Giúp mik với ạ mik cần gấp

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1 were - would you play
2 weren't studying - would have
3 had taken - wouldn't have got
4 would you go - could
5 will you give - is
6 recycle - won't be
7 had heard - wouldn't have gone
8 would you buy - had
9 don't hurry - will miss
10 had phoned - would have given
11 were - wouldn't eat
12 will go - rains
13 had known - would have sent
14 won't feel - swims
15 hadn't freezed - would have gone
1 is explained
2 was stolen
3 will be opened
4 is being closed
5 is going to be built
Câu 3:
a: \(BD=\sqrt{BC^2-DC^2}=4\left(cm\right)\)
b: \(\widehat{A}=180^0-2\cdot70^0=40^0< \widehat{B}\)
nên BC<AC=AB
c: Xét ΔEBC vuông tại E và ΔDCB vuông tại D có
BC chung
\(\widehat{EBC}=\widehat{DCB}\)
Do đó:ΔEBC=ΔDCB
d: Xét ΔOBC có \(\widehat{OBC}=\widehat{OCB}\)
nên ΔOBC cân tại O
Câu 2
a) Thay y = -2 vào biểu thức đã cho ta được:
2.(-2) + 3 = -1
Vậy giá trị của biểu thức đã cho tại y = -2 là -1
b) Thay x = -5 vào biểu thức đã cho ta được:
2.[(-5)² - 5] = 2.(25 - 5) = 2.20 = 40
Vậy giá trị của biểu thức đã cho tại x = -5 là 40
\(\dfrac{2x-3}{5}-x+2\ge\dfrac{x}{3}\)
\(\Leftrightarrow3\left(2x-3\right)-15\left(x+2\right)\ge5x\)
\(\Leftrightarrow6x-9-15x+30\ge5x\)
\(\Leftrightarrow6x-15x-5x\ge9+30\)
\(\Leftrightarrow-14x\ge-21\)
\(\Leftrightarrow x\le\dfrac{21}{14}\le\dfrac{3}{2}\)
-------------|--------]////////////////--->
0 3/2
lâu rồi cũng không nhớ cách làm :v
a: \(\frac{25}{14x^2y}=\frac{25\cdot3\cdot y^4}{14x^2y\cdot3y^4}=\frac{75y^4}{42x^2y^5}\)
\(\) \(\frac{14}{21xy^5}=\frac{14\cdot2\cdot x}{21xy^5\cdot2x}=\frac{28x}{42x^2y^5}\)
b: \(\frac{11}{102x^4y}=\frac{11\cdot y^2}{102x^4y\cdot y^2}=\frac{11y^2}{102x^4y^3}\)
\(\frac{3}{34xy^3}=\frac{3\cdot3\cdot x^3}{34xy^3\cdot3x^3}=\frac{9x^3}{102x^4y^3}\)
c: \(\frac{3x+1}{12xy^4}=\frac{\left(3x+1\right)\cdot3\cdot x}{12xy^4\cdot3x}=\frac{9x^2+3x}{36x^2y^4}\)
\(\frac{y-2}{9x^2y^3}=\frac{\left(y-2\right)\cdot4y}{4y\cdot9x^2y^3}=\frac{4y^2-8y}{36x^2y^4}\)
d: \(\frac{1}{6x^3y^2}=\frac{1\cdot6\cdot y^2}{6x^3y^2\cdot6y^2}=\frac{6y^2}{36x^3y^4}\)
\(\frac{x+1}{9x^2y^4}=\frac{\left(x+1\right)\cdot4\cdot x}{9x^2y^4\cdot4x}=\frac{4x^2+4x}{36x^3y^4}\)
\(\frac{x-1}{4xy^3}=\frac{\left(x-1\right)\cdot9\cdot x^2y}{4xy^3\cdot9x^2y}=\frac{9x^3y-9x^2y}{36x^3y^4}\)
e: \(\frac{3+2x}{10x^4y}=\frac{\left(2x+3\right)\cdot12\cdot y^4}{10x^4y\cdot12y^4}=\frac{24xy^4+36y^4}{120x^4y^5}\)
\(\frac{5}{8x^2y^2}=\frac{5\cdot15\cdot x^2y^3}{8x^2y^2\cdot15x^2y^3}=\frac{75x^2y^3}{120x^4y^5}\)
\(\frac{2}{3xy^5}=\frac{2\cdot40\cdot x^3}{3xy^5\cdot40x^3}=\frac{80x^3}{120x^4y^5}\)
f: \(\frac{4x-4}{2x\left(x+3\right)}=\frac{\left(4x-4\right)\cdot3\cdot\left(x+1\right)}{2x\left(x+3\right)\cdot3\cdot\left(x+1\right)}=\frac{12\left(x-1\right)\left(x+1\right)}{6x\left(x+3\right)\left(x+1\right)}=\frac{12x^2-12}{6x\left(x+3\right)\left(x+1\right)}\)
\(\frac{x-3}{3x\left(x+1\right)}=\frac{\left(x-3\right)\cdot2\cdot\left(x+3\right)}{3x\left(x+1\right)\cdot2\cdot\left(x+3\right)}=\frac{2\left(x^2-9\right)}{6x\left(x+1\right)\left(x+3\right)}\)
g: \(\frac{2x}{\left(x+2\right)^3}=\frac{2x\cdot2x}{2x\left(x+2\right)^3}=\frac{4x^2}{2x\left(x+2\right)^3}\)
\(\frac{x-2}{2x\left(x+2\right)^2}=\frac{\left(x-2\right)\left(x+2\right)}{2x\left(x+2\right)^3}=\frac{x^2-4}{2x\left(x+2\right)^3}\)
h: \(\frac{5}{3x^3-12x}=\frac{5}{3x\left(x^2-4\right)}=\frac{5}{3x\left(x-2\right)\left(x+2\right)}\)
\(=\frac{5\cdot2\cdot\left(x+3\right)}{3x\left(x-2\right)\left(x+2\right)\cdot2\cdot\left(x+3\right)}=\frac{10x+30}{6x\left(x-2\right)\left(x+2\right)\left(x+3\right)}\)
\(\frac{3}{\left(2x+4\right)\left(x+3\right)}=\frac{3}{2\left(x+2\right)\left(x+3\right)}\)
\(=\frac{3\cdot3\cdot x\left(x-2\right)}{3x\left(x-2\right)\cdot2\left(x+2\right)\left(x+3\right)}=\frac{9x^2-18x}{6x\left(x-2\right)\left(x+2\right)\left(x+3\right)}\)