Question

Giải giúp em bài 8 9 10 đi ạ

Answer 1

Bài 8.9.10 của câu 14 hay 15 bạn?

Answer 2

Câu 15:

1: Ta có: \(\dfrac{1}{1-\sqrt{2}}-\dfrac{1}{1+\sqrt{2}}\)

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

\(=-2\sqrt{2}\)

2: Ta có: \(\dfrac{1}{\sqrt{5}+1}+\dfrac{1}{\sqrt{5}-1}\)

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

\(=\dfrac{2\sqrt{5}}{4}=\dfrac{\sqrt{5}}{2}\)

Answer 3

Câu 14:

8: Ta có: \(\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x-2}{x-3\sqrt{x}+2}\)

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

\(=\dfrac{x-4\sqrt{x}+3-2x+4\sqrt{x}+\sqrt{x}-2+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

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

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

9: Ta có: \(\dfrac{3\sqrt{x}+2}{2\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+4}-\dfrac{x-6\sqrt{x}+5}{2x-7\sqrt{x}-4}\)

\(=\dfrac{\left(3\sqrt{x}+2\right)\left(\sqrt{x}+4\right)}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}+\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}-\dfrac{x-6\sqrt{x}+5}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}\)

\(=\dfrac{3x+12\sqrt{x}+2\sqrt{x}+8+2x-\sqrt{x}-2\sqrt{x}+1-x+6\sqrt{x}-5}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}\)

\(=\dfrac{4x+20\sqrt{x}+4}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}\)

Answer 4

Câu 14:

10: Ta có: \(\dfrac{\sqrt{x}+4}{1-7\sqrt{x}}+\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{24\sqrt{x}}{7x-6\sqrt{x}-1}\)

\(=\dfrac{-\left(\sqrt{x}+4\right)\left(\sqrt{x}+1\right)}{\left(7\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}-2\right)\left(7\sqrt{x}-1\right)}{\left(7\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{24\sqrt{x}}{\left(7\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{-x-5\sqrt{x}-4+7x+7\sqrt{x}-14\sqrt{x}-2+24\sqrt{x}}{\left(7\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{6x+12\sqrt{x}-6}{\left(7\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)