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\(53,\sqrt{\left(a-2b\right)^2}\left(a\le2b\right)\)
\(=\left|a-2b\right|=-a+2b\)
\(54,\sqrt{4x^2-4xy+y^2}\left(2x\ge y\right)\)
\(=\sqrt{\left(2x-y\right)^2}=\left|2x-y\right|=2x-y\)
\(55,\sqrt{\left(2x-1\right)^2}\left(x\ge\dfrac{1}{2}\right)\)
\(=\left|2x-1\right|=2x-1\)
\(56,\sqrt{\left(3a-2\right)^2}\left(3a\le2\right)\)
\(=\left|3a-2\right|=-3a+2\)
\(57,\sqrt{\left(6-9x\right)^2}\left(3x\ge2\right)\)
\(=\left|6-9x\right|=-6+9x\)
\(58,\sqrt{25a^2-10a+1}\left(5a\le1\right)\)
\(=\sqrt{\left(5a-1\right)^2}=\left|5a-1\right|=-5a+1\)
\(59,\sqrt{m^2+4mn+4n^2}\left(m\ge-2n\right)\)
\(=\sqrt{\left(m+2n\right)^2}=\left|m+2n\right|=m+2n\)
\(60,\sqrt{9x^2-24xy+16y^2}\left(3x\le4y\right)\)
\(=\sqrt{\left(3x-4y\right)^2}=\left|3x-4y\right|=-3x+4y\)
Bài 3:
53. \(\sqrt{\left(a-2b\right)^2}=\left|a-2b\right|=2b-a\)
54. \(\sqrt{4x^2-4xy+y^2}=\sqrt{\left(2x-y\right)^2}=\left|2x-y\right|=2x-y\)
55. \(\sqrt{\left(2x-1\right)^2}=\left|2x-1\right|=2x-1\)
56. \(\sqrt{\left(3a-2\right)^2}=\left|3a-2\right|=2-3a\)
57. \(\sqrt{\left(6-9x\right)^2}=\left|6-9x\right|=6-9x\)
58. \(\sqrt{25a^2-10a+1}=\sqrt{\left(5a-1\right)^2}=\left|5a-1\right|=1-5a\)
59. \(\sqrt{m^2+4mn+4n^2}=\sqrt{\left(m+2n\right)^2}=\left|m+2n\right|=m+2n\)
60. \(\sqrt{9x^2-24xy+16y^2}=\sqrt{\left(3x-4y\right)^2}=\left|3x-4y\right|=4y-3x\)
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85: =3-căn 2-căn 2+1=4-2căn 2
86: =căn 8+căn 5-căn 8+căn 5=2căn 5
93: =(căn 3+căn 5)(căn 5-căn 2)
=căn 15-căn 6+5-căn 10
94: =(căn 7-căn 3)(căn 7+căn 3)=7-3=4
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1: Khi x=25 thì A=(2*5)/(5+2)=10/7
2: P=A+B
\(=\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{3\sqrt{x}}{\sqrt{x}-2}-\dfrac{5x+4}{x-4}\)
\(=\dfrac{2x-4\sqrt{x}+3x+6\sqrt{x}-5x-4}{x-4}=\dfrac{2\left(\sqrt{x}-2\right)}{x-4}\)
\(=\dfrac{2}{\sqrt{x}+2}\)
3: căn x+2>=2
=>P<=2/2=1
Dấu = xảy ra khi x=0
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a: Xét ΔAHB vuông tại H có HM là đường cao
nên \(AM\cdot AB=AH^2\left(1\right)\)
Xét ΔAHC vuông tại H có HN là đường cao
nên \(AN\cdot AC=AH^2\left(2\right)\)
Từ (1) và (2) suy ra \(AM\cdot AB=AN\cdot AC\)
hay \(\dfrac{AM}{AC}=\dfrac{AN}{AB}\)
Xét ΔAMN vuông tại A và ΔACB vuông tại A có
\(\dfrac{AM}{AC}=\dfrac{AN}{AB}\)
Do đó: ΔAMN\(\sim\)ΔACB
![](https://rs.olm.vn/images/avt/0.png?1311)
\(36,\dfrac{6+2\sqrt{6}}{\sqrt{3}+\sqrt{2}}=\dfrac{\left(6+2\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}=\dfrac{6\sqrt{3}-6\sqrt{2}+6\sqrt{2}-4\sqrt{3}}{\sqrt{3^2}-\sqrt{2^2}}=\dfrac{2\sqrt{3}}{3-2}=2\sqrt{3}\)
\(35,\dfrac{5\sqrt{6}+6\sqrt{5}}{\sqrt{5}+\sqrt{6}}=\dfrac{\sqrt{6}.\sqrt{5}\left(\sqrt{5}+\sqrt{6}\right)}{\sqrt{5}+\sqrt{6}}=\sqrt{30}\)
\(34,\dfrac{6\sqrt{2}-4}{\sqrt{2}-3}\\ =\dfrac{\left(6\sqrt{2}-4\right)\left(\sqrt{2}+3\right)}{\left(\sqrt{2}-3\right)\left(\sqrt{2}+3\right)}\\ =\dfrac{6.2+3.6\sqrt{2}-4\sqrt{2}-12}{\sqrt{2^2}-3^2}\\ =\dfrac{12+18\sqrt{2}-4\sqrt{2}-12}{2-9}\\ =-2\sqrt{2}\)
\(33,\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}=\dfrac{\sqrt{3}.\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}=\sqrt{6}\)
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1:
AC=căn 5^2-3^2=4cm
BH=AB^2/BC=1,8cm
CH=5-1,8=3,2cm
AH=3*4/5=2,4cm
2:
ΔCBA vuông tại B có tan 40=BC/BA
=>BC/10=tan40
=>BC=8,39(m)
ΔCBD vuông tại B có tan D=BC/BD
=>BD=8,39/tan35=11,98(m)