\(\sqrt{a}+\sqrt{b}=m\Leftrightarrow m-\sqrt{a}=\sqrt{b}\Rightarrow m^2-2m\sqrt{a}+a=b\)

\(\Leftrightarrow\sqrt{a}=\frac{m^2+a-b}{2m}\)là số hữu tỉ. 

Tương tự cũng suy ra \(\sqrt{b}\)là số hữu tỉ. 

\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)

\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\frac{a+b+c}{abc}\)

\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)    (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

\(\left(a^2+b^2-2\right)\left(a+b\right)^2+\left(1-ab\right)^2+4ab=0\)

\(\Leftrightarrow\left[\left(a+b\right)^2-2\left(ab+1\right)\right]\left(a+b\right)^2+1+2ab+a^2b^2=0\)

\(\Leftrightarrow\left(a+b\right)^4-2\left(a+b\right)^2\left(ab+1\right)+\left(ab+1\right)^2=0\)

\(\Leftrightarrow\left[\left(a+b\right)^2-\left(ab+1\right)\right]^2=0\)

\(\Leftrightarrow\left(a+b\right)^2-\left(ab+1\right)=0\)

\(\Leftrightarrow ab+1=\left(a+b\right)^2\)

\(\Rightarrow\sqrt{ab+1}=\left|a+b\right|\) là số hữu tỉ (đpcm)

\(A=\sqrt{\dfrac{b^2\left(a-b\right)^2+a^2\left(a-b\right)^2+a^2b^2}{a^2b^2\left(a-b\right)^2}}\)

\(=\sqrt{\dfrac{b^2\left(a^2-2ab+b^2\right)+a^2\left(a^2-2ab+b^2\right)+a^2b^2}{a^2b^2\left(a-b\right)^2}}\)

\(=\sqrt{\dfrac{b^4+a^4-2ab^3-2a^3b+3a^2b^2}{a^2b^2\left(a-b\right)^2}}=\sqrt{\dfrac{\left(b^2+a^2\right)^2-2ab\left(a^2+b^2\right)+a^2b^2}{a^2b^2\left(a-b\right)^2}}\)

\(=\sqrt{\dfrac{\left(b^2+a^2-ab\right)}{a^2b^2\left(a-b\right)^2}}=\left|\dfrac{a^2+b^2-ab}{ab\left(a-b\right)}\right|\)

Do a,b là số hữu tỉ\(\Rightarrow\)\(\left|\dfrac{a^2+b^2-ab}{ab\left(a-b\right)}\right|\) là số hữu tỉ hay A là số hữu tỉ

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 \(a,b,c\in Q\Rightarrow|\left(a+b\right)\left(a+c\right)\left(b+c\right)|\in Q\left(đpcm\right)\)

Câu hỏi của Phạm Quang Dương - Toán lớp 9 - Học toán với OnlineMath

a) Từ giả thiết : \(\dfrac{1}{a}+\dfrac{1}{b}\text{=}\dfrac{1}{c}\)

\(\Rightarrow2ab\text{=}2bc+2ca\)

\(\Rightarrow2ab-2bc-2ca\text{=}0\)

Ta xét : \(\left(a+b-c\right)^2\text{=}a^2+b^2+c^2+2ab-2bc-2ca\)

\(\text{=}a^2+b^2+c^2\)

Do đó : \(A\text{=}\sqrt{a^2+b^2+c^2}\text{=}\sqrt{\left(a+b-c\right)^2}\)

\(\Rightarrow A\text{=}a+b-c\)

Vì a;b;c là các số hữu tỉ suy ra : đpcm

b) Đặt : \(a\text{=}\dfrac{1}{x-y};b\text{=}\dfrac{1}{y-x};c\text{=}\dfrac{1}{z-x}\)

Do đó : \(\dfrac{1}{a}+\dfrac{1}{b}\text{=}\dfrac{1}{c}\)

Ta có : \(B\text{=}\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}\)

Từ đây ta thấy giống phần a nên :

\(B\text{=}a+b-c\)

\(B\text{=}\dfrac{1}{x-y}+\dfrac{1}{y-z}-\dfrac{1}{z-x}\)

Suy ra : đpcm.

Mình bổ sung đề phần b cần phải có điều kiện của x;y;z nha bạn.