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Trả lời:
a, \(A=\frac{\sqrt{x}}{\sqrt{x}-5}-\frac{10\sqrt{x}}{x-25}-\frac{5}{\sqrt{x}+5}\left(ĐK:x\ge0;x\ne25\right)\)
\(=\frac{\sqrt{x}}{\sqrt{x}-5}-\frac{10\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-\frac{5}{\sqrt{x}+5}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-\frac{10\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-\frac{5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+5\right)-10\sqrt{x}-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\frac{x+5\sqrt{x}-10\sqrt{x}-5\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\frac{x-10\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\frac{\left(\sqrt{x}-5\right)^2}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\frac{\sqrt{x}-5}{\sqrt{x}+5}\)
b, Thay x = 9 vào A, ta được:
\(A=\frac{\sqrt{9}-5}{\sqrt{9}+5}=\frac{3-5}{3+5}=\frac{-2}{8}=-\frac{1}{4}\)
c, \(A< \frac{1}{3}\Leftrightarrow\frac{\sqrt{x}-5}{\sqrt{x}+5}< \frac{1}{3}\Leftrightarrow\frac{\sqrt{x}-5}{\sqrt{x}+5}-\frac{1}{3}< 0\)
\(\Leftrightarrow\frac{3\left(\sqrt{x}-5\right)}{3\left(\sqrt{x}+5\right)}-\frac{\sqrt{x}+5}{3\left(\sqrt{x}+5\right)}< 0\)
\(\Leftrightarrow\frac{3\sqrt{x}-15-\sqrt{x}-5}{3\left(\sqrt{x}+5\right)}< 0\)
\(\Leftrightarrow\frac{2\sqrt{x}-20}{3\left(\sqrt{x}+5\right)}< 0\)
\(\Rightarrow2\sqrt{x}-20< 0\) (vì \(3\left(\sqrt{x}+5\right)>0\) )
\(\Leftrightarrow2\sqrt{x}< 20\)
\(\Leftrightarrow\sqrt{x}< 10\)
\(\Leftrightarrow x< 100\)
Vậy \(0\le x< 100\)và \(x\ne25\) là giá trị cần tìm.
a, \(P=\frac{\sqrt{x}+2}{\sqrt{x}+1}-\frac{\sqrt{x}+3}{5-\sqrt{x}}-\frac{3x+4\sqrt{x}-5}{x-4\sqrt{x}-5}\)
\(P=\frac{\sqrt{x}+2}{\sqrt{x}+1}+\frac{\sqrt{x}+3}{\sqrt{x}-5}-\frac{3x+4\sqrt{x}-5}{x-4\sqrt{x}-5}\)
\(P=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-5\right)}+\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-5\right)}-\frac{3x+4\sqrt{x}-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-5\right)}\)
\(P=\frac{x-3\sqrt{x}-10+x+4\sqrt{x}+3-3x-4\sqrt{x}+5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-5\right)}\)
\(P=\frac{-x-3\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-5\right)}\)
\(P=\frac{\left(\sqrt{x}+1\right)\left(-\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-5\right)}=\frac{-\sqrt{x}-2}{\sqrt{x}-5}\)
để P > -2
\(\Rightarrow\frac{-\sqrt{x}-2}{\sqrt{x}-5}>-2\) đoạn này đang chưa nghĩ ra
c, \(P=\frac{-\sqrt{x}-2}{\sqrt{x}-5}\in Z\) \(\Rightarrow-\sqrt{x}-2⋮\sqrt{x}-5\)
=> -căn x + 5 - 7 ⋮ căn x - 5
=> -(căn x - 5) - 7 ⋮ căn x - 5
=> 7 ⋮ x - 5 đoạn này dễ
a, Với \(x\ge0;x\ne25\)thì \(P=\frac{\sqrt{x}+2}{5-\sqrt{x}}\) đoạn này đúng rồi
\(P>-2\)\(\Leftrightarrow\frac{\sqrt{x}+2}{5-\sqrt{x}}>-2\)
\(\Leftrightarrow\frac{\sqrt{x}+2}{5-\sqrt{x}}+2>0\)
\(\Leftrightarrow\frac{12-\sqrt{x}}{5-\sqrt{x}}>0\)
Xét 2 trường hợp cùng âm, cùng dương hoặc "trong trái ngoài cùng"
