a) \(\sqrt{117,5^2-26,5^2-1440}=\sqrt{\left(117,5-26,5\right)\left(117,5+26,5\right)-1440}\)

\(=\sqrt{91.144-1440}=\sqrt{144\left(91-10\right)}=\sqrt{12^2.9^2}=12.9=108\)

b) \(\sqrt{146,5^2-109,5^2+27.256}=\sqrt{\left(146,5-109,5\right)\left(146,5+109,5\right)+27.256}\)

\(=\sqrt{37.256+27.256}=\sqrt{256\left(37+27\right)}=\sqrt{256.64}=\sqrt{16^2.8^2}=16.8=128\)

\(\sqrt{117,5^2-26,5^2}-1440=-202475\)

\(\sqrt{146,5^2-109,5^2+27,256=}-11816494\) 

bạn lê nhat phuong oi sai rồi

1. \(\sqrt{\frac{2ab^2}{162a}}=\sqrt{\frac{b^2}{81}}=\frac{|b|}{9}\)
2. \(2y^2\sqrt{\frac{x^4}{4y^2}}=\frac{2y^2x^2}{-2y}=-yx^2\)

2)

• ​\(\left(\sqrt{2003}+\sqrt{2005}\right)^2=2003+2005+2\sqrt{2003\times2005}\)

\(=4008+2\sqrt{\left(2004-1\right)\left(2004+1\right)}=4008+2\sqrt{2004^2-1}\)

• \(\left(\sqrt{2004}+\sqrt{2004}\right)^2=2004+2004+2\sqrt{2004\times2004}\)

\(=4008+2\sqrt{2004^2}\)

Ta có \(2004^2>2004^2-1\Rightarrow\sqrt{2004^2}>\sqrt{2004^2-1}\Rightarrow4008+2\sqrt{2004^2}>4008+2\sqrt{2004^2-1}\)

Vậy \(2\sqrt{2004}>\sqrt{2003}+\sqrt{2005}\)

1.  a) 108
     b) 128
2.  >

\(a=\sqrt{\left(6,8-3,2\right)\left(6,8+3,2\right)}=\sqrt{3,6\left(10\right)}=\sqrt{36}=6\)

a) \(\sqrt{6,8^2-3,2^2}=\sqrt{\left(6,8-3,2\right)\left(6,8+3,2\right)}\)

=\(\sqrt{3,6.10}=\sqrt{36}=6\)

b)\(\sqrt{21,8^2-18,2^2}=\sqrt{\left(21,8-18,2\right)\left(21,8+18,2\right)}\)

=\(\sqrt{3,6.40}=\sqrt{144}=12\)

c)\(\sqrt{117,5^2-26,5^2-1440}=\sqrt{\left(117,5-26,5\right)\left(117,5+26,5\right)-1440}\)

=\(\sqrt{91.144-1440}=\sqrt{144.81}=\sqrt{144}.\sqrt{81}=108\)

d)\(\sqrt{146,5^2-109,5^2+27.256}\)=\(\sqrt{\left(146,5-109,5\right)\left(146,5+109,5\right)+27.256}\)

=\(\sqrt{37.256+\sqrt{27.256}}=\sqrt{64.256}=\sqrt{64}.\sqrt{256}=128\)

\(P=\left(\frac{2\left(\sqrt{x}+2\right)+\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right).\frac{x+2\sqrt{x}}{2\sqrt{x}}\) điều kiện x >0

\(P=\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}.\frac{x+2\sqrt{x}}{2\sqrt{x}}\)

\(P=\frac{2\sqrt{x}+4+x}{2\sqrt{x}}=1+\frac{4+x}{2\sqrt{x}}.\)

b) P = 3

\(\Leftrightarrow1+\frac{4+x}{2\sqrt{x}}=3\Leftrightarrow\frac{4+x}{2\sqrt{x}}=2\)

\(\Leftrightarrow4+x=4\sqrt{x}\Leftrightarrow4+x-4\sqrt{x}=0\)

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

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

Ngô Văn Tuyên cảm ơn bạn nha. Nhưng cho mình hỏi tí sao bạn lại tách ra thành \(1+\frac{4-x}{2\sqrt{x}}\)

giải thích hộ mình với nhé. Cảm ơn nhiều !!

Bài 22:

1: \(\sqrt{3-\sqrt5}=\frac{\sqrt{6-2\sqrt5}}{\sqrt2}\)

\(=\frac{\sqrt{\left(\sqrt5-1\right)^2}}{\sqrt2}=\frac{\sqrt5-1}{\sqrt2}=\frac{\sqrt{10}-\sqrt2}{2}\)

2: \(\sqrt{7+3\sqrt5}\)

\(=\frac{\sqrt{14+6\sqrt5}}{\sqrt2}\)

\(=\frac{\sqrt{\left(3+\sqrt5\right)^2}}{\sqrt2}=\frac{3+\sqrt5}{\sqrt2}=\frac{3\sqrt2+\sqrt{10}}{2}\)

3: \(\sqrt{9+\sqrt{17}}-\sqrt{9-\sqrt{17}}-2\)

\(=\frac{1}{\sqrt2}\left(\sqrt{18+2\sqrt{17}}-\sqrt{18-2\sqrt{17}}\right)-2\)

\(=\frac{1}{\sqrt2}\left(\sqrt{\left(\sqrt{17}+1\right)^2}-\sqrt{\left(\sqrt{17}-1\right)^2}\right)-2\)

\(=\frac{1}{\sqrt2}\left(\sqrt{17}+1-\sqrt{17}+1\right)-2=\frac{2}{\sqrt2}-2=\sqrt2-2\)

Bài 26:

1: \(\left|3-2x\right|=2\sqrt5\)

=>\(\left[\begin{array}{l}2x-3=2\sqrt5\\ 2x-3=-2\sqrt5\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=3+2\sqrt5\\ 2x=3-2\sqrt5\end{array}\right.\Rightarrow x=\frac{3\pm2\sqrt5}{2}\)

2: \(\sqrt{x^2}=12\)

=>|x|=12

=>x=12 hoặc x=-12

3: \(\sqrt{x^2-2x+1}=7\)

=>\(\sqrt{\left(x-1\right)^2}=7\)

=>|x-1|=7

=>\(\left[\begin{array}{l}x-1=7\\ x-1=-7\end{array}\right.\Rightarrow\left[\begin{array}{l}x=8\\ x=-6\end{array}\right.\)

\(\dfrac{2\sqrt{x}}{\sqrt{x}-2}.\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\)

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

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

\(\dfrac{2\sqrt{x}}{\sqrt{x}-2}.\dfrac{\sqrt{x}+2}{\sqrt{x}-2}=\dfrac{2\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)^2}\) (\(đk:x\ge0;x\ne\sqrt{2}\))

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

\(\)