Bài 2 :

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

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

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

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

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

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

\(=2\left(16-15\right)=2.1=2\)

Bài 1 :

a, ĐKXĐ : \(x\ge0\)

Ta có : \(PT\Leftrightarrow3\sqrt{5x}-4\sqrt{5x}+8\sqrt{5x}=21\)

\(\Leftrightarrow7\sqrt{5x}=21\)

\(\Leftrightarrow\sqrt{5x}=3\)

\(\Leftrightarrow x=\dfrac{9}{5}\left(TM\right)\)

Vậy ...

b, Ta có : \(PT\Leftrightarrow\sqrt{\left(x-5\right)^2}=4\)

\(\Leftrightarrow\left|x-5\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=4\\x-5=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)

Vậy ....

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

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

\(=1.\sqrt{8+2\sqrt{15}}.\left(\sqrt{5}-\sqrt{3}\right)=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}.\left(\sqrt{5}-\sqrt{3}\right)\)

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

1) \(\Leftrightarrow3\sqrt{5x}-4\sqrt{5x}+8\sqrt{5x}=21\)

\(\Leftrightarrow7\sqrt{5x}=21\)

\(\Leftrightarrow\sqrt{5x}=3\)

\(\Leftrightarrow5x=9\)

\(\Leftrightarrow x=\frac{9}{5}\)

2)\(\Leftrightarrow x^2-10x+25=16\)

\(\Leftrightarrow x^2-10x+9=0\)

\(\Leftrightarrow\left(x-9\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-9=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\\x=1\end{cases}}\)

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 ???

1 slot xíu nữa làm :))))) 

8h lên giúp bạn trước rồi giúp mấy bạn khác sau :v

a, nhóm can x vào một nhóm cái trong ngoặc còn lại thì tính ra

\(11\sqrt{5x}=33\)

chia cả hai vế cho 11 căn 5 rồi bình phương hai vế do x>=0

b,sai đề

Làm 1 câu hay 2 câu nhờ :)))))) Tương tự 1 câu câu còn lại tự làm lấy nhé ;v

\(3\sqrt{9x-27}+2\sqrt{25x-75}\)\(-\sqrt{49x-147}=48\)

\(\Leftrightarrow3\sqrt{9\left(x-3\right)}+2\sqrt{25\left(x-3\right)}\)\(-\sqrt{49\left(x-3\right)}=48\)

\(\Leftrightarrow9\sqrt{x-3}+10\sqrt{x-3}-7\sqrt{x-3}\)\(=48\)

\(\Leftrightarrow\sqrt{x-3}=48\)

\(\Leftrightarrow x-3=16\Leftrightarrow x=19\)

P/s câu trên cx tương tự như trên :)))))) Nhưng có điều hơi khác 1 xíu 

a, \(\sqrt{4x^2+20x+25}\) + \(\sqrt{x^2-8x+16}\) = \(\sqrt{x^2+18x+81}\)

⇔ 4x² + 20x + 25 + \(2\sqrt{\left(4x^2+20x+25\right)\left(x^2-8x+16\right)}\) = x² + 18x + 81

⇔ 4x² + 20x + 25 - x² - 18x - 81 + \(2\sqrt{\left(2x+5\right)^2.\left(x-4\right)^2}\) = 0

⇔ 3x² + 2x - 56 + 2.(2x + 5) . (x - 4) = 0

⇔ 3x² + 2x - 56 + (4x + 10) . (x - 4) = 0

⇔ 3x² + 2x - 56 + 4x² - 16x + 10x - 40 = 0

⇔ 7x² - 4x - 96 = 0

x₁ = 4 ( nhận )

x₂ = \(\frac{-24}{7}\) ( nhận )

Vậy: S = {4; \(\frac{-24}{7}\)}

a)

\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+21}=5-2x-x^2\)

\(\Leftrightarrow\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+16}=6-\left(x+1\right)^2\)

\(VT\ge6;VP\le6\Rightarrow VT=VP=6\)

Vậy pt có một nghiệm duy nhất là \(x=-1\)

b)

\(\sqrt{4x^2+20x+25}+\sqrt{x^2-8x+16}=\sqrt{x^2+18x+81}\)

\(\Leftrightarrow\sqrt{\left(2x+5\right)^2}+\sqrt{\left(x-4\right)^2}=\sqrt{\left(x+9\right)^2}\)

\(\Leftrightarrow\left|2x+5\right|+\left|x-4\right|=\left|x+9\right|\)

Lập bảng xét dấu ra nhé ~^o^~

do \(x^2+x+1=x^2+2.\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)

\(\Rightarrow\sqrt{x^2+x+1}>0\forall x\)

voi dk \(x\ge-1\) ta co 

\(x^2+x+1=x^2+2x+1\Rightarrow x=0\)(tm)

b,\(\sqrt{4x^2-20x+25}+2x=5\)

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

    \(\Leftrightarrow\left|2x-5\right|+2x=5\)

th1 \(2x-5\ge0\Leftrightarrow x\ge\frac{5}{2}\) ta co\(2x-5+2x=5\Leftrightarrow4x=10\Rightarrow x=2.5\left(tm\right)\)

th2 \(2x-5< 0\Leftrightarrow x< \frac{5}{2}\) \(5-2x+2x=5\Leftrightarrow5=5\)

\(\Rightarrow\) dung voi moi \(x< \frac{5}{2}\)

kl \(x\le\frac{5}{2}\)

c, \(\left|x-1\right|=4\) \(\Rightarrow\orbr{\begin{cases}x-1=4\left(x\ge1\right)\\x-1=-4\left(x< 1\right)\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\left(tm\right)\\x=-3\left(tm\right)\end{cases}}}\)

d.\(\sqrt{3\left(x^2+2x+1\right)+4}+\sqrt{5\left(x^2+2x+1\right)+16}\)

 =\(\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+16}\ge\sqrt{4}+\sqrt{16}=6\)

ma \(-x^2-2x+5=-\left(x^2+2x+1\right)+6=-\left(x+1\right)^2+6\le6\)

dau = xay ra \(\Leftrightarrow x=-1\)

a) Ta có: \(\sqrt{x-2\sqrt{x-1}}-\sqrt{x-1}=1\)

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

\(\Leftrightarrow\sqrt{x-1}=\sqrt{x-1}+1+1\)(Vô lý)

Vậy: \(S=\varnothing\)

b) Ta có: \(\sqrt{x^4+2x^2+1}=\sqrt{x^2+10x+25}-10x+22\)

\(\Leftrightarrow x^2+1=\left|x+5\right|-10x+22\)

\(\Leftrightarrow\left|x+5\right|=x^2+1+10x-22=x^2+10x-21\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=x^2+10x-21\left(x\ge-5\right)\\-x-5=x^2+10x-21\left(x< -5\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+10x-21-x-5=0\\x^2+10x-21+x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+9x-26=0\\x^2+11x-16=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-9+\sqrt{185}}{2}\\x=\dfrac{-11-\sqrt{185}}{2}\end{matrix}\right.\)