Chứng minh (x-3).(x+5)+20>4 hoặc =4 , dấu = xảy ra khi nào?
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NN
1 tháng 4 2017
c) Áp dụng BĐT cô si cho 2 hai số dương \(a;b\) ta có:
\(a+b\ge2\sqrt{ab}\)
\(\frac{1}{a}+\frac{1}{b}\ge\frac{1}{\sqrt{ab}}\)
\(\Rightarrow\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)\ge4\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
Dấu "=" xảy ra khi \(\Leftrightarrow a=b\)
29 tháng 6 2016
ta có (x-3)(x+5)+ 20
= x^2 +2x - 15 +20
= x^2 + 2x +1 - 16 + 20
= (x+1)^2 - 4
vì \(\left(x+1\right)^2\ge0\)với mọi x
\(\left(x+1\right)^2-4\ge-4\) (cộng cả hai vế với -4)
\(4-\left(x+1\right)^2\le4\) ( nhân cả hai vế với -1 )
30 tháng 6 2016
Giả sử (x-3)(x+5)+20 lớn hơn hoặc bằng 4 với mọi x thuộc R
<=>(x-3)(x+5)+20-4 lớn hơn hoặc bằng 0
<=>X2+2x-15+20-4 lớn hơn hoặc bằng o
<=>x2+2x+1 lớn hơn hoặc bằng 0
<=>(x+1)2 lớn hơn hoặc bằng 0 ( luôn đúng )
Vậy (x-3)(x+5)+20 lớn hơn hoặc bằng 4
(x-3)(x+5)+20 lớn hơn hoặc bằng 4
<=>( x+1)2 =0
Dấu "=" xảy ra khi và chỉ khi x+1=0
<=>x=0-1=-1
a4mb4m-(ambm+1)(a2m b2m+1)(ambm-1)
=a4mb4m-(ambm+1)(ambm-1)(a2mb2m+1)
=a4mb4m-(a2mb2m-1)(a2mb2m+1)
=a4mb4m-a4mb4m +1
=1
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. Dấu "=" xảy ra khi nào?
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 5 2021
Cách đơn giản nhất là sử dụng phép biến đổi tương đương:
BĐT đã cho tương đương:
\(\dfrac{x+y}{xy}\ge\dfrac{4}{x+y}\)
\(\Leftrightarrow\left(x+y\right)^2\ge4xy\)
\(\Leftrightarrow x^2-2xy+y^2\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng)
Vậy BĐT đã cho đúng.
Dấu "=" xảy ra khi x=y
27 tháng 6 2016
câu 1 :a2+ab+ b2/4 +3b2/4=(a+b/2)2 +3b2/2 tong 2 binh phương luôn >=0 dau bang khi ca hai số đó bằng 0. a=0 và b=0
câu 2: a2-ab+ b2/4 +3b2/4=(a-b/2)2 +3b2/2 .a=0 và b=0
