Từ \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Rightarrow\dfrac{ab+bc+ca}{abc}=0\Rightarrow ab+bc+ca=0\)

Và \(a+b+c=\sqrt{2017}\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=2017\)

\(\Rightarrow a^2+b^2+c^2+2\cdot0=2017\)\(\Rightarrow a^2+b^2+c^2=2017\)

Vậy \(a^2+b^2+c^2=2017\)

#5k->Acc_Quỳnh,pls :)

Từ \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Rightarrow\dfrac{ab+bc+ca}{abc}=0\Rightarrow ab+bc+ca=0\)

Và \(a+b+c=\sqrt{2017}\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=2017\)

\(\Rightarrow a^2+b^2+c^2+2=2017\Rightarrow a^2+b^2+c^2=2015\)

Vậy \(a^2+b^2+c^2=2015\)

#5k->Acc_Quỳnh,pls :)

a) điều kiện xác định : \(a>2;a\ne11\)

ta có : \(P=\left(\dfrac{\sqrt{a-2}+2}{3}\right)\left(\dfrac{\sqrt{a-2}}{3+\sqrt{a-2}}+\dfrac{a+7}{11-a}\right):\left(\dfrac{3\sqrt{a-2}+1}{a-3\sqrt{a-2}-2}-\dfrac{1}{\sqrt{a-2}}\right)\)

ta có : \(a+b=\sqrt{2017-a^2}+\sqrt{2017-b^2}\)

\(\Leftrightarrow\left(a+b\right)\left(\sqrt{2017-a^2}-\sqrt{2017-b^2}\right)=b^2-a^2\)

\(\Leftrightarrow b-a=\sqrt{2017-a^2}-\sqrt{2017-b^2}\)

\(\Leftrightarrow2b=2\sqrt{2017-a^2}\Leftrightarrow b^2=2017-a^2\Rightarrow\left(đpcm\right)\)

b, ta giả sử \(a+b=\sqrt{2017-a^2}+\sqrt{2017-b^2}\rightarrow a^2+b^2=2017\)

thật vậy khi ta thay \(a^2+b^2=2017\) vào biểu thức thì thấy thoả mãn

vậy dpcm

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{2018}\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{a+b+c}=0\left(a+b+c=2018\right)\)

\(\Leftrightarrow\dfrac{a+b}{ab}+\dfrac{a+b+c-c}{c\left(a+b+c\right)}=0\)

\(\Leftrightarrow\left[\dfrac{1}{ab}+\dfrac{1}{c\left(a+b+c\right)}\right]\left(a+b\right)=0\)

\(\Leftrightarrow\dfrac{ac+bc+c^2+ab}{abc\left(a+b+c\right)}\times\left(a+b\right)=0\)

\(\Leftrightarrow\dfrac{\left(a+c\right)\left(b+c\right)\left(a+b\right)}{abc\left(a+b+c\right)}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=-c\\b=-c\\a=-b\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}b=2018\\a=2018\\c=2018\end{matrix}\right.\)

\(\Rightarrow P=\dfrac{1}{2018^{2017}}\)

\(1,Q=\dfrac{a^4-2a^2+a^3-2a+a^2-2}{a^4-2a^2+2a^3-4a+a^2-2}\\ Q=\dfrac{\left(a^2-2\right)\left(a^2+a+1\right)}{\left(a^2-2\right)\left(a^2+2a+1\right)}=\dfrac{a^2+a+1}{a^2+2a+1}\)

\(Q=\dfrac{x^2+x+1}{\left(x+1\right)^2}-\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{x^2+x+1-\dfrac{3}{4}x^2-\dfrac{3}{2}x-\dfrac{3}{4}}{\left(x+1\right)^2}+\dfrac{3}{4}\\ Q=\dfrac{\dfrac{1}{4}x^2-\dfrac{1}{2}x+\dfrac{1}{4}}{\left(x+1\right)^2}+\dfrac{3}{4}=\dfrac{\dfrac{1}{4}\left(x-1\right)^2}{\left(x+1\right)^2}+\dfrac{3}{4}\ge\dfrac{3}{4}\\ Q_{min}=\dfrac{3}{4}\Leftrightarrow x=1\)

\(2,\text{Từ GT }\Leftrightarrow\dfrac{ayz+bxz+czy}{xyz}=0\\ \Leftrightarrow ayz+bxz+czy=0\\ \text{Ta có }\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\\ \Leftrightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{zx}{ca}\right)=0\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\cdot\dfrac{cxy+ayz+bzx}{abc}=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\cdot\dfrac{0}{abc}=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)

Lời giải:

Tìm max:

Áp dụng BĐT Bunhiacopxky:

\(A^2=(2x+\sqrt{5-x^2})^2\leq (x^2+5-x^2)(2^2+1)=25\)

\(\Rightarrow A\leq 5\)

Vậy \(A_{\max}=5\Leftrightarrow x=2\)

Tìm min:

ĐKXĐ: \(5-x^2\geq 0\Leftrightarrow -\sqrt{5}\leq x\leq \sqrt{5}\)

Do đó : \(A=2x+\sqrt{5-x^2}\geq 2x\geq -2\sqrt{5}\)

Vậy \(A_{\min}=-2\sqrt{5}\Leftrightarrow x=-\sqrt{5}\)

Bài 2 bạn xem xem có viết nhầm đề bài không nhé.

\(A=\frac{3a}{2a-b}+\frac{3c}{2c-b}-2\)

Chỉ cần cho $b$ càng nhỏ thì giá trị của $A$ càng nhỏ rồi, mà lại không có điều kiện gì của $b$ ?