Cho a+b=1. CM \(\dfrac{a}{b^3-1}+\dfrac{b}{a^3-1}=\dfrac{2.\left(ab-2\right)}{a^2b^2+3}\)
Những câu hỏi liên quan
c/m bất đảng thức :
a)dfrac{a}{3b}+dfrac{bleft(a+bright)}{a^2+ab+b^2}
b)dfrac{a}{b^2}+dfrac{b}{a^2}+dfrac{16}{a+b}ge5left(dfrac{1}{a}+dfrac{1}{b}right)
c)dfrac{a}{2b}+dfrac{2b}{a+b}+dfrac{ab^2}{2left(a^3+2b^3right)}gedfrac{5}{3}
d)dfrac{a}{4b^2}+dfrac{2b}{left(a+bright)^2}gedfrac{9}{4left(a+2bright)}
e)dfrac{2}{a^2+ab+b^2}+dfrac{1}{3b^2}gedfrac{9}{left(a+2bright)^2}
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c/m bất đảng thức :
a)\(\dfrac{a}{3b}+\dfrac{b\left(a+b\right)}{a^2+ab+b^2}\)
b)\(\dfrac{a}{b^2}+\dfrac{b}{a^2}+\dfrac{16}{a+b}\ge5\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
c)\(\dfrac{a}{2b}+\dfrac{2b}{a+b}\)+\(\dfrac{ab^2}{2\left(a^3+2b^3\right)}\ge\dfrac{5}{3}\)
d)\(\dfrac{a}{4b^2}+\dfrac{2b}{\left(a+b\right)^2}\ge\dfrac{9}{4\left(a+2b\right)}\)
e)\(\dfrac{2}{a^2+ab+b^2}+\dfrac{1}{3b^2}\ge\dfrac{9}{\left(a+2b\right)^2}\)
Chứng minh :
a) dfrac{3x}{2y}+dfrac{3}{2}sqrt{dfrac{3}{5}}-sqrt{dfrac{3}{4}}dfrac{3sqrt{x}}{2}.left(dfrac{sqrt{x}}{y}+sqrt{dfrac{3}{5x}}-sqrt{dfrac{1}{3}}right)
b)ab.sqrt{1+dfrac{1}{a^2b^2}}-sqrt{a^2b^2+1}0 , với a ; b 0
c) left(dfrac{3}{a}sqrt{dfrac{a^3}{b}}-dfrac{1}{2}sqrt{dfrac{4}{ab}}-2sqrt{dfrac{b}{a}}right):sqrt{dfrac{1}{ab}}3a-2b-1 với a, b 0
d)left(sqrt{dfrac{16a}{b}}+3sqrt{4ab}-asqrt{dfrac{36b}{a}}+2sqrt{ab}right):left(sqrt{ab}+dfrac{a}{b}sqrt{dfrac{b}{a}}+sqrt{dfrac{a}{b}}rig...
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Chứng minh :
a) \(\dfrac{3x}{2y}+\dfrac{3}{2}\sqrt{\dfrac{3}{5}}-\sqrt{\dfrac{3}{4}}=\dfrac{3\sqrt{x}}{2}.\left(\dfrac{\sqrt{x}}{y}+\sqrt{\dfrac{3}{5x}}-\sqrt{\dfrac{1}{3}}\right)\)
b)\(ab.\sqrt{1+\dfrac{1}{a^2b^2}}-\sqrt{a^2b^2+1}=0\) , với a ; b > 0
c) \(\left(\dfrac{3}{a}\sqrt{\dfrac{a^3}{b}}-\dfrac{1}{2}\sqrt{\dfrac{4}{ab}}-2\sqrt{\dfrac{b}{a}}\right):\sqrt{\dfrac{1}{ab}}=3a-2b-1\) với a, b >0
d)\(\left(\sqrt{\dfrac{16a}{b}}+3\sqrt{4ab}-a\sqrt{\dfrac{36b}{a}}+2\sqrt{ab}\right):\left(\sqrt{ab}+\dfrac{a}{b}\sqrt{\dfrac{b}{a}}+\sqrt{\dfrac{a}{b}}\right)=2\) Với a, b >0
Mọi người giúp tớ với ạ !!!!!! Mình thật sự cần gấp vào ngày mai !!!!
