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

ĐKXĐ : x\(\ge-1\)

\(\Leftrightarrow2\sqrt{x+1}=1+\sqrt{x+6}\)

bình phương 2 vế ta có :

\(4x+4=1+2\sqrt{x+6}+x+6\)

\(3x-3=2\sqrt{x+6}\)

ĐK : \(x>1\)

bình phương 2 vế ta có :

\(9x^2-18x+9=4x+24\Leftrightarrow9x^2-22x-23=0\)\(\Leftrightarrow9x^2-2.11.x+121-144=0\)

\(\Leftrightarrow\left(3x-11\right)^2=144\Leftrightarrow3x-11=\pm12\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-11=12\\3x-11=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{23}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

kết hợp vs đkxđ => x=23/3

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

=\(\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{ab}-1}\)

A=\(\dfrac{x\sqrt{x}+26\sqrt{x}-19}{x+2\sqrt{x}-3}-\dfrac{2\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)

=\(\dfrac{x\sqrt{x}+26\sqrt{x}-19-2\sqrt{x}\left(\sqrt{x}+3\right)+\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

=\(\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

=\(\dfrac{x\sqrt{x}+16\sqrt{x}-x-16}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(x+16\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{x+16}{\sqrt{x}+3}\)

\(\left(2+\sqrt{2}\right)\sqrt{3-\sqrt{8}}\)

=\(2\sqrt{3-\sqrt{8}}+\sqrt{6-2\sqrt{8}}\)=\(\sqrt{12-4\sqrt{8}}+2-\sqrt{2}\)

=\(\sqrt{8}-2+2-\sqrt{2}=\sqrt{8}-\sqrt{2}=\sqrt{2}\)

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

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

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

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

b) ta có : B=3 \(\Leftrightarrow\sqrt{x}-1=3\Leftrightarrow x=16\)

vậy để B=3 thì x=16

n+5 = n-2 + 7

chia hết cho n -2 khi 7 chia hết cho n-2

=> n-2 thuộc U(7) ={1;7}

+n-2 =1 => n =3

+n -2 =7 => n =9

Vậy n =3; n =9

\(\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{\sqrt{4}+\sqrt{3}}+\dfrac{1}{\sqrt{5}+\sqrt{4}}\)

=\(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\sqrt{5}-\sqrt{4}=\sqrt{5}-1\)

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

=\(\sqrt{12-6\sqrt{3}}-\sqrt{4-2\sqrt{3}}\left(2+\sqrt{3}\right)\)

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

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

vậy biểu thức trên là 1 số nguyên

dễ như húp cháo đương nhiên là 20 que rồi 

nhảm nhí

a) D (ĐKXĐ: x\(\ge0,x\ne1\))

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

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

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

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

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

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

=\(\sqrt{x}-x-3=0\Leftrightarrow\left(x-2.\dfrac{1}{2}x+\dfrac{1}{4}\right)-\dfrac{13}{4}=0\)

\(\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{13}{4}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-\dfrac{1}{2}=\dfrac{13}{4}\\\sqrt{x}-\dfrac{1}{2}=-\dfrac{13}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{225}{16}\\x=\dfrac{121}{16}\end{matrix}\right.\)