Question

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Answer 1

Bài 4:

Câu 4.1

A=\(\dfrac{1}{\sqrt{2}-1}-\dfrac{1}{\sqrt{2}+1}\)

A=\(\dfrac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}-\dfrac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}\)

A=\(\dfrac{\sqrt{2}+1}{1}-\dfrac{\sqrt{2}-1}{1}\)

A=\(\sqrt{2}+1-\left(\sqrt{2}-1\right)\)

A=\(\sqrt{2}+1-\sqrt{2}+1\)

A=2

B=\(\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{4-2\sqrt{3}}\)

B=\(\left|\sqrt{3}-2\right|+\sqrt{3-2\sqrt{3}+1}\)

B=\(-\left(\sqrt{3}-2\right)+\sqrt{\left(\sqrt{3}\right)^2-2\sqrt{3}+1^2}\)

B=\(-\sqrt{3}+2+\sqrt{\left(\sqrt{3}-1\right)^2}\)

B=\(-\sqrt{3}+2+\left|\sqrt{3}-1\right|\)

B=\(-\sqrt{3}+2+\left(\sqrt{3}-1\right)\)

B=\(-\sqrt{3}+2+\sqrt{3}-1\)

B=1

Câu 4.2

a, ĐK: x>0; x\(\ne1\)

C=\(\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\)

C=\(\dfrac{x}{\sqrt{x}-1}-\dfrac{\sqrt{x}\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

C=\(\dfrac{x}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}\)

C=\(\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}\)

C=\(\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\)

C=\(\sqrt{x}-1\)

b, Thay x=\(6+2\sqrt{5}\) vào biểu thức trên ta được:

C=\(\sqrt{6+2\sqrt{5}}-1\)

C=\(\sqrt{5+2\sqrt{5}+1}-1\)

C=\(\sqrt{\left(\sqrt{5}\right)^2+2\sqrt{5}+1^2}-1\)

C=\(\sqrt{\left(\sqrt{5}+1\right)^2}-1\)

C=\(\left|\sqrt{5}+1\right|-1\)

C=\(\sqrt{5}+1-1\)

C=\(\sqrt{5}\)