Lời giải:
Áp dụng định lý Pitago:
$BC=\sqrt{AB^2+AC^2}=\sqrt{4,5^2+6^2}=7,5$ (cm)
Áp dụng hệ thức lượng trong tam giác vuông:
$BH.BC=AB^2\Rightarrow BH=\frac{AB^2}{BC}=\frac{4,5^2}{7,5}=2,7$ (cm)
$CH=BC-BH=7,5-2,7=4,8$ (cm)
$AH=AB.AC:BC=4,5.6:7,5=3,6$ (cm)
tìm x khi A=\(\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{-1}{3}\)
1. Thu gọn
a) A=\(\left(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\right)\left(\sqrt{3-2\sqrt{2}}+\sqrt{3+2\sqrt{2}}\right)\)
b) B=\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2-\sqrt{3}}}\)
c) C=\(\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
4. Cho x=\(\sqrt{5}+1\)
Tính P=\(\dfrac{x^4+4x^3+x^2+6x+12}{x^2-2x+12}\)
1.
A=\(\sqrt{1+\dfrac{1}{1^2}+\dfrac{1}{2^2}}+\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+...+\sqrt{1+\dfrac{1}{99^2}+\dfrac{1}{100^2}}\)
1. Cho \(\left(x\sqrt{x^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=2020\)
Tính S=x+y+2020
Tìm x
d, \(\sqrt{x-2\sqrt{x-1}=\sqrt{x-1}-1}\)
e, \(\sqrt{1-12x+36x^2}=5\)
g, \(\sqrt{23+8\sqrt{7}}-\sqrt{7}=4\)
1. Cho\(\left\{{}\begin{matrix}x+y+z=1\\x,y,z>0\end{matrix}\right.\) Tìm GTNN
P=\(\dfrac{1}{16x}+\dfrac{1}{4y}+\dfrac{1}{z}\)
2.4 Rút gọn biểu thức
\(a,\dfrac{3-\sqrt{x}}{x-9}\) ( vs x ≥ 0, x≠ 9)
b, \(\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}\)( vs x ≥ 0 ; x ≠ 9)
c, \(6-2x-\sqrt{9-6x+x^2}\left(x< 3\right)\)
Cm
\(\sqrt{a+4\sqrt{a-2}+2}+\sqrt{a-4\sqrt{a-2}+2}=4\) ( vs 2 ≤ a ≤ 6)
6. √(5+2√6)
7. √(4+2√3)
8. √(4-2√3)
9. √(11-2√30)
10. √(21-4√17)