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}>12\\0\le\sqrt{x}< 5\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x>144\\0\le x< 25\end{cases}}\)
Làm luôn cho đầy đủ =)
b) lấy kết quả rút gọn của câu A ta được
\(P=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}< 1.=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}-1< 0\)
\(P=\frac{x+\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}-1}=\frac{x+2}{\sqrt{x}-1}\)
đề bài cho x>=0 ta suy ra luôn
\(x+2>0\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow x< 1\)
vậy x <1 thì P < 1
\(P=\left(\frac{x+1+\sqrt{x}}{x+1}\right):\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{\sqrt{x}\left(x+1\right)-\left(x+1\right)}\right).\)
\(P=\left(\frac{x+1+\sqrt{x}}{x+1}\right):\left(\frac{1}{\sqrt{x-1}}-\frac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(P=\left(\frac{x+1+\sqrt{x}}{x+1}\right):\left(\frac{x+1-2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(P=\frac{\left(x+\sqrt{x}+1\right)}{\left(x+1\right)}:\frac{\left(\sqrt{x}-1\right)^2}{\left(x+1\right)\left(\sqrt{x}-1\right)}=\frac{\left(x+\sqrt{x}+1\right)}{\left(x+1\right)}.\frac{\left(x+1\right)}{\sqrt{x}-1}\)
\(P=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}\)
a/ \(B=\left(\frac{1}{\sqrt{x}+2}+\frac{7}{x-4}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}-2}-1\right)\)
=> \(B=\left(\frac{1}{\sqrt{x}+2}+\frac{7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\frac{\sqrt{x}-1-\sqrt{x}+2}{\sqrt{x}-2}\right)\)
=> \(B=\frac{\sqrt{x}+5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\frac{1}{\sqrt{x}-2}\)
=> \(B=\frac{\sqrt{x}+5}{\sqrt{x}+2}\)
b/ B>2 <=> \(\frac{\sqrt{x}+5}{\sqrt{x}+2}>2\) <=> \(\sqrt{x}+5>2\sqrt{x}+4\)
<=> \(1>\sqrt{x}\)=> \(-1\le x\le1\)
c/ \(B=\frac{\sqrt{x}+5}{\sqrt{x}+2}=\frac{\sqrt{x}+2+3}{\sqrt{x}+2}=1+\frac{3}{\sqrt{x}+2}\)
Để Bmax thì \(\sqrt{x}+2\) đạt giá trị nhỏ nhất . Do \(\sqrt{x}+2\ge2\)=> Đạt nhỏ nhất khi x=0
Khí đó giá trị lớn nhất của B là: \(1+\frac{3}{2}=\frac{5}{2}\)Đạt được khi x=0
ĐKXĐ của cả A và B : \(\hept{\begin{cases}x\ge0\\x\ne25\end{cases}}\)
\(A=\frac{\sqrt{x}+2}{\sqrt{x}-5}\)
\(B=\frac{x+3\sqrt{x}}{x-25}+\frac{1}{\sqrt{x}+5}\)
\(=\frac{x+3\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}+\frac{\sqrt{x}-5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\frac{x+4\sqrt{x}-5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\frac{x-\sqrt{x}+5\sqrt{x}-5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)+5\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}-5}\)
\(M=\frac{B}{A}=\frac{\frac{\sqrt{x}-1}{\sqrt{x}-5}}{\frac{\sqrt{x}+2}{\sqrt{x}-5}}=\frac{\sqrt{x}-1}{\sqrt{x}-5}\times\frac{\sqrt{x}-5}{\sqrt{x}+2}=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
ĐKXĐ của M : \(\hept{\begin{cases}x\ge0\\x\ne25\end{cases}}\)
\(M\times\left(\sqrt{x}+2\right)\ge3x-3\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+2}\times\left(\sqrt{x}+2\right)\ge3x-3\)( ĐK : \(\hept{\begin{cases}x\ge0\\x\ne25\end{cases}}\))
