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thấy lạ nha , nãy giờ toàn thấy tên Chung Tình Là Tui , yêu tao ư kiếp sau nhé , yêu ông ư tui thà nhảy lầu còn hơn vậy ?
mình nghĩ bài này sai đề,
ĐÚng phải là\(\sqrt[3]{2+\sqrt{3}}\)
( KHÔNG CHẮC NỮA :D )
\(\frac{\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}}{\frac{16}{\frac{16}{x^2}-\frac{8}{x}+1}}\)\(=\frac{\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}}{\left(\frac{4}{x}-1\right)^2}\)
\(\frac{\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}}{\left(\frac{4}{x}-1\right)^2}\)\(=\frac{\sqrt{x-4}+2+\sqrt{x-4}-2}{\left(\frac{4-x}{x}\right)^2}\)
\(=\frac{2\sqrt{x-4}}{\left(\frac{4-x}{x}\right)^2}=\frac{2x^2\sqrt{x-4}}{\left(x-4\right)^2}=\frac{2x^2}{\sqrt{x-4}^3}\)
bài bạn YIM YIM sai nhé, mk làm lại và chỉnh lại đề luôn, bạn tham khảo:
ĐK: \(x>4\)
\(A=\frac{\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}}{\frac{16}{x^2}-\frac{8}{x}+1}\)
\(=\frac{\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}}{\left(1-\frac{4}{x}\right)^2}\)
\(=\frac{\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|}{\left(\frac{x-4}{x}\right)^2}\)
Nếu \(4< x\le8\)thì:
\(A=\frac{\sqrt{x-4}+2+2-\sqrt{x-4}}{\left(\frac{x-4}{x}\right)^2}\)
\(=\frac{4x^2}{\left(x-4\right)^2}\)
Nếu \(x>8\)thì:
\(A=\frac{\sqrt{x-4}+2+\sqrt{x-4}-2}{\frac{\left(x-4\right)^2}{x^2}}=\frac{2x^2}{\sqrt{x-4}^3}\)
a) A= (\(\left(\frac{1+\sqrt{x}}{1+\sqrt{x}}-\frac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x-2}\right)}+\frac{\sqrt{x}+2}{x-2\sqrt{x}-3\sqrt{x}+6}\right)\)
A=\(\left(\frac{1+\sqrt{x}-\sqrt{x}}{1+\sqrt{x}}\right):\left(\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)}\right)\)
A= \(\left(\frac{1}{1+\sqrt{x}}\right):\left(\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{x-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
A=\(\left(\frac{1}{1+\sqrt{x}}\right):\left(\frac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
A=\(\left(\frac{1}{1+\sqrt{x}}\right):\left(\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
A=\(\frac{\sqrt{x}-2}{\sqrt{x}+1}\)
a)ĐKXĐ : x > 0
P = \(\left(\frac{x-1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{\sqrt{x}\left(1+\sqrt{x}\right)}\right)\)
= \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}:\frac{1}{\sqrt{x}}.\left(\sqrt{x}-1+\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
= \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}:\frac{\sqrt{x}-1}{\sqrt{x}}.\left(1-\frac{1}{\sqrt{x}+1}\right)\)
= \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}:\frac{\left(\sqrt{x}-1\right).\sqrt{x}}{\sqrt{x}}\)
= \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}\)
Vậy P = \(\frac{\sqrt{x}+1}{\sqrt{x}}\)
b) x = \(\frac{2}{2+\sqrt{3}}=\frac{2\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=\frac{2.\left(2-\sqrt{3}\right)}{4-3}=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)
\(\Rightarrow\sqrt{x}=\sqrt{3}-1\)
=> P = \(\frac{\sqrt{x}+1}{\sqrt{x}}=\frac{\sqrt{3}-1+1}{\sqrt{3}-1}=\frac{\sqrt{3}}{\sqrt{3}-1}\)
= \(\frac{\sqrt{3}\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3+1}\right)}=\frac{3+\sqrt{3}}{3-1}=\frac{3+\sqrt{3}}{2}\)
c)\(P\sqrt{x}=6\sqrt{x}-3-\sqrt{x-4}\)
\(\Leftrightarrow\frac{\left(\sqrt{x}+1\right)\sqrt{x}}{\sqrt{x}}=6\sqrt{x}-3-\sqrt{x-4}\)
\(\Leftrightarrow\sqrt{x}+1=6\sqrt{x}-3-\sqrt{x-4}\)
\(\Leftrightarrow\sqrt{x-4}=5\sqrt{x-4}\)
Đặt \(\hept{\begin{cases}a=\sqrt{x}\\b=\sqrt{x-4}\end{cases}\Rightarrow a^2+b^2=x-\left(x-4\right)=4}\)
\(\Rightarrow\hept{\begin{cases}a^2-b^2=4\\b=5a-4\end{cases}\Rightarrow\hept{\begin{cases}a^2-\left(5a-4\right)^2=4\left(^∗\right)\\b=5a-4\end{cases}}}\)
Từ (*) <=> a2 -(25a2 -40a + 16 ) =4
<=> -24a2 + 40a - 20 = 0
=> \(\Delta'=-80< 0\)
=> PT vô nghiệm
=> ko tồn tại x thỏa mãn
làm tiếp nè:
\(\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|\)
*)Nếu \(\sqrt{x-1}\)>3<=>x-1>9<=>x>10 thì \(\sqrt{x-1}\)-2>0 \(\sqrt{x-1}\)-3>0
Ta có:|\(\sqrt{x-1}\)-2|+|\(\sqrt{x-1}\)-3|=\(\sqrt{x-1}\)-2+\(\sqrt{x-1}\)-3=2\(\sqrt{x-1}\)-5
*)Nếu 2<\(\sqrt{x-1}\)<3<=>4<x-1<9... làm tiếp đi bận mất rồi
ĐK : \(x\ge1\)
\(A=\sqrt{x+2\sqrt{x-1}}-\sqrt{x+8+6\sqrt{x-1}}\)
\(=\sqrt{x-1+2\sqrt{x-1}}-\sqrt{x-1+6\sqrt{x-1}+9}\)
\(=\sqrt{(\sqrt{x-1}-1)^2}-\sqrt{(\sqrt{x-1}+3)^2}\)
\(=\left|\sqrt{x-1}-1\right|-\left|\sqrt{x-1}+3\right|\)
\(=\hept{\begin{cases}1-\sqrt{x-1}-\sqrt{x-1}-3;1\le x\le2\\\sqrt{x-1}-1-\sqrt{x-1}-3;x>2\end{cases}}\)
\(=\hept{\begin{cases}-2-2\sqrt{x-1};1\le x\le2\\-4;x>2\end{cases}}\)
\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}\) có ĐKXĐ là x>=1
\(=\sqrt{\left(x-1\right)-4\sqrt{x-1}+4}+\sqrt{\left(x-1\right)-6\sqrt{x-1}+9}\)
\(=\sqrt{\sqrt{x-1}^2-4\sqrt{x-1}+2^2}+\sqrt{\sqrt{x-1}^2-6\sqrt{x-1}+3^2}\)
\(=\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}-3\right)^2}\)
\(=\left(\sqrt{x-1}-2\right)+\left(\sqrt{x-1}-3\right)=2\sqrt{x-1}-5\) với x>5
\(=-\left(\sqrt{x-1}-2\right)-\left(\sqrt{x-1}-3\right)=-2\sqrt{x-1}+5\) với x<5