b)CM: \(ab\sqrt{1+\dfrac{1}{a^2b^2}}-\sqrt{a^2b^2+1}=0\)
\(VT=ab\sqrt{\dfrac{a^2b^2+1}{\left(ab\right)^2}}-\sqrt{a^2b^2+1}\)
\(VT=ab\dfrac{\sqrt{a^2b^2+1}}{ab}-\sqrt{a^2b^2+1}\)
\(VT=\sqrt{a^2b^2+1}-\sqrt{a^2b^2+1}\)
\(VT=0=VP\)
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cho biểu thức \(P=\dfrac{a^3-a-2b-\dfrac{b^2}{2}}{\left(1-\sqrt{\dfrac{1}{a}+\dfrac{b}{a^2}}\right)\left(a+\sqrt{a+b}\right)}:\left(\dfrac{a^3+a^2+ab+a^2b}{a^2-b^2}+\dfrac{b}{a-b}\right)\)
xác định điều kiện và rút gọn P
17 tháng 10 2021 lúc 15:03
Cho a,b,c>0 tm: a+b+c=ab+bc+ca
CMR: \(\dfrac{2a-1}{a^2-a+1}+\dfrac{2b-1}{b^2-b+1}+\dfrac{2c-1}{c^2-c+1}=\dfrac{3}{\left(a+b-1\right)\left(b+c-1\right)\left(c+a-1\right)}\)
cho \(7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=6\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)+2017\)
tìm max \(\dfrac{1}{\sqrt{3\left(2a^2+b^2\right)}}+\dfrac{1}{\sqrt{3\left(2b^2+c^2\right)}}+\dfrac{1}{\sqrt{3\left(2c^2+a^2\right)}}\)
Từ \(7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=6\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)+2017\)
\(\Leftrightarrow7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\le6\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)+2017\)\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\le2017\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(T=\dfrac{1}{\sqrt{3\left(2a^2+b^2\right)}}+\dfrac{1}{\sqrt{3\left(2b^2+c^2\right)}}+\dfrac{1}{\sqrt{3\left(2c^2+a^2\right)}}\)
\(=\dfrac{1}{\sqrt{\left(2+1\right)\left(2a^2+b^2\right)}}+\dfrac{1}{\sqrt{\left(2+1\right)\left(2b^2+c^2\right)}}+\dfrac{1}{\sqrt{\left(2+1\right)\left(2c^2+a^2\right)}}\)
\(\le\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}\le\dfrac{1}{9}\left(\dfrac{2^2}{2a}+\dfrac{1^2}{b}\right)+\dfrac{1}{9}\left(\dfrac{2^2}{2b}+\dfrac{1^2}{c}\right)+\dfrac{1}{9}\left(\dfrac{2^2}{2c}+\dfrac{1^2}{a}\right)\)
\(\le\dfrac{1}{9}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)\)\(=\dfrac{1}{3a}+\dfrac{1}{3b}+\dfrac{1}{3c}\le\sqrt{\left(\dfrac{1}{81}+\dfrac{1}{81}+\dfrac{1}{81}\right)\left(\dfrac{9}{a^2}+\dfrac{9}{b^2}+\dfrac{9}{c^2}\right)}\)
\(\le\sqrt{\dfrac{1}{81}\cdot3\cdot9\cdot2017}=\sqrt{\dfrac{2017}{3}}\)
Vậy \(T_{Max}=\sqrt{\dfrac{2017}{3}}\) khi \(a=b=c=\sqrt{\dfrac{3}{2017}}\)
So kimochiii~
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Cho các số thực a,b,c thỏa mãn a>1 , b>\(\dfrac{1}{2}\) , \(c>\dfrac{1}{3}\) và \(\dfrac{1}{a}+\dfrac{2}{2b+1}+\dfrac{3}{3c+2}\ge2\). Tìm GTLN của bt \(P=\left(a-1\right)\left(2b-1\right)\left(3c-1\right)\)
21 tháng 10 2018 lúc 12:10
Xét:
dfrac{c}{a-b}.left(dfrac{a-b}{c}+dfrac{b-c}{a}+dfrac{c-a}{b}right)1+dfrac{c}{a-b}left(dfrac{b-c}{a}+dfrac{c-a}{b}right)1+dfrac{c}{a-b}.dfrac{b^2-bc+ac-a^2}{ab}1+dfrac{c}{a-b}.dfrac{cleft(a-bright)-left(a^2-b^2right)}{ab}1+dfrac{c}{a-b}.dfrac{left(c-a-bright)left(a-bright)}{ab}1+dfrac{c^2-cleft(a+bright)}{ab}1+dfrac{2c^2}{ab}1+dfrac{2c^3}{abc}
CMTT cộng theo vế:
BTCCM3+dfrac{2left(a^3+b^3+c^3right)}{abc}dfrac{6left(a^3+b^3+c^3right)}{3abc}
Mà Khi a+b+c0 thì a^3+b^3+c^33abc ( tự cm,ez)...
