rum em cai help//////

T

thienthanlove20

Bài mấy trang 77 hả bạn********************************************************??????
 
C

cuccuong

tudohap said:
lam ho em bai 26 trang16 (sgk)va bài 14 trang77sgk:confused:
Bấm để xem đầy đủ nội dung ...
bài 26: a) [TEX]\sqrt{25+9}=\sqrt{34}[/TEX] ; [TEX]\sqrt{25}+\sqrt9 = 5+3=8[/TEX]
vì 64>34 \Rightarrow [TEX]\sqrt{64} > \sqrt{34} \Rightarrow 8> \sqrt{34} \Rightarrow \sqrt{25+9}[/TEX]
b) ta có: [TEX](\sqrt{a+b})^2 =a+b[/TEX]
[TEX](\sqrt{a}+\sqrt{b} = a+b+2.\sqrt{ab}[/TEX]
vì a,b>0 \Rightarrow [TEX]2.\sqrt{ab} >0 [/TEX]
\Rightarrow [TEX](\sqrt{a+b})^2 < (\sqrt{a}+\sqrt{b})^2[/TEX]
vì a,b >0 \Rightarrow [TEX](\sqrt{a+b})^2 < (\sqrt{a}+\sqrt{b})^2[/TEX] \Leftrightarrow [TEX]\sqrt{(\sqrt{a+b})^2} < \sqrt{(\sqrt{a}+\sqrt{b})^2} [/TEX] \Leftrightarrow [TEX]|\sqrt{a}+\sqrt{b} > |\sqrt{a+b} \Leftrightarrow \sqrt{a}+\sqrt{b} > \sqrt{a+b}[/TEX]
 
C

cuccuong

tudohap said:
lam ho em bai 26 trang16 (sgk)va bài 14 trang77sgk:confused:
Bấm để xem đầy đủ nội dung ...
vẽ [TEX]\triangle\ ABC[/TEX] vuông tại A. đặt [TEX]\widehat{ABC} =\alpha[/TEX]
a) ta có: (**) [TEX]\frac{sin \alpha}{cos \alpha}=\frac{sinB}{cosB}=\frac{AC}{BC}:\frac{AB}{BC}=\frac{AC}{AB}= tgB = tg \alpha[/TEX]
(**) [TEX]\frac{cos \alpha}{sin \alpha}=\frac{cosB}{sinB}=\frac{AB}{BC}:\frac{AC}{BC}=\frac{AB}{AC}=cotgB=cotg \alpha[/TEX]
(**) [TEX]tg \alpha . cos \alpha = tgB.cosB= \frac{AC}{AB}.\frac{AB}{AC}=1[/TEX]
b) [TEX]sin^2 \alpha + cos^2 \alpha = \frac{AC^2}{BC^2}+\frac{AB^2}{BC^2}=\frac{AC^2+AB^2}{BC^2}=1[/TEX] (theo định lia Pitago trong tam giác vuông)
 
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