Question

GIÚP MÌNH VỚI

Answer 1

$$A = \frac{1}{1 \times 2} + \frac{2}{2 \times 4} + \frac{3}{4 \times 7} + \frac{4}{7 \times 11} + \dots + \frac{10}{46 \times 56} \\ A = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{4} + \frac{1}{4} - \frac{1}{7} + \frac{1}{7} - \frac{1}{11} + \dots + \frac{1}{46} - \frac{1}{56} \\ A = 1 - \frac{1}{56} \\ A = \frac{55}{56}$$

Answer 2

\(D=\sqrt{4+\sqrt7}-\sqrt{4-\sqrt7}\)

Ta thấy \(4+\sqrt7>4-\sqrt7\) nên D > 0

Bình phương hai vế, ta được:

\(D^2=\left(\sqrt{4+\sqrt7}-\sqrt{4-\sqrt7}\right)^2\)

Áp dụng hằng đẳng thức hiệu hai bình phương, có:

\(D^2=\left(\sqrt{4+\sqrt7}\right)^2-2\sqrt{4+\sqrt7}.\sqrt{4-\sqrt7}+\left(\sqrt{4-\sqrt7}\right)^2\)

\(D^2=\left(4+\sqrt7\right)-2\sqrt{\left(4+\sqrt7\right)\left(4-\sqrt7\right)}+\left(4-\sqrt7\right)\)

\(D^2=\left(4+4\right)+\left(\sqrt7-\sqrt7\right)-2\sqrt{4^2-\left(\sqrt7\right)^2}\)

\(D^2=8-2\sqrt{16-7}\)

\(D^2=8-2\sqrt9\)

\(D^2=8-2.3\)

\(D^2=8-6\)

\(D^2=2\)

Mà D > 0

=> \(D=\sqrt2\)

Answer 3

D = √(4 + √7) - √(4 - √7)
D² = (4 + √7) + (4 - √7) - 2√[(4 + √7)(4 - √7)]
D² = 8 - 2√(16 - 7)
D² = 8 - 2√9
D² = 8 - 6 = 2
Vì D > 0 nên D = √2
Kết quả: D = √2

Answer 4

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Answer 5

Ta có: \(D=\sqrt{4+\sqrt7}-\sqrt{4-\sqrt7}\)

\(=\frac{1}{\sqrt2}\left(\sqrt{8+2\sqrt7}-\sqrt{8-2\sqrt7}\right)\)

\(=\frac{1}{\sqrt2}\left(\sqrt{\left(\sqrt7+1\right)^2}-\sqrt{\left(\sqrt7-1\right)^2}\right)\)

\(=\frac{1}{\sqrt2}\left(\sqrt7+1-\sqrt7+1\right)=\frac{2}{\sqrt2}=\sqrt2\)