\(a,\sqrt{1-4a+4a^2}-2a\)

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

\(=1-2a-2a\)

\(=1-4a\)

\(b,x-2y-\sqrt{x^2-4xy+4y^2}\)

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

\(=x-2y-\left(x-2y\right)\)

\(=x-2y-x+2y\)

\(=0\)

\(c,x^2+\sqrt{x^4-8x^2+16}\)

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

\(=x^2+x^2-4\)

\(=2x^2-4\)

Các câu còn lại tương tự nha

\(a,\sqrt{1-4a+4a^2}-2a\)

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

\(=\left(1-2a\right)-2a\)

\(=1-4a\)

\(b,x-2y-\sqrt{x^2-4xy+4y^2}\)

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

\(=x-2y-\left(x-2y\right)\)

\(=x-2y-x+2y\)

\(=0\)

\(c,x^2+\sqrt{x^4-8x^2+16}\)

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

\(=x^2+\left(x^2-4\right)\)

\(=x^2+x^2-4\)

\(=2x^2-4\)

\(d,2x-1-\frac{\sqrt{x^2-10x+25}}{x-5}\)

\(=2x-1-\frac{\sqrt{\left(x-5\right)^2}}{x-5}\)

\(=2x-1-\frac{x-5}{x-5}\)

\(=2x-1-1\)

\(=2x-2\)

\(=2\left(x-1\right)\)

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

\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{x-1-x+4}\)

\(=\dfrac{1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}-2}{3}=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)

b: P=1/4

=>\(\dfrac{\sqrt{x}-2}{3\sqrt{x}}=\dfrac{1}{4}\)

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

=>\(4\sqrt{x}-8-3\sqrt{x}=0\)

=>\(\sqrt{x}=8\)

=>x=64

c: Khi \(x=4+2\sqrt{3}\) thì \(P=\dfrac{\sqrt{4+2\sqrt{3}}-2}{3\cdot\sqrt{4+2\sqrt{3}}}\)

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

a)=1-4a
b) = 2x - 4y
c) = 2x - 2 (nếu x>5)
=2x(nếu x<5)
 

-1.       2x.        2x

a) A= 1 - 4a
b) B=2x-4y
c) C= 2x-2 (nếu x>5)
  = 2x (nếu x< 5)

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

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

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

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

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

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

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

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

1.\(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}=\frac{\left(5+\sqrt{5}\right)\left(5+\sqrt{5}\right)}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}+\frac{\left(5-\sqrt{5}\right)\left(5-\sqrt{5}\right)}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)

\(=\frac{25+10\sqrt{5}+5}{25-5}+\frac{25-10\sqrt{5}+5}{25-5}\)

\(=\frac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{20}\)

\(=\frac{60}{20}=3\)

2.

a) \(\sqrt{45x}-2\sqrt{20x}+2\sqrt{80x}=21\)

ĐK : x ≥ 0

<=> \(\sqrt{5x\cdot9}-2\sqrt{5x\cdot4}+2\sqrt{5x\cdot16}=21\)

<=> \(\sqrt{5x\cdot3^2}-2\sqrt{2^2\cdot5x}+2\sqrt{5x\cdot4^2}=21\)

<=> \(\left|3\right|\sqrt{5x}-2\cdot\left|2\right|\sqrt{5x}+2\cdot\left|4\right|\sqrt{5x}=21\)

<=> \(\sqrt{5x}\cdot\left(3-4+8\right)=21\)

<=> \(\sqrt{5x}\cdot7=21\)

<=> \(\sqrt{5x}=3\)

<=> \(5x=9\)

<=> \(x=\frac{9}{5}\left(tm\right)\)

ơ đang làm lại bấm " Gửi trả lời " ._.

2b) \(\sqrt{x^2-10x+25}=4\)

<=> \(\sqrt{\left(x-5\right)^2}=4\)

<=> \(\left|x-5\right|=4\)

<=> \(\orbr{\begin{cases}x-5=4\\x-5=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\\x=1\end{cases}}\)

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

ĐK : \(\hept{\begin{cases}x>0\\x\ne1\\x\ne4\end{cases}}\)

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

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

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

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

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

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

1/ \(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}=\frac{\sqrt{5}+1}{\sqrt{5}-1}+\frac{\sqrt{5}-1}{\sqrt{5}+1}=\frac{\left(\sqrt{5}+1\right)^2}{5-1}+\frac{\left(\sqrt{5}-1\right)^2}{5-1}\)

\(=\frac{5+2\sqrt{5}+1}{4}+\frac{5-2\sqrt{5}+1}{4}=\frac{12}{4}=3\)

2/ a) \(\sqrt{45x}-2\sqrt{20x}+2\sqrt{80x}=21\Leftrightarrow3\sqrt{5}.\sqrt{x}-4\sqrt{5}.\sqrt{x}+8\sqrt{5}.\sqrt{x}=21\)

\(\Leftrightarrow7\sqrt{5}.\sqrt{x}=21\Leftrightarrow\sqrt{x}=\frac{3}{\sqrt{5}}\Leftrightarrow x=\frac{3}{5}\)

(Bài giải gồm toàn dấu tương đương nên khỏi cần ĐKXĐ ha!! )

b) \(\sqrt{x^2-10x+25}=4\Leftrightarrow\sqrt{\left(x-5\right)^2}=4\Leftrightarrow\left|x-5\right|=4\)

\(\Leftrightarrow\orbr{\begin{cases}x-5=4\\x-5=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=9\\x=1\end{cases}}\)

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

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

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

???? Đề khó hiểu vậy ?? Phải là dấu chia ở giữa mới đúng chứ ??

b) \(A>\frac{1}{6}\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)^2< 18\Leftrightarrow\left(x-2\sqrt{x}\right)\left(x-2\sqrt{x}+1\right)< 18\)

\(\Leftrightarrow\frac{-1-\sqrt{73}}{2}< x-2\sqrt{x}< \frac{-1+\sqrt{73}}{2}\)

??? Bó tay, đề kinh quá ???

\(\frac{\sqrt{3x^2+6xy+3y^2}}{x^2-y^2}\)

<=>\(\frac{\sqrt{3.\left(x+y\right)^2}}{\left(x-y\right).\left(x+y\right)}\)

<=>\(\frac{\sqrt{3}\left|x+y\right|}{\left(x-y\right).\left(x+y\right)}.\)

<=>\(\frac{\sqrt{3}}{x-y}\)