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

\(B=\frac{\sqrt{x}-1}{2+\sqrt{x}}=\frac{\sqrt{5}+1-1}{2+\sqrt{5}+1}=\frac{\sqrt{5}}{\sqrt{5}+3}=\frac{\left(3-\sqrt{5}\right)\sqrt{5}}{\left(3^2-5\right)}=\frac{3\sqrt{5}-5}{4}\)

\(B=\frac{\sqrt{x}-1}{2+\sqrt{x}}=\frac{\sqrt{x}+2-3}{\sqrt{x}+2}=1-\frac{3}{\sqrt{x}+2}\inℤ\Leftrightarrow\frac{3}{\sqrt{x}+2}\inℤ\)

mà \(x\)nguyên nên \(\sqrt{x}+2\inƯ\left(3\right)\)mà \(\sqrt{x}+2\ge2\)nên \(\sqrt{x}+2=3\Leftrightarrow x=1\).

\(a,\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x-1}{\sqrt{x}}.\frac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)^2}{x-1}\)

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

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

\(A=\frac{2\sqrt{x}}{\sqrt{x}}+\frac{2x+2}{\sqrt{x}}\)

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

\(b,A=\frac{2\sqrt{x}+2x+2}{\sqrt{x}}=6\)

\(2\sqrt{x}+2x+2=6\sqrt{x}\)

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

\(\left(\sqrt{2}\sqrt{x}\right)^2-4\sqrt{x}+\sqrt{2}^2=0\)

\(\left(\sqrt{2x}-\sqrt{2}\right)^2=0\)

\(\sqrt{2x}=\sqrt{2}\)

\(x=1\)

Trả lời:

\(\sqrt{x^2-9}=\sqrt{x^2-3x}\)

\(\Leftrightarrow x^2-9=x^2-3x\)

\(\Leftrightarrow x^2-9-\left(x^2-3x\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)-x\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3-x\right)=0\)

\(\Leftrightarrow\left(x-3\right)3=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

Vậy x = 3 là nghiệm của pt.

Ta có:\(\sqrt{n+1}-\sqrt{n}=\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\)

\(2\sqrt{n+1}>\sqrt{n+1}+\sqrt{n}>0\Leftrightarrow\frac{1}{2\sqrt{n+1}}< \frac{1}{\sqrt{n+1}+\sqrt{n}}\)

\(0< 2\sqrt{n}< \sqrt{n+1}+\sqrt{n}\Leftrightarrow\frac{1}{\sqrt{n+1}+\sqrt{n}}< \frac{1}{2\sqrt{n}}\)

Ta có đpcm. 

Đk: \(x\ge3\); \(y\ge5\)

Ta có: \(x+y-2\sqrt{x-3}-4\sqrt{y-5}-3=0\)

<=> \(x-3-2\sqrt{x-3}+1+y-5-4\sqrt{y-5}+4=0\)

<=> \(\left(\sqrt{x-3}-1\right)^2+\left(\sqrt{y-5}-2\right)^2=0\)

<=> \(\hept{\begin{cases}\sqrt{x-3}-1=0\\\sqrt{y-5}-2=0\end{cases}}\) <=> \(\hept{\begin{cases}\sqrt{x-3}=1\\\sqrt{y-5}=2\end{cases}}\) <=> \(\hept{\begin{cases}x=4\\y=9\end{cases}}\)(tm)

Vậy (x;y) = {(4; 9)}

\(A=\frac{x+8}{\sqrt{x}+1}=\frac{x-1+9}{\sqrt{x}+1}=\sqrt{x}-1+\frac{9}{\sqrt{x}+1}=\sqrt{x}+1+\frac{9}{\sqrt{x}+1}-2\)

\(\ge2\sqrt{\left(\sqrt{x}+1\right)\frac{9}{\sqrt{x}+1}}-2=2.3-2=4\)

Dấu \(=\)khi \(\sqrt{x}+1=\frac{9}{\sqrt{x}+1}\Leftrightarrow x=4\).

Vậy \(minA=4\)khi \(x=4\).

\(A=\sqrt{x}-1+\frac{9}{\sqrt{x}+1}>\sqrt{x}-1\)mà \(\sqrt{x}-1\)không có GTLN do đó \(A\)cũng không có GTLN. 

Sửa đề: 6xy³ = 6xy²

Ta có: \(\hept{\begin{cases}8\left(x^3-1\right)+6xy^2=y\left(12x^2+y^2\right)\left(1\right)\\\left(x^2+y-4x\right)\left(x^2+y^2-2x-5\right)=14\left(2\right)\end{cases}}\)

Xét pt (1), ta có: \(8\left(x^3-1\right)+6xy^2=y\left(12x^2+y^2\right)\)

<=> \(8x^3-8+6xy^2-12x^2y-y^3=0\)

<=> \(\left(2x-y\right)^3-8=0\)

<=> \(\left(2x-y-2\right)\left[\left(2x-y\right)^2+2\left(2x-y\right)+4\right]=0\)

