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a: \(BD\cdot CE\cdot BC\)
\(=\dfrac{HB^2}{AB}\cdot\dfrac{HC^2}{AC}\cdot\dfrac{AB\cdot AC}{AH}\)
\(=\dfrac{AH^4}{AH}=AH^3\)
b: \(\dfrac{BD}{CE}=\dfrac{HB^2}{AB}:\dfrac{HC^2}{AC}=\dfrac{HB^2}{AB}\cdot\dfrac{AC}{HC^2}=\dfrac{AB^4}{AB}\cdot\dfrac{AC}{AC^4}=\dfrac{AB^3}{AC^3}\)
1) Áp dụng HTL:
\(\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}=\dfrac{1}{4^2}+\dfrac{1}{\left(4\sqrt{2}\right)^2}=\dfrac{3}{32}\Rightarrow AH=\dfrac{4\sqrt{6}}{3}\left(cm\right)\)
Áp dụng đ/lý Pytago:
\(BC^2=AB^2+AC^2\Rightarrow BC=\sqrt{AB^2+AC^2}=\sqrt{4^2+\left(4\sqrt{2}\right)^2}=4\sqrt{3}\left(cm\right)\)
Bài 2:
a) \(pt\Leftrightarrow\sqrt{\left(2x+1\right)^2}=3\Leftrightarrow\left|2x+1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=3\\2x+1=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
b) \(A=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=2\sqrt{x}.\dfrac{\sqrt{x}+1}{\sqrt{x}}=2\sqrt{x}+2\)
1: Xét ΔABH vuông tại H có HE là đường cao
nên \(AE\cdot AB=AH^2\left(1\right)\)
Xét ΔACH vuông tại H có HF là đường cao
nên \(AF\cdot AC=AH^2\left(2\right)\)
Từ (1) và (2) suy ra \(AE\cdot AB=AF\cdot AC\)
2: \(AE\cdot AB+AF\cdot AC=AH^2+AH^2=2AH^2\)
4: \(4\cdot OE\cdot OF=2OE\cdot2OF=FE\cdot AH=AH^2\)
\(HB\cdot HC=AH^2\)
Do đó: \(4\cdot OE\cdot OF=HB\cdot HC\)
Câu 1:
a)
\(5\sqrt{x}-2=13\Rightarrow 5\sqrt{x}=15\Rightarrow \sqrt{x}=3\)
\(\Rightarrow x=3^2=9\)
b)
\(\sqrt{8x}+7\sqrt{18x}=9-\sqrt{50x}\)
\(\Leftrightarrow \sqrt{4}.\sqrt{2x}+7\sqrt{9}.\sqrt{2x}=9-\sqrt{25}.\sqrt{2x}\)
\(\Leftrightarrow 2\sqrt{2x}+21\sqrt{2x}=9-5\sqrt{2x}\)
\(\Leftrightarrow 28\sqrt{2x}=9\Rightarrow \sqrt{2x}=\frac{9}{28}\)
\(\Rightarrow 2x=(\frac{9}{28})^2\Rightarrow x=\frac{1}{2}.(\frac{9}{28})^2\)
Câu 2:
a) \(Q=\frac{2\sqrt{x}-9}{(\sqrt{x}-2)(\sqrt{x}-3)}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(=\frac{2\sqrt{x}-9}{(\sqrt{x}-2)(\sqrt{x}-3)}-\frac{(\sqrt{x}+3)(\sqrt{x}-3)}{(\sqrt{x}-2)(\sqrt{x}-3)}+\frac{(2\sqrt{x}+1)(\sqrt{x}-2)}{(\sqrt{x}-3)(\sqrt{x}-2)}\)
\(=\frac{2\sqrt{x}-9-(x-9)+(2x-3\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}-3)}\)
\(=\frac{x-\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}-3)}=\frac{(\sqrt{x}+1)(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}-3)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
b) Để \(Q=2\Leftrightarrow \frac{\sqrt{x}+1}{\sqrt{x}-3}=2\Rightarrow \sqrt{x}+1=2\sqrt{x}-6\)
\(\sqrt{x}=7\Rightarrow x=49\)
c)
\(Q\in \mathbb{Z}\Leftrightarrow \frac{\sqrt{x}+1}{\sqrt{x}-3}\in \mathbb{Z}\Rightarrow \sqrt{x}+1\vdots \sqrt{x}-3\)
\(\Leftrightarrow \sqrt{x}-3+4\vdots \sqrt{x}-3\)
\(\Leftrightarrow 4\vdots \sqrt{x}-3\Rightarrow \sqrt{x}-3\in \left\{\pm 1; \pm 2; \pm 4\right\}\)
\(\Rightarrow \sqrt{x}\in \left\{2;4;1; 5; 7\right\}\)
\(\Rightarrow \sqrt{x}\in \left\{4; 16; 1; 25; 49\right\}\)
b: \(BE\cdot CF\cdot BC\)
\(=\dfrac{BH^2}{AB}\cdot\dfrac{CH^2}{AC}\cdot BC\)
\(=\dfrac{AH^4}{AH}=AH^3\)
c: \(\dfrac{BE}{CF}=\dfrac{BH^2}{AB}:\dfrac{CH^2}{AC}=\dfrac{BH^2}{CH^2}\cdot\dfrac{AC}{AB}=\left(\dfrac{AB}{AC}\right)^3\)
Bài 2:
a: \(BC=\sqrt{10^2+8^2}=2\sqrt{41}\left(cm\right)\)
\(AH=\dfrac{8\cdot10}{2\sqrt{41}}=\dfrac{40}{\sqrt{41}}\left(cm\right)\)
\(BH=\dfrac{64}{2\sqrt{41}}=\dfrac{32}{\sqrt{41}}\left(cm\right)\)
\(CH=\dfrac{100}{2\sqrt{41}}=\dfrac{50}{\sqrt{41}}\left(cm\right)\)
b: \(\dfrac{AD}{BD}=\dfrac{AH^2}{AB}:\dfrac{BH^2}{AB}=\dfrac{AH^2}{BH^2}\)