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a) Thay x = 25 vào biểu thức A , ta có
\(A=\frac{5-2}{5-1}=\frac{3}{4}\)
b) \(B=\frac{x-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{4\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B =\frac{x+1+2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B =\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
a, Ta có : \(x=25\Rightarrow\sqrt{x}=5\)
Thay vào biểu thức A ta được :
\(A=\frac{5-2}{5-1}=\frac{3}{4}\)
Vậy với x = 25 thì A = 3/4
b, Với \(x\ge0;x\ne1\)
\(B=\frac{x-5}{x-1}-\frac{2}{\sqrt{x}+1}+\frac{4}{\sqrt{x}-1}\)
\(=\frac{x-5-2\left(\sqrt{x}-1\right)+4\left(\sqrt{x}+1\right)}{x-1}=\frac{x-5-2\sqrt{x}+2+4\sqrt{x}+4}{x-1}\)
\(=\frac{x+1+2\sqrt{x}}{x-1}=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}\pm1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
c, Ta có P = A/B hay \(P=\frac{\sqrt{x}-2}{\sqrt{x}-1}.\frac{\sqrt{x}-1}{\sqrt{x}+1}=\frac{\sqrt{x}-2}{\sqrt{x}+1}\)
\(\sqrt{P}< \frac{1}{2}\)hay \(\sqrt{\frac{\sqrt{x}-2}{\sqrt{x}+1}}< \frac{1}{2}\Rightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}< \frac{1}{4}\)
\(\Leftrightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}-\frac{1}{4}< 0\Leftrightarrow\frac{4\sqrt{x}-8-\sqrt{x}-1}{4\left(\sqrt{x}+1\right)}< 0\)
\(\Rightarrow3\sqrt{x}-9>0\)do \(4\left(\sqrt{x}+1\right)>0\)
\(\Leftrightarrow3\sqrt{x}>9\Leftrightarrow\sqrt{x}>3\Leftrightarrow x>9\)
a, Thay x = 9 vào biểu thức \(A=\frac{\sqrt{x}-2}{\sqrt{x}-1}\) ta được:
\(A=\frac{\sqrt{9}-2}{\sqrt{9}-1}=\frac{\sqrt{3^2}-2}{\sqrt{3^2}-1}=\frac{3-2}{3-1}=\frac{1}{2}\)
Vậy với x = 9 thì \(A=\frac{1}{2}\)
\(b,\left(\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}\right).\frac{\sqrt{x}-1}{x+1}\left(x\ge0;x\ne1\right)\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\frac{\sqrt{x}-1}{x+1}\)
\(=\frac{x-\sqrt{x}+\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}-1}{x+1}\)
\(=\frac{\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+1\right)}\)
\(=\frac{1}{\sqrt{x}+1}\)
a, Thay x=9 ta có
\(A=\frac{\sqrt{9}-2}{\sqrt{9}-1}=\frac{3-2}{3-1}=\frac{1}{2}\)
b,\(B=\left(\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}\right).\frac{\sqrt{x}-1}{x+1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\sqrt{x}-1}{x+1}\)
\(=\frac{x-\sqrt{x}+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\sqrt{x}-1}{x+1}\)
\(=\frac{x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\sqrt{x}-1}{x+1}\)
\(=\frac{1}{\sqrt{x}+1}\)
\(P=\frac{2}{x-1}+\frac{1}{\sqrt{x-1}}+\frac{1}{\sqrt{x}+1}\)
\(P=\frac{2}{ \left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{2+2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{2\left(1+\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{2}{\sqrt{x}-1}\)
\(1,P=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{x-1}\)
\(=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{\left(x+\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{\left(x+\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}-1}\)
\(=\frac{x+2}{x\sqrt{x}-1}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x+2+x-1-x-\sqrt{x}-1}{x\sqrt{x}-1}\)
\(=\frac{x-\sqrt{x}}{x\sqrt{x}-1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
@Mai.T.Loan câu a pha cuối hơi tắt đó nhìn khó hiểu lắm
còn câu b kl sai r nha
\(A=\frac{2}{\sqrt{x}-1}+\frac{2}{\sqrt{x}+1}-\frac{5-\sqrt{x}}{x-1}\)
\(=\frac{2\left(\sqrt{x}+1\right)+2\left(\sqrt{x}-1\right)-5+\sqrt{x}}{x-1}\)
\(=\frac{2\sqrt{x}+2+2\sqrt{x}-2-5+\sqrt{x}}{x-1}=\frac{5\sqrt{x}-5}{x-1}\)
=>m=5;n=5
=>2m+n=10+5=15
Ta có biểu thức:
\(A = \frac{2}{\sqrt{x} - 1} + \frac{2}{\sqrt{x} + 1} - \frac{5 - \sqrt{x}}{x - 1}\)
với \(x \geq 0\), \(x \neq 1\)
🔁 Bước 1: Quy đồng và rút gọn hai phân thức đầu tiên
Ta nhóm hai phân thức đầu:
\(\frac{2}{\sqrt{x} - 1} + \frac{2}{\sqrt{x} + 1} = 2 \left(\right. \frac{1}{\sqrt{x} - 1} + \frac{1}{\sqrt{x} + 1} \left.\right)\)
Quy đồng:
\(\frac{1}{\sqrt{x} - 1} + \frac{1}{\sqrt{x} + 1} = \frac{\left(\right. \sqrt{x} + 1 \left.\right) + \left(\right. \sqrt{x} - 1 \left.\right)}{x - 1} = \frac{2 \sqrt{x}}{x - 1}\)
Vậy:
\(\frac{2}{\sqrt{x} - 1} + \frac{2}{\sqrt{x} + 1} = 2 \cdot \frac{2 \sqrt{x}}{x - 1} = \frac{4 \sqrt{x}}{x - 1}\)
✂️ Bước 2: Ghép với phân thức cuối cùng
Biểu thức A:
\(A = \frac{4 \sqrt{x}}{x - 1} - \frac{5 - \sqrt{x}}{x - 1} = \frac{4 \sqrt{x} - \left(\right. 5 - \sqrt{x} \left.\right)}{x - 1} = \frac{4 \sqrt{x} - 5 + \sqrt{x}}{x - 1} = \frac{5 \sqrt{x} - 5}{x - 1}\) \(A = \frac{5 \left(\right. \sqrt{x} - 1 \left.\right)}{x - 1}\)
Nhớ rằng:
\(x - 1 = \left(\right. \sqrt{x} - 1 \left.\right) \left(\right. \sqrt{x} + 1 \left.\right) \Rightarrow A = \frac{5}{\sqrt{x} + 1}\)
Nhưng đề yêu cầu rút về dạng:
\(A = \frac{m \sqrt{x} - n}{x - 1}\)
Ta quay lại biểu thức vừa tìm:
\(A = \frac{5 \sqrt{x} - 5}{x - 1} = \frac{m \sqrt{x} - n}{x - 1} \Rightarrow m = 5 , n = 5\)
✅ Tính \(2 m + n\)
\(2 m + n = 2 \cdot 5 + 5 = \boxed{15}\)