\(\hept{\begin{cases}2x\left(1+\frac{1}{x^2+y^2}\right)=3\\2y\left(1+\frac{1}{x^2+y^2}\right)=1\end{cases}}\)

\(\hept{\begin{cases}\frac{2x}{2y}=3\\2y\left(1+\frac{1}{x^2+y^2}\right)=1\end{cases}}\)

\(\hept{\begin{cases}x=3y\\2y\left(1+\frac{1}{x^2+y^2}\right)\end{cases}=1}\)

\(2y\left(1+\frac{1}{9y^2+y^2}\right)=1\)

\(2y+\frac{2y}{10y^2}=1\)

\(2y+\frac{1}{5y}-1=0\)

\(10y^2+1-5y=0\)

\(\Delta=\left(-5\right)^2-\left(4.1.10\right)=25-40=-15< 0\)

lên pt vô nghiệm

\(\sqrt{117.5^2-26.5^2-1440}\)

\(\sqrt{5\left(585-130-288\right)}\)

\(\sqrt{5.167}=\sqrt{835}\)

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

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

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

c) \(\sqrt{5}\left(\sqrt{6}+1\right):\frac{\sqrt{2\sqrt{3}+\sqrt{2}}}{\sqrt{2\sqrt{3}-\sqrt{2}}}=\sqrt{5}\left(\sqrt{6}+1\right):\sqrt{\frac{\left(2\sqrt{3}+\sqrt{2}\right)^2}{\left(2\sqrt{3}-\sqrt{2}\right)\left(2\sqrt{3}+\sqrt{2}\right)}}\)

\(=\sqrt{5}\left(\sqrt{6}+1\right):\frac{2\sqrt{3}+\sqrt{2}}{\sqrt{12-2}}=\sqrt{5}\left(\sqrt{6}+1\right)\cdot\frac{\sqrt{10}}{\sqrt{2}\left(\sqrt{6}+1\right)}=\frac{\sqrt{5}.\sqrt{2}.\sqrt{5}}{\sqrt{2}}=5\)

e) \(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\frac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{4-2\sqrt{3}}}\)

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

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

\(a,A=\frac{2}{\sqrt{x}-3}+\frac{2\sqrt{x}}{x-4\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-1}\)

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

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

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

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

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

\(b,A=\frac{\sqrt{x}-3+5}{\sqrt{x}-3}=1+\frac{5}{\sqrt{x}-3}\)

để A nguyên \(5⋮\sqrt{x}-3\)

lập bảng ra đc 

\(x=\left\{2\right\}\)

\(P=\frac{1}{\sqrt{x}-1}\left(x\ge0,x\ne1\right)\)

+ Nếu x ko là số chính phương

=> \(\sqrt{x}\) \(\notin Z\)

=> \(\sqrt{x}-1\notin Z\) 

\(\Rightarrow\frac{1}{\sqrt{x}-1}\notin Z\) ( loại)

+ Nếu x là số chính phương

\(\Rightarrow\sqrt{x}\in Z\)

\(\Rightarrow\sqrt{x}-1\in Z\)

Để P nguyên thì \(1⋮\sqrt{x}-1\) 

Hay \(\sqrt{x}-1\inƯ\left(1\right)=\left\{\pm1\right\}\)

Xét bảng

Vậy ...

\(\sqrt{a+b}< \sqrt{a}+\sqrt{b}\)

\(\Rightarrow\left(\sqrt{a+b}\right)^2< \left(\sqrt{a}+\sqrt{b}\right)^2\)

\(\Rightarrow a+b< a+2\sqrt{ab}+b\)

\(\Rightarrow a+b-a-b< 2\sqrt{ab}\)

\(\Rightarrow0< 2\sqrt{ab}\)

\(\Rightarrow0< \sqrt{ab}\)(luôn đúng với a>0 và b>0)

Vây với a>0 và b>0 thì \(\sqrt{a+b}< \sqrt{a}+\sqrt{b}\)

a) Ta có: \(\Delta'=\left(-m\right)^2+m+1=m^2+m+1=\left(m+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

=> pt luôn có 2 nghiệm phân biệt

Theo hệ thức viet, ta có: \(\hept{\begin{cases}x_1+x_2=2m\\x_1x_2=-m-1\end{cases}}\)

