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#)Giải :
a) \(A=\left(\frac{\sqrt{x}}{2}-\frac{1}{2\sqrt{x}}\right)\left(\frac{x\sqrt{x}}{\sqrt{x}-1}-\frac{x+\sqrt{x}}{\sqrt{x}-1}\right)\)
\(=\frac{x-1}{2\sqrt{x}}\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)^2-\sqrt{x}\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{x-1}{2\sqrt{x}}.\frac{x\sqrt{x}-2x+\sqrt{x}-x\sqrt{x}-2x-\sqrt{x}}{x-1}\)
\(=\frac{-4}{2\sqrt{x}}=-2\sqrt{x}\)
\(P=\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{4-6\sqrt{a}}{1-a}-\frac{-3}{\sqrt{a}+1}\)
ĐK : \(\hept{\begin{cases}a\ge0\\a\ne1\end{cases}}\)
a) \(P=\frac{\sqrt{a}}{\sqrt{a}-1}+\frac{4-6\sqrt{a}}{a-1}+\frac{3}{\sqrt{a}+1}\)
\(=\frac{\sqrt{a}}{\sqrt{a}-1}+\frac{4-6\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{3}{\sqrt{a}+1}\)
\(=\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{4-6\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{3\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\frac{a+\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{4-6\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{3\sqrt{a}-3}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\frac{a+\sqrt{a}+4-6\sqrt{a}+3\sqrt{a}-3}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\frac{a-2\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\frac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\frac{\sqrt{a}-1}{\sqrt{a}+1}\)
Với \(a=4-2\sqrt{3}\)( tmđk \(\hept{\begin{cases}a\ge0\\a\ne1\end{cases}}\))
\(P=\frac{\sqrt{4-2\sqrt{3}}-1}{\sqrt{4-2\sqrt{3}}+1}\)
\(=\frac{\sqrt{3-2\sqrt{3}+1}-1}{\sqrt{3-2\sqrt{3}+1}+1}\)
\(=\frac{\sqrt{\left(\sqrt{3}\right)^2-2\sqrt{3}+1^2}-1}{\sqrt{\left(\sqrt{3}\right)^2-2\sqrt{3}+1^2}+1}\)
\(=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}-1}{\sqrt{\left(\sqrt{3}-1\right)^2}+1}\)
\(=\frac{\left|\sqrt{3}-1\right|-1}{\left|\sqrt{3}-1\right|+1}\)
\(=\frac{\sqrt{3}-1-1}{\sqrt{3}-1+1}=\frac{\sqrt{3}-2}{\sqrt{3}}\)
b) \(P=\frac{\sqrt{a}-1}{\sqrt{a}+1}=\frac{\sqrt{a}+1-2}{\sqrt{a}+1}=1-\frac{2}{\sqrt{a}+1}\)( ĐK \(\hept{\begin{cases}a\ge0\\a\ne1\end{cases}}\))
Để P đạt giá trị nguyên => \(\frac{2}{\sqrt{a}+1}\)nguyên
=> \(2⋮\sqrt{a}+1\)
=> \(\sqrt{a}+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
=> \(\sqrt{a}\in\left\{0;1\right\}\)< đã loại hai trường hợp âm >
=> \(a\in\left\{0\right\}\)< loại trường hợp a = 1 >
Vậy với a = 0 thì P có giá trị nguyên
a) \(M=\frac{x+1+\sqrt{x}}{x+1}:\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{\sqrt{x}\left(x+1\right)-\left(x+1\right)}\right)\)
\(=\frac{x+\sqrt{x}+1}{x+1}:\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{x+\sqrt{x}+1}{x+1}:\frac{x+1-2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\)\(=\frac{x+\sqrt{x}+1}{x+1}.\frac{\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2}=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}\)
b) \(M>3\Rightarrow\frac{x+\sqrt{x}+1}{\sqrt{x}-1}>3\Leftrightarrow\frac{x+\sqrt{x}+1}{\sqrt{x}-1}-3>0\)
\(\Leftrightarrow\frac{x+\sqrt{x}+1-3\left(\sqrt{x}-1\right)}{\sqrt{x}-1}>0\Leftrightarrow\frac{x+\sqrt{x}+1-3\sqrt{x}+3}{\sqrt{x}-1}>0\)\(\Leftrightarrow\frac{x-2\sqrt{x}+4}{\sqrt{x}-1}>0\)
Ta có: \(x-2\sqrt{x}+4=x-2\sqrt{x}+1+3=\left(\sqrt{x}-1\right)+3>0\)\(\Rightarrow\sqrt{x}-1>0\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)
Vậy x>1
c) \(M=7\Rightarrow\frac{x+\sqrt{x}+1}{\sqrt{x}-1}=7\Rightarrow x+\sqrt{x}+1=7\left(\sqrt{x}-1\right)\)
\(\Leftrightarrow x+\sqrt{x}+1=7\sqrt{x}-7\Leftrightarrow x-6\sqrt{x}+8=0\)\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-2=0\\\sqrt{x}-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=2\\\sqrt{x}=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\\x=16\end{cases}\left(tm\right)}}\)
Vậy \(x\in\text{{}4;16\)
\(a,B=\left(\frac{15-\sqrt{x}}{x-25}+\frac{2}{\sqrt{x}+5}\right):\frac{\sqrt{x}+1}{\sqrt{x}-5}\)
\(B=\left(\frac{15-\sqrt{x}+2\sqrt{x}-10}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\frac{\sqrt{x}-5}{\sqrt{x}+1}\)
\(B=\frac{5+\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\frac{\sqrt{x}-5}{\sqrt{x}+1}\)
\(B=\frac{1}{\sqrt{x}+1}\)
\(b,P=A.B=\frac{4\left(\sqrt{x}+1\right)}{25-x}.\frac{1}{\sqrt{x}+1}\)
\(P=\frac{4}{25-x}\)
bổ sung điều kiện cho câu b là x nguyên
\(TH1:x>25< =>P< 0\left(KTM\right)\)
\(TH2:x< 25< =>P>0\)mà x nguyên
\(\frac{4}{25-x}\le4\)
dấu "=" xảy ra khi \(x=24\)
\(< =>MAX:P=4\)
c) Đk: x \(\ge\)0; x \(\ne\)4; x \(\ne\)9
A = \(-\frac{1}{\sqrt{x}-3}\) => -2A = \(\frac{2}{\sqrt{x}-3}\)
Để -2A thuộc Z <=> \(2⋮\sqrt{x}-3\)
<=> \(\sqrt{x}-3\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
Lập bảng:
| \(\sqrt{x}-3\) | 1 | -1 | 2 | -2 |
| x | 8 | 4 (ktm) | 25 | 1 |
Vậy ....
sao biểu thức khi rút gọn xấu vậy bạn ? đề có sai khum :vv, thế tìm x dài lắm bạn ạ
a, Với x > 0 ; \(x\ne1\)
\(M=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{2}{x}-\frac{2-x}{x\sqrt{x}-x}\right)\)
\(=\left(\frac{x+\sqrt{x}+x-\sqrt{x}}{x-1}\right):\left(\frac{2\sqrt{x}-2-2+x}{x\left(\sqrt{x}-1\right)}\right)\)
\(=\left(\frac{2x}{x-1}\right):\left(\frac{x+2\sqrt{x}-4}{x\left(\sqrt{x}-1\right)}\right)=\frac{2x^2}{\left(\sqrt{x}+1\right)\left(x+2\sqrt{x}-4\right)}\)
câu a rút gọn
M = \(\frac{\sqrt{a}-1}{\sqrt{a}}\)