bài 1 cho a,b,c là 3 độ dài 3 cạnh tam giác
CMR \(\frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{a+c-b}>=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
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bài 1 : cho a,b,c là độ dài 3 cạnh tam giác. CMR \(\frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{a+c-b}>=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
cac ban oi giup minh di. minh k biet lam
bài 1 cho a,b,c>0: CMR
a, \(\frac{1}{a}+\frac{1}{b}>\frac{4}{a+b}\)
b, \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right).\left(a+b+c\right)>=9\)
cac ban giup minh di minh k hieu bai nay lam kieu j. minh dang can. cac ban oi lam on giup minh
bai 1 cho a,b,c>0
CMR: \(\frac{a}{1+b-a}+\frac{b}{1+c-b}+\frac{c}{1+a-c}>=1\)
cac ban oi giup minh. minh dang can gap lam. . lam on giup minh di. hu hu
bài 1 cho a,b,c>0. CMR \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}>=\frac{3}{2}\)
cac ban oi giup minh di. toi minh di hoc roi. minh dang can gap lam
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.............
bài 1: cho a,b,c>0 và a+b+c=1
CMR: \(\frac{a}{1+b-a}+\frac{b}{1+c-b}+\frac{c}{1+a-c}>=1\)
bai2 cho a,b,c>0
CMR \(\frac{bc}{a}+\frac{ac}{a}+\frac{ab}{c}>=a+b+c\)
cac ban oi giup minhdi. minh lam chuyen con 2 bai nay minh chua mlam dc. cac ban giup minh nhe. minh dang can lam
Cho a+b=1. Hãy so sánh a4+b4 với \(\frac{1}{8}\)
cac ban oi hay giup minh. bai nay kho ma nhung minh k biet lam. giup minh di minh dang can lam
bài 1. cho a,b,c>0,a+b=1
CMR a, \(\frac{1}{ab}+\frac{1}{a^2+b^2}>=6\)
b, \(\frac{2}{ab}+\frac{3}{a^2+b^2}>=14\)
cac ban oi giup minh chung 2 bđt nay di. minh dang can gap lam
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.