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a: \(BC=\sqrt{21^2+72^2}=75\left(cm\right)\)
b: \(AH=\dfrac{21\cdot72}{75}=20.16\left(cm\right)\)
\(BH=\sqrt{21^2-20.16^2}=5.88\left(cm\right)\)
a, Áp dụng PTG: \(BC=\sqrt{AB^2+AC^2}=75\left(cm\right)\)
b, Áp dụng HTL: \(\left\{{}\begin{matrix}BH=\dfrac{AB^2}{BC}=5,88\left(cm\right)\\AH=\dfrac{AB\cdot AC}{BC}=20,16\left(cm\right)\end{matrix}\right.\)
c, Vì BD là p/g nên \(\dfrac{AD}{DH}=\dfrac{AB}{BH}=\dfrac{25}{7}\Rightarrow AD=\dfrac{25}{7}DH\)
Mà \(AD+DH=AH\Rightarrow\dfrac{32}{7}DH=20,16\Rightarrow DH=4,41\left(cm\right)\)
\(\Rightarrow AD=15,75\left(cm\right)\)
Bài 1:
\(a,A=6\sqrt{2}-6\sqrt{2}+2\sqrt{5}=2\sqrt{5}\\ b,B=\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{3}+\sqrt{2}\\ c,=2\sqrt{3}-6\sqrt{3}+15\sqrt{3}-4\sqrt{3}=7\sqrt{3}\\ d,=1+6\sqrt{3}-\sqrt{3}-1=5\sqrt{3}\\ e,=4\sqrt{2}+\sqrt{2}-6\sqrt{2}+3\sqrt{2}=2\sqrt{2}\)
Bài 2:
\(a,ĐK:x\ge\dfrac{3}{2}\\ PT\Leftrightarrow\sqrt{2x-3}=5\Leftrightarrow2x-3=25\Leftrightarrow x=14\\ b,PT\Leftrightarrow x^2=\sqrt{\dfrac{98}{2}}=\sqrt{49}=7\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=-\sqrt{7}\end{matrix}\right.\\ c,ĐK:x\ge3\\ PT\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}+1\right)=0\\ \Leftrightarrow\sqrt{x-3}=0\left(\sqrt{x+3}+1>0\right)\\ \Leftrightarrow x=3\\ d,ĐK:x\ge1\\ PT\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\\ \Leftrightarrow\sqrt{x-1}=1\Leftrightarrow x=2\left(tm\right)\\ e,PT\Leftrightarrow2x-1=16\Leftrightarrow x=\dfrac{17}{2}\\ f,PT\Leftrightarrow\left|2x-1\right|=\sqrt{3}-1\Leftrightarrow\left[{}\begin{matrix}2x-1=\sqrt{3}-1\\2x-1=1-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{3}}{2}\\x=\dfrac{2-\sqrt{3}}{2}\end{matrix}\right.\)
Bài 3:
\(a,Q=\dfrac{1+5}{3-1}=3\\ b,P=\dfrac{x+\sqrt{x}-6+x-2\sqrt{x}-3-x+4\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ P=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-3}\\ c,M=\dfrac{\sqrt{x}}{\sqrt{x}-3}\cdot\dfrac{3-\sqrt{x}}{\sqrt{x}+5}=\dfrac{-\sqrt{x}}{\sqrt{x}+5}\)
Vì \(-\sqrt{x}\le0;\sqrt{x}+5>0\) nên \(M< 0\)
Do đó \(\left|M\right|>\dfrac{1}{2}\Leftrightarrow M< -\dfrac{1}{2}\Leftrightarrow-\dfrac{\sqrt{x}}{\sqrt{x}+5}+\dfrac{1}{2}< 0\)
\(\Leftrightarrow\dfrac{2\sqrt{x}-\sqrt{x}-5}{2\left(\sqrt{x}+5\right)}< 0\Leftrightarrow\sqrt{x}-5< 0\left(\sqrt{x}+5>0\right)\\ \Leftrightarrow0\le x< 25\)
Bài 4:
