\(a,B=\dfrac{2\cdot4+1}{2+3}=\dfrac{9}{5}\\ b,A=\dfrac{2\sqrt{x}-6-\sqrt{x}+5}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\\ A=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{1}{\sqrt{x}+3}\\ c,P=A:B=\dfrac{1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}+3}{2\sqrt{x}+1}=\dfrac{1}{2\sqrt{x}+1}\\ P=\dfrac{1}{5}\Leftrightarrow2\sqrt{x}+1=5\\ \Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)
b,
A = \(\left(\dfrac{2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{\sqrt{x}-5}{(\sqrt{x}-3)\left(\sqrt{x}+3\right)}\right)\div\dfrac{\sqrt{x}-1}{\sqrt{x}-3}\)
A = \(\dfrac{2\sqrt{x}-6-\sqrt{x}+5}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\div\dfrac{\sqrt{x}-1}{\sqrt{x}-3}\)
A = \(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\times\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\)
A = \(\dfrac{1}{\sqrt{x}+3}\)
Vậy A = \(\dfrac{1}{\sqrt{x}+3}\) với x\(\ge\)0 , x\(\ne\)9
a,
Thay x = 16 vào biểu thức B, ta được :
B = \(\dfrac{2\sqrt{16}+1}{\sqrt{16}+3}\)
B = \(\dfrac{2+1}{3}\)
B = \(\dfrac{3}{3}\) = 1
Với x = 16 thì biểu thức B = 1
c, Đặt P = A : B
\(\Rightarrow\) P = \(\dfrac{1}{\sqrt{x}+3}\div1\)
\(\Rightarrow\) P = \(\dfrac{1}{\sqrt{x}+3}\times1\) = \(\dfrac{1}{\sqrt{x}+3}\)
Ta có : P = \(\dfrac{1}{5}\)
\(\Rightarrow\) \(\dfrac{1}{\sqrt{x}+3}\)= \(\dfrac{1}{5}\)
\(\Leftrightarrow\) \(\dfrac{5}{5\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+3}{5\left(\sqrt{x}+3\right)}\) ĐKXĐ : x\(\ge\)3
\(\Leftrightarrow\) \(\sqrt{x}+3=5\)
\(\Leftrightarrow\sqrt{x}\) = \(5-3\)
\(\Leftrightarrow\) \(\sqrt{x}=2\)
\(\Leftrightarrow\) \(x=4\) (Thoả mãn điều kiện)
Vậy giá trị của x = 4 thì giá trị của P = \(\dfrac{1}{5}\)
