[Toán 9]Bài tập Toán hay :)

H

happytomorrowww

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L

linhhuyenvuong

Bài 1:
Cho a,b,c,d là các số dương. CMR:
[TEX]\frac{a+c}{a+b}+\frac{b+d}{b+c}+[/TEX][tEX]\frac{c+a}{c+d}+\frac{d+b}{d+a}\geq 4[/tEX]

Bài 2:
Giải phương trình:
[TEX]\frac{x^2}{3}+\frac{48}{x^2}-10(\frac{x}{3}-\frac{4}{x})[/TEX]

1,
[TEX]\frac{a+c}{a+b}+\frac{b+d}{b+c}+[/TEX][tEX]\frac{c+a}{c+d}+\frac{d+b}{d+a}=(\frac{a+c}{a+b}+ \frac{c+a}{c+d})+(\frac{b+d}{b+c}+\frac{d+b}{d+a})=\frac{(a+c)(a+b+c+d)}{(a+b)(c+d)}+\frac{(b+d)(a+b+c+d)}{(a+b)(c+d)}[/tEX]


ap dung:[TEX]\frac{1}{xy} \geq\frac{4}{(x+y)^2}[/TEX]

[TEX]\frac{(a+c)(a+b+c+d)}{(a+b)(c+d)}+\frac{(b+d)(a+b+c+d)}{(a+b)(c+d)} \geq\frac{4(a+c)(a+b+c+d)}{(a+b+c+d)^2}+\frac{4(b+d)(a+b+c+d)}{(a+b+c+d)^2} =4[/TEX]
2,x#0
[TEX]\frac{x}{3}-\frac{4}{x}=t[/TEX]
\Rightarrow[TEX]t^2=\frac{x^2}{9}+\frac{16}{x^2}-\frac{8}{3}[/TEX]

\Rightarrow[TEX]3t^2=\frac{x^2}{3}+\frac{48}{x^2}-8[/TEX]

\Rightarrow[TEX]3t^2+8=\frac{x^2}{3}+\frac{48}{x^2}[/TEX]
.........
 
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