tính nhanh giúp mk với ạ mk cảm ơn nhiều
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1 C => hers
2 B => will use
3 D => their
4 A => will be pedaling
5 A => travel
6 C => less
7 D => have they
8 D => than
9 A => worked
10 B => fewer
\(9,=\dfrac{\sqrt{5}+1-\sqrt{5}+1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\dfrac{2}{4}=\dfrac{1}{2}\\ 10,=\dfrac{\sqrt{5}+2+\sqrt{5}-2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}=2\sqrt{5}\\ 11,=\dfrac{8+6\sqrt{2}-8+6\sqrt{2}}{\left(4-3\sqrt{2}\right)\left(4+3\sqrt{2}\right)}=\dfrac{12\sqrt{2}}{-2}=-6\sqrt{2}\\ 12,=\dfrac{2+\sqrt{6}+2-\sqrt{6}}{\left(2-\sqrt{6}\right)\left(2+\sqrt{6}\right)}=\dfrac{4}{-2}=-2\\ 13,=\sqrt{2}-1+\sqrt{2}+3=2\sqrt{2}+2\\ 14,=2-\sqrt{3}+\sqrt{3}-1=1\\ 15,=3-\sqrt{5}+\sqrt{5}-2=1\)
Để A đạt GTLN
=>x2 -2x đạt giá trị dương nhỏ nhất
=>x2-2x=1
=>x2-2x-1=0
=>x=$1-\sqrt{2};\sqrt{2}+1$1−√2;√2+1
Vậy A ko xảy ra GTLN
Để A đạt GTLN
=>x2 -2x đạt giá trị dương nhỏ nhất
=>x2-2x=1
=>x2-2x-1=0
=>x=\(1-\sqrt{2};\sqrt{2}+1\)
Vậy A ko xảy ra GTLN
1/
1. although
2. around
3. where
4. forward
5. and
6. will
2/ 1B 2F 3E 4A 5D 6C
3/
1. daily
2. wearing
3. would do
4. haven't eaten
5. is being carried
6. rises
7. inexperienced
8. equality
4/
1. which -> who
2. pollute -> polluting
5/
1. not to touch that
2. Tom has met is
3. weren't ill
4. may be built
6/
1. environmentalists
2. affect
3. media
4. responsible
7/
1. T
2. T
3. F
4. F
Bài 3:
1, Áp dụng t/c dtsbn:
\(\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{3}=\dfrac{z-x}{3-6}=\dfrac{-21}{-3}=7\\ \Rightarrow\left\{{}\begin{matrix}x=42\\y=28\\z=21\end{matrix}\right.\)
2, Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x+3y-z}{6+15-7}=\dfrac{-14}{14}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-7\end{matrix}\right.\)
Bài 4:
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{4}}=\dfrac{x+y+z}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}=\dfrac{130}{\dfrac{13}{12}}=120\)
Do đó: x=60; y=40; z=30
\(\left|3x+2\right|=\left|4x-3\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=4x-3\\3x+2=3-4x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=-5\\7x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{7}\end{matrix}\right.\)
\(\left|2+3x\right|=\left|4x-3\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2+3x=4x-3\\2+3x=3-4x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{7}\end{matrix}\right.\)
Lời giải:
Gọi tử số là T
\(T=\frac{2-1}{1\times 2}+\frac{4-3}{3\times 4}+\frac{6-5}{5\times 6}+....+\frac{60-59}{59\times 60}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{59}-\frac{1}{60}\)
\(=(1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{59})-(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{60})\)
\(=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{59}+\frac{1}{60})-2\times (\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{60})\)
\(=(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{59}+\frac{1}{60})-(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{30})\)
\(=\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}=\text{mẫu số}\)
Do đó $D=1$