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a: \(P=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b:Sửa đề: 2A
2A=2căn x+5
=>(2căn x+2)/căn x=2căn x+5
=>2x+5căn x-2căn x-2=0
=>2x+3căn x-2=0
=>(căn x+2)(2căn x-1)=0
=>x=1/4
a: \(A=\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)
\(A=\left[1:\left(1-\frac{\sqrt{a}}{1+\sqrt{a}}\right)\right]\left[\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{a\sqrt{a}-a+\sqrt{a}-1}\right]\)
\(=\left[1:\left(\frac{1+\sqrt{a}-\sqrt{a}}{1+\sqrt{a}}\right)\right]\left[\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}\right]\)
\(=\left(1:\frac{1}{1+\sqrt{a}}\right).\frac{a+1-2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\)
\(=\left(\sqrt{a}+1\right).\frac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(a+1\right)}\)
\(=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{a+1}=\frac{a-1}{a+1}\)
Ta có: \(P=\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right)\cdot\left(1-\dfrac{1}{\sqrt{a}}\right)\)
\(=\dfrac{1+\sqrt{a}-1+\sqrt{a}}{-\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
\(=\dfrac{-2\sqrt{a}}{\sqrt{a}\left(\sqrt{a}+1\right)}=\dfrac{-2}{\sqrt{a}+1}\)
\(A=\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\)
\(=\left(\dfrac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+4\sqrt{a}\right)\cdot\dfrac{a-1}{\sqrt{a}}\)
\(=\left(\dfrac{2\cdot2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+4\sqrt{a}\right)\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}\)
\(=\left(\dfrac{4\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+4\sqrt{a}\right)\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}\)
\(=\dfrac{4\sqrt{a}+4\sqrt{a}\cdot\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}\)
\(=4\sqrt{a}\cdot\left(1+\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)\right)\cdot\dfrac{1}{\sqrt{a}}\)
\(=4\left(1+a-1\right)\)
\(=4a\)
Để \(\sqrt{a}>A\) thì \(\sqrt{a}>4a\)
\(\Leftrightarrow a>\sqrt{4a}\left(đk:a\ge0\right)\)
\(\Leftrightarrow a>2\sqrt{a}\)
\(\Leftrightarrow2\sqrt{a}< a\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{a}< a\left(đk:a\ge0\right)\\2\sqrt{a}< a\left(đk:a< 0\right)\end{matrix}\right.\)
A = (15/√x) - (11x + 2√x - 3) - (3√x - 2√x - 1) - (2√x + 3√x - 3)
Tiếp theo, kết hợp các thành phần tương tự:
A = 15/√x - 11x - 2√x + 3 + 3√x - 2√x + 1 - 2√x - 3√x + 3
Đơn giản hóa biểu thức:
A = -11x + 15/√x + 4
Để tìm giá trị lớn nhất của A, ta có thể tìm điểm đạt cực đại của hàm số A(x). Tuy nhiên, để làm điều này, cần biết thêm về giá trị của x.
Sửa đề: (3căn x-2)/căn x-1-(2căn x+3)/(căn x+3)\(A=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}+\dfrac{-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
\(A=\dfrac{-5\sqrt{x}-15+17}{\sqrt{x}+3}==-5+\dfrac{17}{\sqrt{x}+3}< =\dfrac{17}{3}-5=\dfrac{2}{3}\)
Dấu = xảy ra khi x=0
ĐK: a > 0, a khác 1
\(M=\dfrac{a-1}{\sqrt{a}-1}=\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}-1}=\sqrt{a}+1\)
\(N=\dfrac{a-1}{\sqrt{a}+1}=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}+1}=\sqrt{a}-1\)
\(P=\dfrac{a\sqrt{a}-1}{\sqrt{a}-1}=\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}-1}=a+\sqrt{a}+1\)
\(Q=\dfrac{a\sqrt{a}+1}{\sqrt{a}+1}=\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}+1}=a-\sqrt{a}+1\)