tim gtln cua bieu thuc P=\(\frac{\sqrt{x-2017}}{x}\) voi x>=2017
cho x+y=4. tim gtln,gtnn cua bieu thuc: (x-2)y +2017
moi nguoi giup nhanh ho em voi,em can gap lam
tớ hết lượt kết bạn rồi nên bn kết bn vs tớ nha
cho x+y=4.tim GTLN cua bieu thuc A=(x-2).y+2017
\(\frac{2017-2015x}{\sqrt{1-x^2}}\)
tim GTNN cua bieu thuc tren
tim GTNN cua bieu thuc A=x-2017|x - 2017| + |x - 2018| + |x - 2019|
1) Cho bieu thuc: \(B=\left(\frac{\sqrt{x}}{\sqrt{x}+4}+\frac{4}{\sqrt{x}-4}\right):\frac{x+16}{\sqrt{x}+2}\left(x\ge0,x\ne16\right)\)
a) Cho bieu thuc A= \(\frac{\sqrt{x}+4}{\sqrt{x}+2}\) ; voi cac cua bieu thuc A va B da cho, hay tim cac gia tri cua x nguyen de gia tri cua bieu thuc B(A;-1) la so nguyen
Tim GTLN cua bieu thuc sau \(\frac{2}{\frac{-\sqrt{x}}{x+\sqrt{x}+1}}+\sqrt{x}\)
Tim GTLN cua bieu thuc: Q = \(\frac{-\left(\sqrt{x}+2\right)^2}{\sqrt{x}}\)
ta có : (\(\sqrt{x}\)- 2 )\(^2\)\(\ge\)0
\(\Leftrightarrow\)x - 4\(\sqrt{x}\)+ 4 \(\ge\)0
\(\Leftrightarrow\)x - 4\(\sqrt{x}\)+ 4 + 8\(\sqrt{x}\) \(\ge\)8\(\sqrt{x}\)
\(\Leftrightarrow\)(\(\sqrt{x}\)+ 2 )\(^2\)\(\ge\)8\(\sqrt{x}\)
\(\Leftrightarrow\)-(\(\sqrt{x}\)+ 2 )\(^2\)\(\le\)-8\(\sqrt{x}\)
\(\Leftrightarrow\)Q \(\le\)\(\frac{-8\sqrt{x}}{\sqrt{x}}\)= ( - 8 )
Dấu '' = '' xaye ra tại x = 4
tim gia tri nho nhat cua bieu thuc A=(x-1)+(x-2017)
A=|x-1|+|x-2017|
=>A=|x-1|+|2017-x|
Áp dụng bất đẳng thức:|a|+|b| \(\ge\) |a+b|,dấu "=" xảy ra <=> ab \(\ge\) 0
Ta có: A=|x-1|+|2017-x| \(\ge\) |x-1+2017-x|=2016
=>AMin=2016
Dấu "=" xảy ra <=> (x-1)(2017-x) \(\ge\) 0
<=>1 \(\le\)x \(\le\) 2017
Vậy......................
tim GTNN cua bieu thuc C=x3-3x+2017
Sửa đề: Tìm GTNN của \(C=x^2-3x+2017\)
Ta có:
\(C=x^2-3x+2017\)
\(C=\left(x^2-3x+\frac{9}{4}\right)+\frac{3}{4}+2014\)
\(C=\left(x-\frac{3}{2}\right)^2+2014\frac{3}{4}\ge2014\frac{3}{4}\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(x-\frac{3}{2}\right)^2=0\Rightarrow x=\frac{3}{2}\)
Vậy \(Min_C=2014\frac{3}{4}\Leftrightarrow x=\frac{3}{2}\)