Câu 1,2 bạn đã đăng và có lời giải rồi

Câu 3:

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

\(a,\)

\(=\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right):\left(\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)

\(=\left(\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right):\left(\dfrac{3}{3\sqrt{x}+1}\right)\)

\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)

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

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

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

Vậy \(P=\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}\)

\(b,\)Thay \(P=\dfrac{6}{5}\) vào pt, ta có :

\(\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}=\dfrac{6}{5}\)

\(\Leftrightarrow5\left(3\sqrt{x}+1\right)=6\left(3\sqrt{x}-1\right)\)

\(\Leftrightarrow15\sqrt{x}+5-18\sqrt{x}+6=0\)

\(\Leftrightarrow-3\sqrt{x}+11=0\)

\(\Leftrightarrow-3\sqrt{x}=-11\)

\(\Leftrightarrow\sqrt{x}=\dfrac{11}{3}\)

\(\Leftrightarrow x=\left(\dfrac{11}{3}\right)^2\)

\(\Leftrightarrow x=\dfrac{121}{9}\)

Vậy \(x=\dfrac{121}{9}\) thì \(P=\dfrac{6}{5}\)

a. Không giải được\(\sqrt{29}-6\sqrt{6}< 0\)     

b. \(\left(\sqrt{8}-3\sqrt{2}-\sqrt{10}\right)\cdot\sqrt{2}-\sqrt{20}\) 

=\(\left(2\sqrt{2}-3\sqrt{2}-\sqrt{10}\right)\cdot\sqrt{2}-\sqrt{20}\) 

=\(\left(\sqrt{2}-\sqrt{10}\right)\cdot\sqrt{2}-\sqrt{20}\) 

a) Không thể giải vì \(\sqrt{29}-6\sqrt{6}< 0\) 

b) \(\left(\sqrt{8}-3\sqrt{2}-\sqrt{10}\right)\cdot\sqrt{2}-\sqrt{20}\) 

=\(\left(2\sqrt{2}-3\sqrt{2}-\sqrt{10}\right)\cdot\sqrt{2}-\sqrt{20}\) 

=\(\left(-\sqrt{2}-\sqrt{10}\right)\cdot\sqrt{2}-\sqrt{20}\) 

=\(-2-2\sqrt{5}-2\sqrt{5}\) 

=\(-2-4\sqrt{5}\) 

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

Cái này là rút gọn nha ;-; mình quên ko ghi

\(A=\sqrt{24+8\sqrt{5}}+\sqrt{7-4\sqrt{3}}\)

\(=\sqrt{5+2.4\sqrt{5}+16}+\sqrt{4-2.2\sqrt{3}+3}\)

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

\(=|\sqrt{5}+4|+|2-\sqrt{3}|\)

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

\(=\sqrt{5}-\sqrt{3}+8\)

Ko biết đề sai ko?

Cj gì ơi , mặc dù em không biết làm bài của cj e mới có lớp 7 thui 

Nhưng .... e iu cái ảnh 4D trong hình đại diện của cj 

Cj có phải ARMY ko zợ , nếu phải cho e kb nha , ko phải cx dc ạ !!!

Đừng anti tui nhé , mọi người , mơn nhìu !!!

~ HOK TỐT ~

Chỗ bạn tách 24 thành 5 với 16 ấy, 2 số đó cộng lại không bằng 24, mình bấm máy tính ra kết quả khác, còn đề trong vở ghi ra là đúng mà không biết lúc chép vở có sai gì không nữa. Cảm ơn nhiều ạ

a) A= \(\sqrt{2-\sqrt{3}}\) \(\left(\sqrt{6}-\sqrt{2}\right)\)\(\left(2+\sqrt{3}\right)\)

A= \(\sqrt{2-\sqrt{3}}\) . \(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{3}}\) .\(\left(\sqrt{6}-\sqrt{2}\right)\)

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

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

A=\(\sqrt{4+2\sqrt{3}}\) .\(\left(\sqrt{3}-1\right)\)

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

A=\(\left|\sqrt{3}+1\right|\)\(\left(\sqrt{3}-1\right)\)

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

A=3-1

A=2

Vậy A=2

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

a) \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\dfrac{5-\sqrt{5}}{\sqrt{5}}=\sqrt{5}+\sqrt{5}+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}}=\sqrt{5}+\sqrt{5}+\sqrt{5}-1=-1+3\sqrt{5}\)

b) \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(1+\sqrt{3}\right)^2}=\sqrt{\left(2-\sqrt{3}\right)^2}+1+\sqrt{3}=2-\sqrt{3}+1+\sqrt{3}=3\)

a: \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\dfrac{5-\sqrt{5}}{\sqrt{5}}\)

\(=\sqrt{5}+\sqrt{5}+\sqrt{5}-1\)

\(=3\sqrt{5}-1\)

b: \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(\sqrt{3}+1\right)^2}\)

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

=3

???

\(B=\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)+\dfrac{2-2\sqrt{x}}{\sqrt{x}}(x \geq 0,x \neq 1\)
`=((2x+1-x+\sqrtx)/(x\sqrtx-1))(((\sqrtx+1)(x-\sqrtx+1))/(\sqrtx+1)-\sqrtx)+(2-2sqrtx)/sqrtx`
`=((x-\sqrtx+1)/((\sqrtx-1))(x+sqrtx+1)))(x-2\sqrtx+1)-(2\sqrtx-2)/sqrtx`
`=(1/(\sqrtx-1))(\sqrtx-1)^2-(2(\sqrtx-1))/sqrtx`
`=\sqrtx-1-(2(\sqrtx-1))/sqrtx`
`=(x-\sqrtx-2\sqrtx+2)/sqrtx`
`=(x-3sqrtx+2)/sqrtx`

\(B=\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)+\dfrac{2-2\sqrt{x}}{\sqrt{x}}(x \geq 0,x \neq 1\)
`=((2x+1-x+\sqrtx)/(x\sqrtx-1))(((\sqrtx+1)(x-\sqrtx+1))/(\sqrtx+1)-\sqrtx)+(2-2sqrtx)/sqrtx`
`=((x-\sqrtx+1)/((\sqrtx-1))(x+sqrtx+1))))(x-2\sqrtx+1)-(2\sqrtx-2)/sqrtx`
`=(1/(\sqrtx-1))(\sqrtx-1)^2-(2(\sqrtx-1))/sqrtx`
`=\sqrtx-1-(2(\sqrtx-1))/sqrtx`
`=(x-\sqrtx-2\sqrtx+2)/sqrtx`
`=(x-3sqrtx+2)/sqrtx`