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\(\left(4-\sqrt{7}\right)^2=4^2-2\cdot4\cdot\sqrt{7}+7\)
\(=16-8\sqrt{7}+7=23-8\sqrt{7}\)
\(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}-\sqrt{5}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}\)
\(=\left|\sqrt{5}-2\right|-\sqrt{5}\)
\(=\sqrt{5}-2-\sqrt{5}=-2\)
\(\dfrac{\sqrt{4-2\sqrt{3}}}{1+\sqrt{2}}:\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)
\(=\dfrac{\sqrt{3-2\cdot\sqrt{3}\cdot1+1}}{\sqrt{2}+1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}+1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)
\(=\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}=\dfrac{3-1}{2-1}=2\)
\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-\dfrac{6\sqrt{6}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{1}{2}\sqrt{6}-2\sqrt{6}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{1}{2}-2=-\dfrac{3}{2}=-1,5\)
VT tương đương với \(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
\(=\dfrac{\sqrt{1}-\sqrt{2}}{1-2}+\dfrac{\sqrt{2}-\sqrt{3}}{2-3}+...+\dfrac{\sqrt{99}-\sqrt{100}}{99-100}\)
\(=\sqrt{100}-\sqrt{99}+\sqrt{99}-....-\sqrt{3}+\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{1}\) (kiểu do mẫu số nó có kết quả âm nên đảo lại phép)
\(=10-1=9=VP\)
a) a) Biến đổi vế trái thành và làm tiếp.
b) Biến đổi vế trái thành và làm tiếp
Ta có: \(M=\dfrac{3\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+4}{\sqrt{x}+1}-\dfrac{9}{x-\sqrt{x}-2}\)
\(=\dfrac{3\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{2\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3x-3-2x+8-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
Ta có: \(A-1=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-1\)
\(=\dfrac{\sqrt{x}+2-\sqrt{x}-1}{\sqrt{x}+1}\)
\(=\dfrac{1}{\sqrt{x}+1}>0\forall x\) thỏa mãn ĐKXĐ
hay A>1
1, Thay x = 16 vào ta được \(A=\dfrac{4}{4+3}=\dfrac{4}{7}\)
2, \(A+B=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)-3x-9}{x-9}=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{-x+6\sqrt{x}-9}{x-9}=\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{\sqrt{x}-3}{\sqrt{x}+3}=\dfrac{3}{\sqrt{x}+3}\)
Ta có đpcm
1) Thay x=16 vào biểu thức ta có:
\(A=\dfrac{\sqrt{x}}{\sqrt{x+3}}=\dfrac{\sqrt{16}}{\sqrt{16}+3}=\dfrac{4}{4+3}=\dfrac{4}{7}\)
2) \(A+B=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\\ \Rightarrow A+B=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(\Rightarrow A+B=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(\Rightarrow A+B=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(\Rightarrow A+B=\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(\Rightarrow A+B=\dfrac{3}{\sqrt{x}+3}\)
1: Thay x=16 vào A, ta được:
\(A=\dfrac{4}{4+3}=\dfrac{4}{7}\)
