}{12}\)
= \(\dfrac{29}{12}\)
b, \(\dfrac{2}{5}\) x \(\dfrac{1}{2}\) : \(\dfrac{1}{3}\)
= \(\dfrac{1}{5}\) x \(\dfrac{3}{1}\)
= \(\dfrac{3}{5}\)
c, \(\dfrac{2}{9}\) : \(\dfrac{2}{9}\) x \(\dfrac{1}{3}\)
= 1 x \(\dfrac{1}{3}\)
= \(\dfrac{1}{3}\)