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Xét:
\(\dfrac{c}{a-b}.\left(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)=1+\dfrac{c}{a-b}\left(\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)=1+\dfrac{c}{a-b}.\dfrac{b^2-bc+ac-a^2}{ab}=1+\dfrac{c}{a-b}.\dfrac{c\left(a-b\right)-\left(a^2-b^2\right)}{ab}=1+\dfrac{c}{a-b}.\dfrac{\left(c-a-b\right)\left(a-b\right)}{ab}=1+\dfrac{c^2-c\left(a+b\right)}{ab}=1+\dfrac{2c^2}{ab}=1+\dfrac{2c^3}{abc}\)
CMTT cộng theo vế:
\(BTCCM=3+\dfrac{2\left(a^3+b^3+c^3\right)}{abc}=\dfrac{6\left(a^3+b^3+c^3\right)}{3abc}\)
Mà Khi \(a+b+c=0\) thì \(a^3+b^3+c^3=3abc\) ( tự cm,ez)
Vậy \(BTCCM=3+6=9\left(đpcm\right)\)
@Nguyễn Thanh Hằng đọc xong xóa đii nha
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Cho 2 số thực a, b thỏa mãn ab ≠ 0, a ≠ 1, b ≠ 1 và a + b = 1. Tính giá trị của biểu thức
\(P=\dfrac{a}{b^3-1}-\dfrac{b}{a^3-1}+\dfrac{2\left(a-b\right)}{a^2b^2+3}\)
Cho 2 số thực a, b thỏa mãn ab ≠ 0, a ≠ 1, b ≠ 1 và a + b = 1. Tính giá trị của biểu thức
\(P=\dfrac{a}{b^3-1}-\dfrac{b}{a^3-1}+\dfrac{2\left(a-b\right)}{a^2b^2+3}\)
Lời giải:
\(P=\frac{a^4-a-b^4+b}{(b^3-1)(a^3-1)}+\frac{2(a-b)}{a^2b^2+3}\)
\(=\frac{(a^4-b^4)-(a-b)}{a^3b^3-(a^3+b^3)+1}+\frac{2(a-b)}{a^2b^2+3}=\frac{(a-b)[(a+b)(a^2+b^2)-1]}{a^3b^3-[(a+b)^3-3ab(a+b)]+1}+\frac{2(a-b)}{a^2b^2+3}\)
\(=\frac{(a-b)[(a^2+b^2)-(a+b)^2]}{a^3b^3-[1-3ab]+1}+\frac{2(a-b)}{a^2b^2+3}=\frac{-2ab(a-b)}{a^3b^3+3ab}+\frac{2(a-b)}{a^2b^2+3}\)
\(=\frac{-2(a-b)}{a^2b^2+3}+\frac{2(a-b)}{a^2b^2+3}=0\)
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Lời giải:
\(P=\frac{a^4-a-b^4+b}{(b^3-1)(a^3-1)}+\frac{2(a-b)}{a^2b^2+3}\)
\(=\frac{(a^4-b^4)-(a-b)}{a^3b^3-(a^3+b^3)+1}+\frac{2(a-b)}{a^2b^2+3}=\frac{(a-b)[(a+b)(a^2+b^2)-1]}{a^3b^3-[(a+b)^3-3ab(a+b)]+1}+\frac{2(a-b)}{a^2b^2+3}\)
\(=\frac{(a-b)[(a^2+b^2)-(a+b)^2]}{a^3b^3-[1-3ab]+1}+\frac{2(a-b)}{a^2b^2+3}=\frac{-2ab(a-b)}{a^3b^3+3ab}+\frac{2(a-b)}{a^2b^2+3}\)
\(=\frac{-2(a-b)}{a^2b^2+3}+\frac{2(a-b)}{a^2b^2+3}=0\)
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