may ban oi giup minh voi
Rút gọn:
\(\frac{1}{b+2}+\frac{3}{2-b}+\frac{12}{b^2-4}\)
bài1 . cho a>=0, b>=0
CMR:\(\frac{a^3+b^3}{2}>=\left(\frac{a+b}{2}\right)^3\)
cac ban oi giup minh nhe. minh can gap. giup minh di. giup minh di chieu minh di hoc roi
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
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 cho >=0, b>=0
CMR: \(\frac{a^3+b^3}{2}>=\left(\frac{a+b}{2}\right)^3\)
cac ban oi lam on giup minh di chieu minh di hoc roi. nhanh nhe
Ta có: (a-b)2 (a+b)>=0
=> (a-b)(a2-b2) >=0
=> a 3 +b3- a2b -ab2 >=0
=> 3a3+3b3-3a2b-3ab2>=0
=> 4a3+4b3>= (a^3+b^3+3a2b+3ab2)
=> 4a3+ 4b3 >= (a+b)3 => đpcm, tích cho mình nhé
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
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 :Rút gọn:
H=(\(\frac{3}{\sqrt{2}+1}+\frac{14}{2\sqrt{2}-1}-\frac{4}{2-\sqrt{2}}\))(.\(\sqrt{8+2}\))
giup mk vs cac ban oi chieu nay mk phai hk ruj
cho a,b,c >0, a+b=1 CMR
a, \(\frac{1}{ab}+\frac{1}{a^2+b^2}>=6\)
cac ban oi giup minh di. minh dang can gap
\(\frac{1}{ab}+\frac{1}{a^2+b^2}=\frac{2}{2ab}+\frac{1}{a^2+b^2}\ge\frac{\left(\sqrt{2}+1\right)^2}{2ab+a^2+b^2}=\frac{3+2\sqrt{2}}{\left(a+b\right)^2}=3+2\sqrt{2}\)
Xem lại đề.
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