\(\Leftrightarrow\sqrt{x}-1\ge3x-3\)
\(\Leftrightarrow3x-\sqrt{x}-3+1\ge0\)
\(\Leftrightarrow3x-\sqrt{x}-2\ge0\)
\(\Leftrightarrow3x-3\sqrt{x}+2\sqrt{x}-2\ge0\)
\(\Leftrightarrow3\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)\ge0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(3\sqrt{x}+2\right)\ge0\)
Dễ dàng nhận thấy \(3\sqrt{x}+2\ge2>0\forall x\ge0\)
\(\Rightarrow\sqrt{x}-1\ge0\)
\(\Leftrightarrow x\ge1\)
Kết hợp với điều kiện => Với 0 ≤ x ≤ 1 thì thỏa mãn đề bài
a) x = 16 (tm) => A = \(\frac{\sqrt{16}-2}{\sqrt{16}+1}=\frac{4-2}{4+1}=\frac{2}{5}\)
b) B = \(\left(\frac{1}{\sqrt{x}+5}-\frac{x+2\sqrt{x}-5}{25-x}\right):\frac{\sqrt{x}+2}{\sqrt{x}-5}\)
B = \(\frac{\sqrt{x}-5+x+2\sqrt{x}-5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\cdot\frac{\sqrt{x}-5}{\sqrt{x}+2}\)
B = \(\frac{x+3\sqrt{x}-10}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)
B = \(\frac{x+5\sqrt{x}-2\sqrt{x}-10}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)
B = \(\frac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-2}{\sqrt{x}+2}\)
c) P = \(\frac{B}{A}=\frac{\sqrt{x}-2}{\sqrt{x}+2}:\frac{\sqrt{x}-2}{\sqrt{x}+1}=\frac{\sqrt{x}+1}{\sqrt{x}+2}\)
=> \(P\left(\sqrt{x}+2\right)\ge x+6\sqrt{x}-13\)
<=> \(\frac{\sqrt{x}+1}{\sqrt{x}+2}.\left(\sqrt{x}+2\right)-x-6\sqrt{x}+13\ge0\)
<=> \(-x-6\sqrt{x}+13+\sqrt{x}+1\ge0\)
<=> \(-x-5\sqrt{x}+14\ge0\)
<=> \(x+5\sqrt{x}-14\le0\)
<=> \(x+7\sqrt{x}-2\sqrt{x}-14\le0\)
<=> \(\left(\sqrt{x}+7\right)\left(\sqrt{x}-2\right)\le0\)
Do \(\sqrt{x}+7>0\) với mọi x => \(\sqrt{x}-2\le0\)
<=> \(\sqrt{x}\le2\) <=> \(x\le4\)
Kết hợp với Đk: x \(\ge\)0; x \(\ne\)4; x \(\ne\)25
và x thuộc Z => x = {0; 1; 2; 3}
d) M = \(3P\cdot\frac{\sqrt{x}+2}{x+\sqrt{x}+4}\) <=>M = \(3\cdot\frac{\sqrt{x}+1}{\sqrt{x}+2}\cdot\frac{\sqrt{x}+2}{x+\sqrt{x}+4}\)
M = \(\frac{3\sqrt{x}+3}{x+\sqrt{x}+4}=\frac{x+\sqrt{x}+4-x+2\sqrt{x}-1}{\left(x+\sqrt{x}+\frac{1}{4}\right)+\frac{15}{4}}=1-\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{15}{4}}\le1\)(Do \(\left(\sqrt{x}-1\right)^2\ge0\) và \(\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{15}{4}>0\))
Dấu "=" xảy ra <=> \(\sqrt{x}-1=0\) <=> \(x=1\)
Vậy MaxM = 1 khi x = 1
f)\(\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}:\frac{1}{\sqrt{x}-\sqrt{y}}\)
\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}.\left(\sqrt{x}-\sqrt{y}\right)\)
\(=x-y\)
Đkxđ: x≥0, x khác 49
A= \(\frac{2\sqrt{x}}{\sqrt{x}-7}+\frac{x+21\sqrt{x}}{x-49}\)
A=\(\frac{2\sqrt{x}\left(\sqrt{x}+7\right)+x+21\sqrt{x}}{x-49}\)
=\(\frac{3x+35\sqrt{x}}{x-49}\)
B=\(\frac{\sqrt{x}}{x-5}\)
P=A/B=\(\frac{\sqrt{x}\left(3\sqrt{x}+35\right)\left(x-5\right)}{\sqrt{x}\left(x-49\right)}\)
=\(\frac{\left(3\sqrt{x}+35\right)\left(x-5\right)}{\left(x-49\right)}\)
P=1/3
<=>\(\frac{\left(3\sqrt{x}+35\right)\left(x-5\right)}{\left(x-49\right)}=\frac{1}{3}\)
<=>...