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ĐKXĐ: ...
\(A=\left(\frac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-1\right):\left(\frac{25-x+\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}-\frac{\sqrt{x}+3}{\sqrt{x}+5}\right)\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}+5}-\frac{\sqrt{x}+5}{\sqrt{x}+5}\right):\left(\frac{25-x+x-25}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}-\frac{\sqrt{x}+3}{\sqrt{x}+5}\right)\)
\(=\frac{-5}{\left(\sqrt{x}+5\right)}.\frac{\left(\sqrt{x}+5\right)}{-\left(\sqrt{x}+3\right)}=\frac{5}{\sqrt{x}+3}\)
b/ \(B=\frac{x+16}{\sqrt{x}+3}=\sqrt{x}-3+\frac{25}{\sqrt{x}+3}=\sqrt{x}+3+\frac{25}{\sqrt{x}+3}-6\)
\(\Rightarrow B\ge2\sqrt{\frac{\left(\sqrt{x}+3\right).25}{\sqrt{x}+3}}-6=4\)
\(B_{min}=4\) khi \(\left(\sqrt{x}+3\right)^2=25\Rightarrow x=4\)
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
a. P=\(\frac{x-5\sqrt{x}-x+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}:\frac{25-x-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}{\cdot\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}:\frac{-x+9}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{-5}{\sqrt{x}+5}.\frac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{5}{\sqrt{x}+3}\)
b. P=\(\frac{5}{\sqrt{x}+3}\)
P nguyên \(\Leftrightarrow\sqrt{x}+3\inƯ\left(5\right)\Rightarrow\sqrt{x}+3\in\left\{-5;-1;1;5\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{2\right\}\)\(\Rightarrow x=4\)
Vậy x=4 thì P nguyên
a, Với x >= 0 ; x khác 16
\(A=\left(\frac{x+5\sqrt{x}-27+\left(3-\sqrt{x}\right)\left(\sqrt{x}+4\right)}{x-16}\right):\frac{1}{\sqrt{x}+4}\)
\(=\left(\frac{x+5\sqrt{x}-27+3\sqrt{x}+12-x-4\sqrt{x}}{x-16}\right):\frac{1}{\sqrt{x}+4}\)
\(=\left(\frac{4\sqrt{x}-15}{x-16}\right):\frac{1}{\sqrt{x}+4}=\frac{4\sqrt{x}-15}{\sqrt{x}-4}\)
b, Ta có \(B=-2A\Rightarrow\sqrt{x}-4=-\frac{8\sqrt{x}-30}{\sqrt{x}-4}\)
\(\Leftrightarrow x-8\sqrt{x}+16=-8\sqrt{x}+30\Leftrightarrow x-14=0\Leftrightarrow x=14\left(tm\right)\)
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/ ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne25\end{matrix}\right.\)
Thay \(x=9\) vào biểu thức ta có :
\(A=\frac{\sqrt{9}+2}{\sqrt{9}-5}=\frac{3+2}{3-5}=-\frac{5}{2}\)
Vậy....
b/ Ta có :
\(B=\frac{3}{\sqrt{x}+5}+\frac{20-2\sqrt{x}}{x-25}\)
\(=\frac{3}{\sqrt{x}+5}+\frac{20-2\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)
\(=\frac{3\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-5\right)}+\frac{20-2\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)
\(=\frac{3\sqrt{x}-15+20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\frac{\sqrt{x}+5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\frac{1}{\sqrt{x}-5}\)
Vậy...
c/ Ta có :
\(A=B.\left|x-4\right|\)
\(\Leftrightarrow\frac{\sqrt{x}+2}{\sqrt{x}-5}=\frac{1}{\sqrt{x}-5}\left|x-4\right|\)
\(\Leftrightarrow\sqrt{x}+2=\left|x-4\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+2=x-4\\\sqrt{x}+2=4-x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{x}-6=0\\x+\sqrt{x}-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)=0\\\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=9\end{matrix}\right.\)
Vậy...
a) A=
\(\left(\frac{15-\sqrt{x}}{x-25}+\frac{2}{\sqrt{x}+5}\right)\colon\frac{\sqrt{x}+1}{\sqrt{x}-5},x\pm25,x\ge0\), MTC=x-25
\(=\left(\frac{15-\sqrt{x}}{x-25}+\frac{2\left(\sqrt{x}-5\right)}{x-25}\right)\colon\frac{\sqrt{x}+1}{\sqrt{x}-5}\)
=\(\left(\frac{15-\sqrt{x}}{x-25}+\frac{2\sqrt{x}-10}{x-25}\right)\colon\frac{\sqrt{x}+1}{\sqrt{x}-5}\)
=\(\left(\frac{15-\sqrt{x}+2\sqrt{x}-10}{x-25}\right)\colon\frac{\sqrt{x}+1}{\sqrt{x}-5}\)
=\(\frac{5+\sqrt{x}}{x-25}\cdot\frac{\sqrt{x}-5}{\sqrt{x}+1}\)
=\(\frac{5+\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\cdot\frac{\sqrt{x}-5}{\sqrt{x}+1}\)
=\(\frac{1}{\sqrt{x}+1}\)
Bạn thử tham khảo bài của tôi nhé :
b) M=A-B hay
M=\(\frac{1}{\sqrt{x}+1}-\frac{1-\sqrt{x}}{1+\sqrt{x}}\)
đkxd\(x\ge0,x\pm25\)
=\(\frac{1-1+\sqrt{x}}{\sqrt{x}+1}=\frac{\sqrt{x}}{\sqrt{x}+1}\)
Ta có: \(\frac{\sqrt{x}}{\sqrt{x}+1}=1+\frac{-1}{\sqrt{x}+1}\)
Để \(\frac{\sqrt{x}}{\sqrt{x}+1}\) nguyên thì \(\frac{-1}{\sqrt{x+1}}\) nguyên \(hay\sqrt{x}+1\inƯ\left(-1\right)\)
\(\rarr\sqrt{x}+1\in\left.\begin{cases}\\ \end{cases}-1;1\right)\)
+) \(Với\) \(\sqrt{x}+1=-1,\lrArr\sqrt{x}=-2\) (vô lí)
+)\(Với\) \(\sqrt{x}+1=1\lrArr\sqrt{x}=0\rarr x=0\)
Vậy để M nguyên thì \(x=0\)
a)B=√x(√x+5)/x-25 -10√x/x-25 -5(√x-5)/x-25=(x+5√x-10√x-5√x+25)/x-25=(x-10√x+25)/x-25=(√x-5)^2/(√x+5)(√x-5)=√x-5/√x+5
a.
\(B=\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-\dfrac{10\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-\dfrac{5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+5\right)-10\sqrt{x}-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\dfrac{x-10\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\dfrac{\left(\sqrt{x}-5\right)^2}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\dfrac{\sqrt{x}-5}{\sqrt{x}+5}\)
b.
\(B=\dfrac{\sqrt{x}+5-10}{\sqrt{x}+5}=1-\dfrac{10}{\sqrt{x}+5}\ge1-\dfrac{10}{0+5}=-1\)
\(B_{min}=-1\) khi \(x=0\)