Theo bài ra, ta có: \(\hept{\begin{cases}S=2x_1+3x_2+3x_1+2x_2=5\left(x_1+x_2\right)=5.2m=10m\\P=\left(2x_1+3x_2\right)\left(3x_1+2x_2\right)=6x_1^2+13x_1x_2+6x_2^2=6\left(x_1+x_2\right)^2+x_1x_2\end{cases}}\)

\(\hept{\begin{cases}S=10m\\P=6.\left(2m\right)^2-m-1=24m^2-m-1\end{cases}}\)

Hai nghiệm 2x₁ + 3x₂ và 3x₁ + 2x₂ là nghiệm của pt \(x^2-10mx+24m^2-m-1=0\)

b) Theo bài ra, ta có:

\(\left|2x_1+3x_2\right|+\left|3x_1+2x_2\right|=30\)

<=> \(\left(2x_1+3x_2\right)^2+\left(3x_1+2x_2\right)^2+2\left|\left(2x_1+3x_2\right)\left(3x_1+2x_2\right)\right|=900\)

<=> \(\left(2x_1+3x_2+3x_1+2x_2\right)^2-2\left(2x_1+3x_2\right)\left(3x_1+2x_2\right)+2\left|24m^2-m-1\right|=900\)

<=> \(\left(10m\right)^2-2\left(24m^2-m-1\right)+2\left|24m^2-m-1\right|=900\)

<=> \(52m^2+2m+2+2\left|24m^2-m-1\right|=900\)

<=> \(\left|24m^2-m-1\right|=449-26m^2-m\)

<=> \(\orbr{\begin{cases}24m^2-m-1=449-26m^2-m\left(đk:m\ge\frac{1+\sqrt{97}}{48}hoặcx\le\frac{1-\sqrt{97}}{48}\right)\\24m^2-m-1=26m^2+m-449\left(đk:\frac{1-\sqrt{97}}{48}\le x\le\frac{1+\sqrt{97}}{48}\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}50m^2=1\\2m^2+2m-448=0\end{cases}}\)<=> \(\orbr{\begin{cases}m=\pm\frac{1}{5\sqrt{2}}\\m^2+m-224=0\end{cases}}\) (\(\orbr{\begin{cases}m=\frac{1}{5\sqrt{2}}\left(ktm\right)\\m=-\frac{1}{5\sqrt{2}}\left(tm\right)\end{cases}}\))

<=> \(m^2+m-224=0\)(có 2 nghiệm ko thõa mãn -> tự tính)

a) \(\Delta'=m^2+m+1>0\forall m\). Do đó phương trình cho luôn có hai nghiệm phân biệt

Khi đó, theo hệ thức Viet: \(\hept{\begin{cases}x_1+x_2=2m\\x_1x_2=-m-1\end{cases}}\)

Suy ra \(\hept{\begin{cases}5\left(x_1+x_2\right)=10m\\\left(2x_1+3x_2\right)\left(3x_1+2x_2\right)=6\left(x_1+x_2\right)^2+x_1x_2=24m^2-m-1\end{cases}}\)

Áp dụng định lí Viet đảo ta có được phương trình:

\(X^2-10mX+24m^2-m-1=0\left(1\right)\) nhận \(2x_1+3x_2\) và \(3x_1+2x_2\) làm nghiệm.

b) Để \(\left(1\right)\) có nghiệm thì \(100m^2\ge4\left(24m^2-m-1\right)\Leftrightarrow4m^2+4m+4\ge0\left(đ\right)\)

Ta có \(\left|X_1\right|+\left|X_2\right|=30\Leftrightarrow\left(X_1+X_2\right)^2-2X_1X_2+2\left|X_1X_2\right|-900=0\)

\(\Rightarrow100m^2-2\left(24m^2-m-1\right)+2\left|24m^2-m-1\right|+900=0\)

+) Nếu \(24m^2-m-1\ge0\) thì \(100m^2+900=0\Leftrightarrow m=\pm3\)

+) Nếu \(24m^2-m-1< 0\) thì \(4m^2+4m+904=0\)(Vô nghiệm)

Vậy \(m=\pm3.\)