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Ta có :
\(A+B=a\sqrt{a}+\sqrt{ab}+b\sqrt{b}+\sqrt{ab}\)
\(=a\sqrt{a}+b\sqrt{b}+2\sqrt{ab}\)
\(=\)\(\left(\sqrt{a}+\sqrt{b}\right)\left[\left(\sqrt{a}+\sqrt{b}\right)^2-3\sqrt{ab}\right]+2\sqrt{ab}\)
\(A.B=\sqrt{ab}\left(\sqrt{ab+1}\right)+\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\left[\left(\sqrt{a}+\sqrt{b}\right)^2-3\sqrt{ab}\right]\)
Đặt \(\sqrt{a}+\sqrt{b}=x;\)\(\sqrt{ab}=y\)\(\left(x;y\in Q\right)\)thì :
\(A+B=x\left(x^2-3y\right)+2y\)
\(A.B=y\left(y+1\right)+xy\left(x^2-3y\right)\)
\(\Rightarrow\)Các đa thức này là các số hữa tỉ \(\left(đpcm\right)\)
3) Ta có:\(\sqrt{2000}< 2001\)
Áp dụng BĐT AM-GM:
\(\sqrt{1999.\sqrt{2000}}< \sqrt{1999.2001}< \frac{1999+2001}{2}=2000\)
Tương tự ta có:
\(\sqrt{2\sqrt{3\sqrt{4--...\sqrt{1999\sqrt{2000}}}}}< \sqrt{2\sqrt{3\sqrt{4=.\sqrt{1999.2001}}}}< \sqrt{2\sqrt{3\sqrt{4-\sqrt{1998.2000}}}}--< \sqrt{2.4}< 3\)
1)
Với ab + bc + ac = 1 có:
\(a^2+1=a^2+ab+ac+bc=a\left(a+b\right)+c\left(a+b\right)=\left(a+c\right)\left(a+b\right)\)
\(b^2+1=b^2+bc+ca+ab=b\left(b+c\right)+a\left(b+c\right)=\left(a+b\right)\left(b+c\right)\)
\(c^2+1=c^2+bc+ca+ab=c\left(b+c\right)+a\left(b+c\right)=\left(a+c\right)\left(b+c\right)\)
Do đó: \(\sqrt{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}\)
\(=\sqrt{\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(b+a\right)\left(c+a\right)\left(c+b\right)}\)
\(=\sqrt{\left(a+b\right)^2\left(a+c\right)^2\left(b+c\right)^2}\)
\(=|\left(a+b\right)\left(a+c\right)\left(b+c\right)|\)
Vì \(a,b,c\in Q\Rightarrow|\left(a+b\right)\left(a+c\right)\left(b+c\right)|\in Q\left(đpcm\right)\)
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)
\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\frac{a+b+c}{abc}\)
\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\) (do a+b+c = 0)
=> \(B=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{ \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
=> đpcm
a/ \(x+y=a_1+b_1\sqrt{2}+a_2+b_2\sqrt{2}=\left(a_1+a_2\right)+\left(b_1+b_2\right)\sqrt{2}\)
\(xy=\left(a_1+b_1\sqrt{2}\right)\left(a_2+b_2\sqrt{2}\right)=\left(a_1a_2+2b_1b_2\right)+\left(a_1b_2+a_2b_1\right)\sqrt{2}\)
b/ Tương tự câu a.
Ta có \(\Delta=b^2-4ac=\left(a+c\right)^2-4ac=\left(a-c\right)^2\)
\(\Rightarrow x_1=\frac{-b+a-c}{2a};x_2=\frac{-b-a+c}{2a}\in Q.\)
Ta có : \(\left(x+\sqrt{x^2+2017}\right)\left(-x+\sqrt{x^2+2017}\right)=2017\left(1\right)\)
\(\left(y+\sqrt{y^2+2017}\right)\left(-y+\sqrt{y^2+2017}\right)=2017\left(2\right)\)
nhân theo vế của ( 1 ) ; ( 2 ) , ta có :
\(2017\left(-x+\sqrt{x^2+2017}\right)\left(-y+\sqrt{y^2+2017}\right)=2017^2\)
\(\Rightarrow\left(-x+\sqrt{x^2+2017}\right)\left(-y+\sqrt{y^2+2017}\right)=2017\)
rồi bạn nhân ra , kết hợp với việc nhân biểu thức ở phần trên xong cộng từng vế , cuối cùng ta đc :
\(xy+\sqrt{\left(x^2+2017\right)\left(y^2+2017\right)}=2017\)
\(\Leftrightarrow\sqrt{\left(x^2+2017\right)\left(y^2+2017\right)}=2017-xy\)
\(\Leftrightarrow x^2y^2+2017\left(x^2+y^2\right)+2017^2=2017^2-2\cdot2017xy+x^2y^2\)
\(\Rightarrow x^2+y^2=-2xy\Rightarrow\left(x+y\right)^2=0\Rightarrow x=-y\)
A = 2017
( phần trên mk lười nên không nhân ra, bạn giúp mk nhân ra nha :) )
2/ \(\frac{\sqrt{x-2011}-1}{x-2011}+\frac{\sqrt{y-2012}-1}{y-2012}+\frac{\sqrt{z-2013}-1}{z-2013}=\frac{3}{4}\)
\(\Leftrightarrow\frac{4\sqrt{x-2011}-4}{x-2011}+\frac{4\sqrt{y-2012}-4}{y-2012}+\frac{4\sqrt{z-2013}-4}{z-2013}=3\)
\(\Leftrightarrow\left(1-\frac{4\sqrt{x-2011}-4}{x-2011}\right)+\left(1-\frac{4\sqrt{y-2012}-4}{y-2012}\right)+\left(1-\frac{4\sqrt{z-2013}-4}{z-2013}\right)=0\)
\(\Leftrightarrow\left(\frac{x-2011-4\sqrt{x-2011}+4}{x-2011}\right)+\left(\frac{y-2012-4\sqrt{y-2012}+4}{y-2012}\right)+\left(\frac{z-2013-4\sqrt{z-2013}+4}{z-2013}\right)=0\)
\(\Leftrightarrow\frac{\left(\sqrt{x-2011}-2\right)^2}{x-2011}+\frac{\left(\sqrt{y-2012}-2\right)^2}{y-2012}+\frac{\left(\sqrt{z-2013}-2\right)^2}{z-2013}=0\)
Dấu = xảy ra khi \(\sqrt{x-2011}=2;\sqrt{y-2012}=2;\sqrt{z-2013}=2\)
\(\Leftrightarrow x=2015;y=2016;z=2017\)