\(a,A=\dfrac{16+2\cdot4+5}{4-3}=29\\ b,B=\dfrac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ B=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\\ c,P=\dfrac{x+2\sqrt{x}+5}{\sqrt{x}-3}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{x+2\sqrt{x}+5}{\sqrt{x}+1}\\ P=\dfrac{\left(\sqrt{x}+1\right)^2+4}{\sqrt{x}+1}=\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}\\ P\ge2\sqrt{\left(\sqrt{x}+1\right)\cdot\dfrac{4}{\sqrt{x}+1}}=2\sqrt{4}=4\\ P_{min}=4\Leftrightarrow\left(\sqrt{x}+1\right)^2=4\Leftrightarrow\sqrt{x}+1=2\Leftrightarrow x=1\left(tm\right)\)
1, Áp dụng PTG: \(AC=\sqrt{BC^2-AB^2}=8\left(cm\right)\)
Áp dụng HTL: \(\left\{{}\begin{matrix}CH=\dfrac{AC^2}{BC}=6,4\left(cm\right)\\AH=\dfrac{AB\cdot AC}{BC}=4,8\left(cm\right)\end{matrix}\right.\)
\(\sin\widehat{B}=\dfrac{AC}{BC}=\dfrac{4}{5}\approx\sin53^0\\ \Rightarrow\widehat{B}\approx53^0\\ \Rightarrow\widehat{C}\approx90^0-53^0=37^0\)
2,
a, Áp dụng HTL: \(\left\{{}\begin{matrix}AD\cdot AB=AH^2\\AE\cdot AC=AH^2\end{matrix}\right.\Rightarrow AD\cdot AB=AE\cdot AC\)
b, \(AD\cdot AB=AE\cdot AC\Rightarrow\dfrac{AD}{AC}=\dfrac{AE}{AB}\Rightarrow\Delta ABC\sim\Delta AED\left(c.g.c\right)\)
khi bài toán bắt ta chứng minh một hình gì đó mà thiếu một ta hay một đường thẳng...
Bài 5:
a, Áp dụng PTG: \(BC=\sqrt{AB^2+AC^2}=5\left(cm\right)\)
\(\sin B=\dfrac{AC}{BC}=\dfrac{3}{5}\approx\sin37^0\\ \Rightarrow\widehat{B}\approx37^0\\ \Rightarrow\widehat{C}\approx90^0-37^0=53^0\)
b, Áp dụng HTL: \(S_{AHC}=\dfrac{1}{2}AH\cdot HC=\dfrac{1}{2}\cdot\dfrac{AB\cdot AC}{BC}\cdot\dfrac{AC^2}{BC}=\dfrac{1}{2}\cdot\dfrac{12}{5}\cdot\dfrac{9}{5}=\dfrac{54}{25}\left(cm^2\right)\)
c, Vì AD là p/g nên \(\dfrac{DH}{DB}=\dfrac{AH}{AB}\)
Mà \(AC^2=CH\cdot BC\Leftrightarrow\dfrac{HC}{AC}=\dfrac{AC}{BC}\)
Mà \(AH\cdot BC=AB\cdot AC\Leftrightarrow\dfrac{AH}{AB}=\dfrac{AC}{BC}\)
Vậy \(\dfrac{DH}{DB}=\dfrac{HC}{AC}\)
a: Xét (O) có
ΔABC nội tiếp
AB là đường kính
Do đó: ΔACB vuông tại C
=>CB\(\perp\)CA tại C
=>CB\(\perp\)AF tại C
Xét tứ giác BHCF có \(\widehat{BHF}=\widehat{BCF}=90^0\)
nên BHCF là tứ giác nội tiếp
=>B,H,C,F cùng thuộc một đường tròn
\(11,\\ a,=4\cdot5+14:7=20+2=22\\ b,=3\sqrt{2}-12\sqrt{2}+5\sqrt{2}=-4\sqrt{2}\\ c,=\dfrac{3-\sqrt{2}+3+\sqrt{2}}{\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)}=\dfrac{6}{7}\\ 12,\\ a,P=\dfrac{\sqrt{x}+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\\ P=\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{2}{\sqrt{x}+3}\\ b,P=\dfrac{1}{2}\Leftrightarrow\sqrt{x}+3=4\Leftrightarrow x=1\left(tm\right)\)
a: \(=4\cdot5+14:7=20+2=22\)
b: \(=3\sqrt{2}-8\sqrt{2}+5\sqrt{2}=0\)