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1.
\(Q=\left[\frac{\sqrt{x}+2}{(\sqrt{x}+1)^2}-\frac{\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+1)}\right].\sqrt{x}(\sqrt{x}+1)\)
\(=\frac{\sqrt{x}(\sqrt{x}+2)}{\sqrt{x}+1}-\frac{\sqrt{x}(\sqrt{x}-2)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}(\sqrt{x}+2)(\sqrt{x}-1)-\sqrt{x}(\sqrt{x}-2)(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{2x}{x-1}\)
2.
\(A=\left[\frac{\sqrt{x}+2-(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)}-\frac{4\sqrt{x}}{x-4}\right].\frac{x-4}{\sqrt{x}+1}\)
\(=\left(\frac{4}{x-4}-\frac{4\sqrt{x}}{x-1}\right).\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{x-4}.\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{\sqrt{x}+1}\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)
b) Ta có: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{x-2\sqrt{x}}\right)\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{4}{x-4}\right)\)
\(=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
d) Để A>0 thì \(\sqrt{x}-2>0\)
hay x>4
\(a,P=\dfrac{3\left(x+2\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\left(dk:x\ge0,x\ne1\right)\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{3\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{3\sqrt{x}-\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\left(\sqrt{x}+2\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}+2}\\ =\dfrac{2\sqrt{x}+4-\sqrt{x}-1}{\sqrt{x}+2}\\ =\dfrac{\sqrt{x}+3}{\sqrt{x}+2}\)
\(b,x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\)
\(\Rightarrow P=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}+3}{\sqrt{\left(\sqrt{5}-1\right)^2}+2}=\dfrac{\left|\sqrt{5}-1\right|+3}{\left|\sqrt{5}-1\right|+2}=\dfrac{\sqrt{5}-1+3}{\sqrt{5}-1+2}=\dfrac{\sqrt{5}+2}{\sqrt{5}+1}\)
a: \(A=\dfrac{-\left(\sqrt{x}-2\right)}{\sqrt{x}-1}:\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-\left(\sqrt{x}-2\right)^2}{3}\)
Đề bạn gõ sai, mình có sửa lại r nha
\(a,A=\dfrac{1-\sqrt{x}+1}{\sqrt{x}-1}:\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{3}\\ x=5\Leftrightarrow A=\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{3}=\dfrac{5-2\sqrt{5}}{3}\\ c,A=-\dfrac{1}{3}\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=-1\Leftrightarrow x-2\sqrt{x}+1=0\\ \Leftrightarrow\left(\sqrt{x}-1\right)^2=0\Leftrightarrow x=1\left(ktm\right)\Leftrightarrow x\in\varnothing\)
a: \(P=\dfrac{x-1}{\sqrt{x}}:\dfrac{x-1+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
b: \(x=\dfrac{2}{2+\sqrt{3}}=2\left(2-\sqrt{3}\right)=4-2\sqrt{3}\)
Khi x=4-2căn 3 thì \(P=\dfrac{\left(\sqrt{3}-1+1\right)^2}{\sqrt{3}-1}=\dfrac{3}{\sqrt{3}-1}=\dfrac{3\sqrt{3}+3}{2}\)
a) Ta có: \(P=\left(\dfrac{1}{\sqrt{x}-\sqrt{x-1}}-\dfrac{x-3}{\sqrt{x-1}-\sqrt{2}}\right)\left(\dfrac{2}{\sqrt{2}-\sqrt{x}}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}\right)\)
\(=\left(\dfrac{\sqrt{x}+\sqrt{x-1}}{x-\left(x-1\right)}-\dfrac{\left(\sqrt{x-1}-\sqrt{2}\right)\left(\sqrt{x-1}+\sqrt{2}\right)}{\sqrt{x-1}-\sqrt{2}}\right)\cdot\left(\dfrac{2}{\sqrt{2}-\sqrt{x}}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\right)\)
\(=\left(\sqrt{x}+\sqrt{x-1}-\sqrt{x-1}-\sqrt{2}\right)\cdot\left(\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\right)\)
\(=\left(\sqrt{x}-\sqrt{2}\right)\cdot\dfrac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{-\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\left(\sqrt{x}-\sqrt{2}\right)\cdot\dfrac{\sqrt{x}-\sqrt{2}}{-\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{2}-\sqrt{x}}{\sqrt{x}}\)
b) Ta có: \(x=3-2\sqrt{2}\)
\(=2-2\cdot\sqrt{2}\cdot1+1\)
\(=\left(\sqrt{2}-1\right)^2\)
Thay \(x=\left(\sqrt{2}-1\right)^2\) vào biểu thức \(P=\dfrac{\sqrt{2}-\sqrt{x}}{\sqrt{x}}\), ta được:
\(P=\dfrac{\sqrt{2}-\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(\sqrt{2}-1\right)^2}}\)
\(=\dfrac{\sqrt{2}-\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\)
\(=\dfrac{\sqrt{2}-\sqrt{2}+1}{\sqrt{2}-1}\)
\(=\dfrac{1}{\sqrt{2}-1}\)
\(=\sqrt{2}+1\)
Vậy: Khi \(x=3-2\sqrt{2}\) thì \(P=\sqrt{2}+1\)
Phần a,b,c bạn có thể tham khảo bài bên dưới.
