a: Qua N, kẻ tia NA nằm giữa hai tia NG và NK sao cho NA//EG

NA//EG

=>\(\hat{EGN}=\hat{GNA}\) (hai góc so le trong)

=>\(\hat{GNA}=50^0\)

Ta có: tia NA nằm giữa hai tia NG và NK

=>\(\hat{GNA}+\hat{KNA}=\hat{GNK}\)

=>\(\hat{KNA}=120^0-50^0=70^0\)

Ta có: \(\hat{KNA}=\hat{K}\left(=70^0\right)\)

mà hai góc này là hai góc ở vị trí so le trong

nên NA//MK

mà NA//EG

nên EG//MK

b: EG//MK

EM⊥MK

Do đó: EM⊥ EG

Bổ sung vào hình vẽ là điểm K ở chỗ góc 50 độ nhé

a: Ta có: \(\hat{HEK}+\hat{EKC}=130^0+50^0=180^0\)

mà hai góc này là hai góc ở vị trí trong cùng phía

nên HE//CK

=>x'x//y'y

b: x'x//y'y

HC⊥x'x

Do đó: HC⊥y'y

=>\(\hat{BCy}+\hat{BCH}=90^0\)

=>\(\hat{BCy}=90^0-60^0=30^0\)

c: Qua B, kẻ tia BM nằm giữa hai tia BA và BC sao cho BM//xx'//y'y

BM//y'y

=>\(\hat{MBC}=\hat{BCy}\) (hai góc so le trong)

=>\(\hat{MBC}=30^0\)

Ta có: \(\hat{MBC}+\hat{MBA}=\hat{ABC}\)

=>\(\hat{ABM}=90^0-30^0=60^0\)

BM//AE
=>\(\hat{ABM}+\hat{BAx^{\prime}}=180^0\) (hai góc trong cùng phía)

=>\(\hat{x^{\prime}AB}=180^0-60^0=120^0\)

ta có:

\(-\frac{x}{2}=\frac{-y}{4}=\frac{6}{-8}\)

=>\(\frac{-x}{2}=\frac{6}{-8}\)

=>-8.(-x)=6.2

=>8x=12

=>x=3/2

lại có:

\(\frac{-y}{4}=\frac{6}{-8}\)

=>-8.(-y)=6.4

=>8y=24

=>y=3

Vậy x=3/2; y=3

làm như thế này mới nhanh

\(\frac{-x}{2}=\frac{-y}{4}=\frac{6}{-8}\)

=>\(\frac{-x}{2}=-y=\frac{6.4}{-8}\)

=>\(\frac{-x}{2}=-y=-3\)

=>\(\frac{-x}{2}=\frac{-3}{1}=>-x=\frac{-3.2}{1}=-6\)

=>x=-6 ;y=-3

\(a,A=0,2\left(5x-1\right)-\dfrac{1}{2}\left(\dfrac{2}{3}x+4\right)+\dfrac{2}{3}\left(3-x\right)\)

\(=x-0,2-\dfrac{1}{3}x-2+2-\dfrac{2}{3}x\)

\(=\left(-0,2-2+2\right)+\left(x-\dfrac{1}{3}x-\dfrac{2}{3}x\right)\)

\(=-0,2\)

\(b,B=\left(x-2y\right)\left(x^2+2xy+4y^2\right)-\left(x^3-8y^3+10\right)\)

\(=x^3-8y^3-x^3+8y^3-10\)

\(=-10\)

\(c,C=4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)-4x\)

\(=4\left(x^2+2x+1\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)-4x\)

\(=4x^2+8x+4+4x^2-4x+1-8x^2+8-4x\)

\(=13\)

a) \(A=0,2\left(5x-1\right)-\dfrac{1}{2}\left(\dfrac{2}{3}x+4\right)+\dfrac{2}{3}\left(3-x\right)\)

\(A=x-\dfrac{1}{5}-\dfrac{1}{3}x-2+2-\dfrac{2}{3}x\)

\(A=\left(x-\dfrac{1}{3}x-\dfrac{2}{3}x\right)-\left(\dfrac{1}{5}+2-2\right)\)

\(A=-\dfrac{1}{5}\)

Vậy: ...

b) \(B=\left(x-2y\right)\left(x^2+2xy+4y^2\right)-\left(x^3-8y^3+10\right)\)

\(B=\left[x^3-\left(2y\right)^3\right]-\left[x^3-\left(2y\right)^3\right]-10\)

\(B=-10\)

Vậy: ...

c) \(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x+1\right)\left(x-1\right)-4x\)

\(=4\left(x^2+2x+4\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)-4x\)

\(=4x^2+8x+4+4x^2-4x+1-8x^2+8-4x\)

\(=\left(4x^2+4x^2-8x^2\right)+\left(8x-4x-4x\right)+\left(4+1+8\right)\)

\(=13\)

Vậy:...

Kết quả là -41/18

Đúng đó cho mk nha!!!!!!!!!!!

19/18 NHA

NHỚ TK CHO MK NHA

jimmmmmmmmmmmmmmmmmmmmmmmmmmm

he he he he he he

\(x=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)

\(\Rightarrow x^3=9+4\sqrt{5}+9-4\sqrt{5}+3\sqrt[3]{\left(9+4\sqrt[]{5}\right)\left(9-4\sqrt{5}\right)}\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\)

\(=18+3\sqrt{81-80}.x=18+3x\)\(\Rightarrow x^3-3x=18\left(1\right)\)

\(y=\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\)

\(\Rightarrow y^3=3+2\sqrt{2}+3-2\sqrt{2}+3\sqrt[3]{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)\)

\(=6+3\sqrt[3]{9-8}.y=6+3y\)\(\Rightarrow y^3-3y=6\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow P=x^3+y^3-3\left(x+y\right)+1996=x^3-3x+y^3-3y+1996\)

\(=18+6+1996=2020\)