ĐKXĐ: x>=1

\(PT\Leftrightarrow8\sqrt{x-1}+7\sqrt{x-1}-\sqrt{x-1}=46\)

=>\(14\sqrt{x-1}=46\)

=>\(\sqrt{x-1}=\dfrac{23}{7}\)

=>\(x-1=\dfrac{529}{49}\)

=>\(x=\dfrac{578}{49}\)

2\(\sqrt{1\text{6}x-1\text{6}}\)+\(\sqrt{49x-49}\)-\(\sqrt{x-1}\)=46

<=>2.4.4\(\sqrt{x-1}\)+7.7\(\sqrt{x-1}\)-\(\sqrt{x-1}\)=46

<=> 32\(\sqrt{x-1}\)+49\(\sqrt{x-1}\)-\(\sqrt{x-1}\)=46

<=> (32+49-1)\(\sqrt{x-1}\)=46

<=> 80\(\sqrt{x-1}\)=46

<=> \(\sqrt{x-1}\)=\(\dfrac{23}{40}\)

<=> x-1=\(\dfrac{529}{1\text{6}00}\)

<=> x=\(\dfrac{2159}{1\text{6}00}\)

<=>x=1,35

\(N=x+2y-\sqrt{2x-1}-5\sqrt{4y-3}+13\)

\(2N=2x+4y-2\sqrt{2x-1}-10\sqrt{4y-3}+26\)

\(=\left(2x-1-2\sqrt{2x-1}+1\right)+\left(4y-3-10\sqrt{4y-3}+25\right)+4\)

\(=\left(\sqrt{2x-1}-1\right)^2+\left(\sqrt{4y-3}-5\right)^2+4\ge4\)

\(\sqrt{x-1}\) - \(\sqrt{9x-9}\) + \(\sqrt{16x-16}\) = 4 (đk \(x\ge\)1)

\(\sqrt{x-1}-\) \(\sqrt{9\left(x-1\right)}\) + \(\sqrt{16\left(x-1\right)}\) = 4

\(\sqrt{x-1}\) - 3\(\sqrt{x-1}\) + \(4\sqrt{x-1}\) = 4  

 \(\sqrt{x-1}\)( 1 - 3 + 4 ) = 4

  \(\sqrt{x-1}\) . 2 = 4

  \(\sqrt{x-1}\) = 4 : 2

  \(\sqrt{x-1}\) = 2

   \(x-1\) =4

  \(x=4+1\)

 \(x=5\) (thỏa mãn)

Vậy \(x\) = 5 

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

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

\(=\dfrac{\left(a+1\right)\cdot\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\)

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

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

\(=\dfrac{\left(a+1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\)

a: ĐKXĐ: \(x>0\)

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

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

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

ĐK: \(x,y\ge0\)

\(hpt\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x}-2\sqrt{y}=-2\left(1\right)\\4\sqrt{x}+2\sqrt{y}=2\end{matrix}\right.\)

Cộng vế theo vế 2 phương trình ta được: \(7\sqrt{x}=0\Leftrightarrow x=0\)

Khi đó \(\left(1\right)\Leftrightarrow-2\sqrt{y}=-2\Leftrightarrow y=1\)

Vậy hệ đã cho có nghiệm \(\left(x;y\right)=\left(0;1\right)\)

ĐKXĐ: \(x\ge0;y\ge0\)

\(\left\{{}\begin{matrix}3\sqrt{x}-2\sqrt{y}=-2\\2\sqrt{x}+\sqrt{y}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x}-2\sqrt{y}=-2\\4\sqrt{x}+2\sqrt{y}=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}7\sqrt{x}=0\\2\sqrt{x}+\sqrt{y}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2\sqrt{0}+\sqrt{y}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\0+\sqrt{y}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\\sqrt{y}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\) (TM)

Vậy...

\(\left\{{}\begin{matrix}3\sqrt{x}-2\sqrt{y}=-2\\2\sqrt{x}+\sqrt{y}=1\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}3\sqrt{x}-2\sqrt{y}=-2\\4\sqrt{x}+2\sqrt{y}=2\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}7\sqrt{x}=0\\2\sqrt{x}+\sqrt{y}=1\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}\sqrt{x}=0\\2.0+\sqrt{y}=1\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=0\\\sqrt{y}=1\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}x=0\\y=^+_11\end{matrix}\right.\)

Coi như bước trên bạn đã làm đúng, giải pt vô tỉ thôi nhé:

TH1: \(x=y\)

\(\Rightarrow x^2+x+2=\sqrt{5x+5}+\sqrt{3x+2}\)

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

\(\Leftrightarrow x^2-x-1+\dfrac{x^2-x-1}{x+1+\sqrt{3x+2}}+\dfrac{x^2-x-1}{x+2+\sqrt{5x+5}}=0\)

TH2: \(x=4y+3\)

Đây là trường hợp nghiệm ngoại lai, lẽ ra phải loại (khi bình phương lần 2 phương trình đầu, bạn quên điều kiện nên ko loại trường hợp này)