Giúp tuoiii vs 😢
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Bài 1:
a: \(A=8\sqrt{2}-15\sqrt{2}-8\sqrt{2}=-15\sqrt{2}\)
b: \(B=-\sqrt{7}\)
Bài 1:
\(A=8\sqrt{2}-15\sqrt{2}-8\sqrt{2}=-15\sqrt{2}\\ B=\dfrac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}=-\sqrt{7}\\ C=\sqrt{\left(\sqrt{2}+\sqrt{5}\right)^2}-\sqrt{5}=\sqrt{2}+\sqrt{5}-\sqrt{5}=\sqrt{2}\)
Bài 2:
\(a,\Leftrightarrow x+2=9\Leftrightarrow x=7\\ b,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-6\sqrt{x+5}=-16\\ \Leftrightarrow\sqrt{x+5}=\dfrac{-16}{-4}=4\\ \Leftrightarrow x+5=16\\ \Leftrightarrow x=11\left(tm\right)\)
Bài 4:
Gọi góc đó là \(\alpha\) thì \(\sin\alpha=\dfrac{6}{8}\approx\sin49^0\Leftrightarrow\alpha\approx49^0\)
Vậy góc tạo bởi thang và mặt đất xấp xỉ 49 độ
11. A
12. B
13. C
14. D
15. A
16. A
17. C
18. B
19. A
20. D
21. D
22. C
23. C
24. B
25. B
d: ta có: \(C=3+3^3+3^5+...+3^{1991}\)
\(=3\left(1+3^2+3^4\right)+...+3^{1987}\left(1+3^2+3^4\right)\)
\(=91\cdot\left(3+...+3^{1987}\right)⋮13\)
c) \(\left(\dfrac{\sqrt{x}}{2\sqrt{x}-2}-\dfrac{\sqrt{x}}{2\sqrt{x}+2}\right):\dfrac{\sqrt{x}}{x+2\sqrt{x}+1}\)
= \(\left(\dfrac{\sqrt{x}}{2\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}}{2\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}}{\sqrt{x}^2+2\sqrt{x}+1^2}\)
= \(\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\) \(:\dfrac{\sqrt{x}}{\left(\sqrt{x}+1\right)^2}\)
= \(\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)-\sqrt{x}\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\) \(.\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
= \(\dfrac{2\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
= \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
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