làm hộ mình bài ạ
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1 She failed to do everything she was supposed to do
2 There is no likelihood that anyone will apply for this post
3 He succeeded in finishing the English test in time
4 There is no likelihood that the situation will improve in the near future
5 There is a chance that you will get what you want there
6 Don't check your answers before answering all the questions
7 I won't give an opinion before thinking this through
8 The judge will decide after hearing all the evidence
9 I won't be able to have a holiday until I save enough money
10 There is little likelihood that my sister will win the first prize in the singing contest
11 He made no reference of his new job when I last spoke to him
1 She failed to do everything she was supposed to do
2 There is no likelihood that anyone will apply for this post
3 He succeeded in finishing the English test in time
4 There is no likelihood that the situation will improve in the near future
5 There is a chance that you will get what you want there
6 Don't check your answers before answering all the questions
7 I won't give an opinion before thinking this through
8 The judge will decide after hearing all the evidence
9 I won't be able to have a holiday until I save enough money
10 There is little likelihood that my sister will win the first prize in the singing contest
11 He made no reference of his new job when I last spoke to him
Câu 2:
a) Ta có: \(-7x+21< 0\)
\(\Leftrightarrow-7x< -21\)
hay x>3
Vậy: S={x|x>3}
Câu 2:
b) Ta có: x<y
nên -x>-y
\(\Leftrightarrow-x+2021>-y+2021\)
mà \(-y+2021>-y+2020\)
nên -x+2021>-y+2020
hay 2021-x>2020-y
C1
A. baseball - football - volleyball - basketball - ice hockey
B. judo - gymnastics
C. skiing - ice-skating - horse-riding
1) Vì x=25 thỏa mãn ĐKXĐ nên Thay x=25 vào biểu thức \(A=\dfrac{\sqrt{x}-2}{x+1}\), ta được:
\(A=\dfrac{\sqrt{25}-2}{25+1}=\dfrac{5-2}{25+1}=\dfrac{3}{26}\)
Vậy: Khi x=25 thì \(A=\dfrac{3}{26}\)
2) Ta có: \(B=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}+\dfrac{2x+8\sqrt{x}-6}{x-\sqrt{x}-2}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}+\dfrac{2x+8\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-5\sqrt{x}+6+2x+8\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3x+3\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3\sqrt{x}}{\sqrt{x}-2}\)
Câu IV:
1) Xét tứ giác BFEC có
\(\widehat{BFC}=\widehat{BEC}\left(=90^0\right)\)
\(\widehat{BFC}\) và \(\widehat{BEC}\) là hai góc cùng nhìn cạnh BC
Do đó: BFEC là tứ giác nội tiếp(Dấu hiệu nhận biết tứ giác nội tiếp)
hay B,F,E,C cùng nằm trên 1 đường tròn(đpcm)
bài gì
bài gì ,ko cho bài sao làm ,bạn cần mình báo cáo ko
Cứ bảo giúp mình làm xong ko cho đề
THEO các bạn có nên báo cáo ko
CÓ : k đúng
Ko:k sai