Lời giải:

Đặt mẫu số của $B$ là $M$.

Từ \(2018x^3=2019y^3=2020z^3\)

\(\Rightarrow \sqrt[3]{2018}x=\sqrt[3]{2019}y=\sqrt[3]{2020}z=\frac{\sqrt[3]{2018}}{\frac{1}{x}}=\frac{\sqrt[3]{2019}}{\frac{1}{y}}=\frac{\sqrt[3]{2020}}{\frac{1}{z}}=\frac{\sqrt[3]{2018}+\sqrt[3]{2019}+\sqrt[3]{2020}}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}\)

\(=\frac{\sqrt[3]{2018}+\sqrt[3]{2019}+\sqrt[3]{2020}}{8}=\frac{M}{8}\)

\(\Rightarrow \left\{\begin{matrix} x=\frac{M}{8\sqrt[3]{2018}}\\ y=\frac{M}{8\sqrt[3]{2019}}\\ z=\frac{M}{8\sqrt[3]{2020}}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} 2018x^2=\frac{\sqrt[3]{2018}M^2}{64}\\ 2019y^2=\frac{\sqrt[3]{2019}M^2}{64}\\ 2020z^2=\frac{\sqrt[3]{2020}M^2}{64}\end{matrix}\right.\)

\(\Rightarrow 2018x^2+2019y^2+2020z^2=\frac{M^2(\sqrt[3]{2018}+\sqrt[3]{2019}+\sqrt[3]{2020})}{64}=\frac{M^3}{64}\)

\(\Rightarrow B=\frac{\sqrt[3]{\frac{M^3}{64}}}{M}=\frac{M}{4M}=\frac{1}{4}\)

\(x=\frac{1}{2}\sqrt{\sqrt{2}+\frac{1}{8}}-\frac{1}{8}\sqrt{2}\)

\(\Leftrightarrow x+\frac{\sqrt{2}}{8}=\frac{1}{2}\sqrt{\sqrt{2}+\frac{1}{8}}\)

\(\Leftrightarrow\left(x+\frac{\sqrt{2}}{8}\right)^2=\frac{1}{4}\left(\sqrt{2}+\frac{1}{8}\right)\)

\(\Leftrightarrow x^2+\frac{x\sqrt{2}}{4}+\frac{1}{32}=\frac{\sqrt{2}}{4}+\frac{1}{32}\)

\(\Leftrightarrow x^2+\frac{x\sqrt{2}}{4}-\frac{\sqrt{2}}{4}=0\)

\(\Leftrightarrow4x^2+x\sqrt{2}-\sqrt{2}=0\)(1)

\(\Leftrightarrow x\sqrt{2}=\sqrt{2}-4x^2\)

\(\Leftrightarrow x=1-2x^2\sqrt{2}\)

Thay vào M ta sẽ được

\(M=x^2+\sqrt{x^4+1-2x^2\sqrt{2}+1}\)

     \(=x^2+\sqrt{\left(x^2-\sqrt{2}\right)^2}\)

     \(=x^2+\left|x^2-\sqrt{2}\right|\)

Từ \(\left(1\right)\Rightarrow\sqrt{2}-x\sqrt{2}=4x^2\ge0\)

           \(\Leftrightarrow\sqrt{2}\left(1-x\right)\ge0\)

           \(\Leftrightarrow x\le1\)

           \(\Leftrightarrow x^2\le1< \sqrt{2}\)

           \(\Rightarrow\left|x^2-\sqrt{2}\right|=\sqrt{2}-x^2\)

Khi đó \(M=x^2+\left|x^2-\sqrt{2}\right|=x^2-\sqrt{2}+x^2=\sqrt{2}\)

|N|

ĐKXĐ: ...

\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{2}{x}-\frac{2-x}{x\left(\sqrt{x}+1\right)}\right)\)

\(=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{2\left(\sqrt{x}+1\right)-2+x}{x\left(\sqrt{x}+1\right)}\right)\)

\(=\frac{\left(x+2\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{x\left(\sqrt{x}+1\right)}{\left(x+2\sqrt{x}\right)}=\frac{x}{\sqrt{x}-1}\)

\(x=\frac{2}{2-\sqrt{3}}=\frac{4}{4-2\sqrt{3}}=\left(\frac{2}{\sqrt{3}-1}\right)^2\)

\(\Rightarrow P=\frac{\frac{2}{2-\sqrt{3}}}{\frac{2}{\sqrt{3}-1}-1}=\frac{\frac{2}{2-\sqrt{3}}}{\frac{3-\sqrt{3}}{\sqrt{3}-1}}=\frac{2}{2\sqrt{3}-3}\)

\(\sqrt{P}\) xác định khi \(x>1\)

Khi đó: \(\sqrt{P}=\sqrt{\frac{x}{\sqrt{x}-1}}=\sqrt{\frac{x}{\sqrt{x}-1}-4+4}=\sqrt{\frac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-1}+4}\ge2\)

\(\sqrt{P}_{min}=2\) khi \(x=4\)

Lời giải:

Từ \(2\sqrt{x}-\sqrt{y}=0\Rightarrow \sqrt{y}=2\sqrt{x}\)

Thay vào biểu thức $M$:

\(M=x-\sqrt{y}+2019=x-2\sqrt{x}+2019=(x-2\sqrt{x}+1)+2018\)

\(=(\sqrt{x}-1)^2+2018\geq 0+2018=2018\)

Dấu "=" xảy ra khi \(\sqrt{x}-1=0\Leftrightarrow x=1\)

Vậy \(M_{\min}=2018\Leftrightarrow x=1\)

a) DK : x > 0; x khác 1

 \(P=\sqrt{x}\left(\sqrt{x}-1\right)-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)

\(=x-\sqrt{x}+1\)

c )  \(Q=\frac{2\sqrt{x}}{P}=\frac{2\sqrt{x}}{x-\sqrt{x}+1}\)

<=> \(xQ-\left(Q+2\right)\sqrt{x}+Q=0\)(1)

TH1: Q = 0 => x = 0 loại

TH2: Q khác 0

(1) là phương trình bậc 2 với tham số Q ẩn x.

(1) có nghiệm <=> \(\left(Q+2\right)^2-4Q^2\ge0\)

<=> \(-3Q^2+4Q+4\ge0\)

<=> \(-\frac{2}{3}\le Q\le2\)

Vì Q nguyên và khác 0 nên Q =  1 hoặc Q = 2

Với Q = 1 => \(x-3\sqrt{x}+1=0\)

<=> \(\sqrt{x}=\frac{3}{2}\pm\frac{\sqrt{5}}{2}\)----> Tìm được x 

Với Q = 2 => \(2x-4\sqrt{x}+1=0\Leftrightarrow\sqrt{x}=1\pm\frac{1}{\sqrt{2}}\)-----> tìm đc x.

Tự làm tiếp nhé! Kiểm tra lại đề bài câu b.