\(g,\)\(x+\frac{2}{3}+\frac{2}{5}+\frac{2}{5\cdot9}+\frac{2}{9\cdot13}+........+\frac{2}{101\cdot105}=100\)
Lam on giup minh voi, minh dang can gap lam !!!
thank you very much ><
Tìm a,b,c sao cho
a, \(\frac{1}{x\left(x+1\right).\left(x+2\right)}=\frac{a}{x}+\frac{b}{x+1}+\frac{c}{x+2}\)
cac ban oi giup minh di. minh dang can gap lam. lam on
tick đi giải cho
đáp án là a=4.............
Chung minh bat dang thuc sau: \(\frac{x^2}{y^2}\)+\(\frac{y^2}{x^2}\)+4>=3(\(\frac{x}{y}\)+\(\frac{y}{x}\))
Giup mk voi . mk can gap lam. Neu thich thi ket ban voi mk nhe
Tim x, biet :
\(a,\frac{15}{2}-\left(\frac{x}{2}-\frac{3}{4}\right)=\frac{5}{26}\)
\(b,\frac{5}{6}+\frac{5}{18}+\frac{5}{36}+\frac{5}{60}+..........+\frac{5}{29700}-\frac{x}{29700}=1000\)
Lam giup minh voi, minh dang can gap !!!
\(a,\frac{15}{2}-\left(\frac{x}{2}-\frac{3}{4}\right)=\frac{5}{26}\)
\(\frac{x}{2}-\frac{3}{4}=\frac{15}{2}-\frac{5}{26}\)
\(\frac{x}{2}-\frac{3}{4}=39\)
\(\frac{x}{2}=39+\frac{3}{4}\)
\(\frac{x}{2}=\frac{159}{4}\)
\(\Rightarrow\frac{2.x}{4}=\frac{159}{4}\)
\(\Rightarrow2.x=159\)
\(\Rightarrow x=159:2=\frac{159}{2}\)
cho x>=y>=z>0.chứng minh \(\frac{x^2y}{z}+\frac{y^2z}{x}+\frac{z^2x}{y}>=x^2+y^2+z^2\)
minh dang can gap lam ai giup minh vs
Bài 1: Cho A=\(\left(\frac{1}{3}+\frac{3}{x^2-3x}\right):\left(\frac{x^2}{27-3x^2}+\frac{1}{x+3}\right)\)
a, Rút gọn A
b, Tìm x để A<-1
cac ban oi giup minh di minh dang can gap lam. ai giup minh hên nhat nam 2016
a. \(A=\left[\frac{1}{3}+\frac{3}{x.\left(x-3\right)}\right]:\left[\frac{x^2}{3.\left(9-x^2\right)}+\frac{1}{x+3}\right]\)
\(=\left[\frac{x.\left(x-3\right)}{3.x.\left(x-3\right)}+\frac{3.3}{x\left(x-3\right).3}\right]:\left[\frac{x^2}{3.\left(3-x\right)\left(3+x\right)}+\frac{1}{x+3}\right]\)
\(=\left[\frac{x^2-3x+9}{3x.\left(x-3\right)}\right]:\left[\frac{x^2}{3.\left(3-x\right)\left(3+x\right)}+\frac{\left(3-x\right).3}{\left(x+3\right).\left(3-x\right).3}\right]\)
\(=\frac{x^2-3x+9}{3x.\left(x-3\right)}:\left[\frac{x^2+9-3x}{3.\left(3-x\right)\left(3+x\right)}\right]\)
\(=\frac{x^2-3x+9}{3x.\left(x-3\right)}.\frac{3.\left(3-x\right)\left(3+x\right)}{x^2-3x+9}\)
\(=\frac{-\left(x-3\right)\left(3+x\right)}{x-3}=-\left(3+x\right)\)
b. Để A < -1 thì:
-(3+x) < -1
=> -3 - x < -1
=> x < -3 - (-1) = -2
Vậy x < -2 thì A < -1.
Bài 1: Tìm x biết
\(\frac{\left(2009-x\right)^2+\left(2009-x\right).\left(x-2010\right)+\left(x-2010\right)^2}{\left(2009-x\right)^2-\left(2009-x\right).\left(x-2010\right)+\left(x-2010\right)^2}=\frac{19}{49}\)
cac ban oi giup minh di. minh dang can gap lam. chieu minh di hoc roi. lam on
Lời giải của mình ở đây nha bạn!
http://olm.vn/hoi-dap/question/424173.html
\(S=\left\{\frac{4023}{2};\frac{4015}{2}\right\}\)
S=\(\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+..........+\frac{2}{93\cdot95}\)
nhanh nha minh dang can gap do
\(S=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{93.95}\)
\(S=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)
\(S=\frac{1}{5}-\frac{1}{95}\)
\(S=\frac{18}{95}\)
Vậy \(S=\frac{18}{95}\)
Giải
\(S=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)
\(=\frac{1}{5}-\frac{1}{95}\)
\(=\frac{18}{95}\)
Vậy S=\(\frac{18}{95}\)
may bn giai giup minh bai hinh luon voi
Tìm x biết :
\(\frac{x}{2}-\frac{2}{3}\left(3x-2\right)=\left(-\frac{1}{2}\right)^3\)
Ai biet giai thi giup minh voi minh like cho minh dang can gap .
\(\frac{3}{4}x\left(\frac{8}{3}-\frac{4}{5}\right)+\frac{3}{4}-\left(\frac{4}{5}-\frac{2}{3}\right)\)
Giup minh lam bai nay voi
Mình cần giúp. Ai đó có thể giúp mình đc ko zậy