Trả lời:

\(\frac{a-\sqrt{a}}{1-a}\)

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

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

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

Trả lời:

Bài 1:

a, \(3\sqrt{2}-4\sqrt{18}+2\sqrt{32}\)

\(=3\sqrt{2}-4\sqrt{3^2.2}+2\sqrt{4^2.2}\)

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

\(=-\sqrt{2}\)

b, \(\sqrt{\left(2+\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\)

\(=\left|2+\sqrt{5}\right|+\sqrt{5-6\sqrt{5}+9}\)

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

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

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

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

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

c, \(\frac{2}{\sqrt{3}-1}+\frac{2}{\sqrt{3}+1}\)

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

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

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

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

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

Bài 2:

a, \(\sqrt{2x+1}-3=1\left(ĐK:x\ge-\frac{1}{2}\right)\)

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

\(\Leftrightarrow2x+1=16\)

\(\Leftrightarrow2x=15\)

\(\Leftrightarrow x=\frac{15}{2}\left(tm\right)\)

Vậy x = 15/2 

Trả lời:

a, \(5\sqrt{x}-2=13\left(ĐK:x\ge0\right)\)

\(\Leftrightarrow5\sqrt{x}=15\)

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

\(\Leftrightarrow x=9\left(tm\right)\)

Vậy x = 9 

b, \(2\sqrt{8x}+7\sqrt{18x}=9-\sqrt{50x}\left(ĐK:x\ge0\right)\)

\(\Leftrightarrow2\sqrt{2^2.2x}+7\sqrt{3^2.2x}=9-\sqrt{5^2.2x}\)

\(\Leftrightarrow4\sqrt{2x}+21\sqrt{2x}=9-5\sqrt{2x}\)

\(\Leftrightarrow4\sqrt{2x}+21\sqrt{2x}+5\sqrt{2x}=9\)

\(\Leftrightarrow30\sqrt{2x}=9\)

\(\Leftrightarrow\sqrt{2x}=\frac{3}{10}\)

\(\Leftrightarrow2x=\frac{9}{100}\)

\(\Leftrightarrow x=\frac{9}{200}\left(tm\right)\)

Vậy x = 9/200 

TL

HT:

TL:

\(\sqrt{x+1}\) \(+\sqrt{x-1}\)

~HT~

Vì OH vuông với AB => H là trung điểm 

=> AH = HB = AB/2 = 12/2 = 6 cm 

Theo định lí Pytago tam giác AHO vuông tại H ta được : 

\(AO=\sqrt{AH^2+OH^2}=\sqrt{64+36}=10\)cm 

hay R = 10 cm 

à bạn ơi

mình xin lỗi giải được

3x^2+4(x+3)-2\(\sqrt{14x^2-7}\)

<=> \(2x^2-1+2\sqrt{14x^2-7}+x^2+4x+6\)

<=>\(\left(\sqrt{2x^2-1}+\sqrt{7}\right)^2+x^2+4x+6\)

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

\(=>A>=2\)

\(dấu"="xảyrakhĩ=-2\)

có vấn đề ở chỗ 2 can 14x^2-7 hay sao ấy

Đặt y = ax + 4 (d) ; y = -3x (d1) 

d // d1 <=> \(\hept{\begin{cases}a=-3\\4\ne0\end{cases}}\Leftrightarrow a=-3\)

b, Thay x = 1 ; y = 6 vào (d) ta được : \(a+4=6\Leftrightarrow a=2\)