<=> \(2x-y-2=0\)(vì (2x - y)² + 2(2x - y) + 4 = (2x - y + 1)² + 3 > 0)

<=> \(x-1=\frac{y}{2}\)

Xét pt (2) \(\left(x^2+y-4x\right)\left(x^2+y^2-2x-5\right)=14\)

<=> \(\left[\left(x-1\right)^2+y-2x-1\right].\left[\left(x-1\right)^2+y^2-6\right]=14\)

<=> \(\left(\frac{y^2}{4}+y-y-2-1\right)\left(\frac{y^2}{4}+y^2-6\right)=14\)

<=> \(\left(\frac{y^2}{4}-3\right)\left(\frac{5y^2}{4}-6\right)=14\)

<=> \(\left(y^2-12\right)\left(5y^2-24\right)=224\)

<=> \(\left(y^2-12\right).\left[5\left(y^2-12\right)+36\right]=224\)

<=> \(5\left(y^2-12\right)^2+36\left(y^2-12\right)-224=0\)

<=> \(5\left(y^2-12\right)^2-20\left(y^2-12\right)+56\left(y^2-12\right)-224=0\)

<=> \(\left(y^2-12-20\right)\left(5y^2-60+56\right)=0\)

<=> \(\left(y^2-32\right)\left(5y^2-4\right)=0\)

<=> \(\orbr{\begin{cases}y=\pm4\sqrt{2}\\y=\pm\frac{2}{\sqrt{5}}\end{cases}}\) 

\(y=4\sqrt{2}\) => \(x=\frac{y-2}{2}=\frac{4\sqrt{2}-2}{2}=2\sqrt{2}-1\)

\(y=-4\sqrt{2}\) => \(x=\frac{-4\sqrt{2}-2}{2}=-2\sqrt{2}-1\)

\(y=\frac{2}{\sqrt{5}}\) => \(x=\frac{\frac{2}{\sqrt{5}}-2}{2}=\frac{1}{\sqrt{5}}-1\)

\(y=-\frac{2}{\sqrt{5}}\)=> \(y=\frac{-\frac{2}{\sqrt{5}}-2}{2}=-\frac{1}{\sqrt{5}}-1\)

Ta có: \(\hept{\begin{cases}x^2+y^2+xy+2y+x=2\left(1\right)\\2x^2-y^2-2y-2=0\left(2\right)\end{cases}}\)

<=> \(3x^2+xy+x-4=0\)

<=> \(x\left(y+1\right)=4-3x^2\) 

<=> \(y+1=\frac{4-3x^2}{x}\)

Khi đó,  pt (2) <=> \(2x^2-1-\left(y+1\right)^2=0\)

<=> \(2x^2-1-\left(\frac{4-3x^2}{x}\right)^2=0\)

<=> \(2x^2-1-\frac{9x^4-24x^2+16}{x^2}=0\)

<=> \(2x^4-x^2-9x^4+24x^2-16=0\)

<=> \(7x^4-23x^2+16=0\)

<=>> \(7x^4-7x^2-16x^2+16=0\)

<=> \(\left(x^2-1\right)\left(7x^2-16\right)=0\)

<=> \(\orbr{\begin{cases}x=\pm1\\x=\pm\frac{4}{\sqrt{7}}\end{cases}}\)

Với x = 1 => \(y=\frac{4-3.1^2}{1}-1=0\)

(còn lại tt)

\(\hept{\begin{cases}x^2+y^2+xy+2y+x=2\left(1\right)\\2x^2-y^2-2y-2=0\left(2\right)\end{cases}}\)

Lấy \(3\left(2\right)-\left(1\right)\)ta được: 

\(3\left(2x^2-y^2-2y-2\right)-\left(x^2+y^2+xy+2y+x\right)=-2\)

\(\Leftrightarrow5x^2-4y^2-8y-4-xy-x=0\)

\(\Leftrightarrow\left(x-y-1\right)\left(5x+4y+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=y+1\\x=-\frac{4y+4}{5}\end{cases}}\)

Từ đây bạn thế vào (1) hoặc (2) và giải phương trình bậc hai thu được các nghiệm của hệ phương trình. 

Đáp án các nghiệm là: \(\left(-1,-2\right),\left(1,0\right),\left(-\frac{4}{\sqrt{7}},\frac{5}{\sqrt{7}}-1\right),\left(\frac{4}{\sqrt{7}},-\frac{5}{\sqrt{7}}-1\right)\).

\(x=6-2\sqrt{5}\)

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

\(x= \left(\sqrt{5}-1\right)^2\)

thay vào P ta đc:

\(P=\frac{2\sqrt{\left(\sqrt{5}-1\right)^2}-1}{\sqrt{\left(\sqrt{5}-1\right)^2}+1}\)

\(P=\frac{2\left(\sqrt{5}-1\right)-1}{\sqrt{5}-1+1}\)

\(P=\frac{2\sqrt{5}-2-1}{\sqrt{5}}\)

\(P=\frac{2\sqrt{5}-3}{\sqrt{5}}\)

\(P=\frac{10-3\sqrt{5}}{5}\)