\(1,\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)

Để \(\frac{\sqrt{x}+1}{\sqrt{x}-3}\in Z\Rightarrow\frac{4}{\sqrt{x}-3}\in Z\)

\(\Rightarrow\sqrt{x}-3\in\left(1;4;-1;-4\right)\)

\(\Rightarrow\sqrt{x}\in\left(4;7;2;-1\right)\)

\(\Rightarrow\sqrt{x}=4\Leftrightarrow x=2\)

\(4,A=x+\sqrt{x}+1\)

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

\(A=\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{3}{4}\)

\(\Rightarrow A\ge\frac{3}{4}.\left(\sqrt{x}+\frac{1}{2}\right)^2\ge0\)

Dấu "=" xảy ra khi :

\(\sqrt{x}+\frac{1}{2}=0\Leftrightarrow\sqrt{x}=-\frac{1}{2}\)

Vậy Min A = 3/4 khi căn x = -1/2

đè hinh như là 6\(\sqrt{x}\) nhi bạn

điều kiện \(x\ge0\)và x khác 1/4

Q= \(\frac{3\sqrt{x}+2}{2\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+4}-\frac{x-6\sqrt{x}+5}{2x+7\sqrt{x}-4}=\frac{3x+14\sqrt{x}+8+2x-3\sqrt{x}+1-x+6\sqrt{x}-5}{2x+7\sqrt{x}-4}\)

=\(\frac{4x+17\sqrt{x}+4}{2x+7\sqrt{x}-4}\)

đề Q>1/2 thì \(\frac{4x+17\sqrt{x}+4}{2x+7\sqrt{x}-4}>\frac{1}{2}\)

<=> \(8x+34\sqrt{x}+8>2x+7\sqrt{x}-4\)<=> \(6x+27\sqrt{x}+12>0\) với mọi x>=0

vậy Q>1/2 khi x>=0 và x khác 1/4

a. ĐK \(\hept{\begin{cases}a\ge0\\a\ne4\\a\ne9\end{cases}}\)

P=\(\frac{2\sqrt{a}-9-\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)+\left(2\sqrt{a}+1\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)

\(=\frac{2\sqrt{a}-9-a+9+2a-4\sqrt{a}+\sqrt{a}-2}{\left(\sqrt{a}-3\right)\left(\sqrt{a}-2\right)}\)

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

b. P = \(\frac{\sqrt{a}+1}{\sqrt{a}-3}=1+\frac{4}{\sqrt{a}-3}\)

P nguyên \(\sqrt{a}-3\inƯ\left(4\right)\Rightarrow\sqrt{a}-3\in\left\{-4;-2;-1;1;2;4\right\}\)

\(\Rightarrow\sqrt{a}\in\left\{1;2;4;5;7\right\}\Rightarrow a\in\left\{1;4;16;25;49\right\}\)

c. \(P< 1\Rightarrow P-1< 0\Rightarrow\frac{\sqrt{a}+1-\sqrt{a}+3}{\sqrt{a}-3}< 0\Rightarrow\frac{4}{\sqrt{a}-3}< 0\)

\(\Rightarrow0\le a< 9\)và \(a\ne4\)

bn tham khảo câu hỏi tương tự nha!

hok tốt!

a) \(ĐKXĐ:x\ne4;x\ne9\)

b) \(A=\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)

        \(=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)

         \(=\frac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

          \(=\frac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{-\sqrt{x}+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

c) Ta có: \(A=\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\left(\sqrt{x}-3\right)+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)

\(\Rightarrow\sqrt{x}-3\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\) (ĐK: x thuộc Z)

Vậy để A thuộc Z khi x = {2;\(\sqrt{2};\sqrt{5};\sqrt{1};\sqrt{7}\) }

ĐKXĐ: \(x\ge0;x\ne\left\{4;9\right\}\)

\(A=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\frac{2\sqrt{x}-9+2x-3\sqrt{x}-2-x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)

\(x=7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2\Rightarrow\sqrt{x}=2-\sqrt{3}\)

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

\(A=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)

Để A nguyên \(\Rightarrow\sqrt{x}-3=Ư\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\)

\(\Rightarrow\sqrt{x}=\left\{-1\left(ktm\right);1;2;4;5;7\right\}\)

\(\Rightarrow x=\left\{1;4\left(ktm\right);16;25;49\right\}\)

em phai khong biet

moi hoc lop 6 thoi anh a