Help me
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
DT
Đỗ Thanh Hải
CTVVIP
14 tháng 4 2021
II
1 driver
2 participants
3 useless
4 electrician
5 building
6 variety
7 creative
8 poisonous
III
1 B => happy
2 C => between
3 B => seriously
4 A => excited
5C => to spending
6 B => ought to be
7 C => for
8 her => herself
IV
1 to buy
2 has been
3 will be going
4 to work
5 saw
6 to lock
7 don't know
8 are wasting
DT
Đỗ Thanh Hải
CTVVIP
14 tháng 4 2021
D Reading
1 T
2 F
3 T
4 F
II
1 to
2 Most
3 However
4 use
III
1 In Milton Keynes, a town about 70 kilometres northwest of London.
2 They can select their terminal, read the news, check e-mails or play video games.
3 No, they aren't
4 Yes, they are
NT
0
18 tháng 10 2021
Bài 3:
a: Theo đề, ta có hệ phương trình:
\(\left\{{}\begin{matrix}4a+b=3\\2a+b=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a=4\\2a+b=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=-1-2a=-5\end{matrix}\right.\)
DC
16 tháng 4 2022
= 28/15 . 3/4 - ( 11/20 + 1/4 ) : 7/3
= 28/15 . 3/4 - 4/5 : 7/3
= 7/5 - 12/35
= 37/35
KM
16 tháng 4 2022
= \(\dfrac{28}{15}\) . \(\dfrac{3}{4}\) - (\(\dfrac{11}{20}\) + \(\dfrac{1}{4}\)) : \(\dfrac{7}{3}\)
= \(\dfrac{7}{5}\) - (\(\dfrac{11}{20}\) + \(\dfrac{5}{20}\)) : \(\dfrac{7}{3}\)
= \(\dfrac{7}{5}\) - \(\dfrac{16}{20}\) : \(\dfrac{7}{3}\)
= \(\dfrac{7}{5}\) - \(\dfrac{16}{20}\) x \(\dfrac{3}{7}\)
= \(\dfrac{7}{5}\) - \(\dfrac{12}{35}\)
= \(\dfrac{49}{35}\) - \(\dfrac{12}{35}\)
= \(\dfrac{37}{35}\)
9 tháng 7 2021
-13/25x9/17+13/17x8/25+(-13/25)
=-13/25x9/17+8/17x13/25+(-13/25)
=13x(-9/17+8/17-1)
=13x(-18/17)=-234
9 tháng 7 2021
a) Ta có: \(\dfrac{-13}{25}\cdot\dfrac{9}{17}+\dfrac{13}{17}\cdot\dfrac{8}{25}+\dfrac{-13}{25}\)
\(=\dfrac{-13}{25}\cdot\dfrac{9}{17}-\dfrac{-13}{25}\cdot\dfrac{8}{17}+\dfrac{-13}{25}\)
\(=\dfrac{-13}{25}\left(\dfrac{9}{17}-\dfrac{8}{17}+1\right)\)
\(=\dfrac{-13}{25}\cdot\dfrac{18}{17}\)
\(=-\dfrac{234}{425}\)
Bài 1.1 TXĐ: $D=\mathbb{R}$
Bài 1.2:
ĐKXĐ: $4x^2+5x-9\neq 0$
$\Leftrightarrow (x-1)(4x+9)\neq 0$
$\Leftrightarrow x\neq 1$ và $x\neq \frac{-9}{4}$
Vậy TXĐ: $D=\mathbb{R}\setminus \left\{1; \frac{-9}{4}\right\}$
Bài 1.3
ĐKXĐ: $x^2+2x+5\neq 0$
$\Leftrightarrow (x+1)^2+4\neq 0$
$\Leftrightarrow (x+1)^2\neq -4\Leftrightarrow x\in\mathbb{R}$
Vậy TXĐ: $D=\mathbb{R}$
Bài 1.4
ĐKXĐ: \(\left\{\begin{matrix} x+4\geq 0\\ x-2\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq -4\\ x\neq 2\end{matrix}\right.\)
Vậy TXĐ: $D=[-4;+\infty)\setminus \left\{2\right\}$