dạng 1
bài 2: đkxđ: \(x \ge 0; x \neq 1\)
a. thay x = 25 vào biểu thức A ta được:
\(A=\frac{\sqrt{25}-2}{\sqrt{25}+1}=\frac{5-2}{5+1}=\frac36=\frac12\)
b. \(B = \frac{1}{\sqrt{x}+1} - \frac{\sqrt{x}}{1-\sqrt{x}} + \frac{2}{x-1}\)
\(B = \frac{1}{\sqrt{x}+1} + \frac{\sqrt{x}}{\sqrt{x}-1} + \frac{2}{(\sqrt{x}-1)(\sqrt{x}+1)}\)
\(B = \frac{\sqrt{x}-1 + \sqrt{x}(\sqrt{x}+1) + 2}{(\sqrt{x}-1)(\sqrt{x}+1)}\)
\(B = \frac{\sqrt{x}-1 + x + \sqrt{x} + 2}{(\sqrt{x}-1)(\sqrt{x}+1)}\)
\(B = \frac{x + 2\sqrt{x} + 1}{(\sqrt{x}-1)(\sqrt{x}+1)}\)
\(B = \frac{(\sqrt{x}+1)^2}{(\sqrt{x}-1)(\sqrt{x}+1)}\)
\(B = \frac{\sqrt{x}+1}{\sqrt{x}-1} \quad (\text{đpcm})\)
\(c.P=A:B\)
\(P = \frac{\sqrt{x}-2}{\sqrt{x}+1} : \frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(P = \frac{\sqrt{x}-2}{\sqrt{x}+1} \cdot \frac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(P = \frac{(\sqrt{x}-2)(\sqrt{x}-1)}{(\sqrt{x}+1)^2}\)
vì P = 5/2 nên: \(\frac{(\sqrt{x}-2)(\sqrt{x}-1)}{(\sqrt{x}+1)^2} = \frac{5}{2}\)
\(\Rightarrow2(\sqrt{x}-2)(\sqrt{x}-1)=5(\sqrt{x}+1)^2\)
\(\Rightarrow2(x-3\sqrt{x}+2)=5(x+2\sqrt{x}+1)\)
\(\Rightarrow2x-6\sqrt{x}+4=5x+10\sqrt{x}+5\)
\(\Rightarrow3x+16\sqrt{x}+1=0\)
vì \(x \ge 0 \Rightarrow \sqrt{x} \ge 0\) nên \(3x+16\sqrt{x}+1\ge1>0\quad\left(\forall x\ge0\right)\)
⇒ phương trình trên vô nghiệm
vậy không có giá trị x thoả P = 5/2
bài 3: đkxđ: \(x \ge 0; x \neq 1; x \neq 4\)
thay x = 9 vào biểu thức A ta được: \(A=\frac{4\sqrt{9}}{\sqrt{9}-1}=\frac{4 \cdot3}{3 - 1}=\frac{12}{2}=6\)
vậy khi x = 9 thì A = 6
\(B = \frac{\sqrt{x}+5}{\sqrt{x}+2} - \frac{1}{2-\sqrt{x}} + \frac{12}{x-4}\)
\(B = \frac{\sqrt{x}+5}{\sqrt{x}+2} + \frac{1}{\sqrt{x}-2} + \frac{12}{(\sqrt{x}-2)(\sqrt{x}+2)}\)
\(B = \frac{(\sqrt{x}+5)(\sqrt{x}-2) + 1 \cdot (\sqrt{x}+2) + 12}{(\sqrt{x}-2)(\sqrt{x}+2)}\)
\(B = \frac{x - 2\sqrt{x} + 5\sqrt{x} - 10 + \sqrt{x} + 2 + 12}{(\sqrt{x}-2)(\sqrt{x}+2)}\)
\(B = \frac{x + 4\sqrt{x} + 4}{(\sqrt{x}-2)(\sqrt{x}+2)}\)
\(B = \frac{(\sqrt{x}+2)^2}{(\sqrt{x}-2)(\sqrt{x}+2)}\)
\(B = \frac{\sqrt{x}+2}{\sqrt{x}-2}\)
c. ta có \(P = A \cdot B\)
\(\Rightarrow P=\frac{4\sqrt{x}}{\sqrt{x}-1}\cdot\frac{\sqrt{x}+2}{\sqrt{x}-2}\)
\(\Rightarrow P=\frac{4\sqrt{x}(\sqrt{x}+2)}{(\sqrt{x}-1)(\sqrt{x}-2)}\)
vì P = 2 nên \(\frac{4\sqrt{x}(\sqrt{x}+2)}{(\sqrt{x}-1)(\sqrt{x}-2)} = 2\)
\(\Rightarrow2\sqrt{x}(\sqrt{x}+2)=(\sqrt{x}-1)(\sqrt{x}-2)\)
\(\Rightarrow2x+4\sqrt{x}=x-3\sqrt{x}+2\)
\(\Rightarrow x+7\sqrt{x}-2=0\)
đặt \(t = \sqrt{x}\) với \(t\ge0,t\neq1,t\neq2\)
ta có: \(t^2 + 7t - 2 = 0\)
\(\Delta = 7^2 - 4 \cdot 1 \cdot (-2) = 49 + 8 = 57 > 0\)
\(\Rightarrow\begin{cases}t_1=\frac{-7 + \sqrt{57}}{2}\left(N\right)\\ t_2=\frac{-7 - \sqrt{57}}{2}<0\left(L\right)\end{cases}\)
\(x=t^2=\left(\frac{\sqrt{57}-7}{2}\right)^2=\frac{57 - 14\sqrt{57} + 49}{4}\)
\(=\frac{106 - 14\sqrt{57}}{4}=\frac{53 - 7\sqrt{57}}{2}\)
