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![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=\dfrac{x^3+2x}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2}{x^2+x+1}-\dfrac{1}{x-1}=\dfrac{x^3+2x+2x-2-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^3+3x-3}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^3+3}{\left(x^2+x+1\right)}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=\dfrac{x^3+2x+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^3-x^2+3x-3}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+3}{x^2+x+1}\)
b: \(=\dfrac{x^2-2x-3+x^2+2x-3+2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{2x-6}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x+3}\)
c: \(=\dfrac{6-7+x}{3\left(x-1\right)}=\dfrac{x-1}{3\left(x-1\right)}=\dfrac{1}{3}\)
d: \(=\dfrac{x^3+2x+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^3-x^2+3x-3}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+3}{x^2+x+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1 is explained
2 was stolen
3 will be opened
4 is being closed
5 is going to be built
a) x2y - xy2 - 5x + 5y = xy(x - y) - 5(x - y) = (xy - 5)(x - y)
b) x2 + 8x - y2 + 16 = (x2 + 8x + 16) - y2 = (x + 4)2 - y2 = (x + y + 4)(x - y + 4)
c) x2 - 7x + 10 = x2 - 2x - 5x + 10 = x(x - 2) - 5(x - 2) = (x - 5)(x - 2)