Ngũ giác đều chứ nhỉ ._?

Các vector bằng nhau: `\vec(AB) =\vec(ED) ; \vec(BA)=\vec(DE) ; \vec(BC)=\vec(FE) ; \vec(CB)=\vec(EF) ; \vec(AF)=\vec(CD) ; \vec(FA)=\vec(DC) ; \vec(AO)=\vec(OD) ; \vec(OA) =\vec(DO) ; \vec(BO)=\vec(OE) ; \vec(OB)=\vec(EO) ; \vec(FO)=\vec(OC) ; \vec(OF)=\vec(CO)`

`=(\sqrt5-2)/(\sqrt5^2-2^2)+(\sqrt5+2)/(\sqrt5^2-2^2)`

`=\sqrt5-2+\sqrt5+2`

`=2\sqrt5`

`A=1/(2.3) + 1/(3.4) +........ +1/(99.100)`

`=1/2-1/3+1/3-1/4+......+1/99-1/100`

`=1/2-1/100`

`=49/100`

a) `27 .75 + 25.27 -150`

`=27 xx (75 + 25)-150`

`=27 xx 100-150`

`=2700-150`

`=2550`

b) `142-[50-(2^3 .10-2^3 .5)]`

`=142-[50-2^3 .(10-5)]`

`=142-(50-8 .5)`

`=142-(50-40)`

`=142-10`

`=132`

c) `375:{32-[4+(5.3^2-42)]}-14`

`=375:{32-[4+(45-42)]}-14`

`=375:[32-(4+3)]-14`

`=375: 25 -14`

`=15-14`

`=1`

Ta có: `\hatA+\hatB+\hatC=180^@`

`<=>60^@+\hatB+50^@=180^@`

`<=> \hatB=70^@`

`=>\hat(ABD)=\hat(CBD) = \hatB :2=35^@`

Xét `\Delta ABD`: `\hatA+\hat(ABD)+\hat(ADB)=180^@`

`<=> 60^@+35^@+\hat(ADB)=180^@`

`<=> \hat(ADB)=85^@`

Xét `\DeltaCBD`: `\hat(CDB)+\hat(CBD)+\hatC=180^@`

`<=>\hat(CDB)+35^@+50^@=180^@`

`<=>\hat(CDB)=95^@`

b) `sin^2 3x=1`

`<=> (1-cos6x)/2=1`

`<=> 1-cos6x=2`

`<=> cos6x=-1`

`<=> 6x=π +k2π`

`<=>x=π/6 +k π/3 ( k \in ZZ)`

c) `tan^2 2x=3`

`<=> (1-cos4x)/(1+cos4x)=3`

`<=> 1-cos4x=3+3cos4x`

`<=>cos4x = -1/2`

`<=>4x= \pm (2π)/3 +k2π`

`<=>x =  \pm π/6 + k π/2 (k \in ZZ)`

a) `ĐK: x >= 0 ;x \ne 1`

\(A=\left(\sqrt{x}-\dfrac{x+2}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}-4}{x-1}\right)\\ =\dfrac{x-\sqrt{x}-x-2}{\sqrt{x}-1}:\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{-\sqrt{x}-2}{\sqrt{x}-1}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{x-4}\\ =\dfrac{-\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{-\sqrt{x}-1}{\sqrt{x}-2}\)

b) 

Khi `x=9 => A=(-\sqrt9-1)/(\sqrt9-2)=-4`

Khi `x=1/4 => A=(-\sqrt(1/4)-1)/(\sqrt(1/4)-2)=1`

c) `A<1/2 <=> (-\sqrtx-1)/(\sqrtx-2) <1/2`

`<=>(-\sqrtx-1-2\sqrtx+4)/(\sqrtx-2)<0`

`<=>(-3\sqrtx+3)/(\sqrtx-2)<0`

`<=>(\sqrtx-1)/(\sqrtx-2)>0`

TH1: `{(\sqrtx-1>0),(\sqrtx-2>0):} <=> x>4`

TH2: `{(\sqrtx-1<0),(\sqrtx-2<0):} <=>x<1`

Vậy `0<= x <1 \vee x>4` thỏa mãn.

`(\sqrtx+1)/(\sqrtx-3)` là số nguyên 

`<=> \sqrtx +1 vdots \sqrtx-3`

`<=> (\sqrtx-3)+4 vdots \sqrtx-3`

`<=> \sqrtx-3 \in {-4;4;-2;2;-1;1}`

`<=> \sqrtx \in {7;1;5;2;4}`

`<=> x \in {49;1;25;4;16}`

Ta có: `Cx////AB=>` \(\left\{{}\begin{matrix}\widehat{BCx}=\widehat{B}\left(\text{so le trong}\right)\\\widehat{DCx}=\widehat{A}\left(\text{đồng vị}\right)\end{matrix}\right.\)

Mà `\hatA=\hatB` (GT)

`=> \hat(BCx)=\hat(DCx)`

`=> Cx` là phân giác `\hat(DCB)`.

\(\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\left(\dfrac{x+1}{3x}-x-1\right)\right]:\dfrac{x-1}{x}\\ =\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\left(\dfrac{x+1}{3x}-\left(x+1\right)\right)\right].\dfrac{x}{x-1}\\ =\left[\dfrac{2}{3x}-\dfrac{2}{x+1}.\left(x+1\right)\left(\dfrac{1}{3x}-1\right)\right].\dfrac{x}{x-1}\\ =\left[\dfrac{2}{3x}-2\left(\dfrac{1}{3x}-1\right)\right].\dfrac{x}{x-1}\\ =\left[2\left(\dfrac{1}{3x}-\dfrac{1}{3x}+1\right)\right].\dfrac{x}{x-1}\\ =\dfrac{2x}{x-1}\)