Một cách khác nhé!

Đặt a=2014, b=2015 => b-a=1

Khi đó: \(Q=\sqrt{a^2+a^2b^2+b^2}=\sqrt{\left(b-a\right)^2+a^2b^2+2ab}=\sqrt{a^2b^2+2ab+1}=\sqrt{\left(ab+1\right)^2}\)

\(=ab+1=2014.2015+1=4058211\)

Đặt \(2014=a\) thì ta có:

\(Q=\sqrt{a^2+a^2.\left(a+1\right)^2+\left(a+1\right)^2}\)

\(=\sqrt{a^4+2a^3+3a^2+2a+1}\)

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

Vậy Q là số nguyên

a. ĐKXĐ : \(\orbr{\begin{cases}x\ge0\\1-\sqrt{x}\ne0\end{cases}}\)<=> \(\orbr{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

b. \(P=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(\Leftrightarrow P=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(\Leftrightarrow P=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(\Leftrightarrow P=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(\Leftrightarrow P=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(\Leftrightarrow P=\frac{\left(\sqrt{x}-1\right)\left(2-5\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

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

là bằng 2 phần 3 phải ko

a. ĐKXĐ : \(\hept{\begin{cases}x\ge0\\y\ge0\\y-x\ne0\end{cases}}\)<=> \(\hept{\begin{cases}x\ge0\\y\ge0\\x\ne y\end{cases}}\)

b. \(R=\left(\frac{x-y}{\sqrt{x}-\sqrt{y}}+\frac{\sqrt{x^3}-\sqrt{y^3}}{y-x}\right):\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

\(\Leftrightarrow R=\left(\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}+\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{y-x}\right):\frac{x-\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\)

\(\Leftrightarrow R=\left(\sqrt{x}+\sqrt{y}-\frac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right):\frac{x-\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\)

\(\Leftrightarrow R=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}.\frac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)

\(\Leftrightarrow R=\frac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{x-\sqrt{xy}+y}\)

\(\Leftrightarrow R=\frac{\sqrt{xy}}{x-\sqrt{xy}+y}\)

c. Với \(\hept{\begin{cases}x\ge0\\y\ge0\\x\ne y\end{cases}}\)thì \(\sqrt{xy}\ge0\)  ( 1 )

Ta có : \(x-\sqrt{xy}+y=\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}\)

Mà \(\orbr{\begin{cases}\left(\sqrt{x}-\sqrt{y}\right)^2\ge0\\\left(1\right)\end{cases}}\)=> \(x-\sqrt{xy}+y\ge0\)( 2 )

Từ ( 1 ) và ( 2 ) => \(R\ge0\) ( Đpcm )

ĐK: \(x\ge0;x\ne1\)

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

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

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

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

Để  \(P=\sqrt{x}\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}-1}=\sqrt{x}\Leftrightarrow\sqrt{x}+1=\sqrt{x}\left(\sqrt{x}-1\right)\)\(\sqrt{x}+1\Leftrightarrow x-\sqrt{x}\Leftrightarrow-x+2\sqrt{x}+1=0\)

\(\Leftrightarrow-\left(x-2\sqrt{x}+1\right)+2=0\Leftrightarrow\left(\sqrt{x}-1\right)^2=2\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-1=\sqrt{2}\\\sqrt{x}-1=-\sqrt{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=\sqrt{2}+1\\\sqrt{x}=-\sqrt{2}+1\end{cases}\Leftrightarrow}x=3\pm2\sqrt{2}}\)

b) Với \(x>1\)thì \(P>0\)

Ta dễ thấy \(P=\frac{\sqrt{x}+1}{\sqrt{x}-1}>1\)

Ta có: \(P>0;P>1\)\(\Rightarrow P\left(P-1\right)>0\Leftrightarrow P^2>P\Leftrightarrow P>\sqrt{P}\)

\(tacó 18-8\sqrt{2}=\left(\sqrt{2}-4\right)^2 \))  (phân tích theo HĐt)
suy ra  \(\sqrt{6-2\sqrt{2}+\sqrt{12}+4-\sqrt{2}}\)( vì 4 > căn 2)
          RG ta đc
   \(\sqrt{10-3\sqrt{2}+2\sqrt{3}}\)
 {   \(\sqrt{10-\sqrt{6}\left(\sqrt{2}+\sqrt{3}\right)}\)bỏ bước này cx đc }
bn nên xem lại đề vì k bài nào kêu tính mà ra KQ nhìu căn như w
  nhớ cho mik nha ~!!!

ĐKXĐ : \(x\ge\pm5\)

\(\sqrt{x-5}-3\sqrt{x^2-25}=0\)

\(\Leftrightarrow\sqrt{x-5}\left(1-3\sqrt{x+5}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-5}=0\\1-3\sqrt{x+5}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\3\sqrt{x+5}=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\\sqrt{x+5}=\frac{1}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x+5=\frac{1}{9}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=5\\x=-\frac{44}{9}\end{cases}\left(tm\right)}\)

Vậy ....

đk: \(x\ge5\)

Ta có: \(\sqrt{x-5}-3\sqrt{x^2-25}=0\)

\(\Leftrightarrow\sqrt{x-5}=3\sqrt{x^2-25}\)

\(\Leftrightarrow x-5=9\left(x^2-25\right)\)

\(\Leftrightarrow9x^2-x-220=0\)

\(\Leftrightarrow\left(x-5\right)\left(9x+44\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\left(tm\right)\\x=-\frac{44}{9}\left(ktm\right)\end{cases}}\)

Vậy x = 5

x= √5+√13+√5+√13=√5+√13+√5+√16=

 = √5+√13+√5+4=√5+√13+√9=√5+√13+3

x= \(\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{13}}}}=\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{16}}}}=\)

 = \(\sqrt{5+\sqrt{13+\sqrt{5+4}}}=\sqrt{5+\sqrt{13+\sqrt{9}}}=\)\(\sqrt{5+\sqrt{13+3}}\)

= \(\sqrt{5+\sqrt{16}}=\sqrt{5+4}=\sqrt{9}=3\)

Gọi \(A=\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{7}}+...+\frac{1}{\sqrt{97}+\sqrt{99}}\)

\(\Rightarrow2A=\frac{2}{\sqrt{1}+\sqrt{3}}+\frac{2}{\sqrt{5}+\sqrt{7}}+...+\frac{2}{\sqrt{97}+\sqrt{99}}\)

\(=\frac{\left(\sqrt{3}\right)^2-\left(\sqrt{1}\right)^2}{\sqrt{3}+\sqrt{1}}+...+\frac{\left(\sqrt{99}\right)^2-\left(\sqrt{97}\right)^2}{\sqrt{99}+\sqrt{97}}\)

\(=\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+...+\sqrt{99}-\sqrt{97}\)

\(=\sqrt{99}-1\)

Vậy \(A=\frac{\sqrt{99}-1}{2}=\frac{2\sqrt{99}-2}{4}>\frac{9}{4}\)