Toán

a: Xét ΔAKC và ΔPKB có

KA=KP

góc AKC=góc PKB

KC=KB

=>ΔAKC=ΔPKB

=>góc KAC=góc KPB

=>AC//PB

b: Xét ΔABP và ΔNAM có

AB=NA

BP=AM

AP=MN

=>ΔABP=ΔNAM

AE=5-3=2cm

CE=căn 3^2+8^2=căn 73(cm)

Xét ΔFDC có AE//DC

nên AE/DC=FA/FD=FE/FC

=>FA/(FA+8)=FE/(FE+căn 73)=2/5

=>5FA=2FA+16 và 5FE=2FE+2căn 73

=>FA=16/3; FE=2*căn 73/3

S BEF=S BAF-S EAF

=1/2*5*16/3-1/2*2*16/3

=8cm²

S BCF=8+1/2*8*3=20cm²

\(199^{20}=\left(199^4\right)^5\)

\(2003^{15}=\left(2003^3\right)^5\)

mà \(199^4< 2003^3\)

nên \(199^{20}< 2003^{15}\)

`1/6 + 3/4 + 1/3`

`= 2/12 + 9/12 + 4/12 `

`= (2+9+4)/12`

` = 15/12 `

`= 5/4`

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`3/5 + 3/4 + 4/20`

`= 12/20 + 15/20 + 4/20`

`= (12+15+4)/20`

`= 31/20`

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`9/2 - 2/3 - 3/4`

`= 54/12 - 8/12 -9/12`

`= (54-8-9)/12`

`= 37/12`

\(a.A=\left(x+y\right)^2-\left(x-y\right)^2=x^2+2xy+y^2-x^2+2xy-y^2\)

\(=x^2-x^2+2xy+2xy+y^2-y^2=4xy.\)

\(b.B=\left(x+2\right)^2+2\left(x+2\right)\left(x-3\right)+\left(x-3\right)^2\)

\(=x^2+4x+4+2x^2-2x-12+x^2-6x+9\)

\(=x^2+2x^2+x^2+4x-2x-6x+4-12+9\)

\(=4x^2-4x+1.\)

\(c.C=\left(x+2\right)^2-2\left(x+2\right)\left(x+3\right)+\left(x+3\right)^2\)

\(=x^2+4x+4-2x^2-10x-12+x^2+6x+9\)

\(=x^2-2x^2+x^2+4x-10x+6x+4-12+9\)

\(=1.\)

\(=\dfrac{2a+3b}{\sqrt{2a}+\sqrt{3b}}+\dfrac{\left(\sqrt{2a}-\sqrt{3b}\right)\left(\sqrt{4a^2}+\sqrt{2a\cdot3b}+\sqrt{9b^2}\right)}{3b-2a}\)

\(=\dfrac{\left(2a+3b\right)\left(\sqrt{2a}-\sqrt{3b}\right)-\left(\sqrt{2a}-\sqrt{3b}\right)\left(2a+\sqrt{6ab}+3b\right)}{2a-3b}\)

\(=\dfrac{\left(\sqrt{2a}-\sqrt{3b}\right)\cdot\left(-\sqrt{6ab}\right)}{2a-3b}=\dfrac{-\sqrt{6ab}}{\sqrt{2a}+\sqrt{3b}}\)

\(=4\cdot\dfrac{5}{2}\cdot\sqrt{x}-\dfrac{8}{3}\cdot\dfrac{3}{2}\cdot\sqrt{x}-\dfrac{4}{3x}\cdot\dfrac{\left|3x\right|}{8}\)

\(=10\sqrt{x}-4\sqrt{x}-\dfrac{1}{2}=6\sqrt{x}-\dfrac{1}{2}\)

|3 - x| = -4y² + 20y - 25

<=> |3 - x| = -(2y - 5)²

mà -(2y - 5)² ≤ 0 vs mọi x; |3 - x| ≥ 0 vs mọi x

=> 3 - x = 2y - 5 = 0

<=> x = 3 ; y = 5/2

1: Xét ΔAHC vuông tại H có tanC=AH/HC

\(\Leftrightarrow HC=6:tan30^0=6\sqrt{3}\left(cm\right)\)

\(HB=\dfrac{6^2}{\left(6\sqrt{3}\right)^2}=\dfrac{1}{3}\left(cm\right)\)

\(BC=BH+CH=6\sqrt{3}+\dfrac{1}{3}\left(cm\right)\)

2: Xét ΔBAH vuông tại H có cos BAH=AH/AB

nên AH/24=1/2

=>AH=12(cm)

29,9.30,1 = 899,99

31,8² − 2.31,8.21,8 + 21,8² = (31,8 + 21,8)² = 53,6² = 2872,96

37² + 2.37.13 + 13² = (37 + 13)² = 50² = 2500

51,7² − 2.51.31,7 + 31,7² = (51,7 + 31,7)² = 83,4² = 6955,56

37.43 = 1591

20,1.19,9 = 399,99