GIÚP MIK B2 Ạ MIK ĐANG CẦN GẤP LẮM
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1 had stayed
2 were
3 arrive
4 would have bought
5 would go
6 comes
7 had thought
8 gets
9 will become
10 had known
11 hurries
12 would change
13 would have trusted
14 doesn't study
15 weren't
Bài 34:
a: =>x+28=0
=>x=-28
b: =>27-x=0 hoặc x+9=0
=>x=27 hoặc x=-9
c: =>x(x-43)=0
=>x=0 hoặc x=43
Bài 1:
a: 4/9=4/9
5/3=15/9
b: 3/14=-18/84
-5/6=-70/84
c: -2/25=-14/175
-3/35=-15/175
d: -7/20=-35/100
-11/25=-44/100
e: -25/75=-1/3=-5/15
-12/30=-2/5=-6/15
c: \(=\dfrac{-27\cdot100}{-30}=\dfrac{2700}{30}=90\)
Câu 3:
a: Ta có: \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)\)
\(=6x^2-2x-6x^2-2x+18x+6\)
=14x+6
b: Ta có: \(2x\left(x+7\right)-3x\left(x+1\right)\)
\(=2x^2+14x-3x^2-3x\)
\(=-x^2+11x\)
Câu 2:
a: Ta có: \(\left(-8x^5+12x^3-16x^2\right):4x^2\)
\(=-8x^5:4x^2+12x^3:4x^2-16x^2:4x^2\)
\(=-2x^3+3x-4\)
b: Ta có: \(\left(12x^3y^3-18x^2y+9xy^2\right):6xy\)
\(=12x^3y^3:6xy-18x^2y:6xy+9xy^2:6xy\)
\(=2x^2y^2-3x+\dfrac{3}{2}y\)
c: Ta có: \(\dfrac{x^3-11x^2+27x-9}{x-3}\)
\(=\dfrac{x^3-3x^2-8x^2+24x+3x-9}{x-3}\)
\(=x^2-8x+3\)
d: Ta có: \(\dfrac{6x^4-13x^3+7x^2-x-5}{3x+1}\)
\(=\dfrac{6x^4+2x^3-15x^3-5x^2+12x^2+4x-5x-\dfrac{5}{3}-\dfrac{10}{3}}{3x+1}\)
\(=2x^3-5x^2+4x-\dfrac{5}{3}-\dfrac{\dfrac{10}{3}}{3x+1}\)