thèn này ko làm mà lôi BTVN ra hỏi lmj z ?
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Bài 1: Căn bậc hai
@@@ Các câu hỏi tương tự
1/ cho a,b,c thỏa ab+bc+cage11
c/m sqrt[3]{a^2+3}+dfrac{7}{5sqrt[3]{14}}sqrt[3]{b^2+3}+dfrac{sqrt[3]{9}}{5}sqrt[3]{c^2+3}gedfrac{23}{5sqrt[3]{2}}
2)cho a,b,c dương thỏa a+b+c3
c/m left(a^3+b^3+c^3right)left(a^2-b^2right)left(b^2-c^2right)left(c^2-a^2right)ledfrac{729sqrt{3}}{8}
p/s: cách của mik đa phần dùng cô-si (I need another way!!)
Đọc tiếp
1/ cho a,b,c thỏa \(ab+bc+ca\ge11\)
c/m \(\sqrt[3]{a^2+3}+\dfrac{7}{5\sqrt[3]{14}}\sqrt[3]{b^2+3}+\dfrac{\sqrt[3]{9}}{5}\sqrt[3]{c^2+3}\ge\dfrac{23}{5\sqrt[3]{2}}\)
2)cho a,b,c dương thỏa a+b+c=3
c/m \(\left(a^3+b^3+c^3\right)\left(a^2-b^2\right)\left(b^2-c^2\right)\left(c^2-a^2\right)\le\dfrac{729\sqrt{3}}{8}\)
p/s: cách của mik đa phần dùng cô-si (I need another way!!)
cho a,b,c\(\le\dfrac{3}{2}\)
Tìm giá trị nhỏ nhất của
\(A=\left(3+\dfrac{1}{a}+\dfrac{1}{b}\right)\left(3+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(3+\dfrac{1}{c}+\dfrac{1}{a}\right)\)
8 tháng 4 2021 lúc 21:00
Cho abc=1
CMR\(\dfrac{a+3}{\left(a+1\right)^2}+\dfrac{b+3}{\left(b+1\right)^2}+\dfrac{c+3}{\left(c+1\right)^2}\ge3\)
1) gpt \(x^2+3x\sqrt{\dfrac{x^2+1}{x}}=10x-1\)
2) ghpt \(\left\{{}\begin{matrix}x^2+y^2+2\left(x+y\right)=6\\xy\left(x+2\right)\left(y+2\right)=9\end{matrix}\right.\)
3) cho a,b,c dương thỏa abc=1
CMR \(\dfrac{2}{a^2\left(b+c\right)}+\dfrac{2}{b^2\left(c+a\right)}+\dfrac{2}{c^2\left(a+b\right)}\ge3\)
ta có sqrt{left(1+a^3right)left(1+b^3right)}sqrt{left(1+aright)left(a^2-a+1right)}.sqrt{left(1+bright)left(b^2-b+1right)}
Mà sqrt{left(a+1right)left(a^2-a+1right)}ledfrac{a+1+a^2-a+2}{2}dfrac{a^2+2}{2}
Tương tự thì sqrt{left(1+a^3right)left(1+b^3right)}ledfrac{left(a^2+2right)left(b^2+2right)}{4}Rightarrowdfrac{a^2}{sqrt{left(1+a^3right)left(1+B^3right)}}gedfrac{4a^2}{left(a^2+2right)left(b^2+2right)}...
Đọc tiếp
ta có \(\sqrt{\left(1+a^3\right)\left(1+b^3\right)}=\sqrt{\left(1+a\right)\left(a^2-a+1\right)}.\sqrt{\left(1+b\right)\left(b^2-b+1\right)}\)
Mà \(\sqrt{\left(a+1\right)\left(a^2-a+1\right)}\le\dfrac{a+1+a^2-a+2}{2}=\dfrac{a^2+2}{2}\)
Tương tự thì \(\sqrt{\left(1+a^3\right)\left(1+b^3\right)}\le\dfrac{\left(a^2+2\right)\left(b^2+2\right)}{4}\Rightarrow\dfrac{a^2}{\sqrt{\left(1+a^3\right)\left(1+B^3\right)}}\ge\dfrac{4a^2}{\left(a^2+2\right)\left(b^2+2\right)}\)
=\(\dfrac{4a^2\left(c^2+2\right)}{\left(a^2+2\right)\left(b^2+2\right)\left(c^2+2\right)}\)
Tương tự rồi + vào, ta có
...\(\ge4\dfrac{a^2\left(c^2+2\right)+b^2\left(a^2+2\right)+c^2\left(b^2+2\right)}{\left(a^2+2\right)\left(b^2+2\right)\left(c^2+2\right)}\)
ta cần chứng minh \(3\left[a^2\left(c^2+2\right)+b^2\left(a^2+2\right)+c^2\left(b^2+2\right)\right]\ge\left(a^2+2\right)\left(b^2+2\right)\left(c^2+2\right)\)
đến đây nhân tung ra và dùng cô-si tiếp
bài 1 cho a,b,c\(\ge1\)
chứng minh rằng
a(b+c)+b(c+a)+c(a+b)+2\(\left(\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}+\dfrac{1}{1+c^2}\right)\ge9\)
bài 2 cho x,y,z>0 thỏa mãn \(x^3+y^3+z^3=3\)
Tìm min của A=3xy+3yz+3zx-xyz
cho a,b,c>0. chứng minh rằng
\(\dfrac{1}{a\left(b+1\right)}+\dfrac{1}{b\left(c+1\right)}+\dfrac{1}{c\left(a+1\right)}\ge\dfrac{3}{abc+1}\)
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}
Đọc tiếp
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)\(\dfrac{2}{a^2+ab+b^2}+\dfrac{1}{3b^2}\ge\dfrac{9}{\left(a+2b\right)^2}\)
a, \(A=\left(\sqrt{2}+1\right)[\left(\sqrt{2}\right)^2+1][(\sqrt{2})^4+1][\left(\sqrt{2}\right)^8+1][1\left(\sqrt{2}\right)^{16}+1]\)
b, \(B=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2019}+1\sqrt{2020}}\)
c,\(C=^3\sqrt[]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}}\)