cho \(\left(a+b+c\right).\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1\)
Tính giá trị biểu thức M=(a23+b23).(b5+c5).(a2015+b2015)
moi nguoi oi giup minh 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 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
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.
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: 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
\(S=\left\{\frac{4023}{2};\frac{4015}{2}\right\}\)
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: chứng minh BĐT
a,\(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}<=\frac{a+b+c}{2}\left(a,b,c>0\right)\)
cac oi giup minh di. minh can lam roi
BAi 1
a) hay so sanh
\(\frac{2014}{2015}+\frac{2015}{2014}\)va \(\frac{666665}{333333}\)
b)tinh
A=\(\left(1-\frac{1}{28}\right)\left(1-\frac{1}{36}\right)\left(1-\frac{1}{45}\right)...\left(1-\frac{1}{1326}\right)\)
cac ban lam nhanh nhe minh dang can gap xong minh tick cho
A)=vậy\(\frac{2014}{2015}+\frac{2015}{2014}>\frac{666665}{333333}.\)
bạn nhé
Cho a3+b3+c3 = 3abc và a +b +c khác 0
a) Tính giá trị biểu thức \(\frac{a^2+b^2+c^2}{_{\left(a+b+c\right)^2}}\)
b)Chứng minh : P=\(\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{c}+\frac{1}{a}\right)=\frac{8}{abc}\)
Ta có:
\(a^3+b^3+c^3=3abc=>a^3+b^3+c^3-3abc=0\)
\(=>\left(a+b\right)^3-3a^2b-3ab^2+c^3-3abc=0\)
\(=>\left[\left(a+b\right)^3+c^3\right]-3a^2b-3ab^2-3abc=0\)
\(=>\left[\left(a+b\right)^3+c^3\right]-3ab\left(a+b+c\right)=0\)
\(=>\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
\(=>\left(a+b+c\right)\left(a^2+2ab+b^2-ca-bc+c^2-3ab\right)=0\)
\(=>\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
Vì a3+b3+c3=3abc và a+b+c khác 0
=>\(a^2+b^2+c^2-ab-bc-ca=0\)
\(=>2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(=>\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(=>\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Tổng 3 số không âm = 0 <=> chúng đều = 0
\(< =>\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}< =>a=b=c}\)
Vậy \(\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{1}{3}\)
\(\)
Ta có ; \(a^3+b^3+c^3=3abc\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-bc-ac\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\frac{a+b+c}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
Vì \(a+b+c\ne0\) nên ta có \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\Leftrightarrow a=b=c\)
a) Thay a = b = c vào biểu thức được : \(\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)
b) Thay a = b = c vào P : \(P=\frac{2}{a}.\frac{2}{b}\frac{2}{c}=\frac{8}{abc}\)
b,
Ta có: \(P=\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{c}+\frac{1}{a}\right)=\frac{a+b}{ab}.\frac{b+c}{bc}.\frac{a+c}{ca}\)
Vì a=b=c (chứng minh tương tự câu a)
=>a+b=2b
b+c=2c
a+c=2a
Do đó \(P=\frac{2b}{ab}.\frac{2c}{bc}.\frac{2a}{ca}=\frac{\left(2.2.2\right).abc}{\left(abc\right)^2}=\frac{8}{abc}\left(đpcm\right)\)