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![](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)
![](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
![](https://rs.olm.vn/images/avt/0.png?1311)
![](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)
\(\dfrac{2x-3}{5}-x+2\ge\dfrac{x}{3}\)
\(\Leftrightarrow3\left(2x-3\right)-15\left(x+2\right)\ge5x\)
\(\Leftrightarrow6x-9-15x+30\ge5x\)
\(\Leftrightarrow6x-15x-5x\ge9+30\)
\(\Leftrightarrow-14x\ge-21\)
\(\Leftrightarrow x\le\dfrac{21}{14}\le\dfrac{3}{2}\)
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0 3/2
lâu rồi cũng không nhớ cách làm :v
a: A(x)+B(x)
\(=-3x^3+5x^2+4x+1+3x^3+6x^2-8x+9\)
\(=11x^2-4x+10\)
A(x)-B(x)
\(=-3x^3+5x^2+4x+1-3x^3-6x^2+8x-9\)
\(=-6x^3-x^2+12x-8\)
b: C(x)+D(x)
\(=-x^3+5x^2+5x-\dfrac{3}{4}+4x^3-5x^2-3x-\dfrac{1}{4}\)
\(=3x^3+2x-1\)
C(x)-D(x)
\(=-x^3+5x^2+5x-\dfrac{3}{4}-4x^3+5x^2+3x+\dfrac{1}{4}\)
\(=-5x^3+10x^2+8x-\dfrac{1}{2}\)
c: E(x)+F(x)
\(=3x^3+7x^2+5x-8+3x^3+7x^2-9x+1\)
\(=6x^3+14x^2-4x-7\)
E(x)-F(x)
\(=3x^3+7x^2+5x-8-3x^3-7x^2+9x-1\)
\(=14x-9\)
d: G(x)+H(x)
\(=5x^4-6x^3-3x^2-2x+8+x^4+3x^2-3x-5\)
\(=6x^4-6x^3-5x+3\)
G(x)-H(x)
\(=5x^4-6x^3-3x^2-2x+8-x^4-3x^2+3x+5\)
\(=4x^4-6x^3-6x^2+x+13\)
e: I(x)+J(x)
\(=5x^4-2x^3-6x^2+7x+6+2x^3+3x^2-7x-5\)
\(=5x^4-3x^2+1\)
I(x)-J(x)
\(=5x^4-2x^3-6x^2+7x+6-2x^3-3x^2+7x+5\)
\(=5x^4-4x^3-9x^2+14x+11\)
f: K(x)+L(x)
\(=4x^4+3x^3+5x^2-2x+6-4x^4-3x^3-4x^2+2x-9\)
\(=x^2-3\)
K(x)-L(x)
\(=4x^4+3x^3+5x^2-2x+6+4x^4+3x^3+4x^2-2x+9\)
\(=8x^4+6x^3+9x^2-4x+15\)
g: M(x)+N(x)
\(=-5x^4+4x^3-5x^2-\dfrac{1}{2}x-19+6x^4-4x^3+3x^2+\dfrac{1}{2}x-20\)
\(=x^4-2x^2-39\)
M(x)-N(x)
\(=-5x^4+4x^3-5x^2-\dfrac{1}{2}x-19-6x^4+4x^3-3x^2-\dfrac{1}{2}x+20\)
\(=-11x^4+8x^3-8x^2-x+1\)
h:
\(O\left(x\right)=x^5+x^3-4x-x^5+3x+7\)
\(=\left(x^5-x^5\right)+x^3+\left(-4x+3x\right)+7\)
\(=x^3-x+7\)
\(P\left(x\right)=3x^2-x^3+8x-3x^2-14\)
\(=-x^3+\left(3x^2-3x^2\right)+8x-14=-x^3+8x-14\)
O(x)+P(x)
\(=x^3-x+7-x^3+8x-14\)
\(=7x-7\)
O(x)-P(x)
\(=x^3-x+7+x^3-8x+14\)
\(=2x^3-9x+21\)