Giải hộ e bài7 vs ah
Giúp e bài7 vs ah
VII hả e, e chụp thế này sao nhìn rõ đc
Giải hộ e bài 3,5 vs ah
III
1 appreciated
2 worried
3 tense
4 confident
5 delighted
6 frustrated
7 calm
8 relaxed
9 self-disciplined
10 depressed
V
1 had resolved
2 has taken
3 Did you say
4 not to take
5 wanted
6 is being repaired
7 taking
8 to think
9 had worked
10 was washing - dropped
11 will find
12 faces
13 turned
14 to give
15 will empathise
Mn giải chi tiết hộ e vs ah
Đặt \(A=\dfrac{1}{\sqrt[3]{a+7b}}+\dfrac{1}{\sqrt[3]{b+7c}}+\dfrac{1}{\sqrt[3]{c+7a}}\)
\(A=\dfrac{\sqrt[3]{64}}{\sqrt[3]{8.8\left(a+7b\right)}}+\dfrac{\sqrt[3]{64}}{\sqrt[3]{8.8\left(b+7c\right)}}+\dfrac{\sqrt[3]{64}}{\sqrt[3]{8.8\left(c+7a\right)}}\)
\(\ge\dfrac{4}{\dfrac{8+8+a+7b}{3}}+\dfrac{4}{\dfrac{8+8+b+7c}{3}}+\dfrac{4}{\dfrac{8+8+c+7a}{3}}\ge\dfrac{\left(2+2+2\right)^2}{\dfrac{8+8+a+7b+8+8+b+7c+8+8+c+7a}{3}}\)
\(=\dfrac{36.3}{8\left(a+b+c\right)+48}=\dfrac{3}{2}\)
Vậy \(A_{min}=\dfrac{3}{2}\Leftrightarrow a=b=c=1\)
Giải hộ mk vs ah
1 Your sister is interested in doing DIY in her free time, isn't she?
2 Unless that girl study harder, she will fail the exam
3 My father has given up smoking since last year
4 Mary wanted to know what Peter would give his brother on his birthday the day after
1. Your sister is interested in doing DIY in her free time, isn't she?
2. Unless that girl studies harder, she'll fail the exam.
3. My father has given up smoking since last year.
4. Mary wanted to know what Peter would give his brother on his birthday the next day.
Giải hộ mik bài hình vs ah
Cho tam giác ABC vuông tại A.Kẻ AH vuông góc với BC (H€BC).Tia phân giác góc HAC cắt cạnh BC ở D và tia phân giác HAB cắt cạnh BC ở E. Chứng minh rằng AB + AC = BD + EC. Mn giải hộ e câu này vs ạk:)))
Giải hộ e vs
Giải hộ e vs ạ. E cảm ơn nhiều
Câu 1: C
Câu 2: B
Câu 3: A
Câu 4: C
Câu 5: B
Câu 6: A
Câu 7: B
Câu 8: A
Câu 9: D
Câu 10: B
Câu 11: B
Câu 12: D
Câu 13: A
Câu 14: B
Câu 15: B
Câu 16: C
Câu 17: C
Câu 18: \(\left\{{}\begin{matrix}a:D\\b:B\end{matrix}\right.\)
Giải hộ e vs ạ
\(\)với \(x\ge0,x\ne4\)
\(=>B=\left[\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\right]:\dfrac{3}{\sqrt{x}+1}\)
\(B=\left[\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\right].\dfrac{\sqrt{x}+1}{3}\)
\(B=\dfrac{3.\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)3}=\dfrac{1}{\sqrt{x}-2}\)
\(=>P=\dfrac{18}{A.B}=\dfrac{18}{\dfrac{x-4}{\sqrt{x}-1}.\dfrac{1}{\sqrt{x}-2}}=\dfrac{18}{\dfrac{\sqrt{x}+2}{\sqrt{x}-1}}\)
\(=\dfrac{18\left(\sqrt{x}-1\right)}{\sqrt{x}+2}\)\(=\dfrac{18\sqrt{x}-18}{\sqrt{x}+2}=\dfrac{18\sqrt{x}}{\sqrt{x}+2}-\dfrac{18}{\sqrt{x}+2}\ge-9\)
dấu"=" xảy ra<=>x=0(tm)
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