Phần d.
ĐKXĐ: $x\geq 0; x\neq 4$
$A>5\Leftrightarrow \frac{x+9}{2\sqrt{x}}>5$ ($x> 0$)
$\Leftrightarrow x+9> 10\sqrt{x}$
$\Leftrightarrow x-10\sqrt{x}+9>0$
$\Leftrightarrow (\sqrt{x}-1)(\sqrt{x}-9)>0$
\(\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} \sqrt{x}-1>0\\ \sqrt{x}-9>0\end{matrix}\right.\\ \left\{\begin{matrix} \sqrt{x}-1<0\\ \sqrt{x}-9<0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} x>1\\ x>81\end{matrix}\right.\\ \left\{\begin{matrix} 0\leq x< 1\\ 0\leq x< 81\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow \left[\begin{matrix} x>81\\ 0\leq x< 1\end{matrix}\right.\)
Kết hợp với đkxđ suy ra $x>81$ hoặc $0< x< 1$
a
Với: x \(\ge0,x\) \(\ne4\) có:
\(A=\left(\dfrac{x-\sqrt{x}+7}{x-4}+\dfrac{\sqrt{x}+2}{x-4}\right):\left(\dfrac{\left(\sqrt{x}+2\right)^2}{x-4}-\dfrac{\left(\sqrt{x}-2\right)^2}{x-4}-\dfrac{6\sqrt{x}}{x-4}\right)\)
\(=\left(\dfrac{x-\sqrt{x}+7+\sqrt{x}+2}{x-4}\right):\left(\dfrac{x+4\sqrt{x}+4}{x-4}-\dfrac{x-4\sqrt{x}+4}{x-4}-\dfrac{6\sqrt{x}}{x-4}\right)\)
\(=\left(\dfrac{x+9}{x-4}\right):\left(\dfrac{x+4\sqrt{x}+4-x+4\sqrt{x}-4-6\sqrt{x}}{x-4}\right)\)
\(=\left(\dfrac{x+9}{x-4}\right):\left(\dfrac{2\sqrt{x}}{x-4}\right)\)
\(=\dfrac{\left(x+9\right)\left(x-4\right)}{2\sqrt{x}\left(x-4\right)}=\dfrac{x+9}{2\sqrt{x}}\)
b
Giải \(x^2-5x+4=0\)
Nhẩm nghiệm: a + b + c = 0 (1 - 5 + 4 = 0)
\(\Rightarrow x_1=1;x_2=\dfrac{c}{a}=\dfrac{4}{1}=4\)
Thay x = 1 vào A:
\(A=\dfrac{1+9}{2\sqrt{1}}=\dfrac{10}{2}=5\)
Thay x = 4 vào A:
\(A=\dfrac{4+9}{2.\sqrt{4}}=\dfrac{13}{2.2}=\dfrac{13}{4}\)
c
ĐK: x > 0
\(A=0\Leftrightarrow\dfrac{x+9}{2\sqrt{x}}=0\)
=> \(x+9=0\Rightarrow x=-9\) (không thỏa mãn)
Vậy không xác định được giá trị x
d
ĐK: x > 0
\(A>5\Leftrightarrow\dfrac{x+9}{2\sqrt{x}}>5\)
\(\Leftrightarrow x+9>5.2\sqrt{x}\Leftrightarrow x+9>10\sqrt{x}\)
\(\Leftrightarrow\left(x+9\right)^2>\left(10\sqrt{x}\right)^2=100x\)
<=> \(x^2+18x+81-100x>0\)
<=> \(x^2-82x+81>0\)
<=> \(x^2-81x-x+81>0\)
<=> \(x\left(x-81\right)-\left(x-81\right)>0\)
<=> \(\left(x-1\right)\left(x-81\right)>0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x-1>0\\x-81>0\end{matrix}\right.\\\left[{}\begin{matrix}x-1< 0\\x-81< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>1\\x>81\end{matrix}\right.\\\left[{}\begin{matrix}x< 1\\x< 81\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>81\\x< 81\end{matrix}\right.\)
Vậy để A > 5 thì x > 81 và 0 < x < 81
a: \(A=\left(\dfrac{x+4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}-1+\sqrt{x}+1}{x-1}\)
\(=\dfrac{x+4\sqrt{x}+4-x-2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{x-1}{2\sqrt{x}}\)
\(=\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{2\sqrt{x}+2}{\sqrt{x}}\)
c: 2x-3căn x-5=0
=>2x-5căn x+2căn x-5=0
=>2căn x-5=0
=>x=25/4
Khi x=25/4 thì \(A=\dfrac{2\cdot\dfrac{5}{4}+2}{\dfrac{5}{4}}=\dfrac{18}{5}\)