`a)\sqrt{45x}-2\sqrt{20x}+2\sqrt{80x}=21`    `ĐK: x >= 0`

`<=>3\sqrt{5x}-4\sqrt{5x}+8\sqrt{5x}=21`

`<=>7\sqrt{5x}=21`

`<=>\sqrt{5x}=3`

`<=>5x=9<=>x=9/5` (t/m).

`b)\sqrt{x^2-10x+25}=4`

`<=>\sqrt{(x-5)^2}=4`

`<=>|x-5|=4`

`<=>[(x-5=4),(x-5=-4):}`

`<=>[(x=9),(x=1):}`

Lời giải:

$(4+\sqrt{15})(\sqrt{10}-\sqrt{6})\sqrt{4-\sqrt{15}}$

$=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}$

$=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}$

$=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})(\sqrt{5}-\sqrt{3})$

$=(4+\sqrt{15})(8-2\sqrt{15})=2(4+\sqrt{15})(4-\sqrt{15})$

$=2(4^2-15)=2$ (đpcm)

ai giúp vs

Bài 1

a) Đặt VT = A

<=> \(2\sqrt{2}A=\left(8+2\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{8-2\sqrt{15}}\)

<=> \(2\sqrt{2}A=\left(\sqrt{5}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right).\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

<=> \(2A=\left(\sqrt{5}+\sqrt{3}\right)^2.\left(\sqrt{5}-\sqrt{3}\right)^2\)

<=> 2A = \(\left(5-3\right)^2=4\)

<=> A = 2

b) Đặt VT = B

<=> \(2\sqrt{2}B=\left(10+2\sqrt{21}\right).\left(\sqrt{14}-\sqrt{6}\right)\sqrt{10-2\sqrt{21}}\)

<=> \(2\sqrt{2}B=\left(\sqrt{7}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{7}-\sqrt{3}\right).\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)

<=> \(2B=\left(\sqrt{7}+\sqrt{3}\right)^2.\left(\sqrt{7}-\sqrt{3}\right)^2=\left(7-3\right)^2=16\)

<=> B = 8 

Bài 2

Đặt VT = A

<=> A² = \(\dfrac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}{2}\)

<=> A² = \(\dfrac{2\sqrt{5}+2\sqrt{5-4}}{2}=\dfrac{2\sqrt{5}+2}{2}=\sqrt{5}+1\)

<=> \(A=\sqrt{\sqrt{5}+1}\)

\(=\left(\dfrac{\sqrt{10}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\dfrac{\sqrt{6^2}}{\sqrt{6}}\right)\sqrt{4+\sqrt{15}}\)

\(=\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4+\sqrt{15}}\)

\(=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{5+2\sqrt{3}\sqrt{5}+3}\)

\(=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)=5-3=2\)

 \(VT\Leftrightarrow\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4+\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8+2\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)=5-3=2=VP\left(dpcm\right)\)

\(\left(\dfrac{\sqrt{30}-\sqrt{20}}{\sqrt{3}-\sqrt{2}}-\dfrac{6}{\sqrt{6}}\right)\sqrt{4+\sqrt{15}}=\left(\dfrac{\sqrt{10}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\sqrt{6}\right)\left(4+\sqrt{15}\right)\)

\(=\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4+\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8+2\sqrt{15}}\)

\(=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)\)

\(=5-3=2\)

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 ???

a, \(VT=\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\)

\(=\frac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)\left(\sqrt{20}-2\right)}{2}\)

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

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

\(=\left(\sqrt{5}-1\right)^2\left(3+\sqrt{5}\right)=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)

\(=18-6\sqrt{5}+6\sqrt{5}-10=8=VP\)

b, \(VT=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{5-2\sqrt{5}\sqrt{3}+3}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\)

\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

\(=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)\)

\(=2\left(16-15\right)=2=VP\)