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a) Ta có: \(A=\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\sqrt{x}\right)\cdot\left(\frac{1-\sqrt{x}}{1-x}\right)^2\)
\(=\left(\frac{1-x\sqrt{x}+\sqrt{x}\left(1-\sqrt{x}\right)}{1-\sqrt{x}}\right)\cdot\left(\frac{1}{1+\sqrt{x}}\right)^2\)
\(=\frac{1-x\sqrt{x}+\sqrt{x}-x}{1-\sqrt{x}}\cdot\frac{1}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{-\left(x-1\right)\left(-1-\sqrt{x}\right)}{1-\sqrt{x}}\cdot\frac{1}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{\left(1+\sqrt{x}\right)\cdot\left(-1-\sqrt{x}\right)}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{-1\cdot\left(1+\sqrt{x}\right)^2}{\left(1+\sqrt{x}\right)^2}=-1\)
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
Ta có: \(B=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)+5\left(\sqrt{x}+1\right)+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x+2\sqrt{x}-3+5\sqrt{x}+5+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+6}{\sqrt{x}-1}\)
do đó \(P=\frac{\sqrt{x}-1}{\sqrt{x}+1}.\frac{\sqrt{x}-6}{\sqrt{x}-1}=\frac{\sqrt{x}-6}{\sqrt{x}+1}=1-\frac{7}{\sqrt{x}+1}\)
Vì \(x\ge0\Rightarrow0< \frac{7}{\sqrt{x}+1}\le7\)
Để P nguyên thì \(\frac{7}{\sqrt{x}+1}\in Z\)
do đó \(\frac{7}{\sqrt{x}+1}\in\left\{1,2,3,4,5,6,7\right\}\)
Đến đây xét từng TH là ra
rút gọn B ta có B=\(\frac{\sqrt{x}+6}{\sqrt{x}-1}\)\(\Rightarrow\)\(AB=\frac{\sqrt{x}+6}{\sqrt{x}+1}\in Z\)
=\(1+\frac{5}{\sqrt{x}+1}\)
Vì 1\(\in Z\) nên để P thuộc Z thì \(\frac{5}{\sqrt{x}+1}\in Z\)
\(\Rightarrow\left(\sqrt{x}+1\right)\inƯ\left(5\right)=\pm1;\pm5\)
Đến đây thì ez rồi
Lời giải:
a)
\(A=\frac{\sqrt{3}-1+\sqrt{3}+1}{(\sqrt{3}+1)(\sqrt{3}-1)}+2-\sqrt{3}=\frac{2\sqrt{3}}{3-1}+2-\sqrt{3}=\sqrt{3}+2-\sqrt{3}=2\)
b)
\(B=\left(\frac{1}{\sqrt{x}(\sqrt{x}-1)}+\frac{\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)}\right):\frac{\sqrt{x}}{(\sqrt{x}-1)^2}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}.(\sqrt{x}-1)}.\frac{(\sqrt{x}-1)^2}{\sqrt{x}}=\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{x}=\frac{x-1}{x}\)
\(A=\frac{\left(1+\sqrt{x}\right)^2-4\sqrt{x}}{\sqrt{x}-1}\) \(\left(x\ge0;x\ne1\right)\)
\(A=\frac{x+2\sqrt{x}+1-4\sqrt{x}}{\sqrt{x}-1}=\frac{x-2\sqrt{x}+1}{\sqrt{x}-1}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}=\sqrt{x}-1\)
và \(B=\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{2}}+\frac{2+\sqrt{2}}{\sqrt{x}+1}\)
\(B=\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{2}}+\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)
\(B=\sqrt{3}+2+\frac{1}{\sqrt{3}-\sqrt{2}}+\sqrt{2}\)
\(B=\sqrt{3}+\sqrt{2}+\frac{1}{\sqrt{3}-\sqrt{2}}+2\)
\(B=\frac{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)+1}{\sqrt{3}-\sqrt{2}}+2\)
\(B=\frac{3-2+1}{\sqrt{3}-\sqrt{2}}+2\)
\(B=\frac{2}{\sqrt{3}-\sqrt{2}}+2\)
để A = B thì \(\sqrt{x}-1\)= \(\frac{2}{\sqrt{3}-\sqrt{2}}+2\)
\(\sqrt{x}=\frac{2}{\sqrt{3}-\sqrt{2}}+3\)
\(\sqrt{x}=\frac{2\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+3\)
\(\sqrt{x}=2\sqrt{3}+2\sqrt{2}+3\)
tới bước này tui bí :(( mong các bạn giỏi khác giúp bạn :D
b) \(A=\sqrt{x}+1\)
Để A \(\in\)Z \(\Leftrightarrow\sqrt{x}+1\in Z\)\(\Leftrightarrow\sqrt{x}\in Z\)
\(\Leftrightarrow\sqrt{x}=a\left(a\in Z;a\ge0\right)\)\(\Leftrightarrow x=a^2\)
vậy x là bình phương 1 số tự nhiên thì A thuộc Z
giúp mk vs