Ta co :\(\dfrac{1}{f\left(x\right)}=\) \(x^4-x^2+1=x^4-2.\dfrac{1}{2}x^2+\dfrac{1}{4}+\dfrac{3}{4}\)
= \(\left(x^2-\dfrac{1}{4}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
=> f(x) ≤ \(\dfrac{4}{3}\)
Vay max f(x) =\(\dfrac{4}{3}\)
tim gia tri lon nhat
cua bieu tthuc \(M=\dfrac{x^2}{x^4+x^2+1}\)
gia tri nho nhat cua bieu thuc\(\dfrac{^{^{ }}^{ }x^2+1}{x^2-1^{ }}\)
Cho \(xy+\sqrt{\left(1+x\right)^2\left(1+y\right)^2}=\sqrt{2017}\)
Tinh gia tri cua bieu thuc: \(P=x+\sqrt{1+y^2}+y\sqrt{1+x^2}\)
Giup mk!!!
cho a c la cac so thuc khong am thoa man dieu kien a+b+c=3 tim gia tri lon nhat cua bieu thucP=a\b3+1+b\c3+1+c\a3+1
cho x,y>0 va \(x+y\le1.\)
tim GTNN cua bieu thuc \(A=\dfrac{1}{x^2+y^2}+\dfrac{1}{xy}\)
cho a=4+\(\sqrt{5}\) va b=4-\(\sqrt{5}\) tinh gia tri cua bieu thuc
A=\(\left(a^{2019}-8a^{2018}+11a^{2017}\right)\)+\(\left(b^{2019}-8b^{2018}+11b^{2017}\right)\)
F = \(\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\left(\dfrac{1}{x}+\dfrac{1}{y}\right).\dfrac{1}{x+y+2\sqrt{xy}}+\dfrac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}.\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)\right]\)
tim gtnn cua bt A
A= \((x-1)^4+(x-3)^4+6(x-1)^2\left(x-3\right)^2\)
Rut gon bieu thuc sau:
\(A=\left(\dfrac{a-\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}+1}{a+\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{a}\right)\)
