HÔM NAY MÌNH PHẢI NỘP RỒI .MK CẦN GẤP!!
Tìm x; y là số nguyên tố biết:49 × X + 11 × Y = 224
Các bạn giúp mk nha.Cảm ơn nhiều.
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Ingredients:
Directions:
In a 2 quart saucepan bring broth, ginger, star anise, and cinnamon to a boil. Reduce heat and simmer for 15 minutes.
With a very sharp knife cut sirloin across the grain into very thin slices.
In a large bowl soak noodles in hot water to cover 15 minutes, or until softened and pliable.
While noodles are soaking, bring a kettle of salted water to a boil for noodles. Drain noodles in a colander and cook in boiling water, stirring 45 seconds, or until tender. Drain noodles in a colander. Set aside.
Strain broth into saucepan and bring to a boil. Stir in fish sauce, salt and pepper. Add sirloin and sprouts and cook 30 to 45 seconds, or until sirloin changes color. Skim any froth from soup.
To serve, divide noodles into 4 bowls. Ladle soup over noodles. Sprinkle scallion greens, cilantro, chilies and basil over soup and serve with lime wedges.
Bài 1:
1) Kẻ tia Cx//AB//DE
Ta có: Cx//AB
\(\Rightarrow\widehat{BAC}+\widehat{ACx}=180^0\)(2 góc trong cùng phía)
\(\Rightarrow\widehat{ACx}=180^0-\widehat{BAC}=180^0-140^0=40^0\)
Ta có: Cx//DE
\(\Rightarrow\widehat{xCD}+\widehat{CDE}=180^0\)( 2 góc trong cùng phía)
\(\Rightarrow\widehat{xCD}=180^0-\widehat{CDE}=180^0-150^0=30^0\)
\(\Rightarrow\widehat{ACD}=\widehat{ACx}+\widehat{xCD}=40^0+30^0=70^0\)
2) Ta có AB//DE(gt)
Mà DE⊥MN
=> AB⊥MN =>\(\widehat{AMN}=90^0\Rightarrow\dfrac{1}{2}\widehat{AMN}=45^0\Rightarrow\widehat{AMP}=45^0\) (do MP là tia phân giác \(\widehat{AMN}\))
Ta có AB//DE
=> \(\widehat{AMP}+\widehat{DPM}=180^0\) (2 góc trong cùng phía)
\(\Rightarrow\widehat{DPM}=180^0-\widehat{AMP}=180^0-45^0=135^0\)
\(1,\\ a,\left\{{}\begin{matrix}AC\perp AB\\BD\perp AB\end{matrix}\right.\Rightarrow AC//BD\\ b,AC//BD\Rightarrow\widehat{D_2}=\widehat{C_1}=57^0\left(đồng.vị\right)\\ \widehat{D_2}+\widehat{D_1}=180^0\left(kề.bù\right)\Rightarrow\widehat{D_1}=180^0-57^0=123^0\\ c,AC//BD\Rightarrow\widehat{D_1}=\widehat{C_1}=123^0\left(đồng.vị\right)\)
\(2,\\ \widehat{DAB}+\widehat{ABE}=50^0+130^0=180^0\)
Mà 2 góc này ở vị trí TCP nên AD//BE (1)
\(\widehat{EBC}+\widehat{BCG}=140^0+40^0=180^0\)
Mà 2 góc này ở vị trí TCP nên BE//CG (2)
Từ (1)(2) ta được AD//CG
<=> 2x + 12 = 3x - 21
<=> 2x - 3x = -21 - 12
<=> -x = -33
<=> x = 33