Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
dkxd \(x\ge4\)
A=\(\sqrt{x-4+4\sqrt{x-4}+4}\) +\(\sqrt{x-4-4\sqrt{x-4}+4}\)
=\(\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\)
th1 \(\sqrt{x-4}\ge2\Leftrightarrow x\ge8\)
ta co\(\sqrt{x-4}+2+\sqrt{x-4}-2=2\sqrt{x-4}\)
th2 \(4\le x< 8\)
ta co \(\sqrt{x-4}+2+2-\sqrt{x-4}=4\)
mọi ng ơi mk viết thiếu dấu ngoặc nha.thiếu ngoặc lownns nha. đóng ngoắc ở trước dấu chia
ĐKXĐ X >= 4
\(y=\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}\)
\(=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}\)
\(=\sqrt{x-4}+2+l\sqrt{x-4}-2l\)
(+) \(l\sqrt{x-4}-2l=\sqrt{x-4}-2\) khi x>= 8
(+) \(l\sqrt{x-4}-2l=2-\sqrt{x-4}\) khi x<= 8
Với x >=8 => y = \(\sqrt{x-4}+2+\sqrt{x-4}-2=2\sqrt{x-4}\)
Với \(x<=8\Rightarrow y=\sqrt{x-4}+2+2-\sqrt{x-4}=4\)
\(a,ĐKXĐ:x\ne\sqrt{2};-\sqrt{2};x\ne4\)
\(P=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)
\(P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{x-4}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{x-4}+\frac{-2-5\sqrt{x}}{x-4}\)
\(P=\frac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{x-4}\)
\(P=\frac{3x-6\sqrt{x}}{x-4}\)
\(b;\)Để P<2
\(\Rightarrow3x-6\sqrt{x}< 2x-8\)
\(\Rightarrow3x-2x< -8+6\sqrt{x}\)
\(\Rightarrow x-6\sqrt{x}< -8\)
\(\Rightarrow\sqrt{x}\left(\sqrt{x}-6\right)< 8\)
Tìm x là xong
a) \(P=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)\(\left(ĐKXĐ:x>4\right)\)
\(P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{-2-5\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(P=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3\sqrt{x}}{\sqrt{x}+2}\)
b) Ta có : \(P< 2\Leftrightarrow\frac{3\sqrt{x}}{\sqrt{x}+2}< 2\)
\(\Leftrightarrow\frac{3\sqrt{x}}{\sqrt{x}+2}-2< 0\)
\(\Leftrightarrow\frac{3\sqrt{x}}{\sqrt{x}+2}-\frac{2\sqrt{x}+4}{\sqrt{x}+2}< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-4}{\sqrt{x}+2}< 0\)
Mà \(\sqrt{x}-4< \sqrt{x}+2\)
\(\Rightarrow\hept{\begin{cases}\sqrt{x}-4< 0\\\sqrt{x}+2>0\end{cases}\Leftrightarrow}\hept{\begin{cases}\sqrt{x}< 4\\\sqrt{x}>-2\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 16\\x>4\end{cases}}\Leftrightarrow4< x< 16\)
Vậy ...
Kết quả rút gọn: \(P=\frac{\sqrt{x}+2}{\sqrt{x}-1}\)
\(M=\frac{x+12}{\sqrt{x}-1}.\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{x+12}{\sqrt{x}+2}\)
\(M=\frac{x-4+16}{\sqrt{x}+2}=\sqrt{x}-2+\frac{16}{\sqrt{x}+2}=\left(\sqrt{x}+2+\frac{16}{\sqrt{x}+2}\right)-4\)
Âp dụng BĐT AM-GM cho 2 số không âm ta có:
\(M\ge2\sqrt{\left(\sqrt{x}+2\right).\frac{16}{\sqrt{x}+2}}-4=2.4-4=4\)
Vậy min M =4. Dấu bằng xảy ra \(\Leftrightarrow\left(\sqrt{x}+2\right)^2=16\Leftrightarrow\sqrt{x}+2=4\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)
\(P=\left(\frac{3}{x-1}+\frac{1}{\sqrt{x}+1}\right):\frac{1}{\sqrt{x}+1}\) \(ĐKXĐ:x\ne1\)
\(P=\left(\frac{3}{x-1}+\frac{\sqrt{x}-1}{x-1}\right):\frac{1}{\sqrt{x}+1}\)
\(P=\frac{\sqrt{x}+2}{x-1}.\left(\sqrt{x}+1\right)\)
\(P=\frac{\sqrt{x}+2}{\sqrt{x}-1}\)
b) theo câu a) \(P=\frac{\sqrt{x}+2}{\sqrt{x}-1}\) với \(ĐKXĐ:x\ne1\)
theo bài ra \(P=\frac{5}{4}\)thì \(\Leftrightarrow\frac{\sqrt{x}+2}{\sqrt{x}-1}=\frac{5}{4}\)
\(\Leftrightarrow\left(\sqrt{x}+2\right).4=\left(\sqrt{x}-1\right).5\)
\(\Leftrightarrow4\sqrt{x}+8=5\sqrt{x}-5\)
\(\Leftrightarrow-\sqrt{x}+13=0\)
\(\Leftrightarrow-\sqrt{x}=-13\)
\(\Leftrightarrow\sqrt{x}=13\)
\(\Leftrightarrow x=169\)
vậy \(x=169\)khi \(P=\frac{5}{4}\)