Chương I - Căn bậc hai. Căn bậc ba
Các câu hỏi tương tự
Cho a,b,c la cac so duong thoa man dieu kien
a+b+c=1
Tim GTNN :
A=\(\dfrac{\left(1+a\right)\left(1+b\right)\left(1+c\right)}{\left(1-a\right)\left(1-b\right)\left(1-c\right)}\)
cho a, b, c la cac so thuc duong thoa man a + b + c =abc chung minh rang :
\(\frac{1}{a^2\left(1+bc\right)}+\frac{1}{b^2\left(1+ac\right)}+\frac{1}{c^2\left(1+ab\right)}\le\frac{1}{4}\)
Tim cac so nguyen duong a, b, c thoa man: \(\left\{{}\begin{matrix}\sqrt{a-b+c}=\sqrt{a}-\sqrt{b}+\sqrt{c}\\\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\end{matrix}\right.\)
cho a b c la cac so thuc ko am thoa man a+b+c=3. tim GTLN cua K=\(\sqrt{12a+\left(b-c\right)^2}+\sqrt{12b+\left(a-c\right)^2}+\sqrt{12c+\left(a-b\right)^2}\)
cho 3 so thuc duong a, b, c thoa man 1/a+1/c=2/b. tim GTNN cua (a+b)/(2a-b)+(b+c)(/2c-b)
Cho x,y, z la cac so duong thoa man dieu kien x+y+z=a
tim GTNN : Q=\(\left(1+\dfrac{a}{x}\right)\left(1+\dfrac{a}{y}\right)\left(1+\dfrac{a}{z}\right)\)
Cho x,y,z la cac so duong thoa man dieu kien x+y+z > hoac = 12
GTNN: P =\(\dfrac{x}{\sqrt{y}}+\dfrac{y}{\sqrt{z}}+\dfrac{z}{\sqrt{x}}\)
1. tim x biet :
a, (x-2)(x+3) > 2x\(^2\) -x -5
b, x( x-5) > x-4
2. cho 2 so x va y thoa man : x+y = 7 va xy=2 . khong tinh x va y , hay tinh gia tri cua bieu thuc A= x - y ( biet x< y)
Cho a, b, c > 0 thoa man a + b + c = 3.
Tim GTNN : \(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}+\dfrac{1}{a^2+b^2+c^2}\)