ai nay dung kinh nghiem la chinh

cau a)

ta thay \(10+6\sqrt{3}=\left(1+\sqrt{3}\right)^3\)

\(6+2\sqrt{5}=\left(1+\sqrt{5}\right)^2\)

khi do \(x=\frac{\sqrt[3]{\left(\sqrt{3}+1\right)^3}\left(\sqrt{3}-1\right)}{\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{5}}\)

\(x=\frac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{1+\sqrt{5}-\sqrt{5}}\)

\(x=\frac{3-1}{1}=2\)

suy ra 

x^3-4x+1=1

A=1^2018

A=1

b)

ta thay

\(7+5\sqrt{2}=\left(1+\sqrt{2}\right)^3\)

khi do 

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

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

x=2

thay vao

x^3+3x-14=0

B=0^2018

B=0

ĐK: \(x-9\ne0\Rightarrow x\ne9\)

\(\sqrt{x}\ge0\Rightarrow x\ge0\)

\(x+\sqrt{x}-6\ne0\Rightarrow x+3\sqrt{x}-2\sqrt{x}-6\ne0\Rightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)\ne0\)

\(\Rightarrow\sqrt{x}-2\ne0\Rightarrow\sqrt{x}\ne2\Rightarrow x\ne4\)

ĐKXĐ: \(x\ge0;x\ne4;x\ne9\)

\(A=\left(\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{1}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\left(\frac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\left(\frac{1+\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)

\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\frac{1+x-9-x+4\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{\sqrt{x}}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{4\sqrt{x}-12}\)

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

2, Với \(x=\frac{25}{16}\)\(\Rightarrow\sqrt{x}=\sqrt{\frac{25}{16}}=\frac{5}{4}\)

\(A=\frac{\frac{5}{4}\left(\frac{5}{4}-2\right)}{4\left(\frac{5}{4}-3\right)}=\frac{5}{4}.\left(-\frac{3}{4}\right):4\left(-\frac{7}{4}\right)=-\frac{15}{16}:-7=\frac{15}{112}\)

\(\orbr{\begin{cases}\orbr{\begin{cases}\\\end{cases}}\\\end{cases}}\)\(\orbr{\begin{cases}\orbr{\begin{cases}\sqrt{x}-2< 0\\\sqrt{x}-3>0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}< 2\\\sqrt{x}>3\end{cases}}\Rightarrow\orbr{\begin{cases}x< 4\\x>9\end{cases}}}\\\orbr{\begin{cases}\sqrt{x}-2>0\\\sqrt{x}-3< 0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}>2\\\sqrt{x}< 3\end{cases}\Rightarrow\orbr{\begin{cases}x>4\\x< 9\end{cases}}}}\end{cases}}\)

\(\sqrt{6-2\sqrt{7}}< \sqrt{6+2\sqrt{7}}\Rightarrow x< 0\)

\(x^2=12-2\sqrt{\left(6-2\sqrt{7}\right)\left(6+2\sqrt{7}\right)}=12-4\sqrt{2}\)

\(\Rightarrow x=-2\sqrt{3-\sqrt{2}}\)

Bạn tự thay vào Q tính (kết quả rất xấu)

\(x=\dfrac{1}{\sqrt{2}}\left(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\right)\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\right)=\sqrt{6}\)

\(y=\sqrt{\left(\sqrt{6}-1\right)^2}=\sqrt{6}-1\)

\(\Rightarrow x-y=1\Rightarrow P=1\)

\(B=x-2020-\sqrt{x-2020}+\dfrac{1}{4}+\dfrac{8079}{4}\)

\(B=\left(\sqrt{x-2020}-\dfrac{1}{2}\right)^2+\dfrac{8079}{4}\ge\dfrac{8079}{4}\)

\(B_{min}=\dfrac{8079}{4}\) khi \(x=\dfrac{8081}{4}\)

1. 

= -(13 + 3 căn7 ) / 2  +  -(7 + 3 căn7 ) / 2 

=  -7 + 3 căn7

`a)P=(x^2+sqrtx)/(x-sqrtx+1)-(2x+sqrtx)/sqrtx`

`P=(sqrtx(sqrtx+1)(x-sqrtx+1))/(x-sqrtx+1)-(sqrtx(2sqrtx+1))/sqrtx`

`P=x+sqrtx-2sqrtx-1`

`P=x-sqrtx-1`

a: Ta có: \(P=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}\)

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

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

Bài 1:

\(\frac{x-9}{\sqrt{x}+3}+\frac{2\sqrt{x}-6}{\sqrt{x}-3}=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}+\frac{2\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\)

\(=\sqrt{x}-3+2=\sqrt{x}-1\)

Bài 2:

a) Không rõ đề

b) \(\sqrt{x^2-6x+9}=\sqrt{4+2\sqrt{3}}\)

\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=\sqrt{\left(\sqrt{3}+1\right)^2}\)

\(\Leftrightarrow\left|x-3\right|=\sqrt{3}+1\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=\sqrt{3}+1\\x-3=-\sqrt{3}-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4+\sqrt{3}\\x=2-\sqrt{3}\end{cases}}\)