Question

1) CMR : \(2^{1975}+5^{2010}⋮3\)

2) CMR nếu \(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=1\) thì \(x\sqrt{1+y^2}+y\sqrt{1+x^2}=0\)

3) cho a,b,c dương . CM \(\sqrt{\dfrac{2}{a}}+\sqrt{\dfrac{2}{b}}+\sqrt{\dfrac{2}{c}}\le\sqrt{\dfrac{a+b}{ab}}+\sqrt{\dfrac{b+c}{bc}}+\sqrt{\dfrac{c+a}{ca}}\)

p/s : đề GIa lai nhé mik hỏi cách làm khác thui, sắp thi tỉnh oy =)

Answer 1

bài BĐT cuối làm r` mà ko nhớ ở đâu, bn vào đây tìm lại hộ mình Here nhân tiện ở đây cũng có 1 số bài BĐT+HPT+GPT hay lắm đấy chịu khó tìm nhé ko tìm dc bảo mình :v

Answer 2

\(2^{1975}\) + 1 \(⋮\) (2+1) = 3 ; \(5^2\) = 25 \(\equiv\) 1 (mod3) \(\Rightarrow\) \(\left(5^2\right)^{1005}\) \(\equiv\)1 (mod3) \(\Rightarrow\) \(5^{2010}\) - 1 \(⋮\) 3 \(\Rightarrow\) điều phải chứng minh

Answer 3

e làm cách nầy dk k

we đặt \(\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)\rightarrow\left(x;y;z\right)\)

cần chứng minh \(\sqrt{2x}+\sqrt{2y}+\sqrt{2z}\le\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\)

ta có: \(\left(\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\right)^2=2\left(x+y+z\right)+2\left[\sqrt{\left(x+y\right)\left(y+z\right)}+\sqrt{\left(y+z\right)\left(z+x\right)}+\sqrt{\left(z+x\right)\left(x+y\right)}\right]\)

Áp dụng BĐT bunyakovsky:\(\sqrt{\left(x+y\right)\left(y+z\right)}\ge\sqrt{xy}+\sqrt{yz}\)

thiết lập tương tự , ta có: \(2\sum\sqrt{\left(x+y\right)\left(y+z\right)}\ge4\left(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\right)\)

\(\Rightarrow VT^2\ge2\left(x+y+z+2\sqrt{xy}+2\sqrt{yz}+2\sqrt{xz}\right)=2\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2=VF^2\)

do đó \(\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\ge\sqrt{2x}+\sqrt{2y}+\sqrt{2z}\)

dấu = xảy ra khi x=y=z hay a=b=c

Answer 4

Ta đã biết: \(\left(\dfrac{1}{\sqrt{a}}-\dfrac{1}{\sqrt{b}}\right)^2\ge0\Leftrightarrow\dfrac{2}{\sqrt{ab}}\le\dfrac{1}{a}+\dfrac{1}{b}\)

\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{\sqrt{ab}}\le2\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)

\(\Leftrightarrow\left(\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}\right)^2\le2\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)

\(\Leftrightarrow\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}\le\sqrt{2\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}\).Tương tự ta có:

\(\dfrac{1}{\sqrt{b}}+\dfrac{1}{\sqrt{c}}\le\sqrt{2\left(\dfrac{1}{b}+\dfrac{1}{c}\right)};\dfrac{1}{\sqrt{c}}+\dfrac{1}{\sqrt{a}}\le\sqrt{2\left(\dfrac{1}{c}+\dfrac{1}{a}\right)}\)

Cộng theo vế các BĐT trên ta có:

\(2\left(\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}+\dfrac{1}{\sqrt{c}}\right)\leq \sqrt{2}\left[\sqrt{\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}+\sqrt{\left(\dfrac{1}{b}+\dfrac{1}{c}\right)}+\sqrt{\left(\dfrac{1}{c}+\dfrac{1}{a}\right)}\right]\)

\(\sqrt{\dfrac{2}{a}}+\sqrt{\dfrac{2}{b}}+\sqrt{\dfrac{2}{c}}\leq\sqrt{\dfrac{a+b}{ab}}+\sqrt{\dfrac{b+c}{bc}}+\sqrt{\dfrac{c+a}{ca}}\)