Anh chị ơi giúp e làm bài này với ạ, e cần gấp ý ạ:(( E cảm ơn các anh chị nhiều ạ<3
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15 tháng 11 2021
\(a,=x^2+x+4x+4=\left(x+1\right)\left(x+4\right)\\ b,=x^2+2x-3x-6=\left(x-3\right)\left(x+2\right)\\ c,=x^2-2x-3x+6=\left(x-2\right)\left(x-3\right)\\ d,=3\left(x^2-2x+5x-10\right)=3\left(x-2\right)\left(x+5\right)\\ e,=-3x^2+6x-x+2=\left(x-2\right)\left(1-3x\right)\\ f,=x^2-x-6x+6=\left(x-1\right)\left(x-6\right)\\ h,=4\left(x^2-3x-6x+18\right)=4\left(x-3\right)\left(x-6\right)\\ i,=3\left(3x^2-3x-8x+5\right)=3\left(x-1\right)\left(3x-8\right)\\ k,=-\left(2x^2+x+4x+2\right)=-\left(2x+1\right)\left(x+2\right)\\ l,=x^2-2xy-5xy+10y^2=\left(x-2y\right)\left(x-5y\right)\\ m,=x^2-xy-2xy+2y^2=\left(x-y\right)\left(x-2y\right)\\ n,=x^2+xy-3xy-3y^2=\left(x+y\right)\left(x-3y\right)\)
MA
0
MA
0
MA
1
15 tháng 11 2021
a) ĐKXĐ: \(\left\{{}\begin{matrix}2x+3\ne0\\2x+1\ne0\\\left(2x+3\right)\left(2x+1\right)\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-\dfrac{3}{2}\\x\ne-\dfrac{1}{2}\\\left(2x+3\right)\left(2x+1\right)\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-\dfrac{3}{2}\\x\ne-\dfrac{1}{2}\end{matrix}\right.\)
b) \(\Rightarrow P=\dfrac{2\left(2x+1\right)+3\left(2x+3\right)-6x-5}{\left(2x+3\right)\left(2x+1\right)}\)
\(\Rightarrow P=\dfrac{4x+2+6x+9-6x-5}{\left(2x+3\right)\left(2x+1\right)}\)
\(\Rightarrow P=\dfrac{4x+6}{\left(2x+3\right)\left(2x+1\right)}\)
\(\Rightarrow P=\dfrac{2\left(2x+3\right)}{\left(2x+3\right)\left(2x+1\right)}\)
\(\Rightarrow P=\dfrac{2}{2x+1}\)
c) \(P=-1\Rightarrow\dfrac{2}{2x+1}=-1\\ \Rightarrow2=-2x-1\\ \Rightarrow2x=-3\\ \Rightarrow x=-\dfrac{3}{2}\)
MA
1
HV
7 tháng 12 2021
1 were you doing
2 were having - rang
3 takes - is celebrated
4 was formed
5 have lost - haven't found
6 is held to worship
7 skating
8 getting up
9 reading - doing
10 has been built
11 swimming - feel
12 were watching - failed
13 has worked - graduated
14 have been invited
15 will be discussing
16 decided not to stay
17 to pass - testing
18 not to phone
19 doing
20 to stay - do
MA
3
28 tháng 10 2021
a) \(A=\dfrac{\left(2x^2+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}\)
\(=\dfrac{2\left(x-2\right)}{x+2}\)
Thay \(x=\dfrac{1}{2}\) vào A ta được:
\(A=\dfrac{2\cdot\left(\dfrac{1}{2}-2\right)}{\dfrac{1}{2}+2}=\dfrac{-3}{\dfrac{5}{2}}=-\dfrac{6}{5}\)
28 tháng 10 2021
b) \(B=\dfrac{x^3-x^2y+xy^2}{x^3+y^3}=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}\)
Thay \(x=-5,y=10\) vào B ta đc:
\(B=\dfrac{-5}{-5+10}=-1\)
MA
1
8 tháng 11 2021
a: Xét tứ giác MIPC có
K là trung điểm của MP
K là trung điểm của IC
Do đó: MIPC là hình bình hành
mà MI=PI
nên MIPC là hình thoi
MA
0
MA
2
11 tháng 11 2021
\(a,=\left(6x+1-6x+1\right)^2=4\\ b,=3x^2-6x-5x+5x^2-8x^2-24=-11x-24\\ c,=14x^2+x-3-5x^2-18x+8-9x^2+17x=5\\ d,=6x^2+43x-40-6x^2-7x+3-36x+27=-10\)
11 tháng 11 2021
a) \(=\left(6x+1\right)^2-2\left(6x+1\right)\left(6x-1\right)+\left(6x-1\right)^2=\left(6x+1-6x+1\right)^2=2^2=4\)
b) \(=3x^2-6x-5x+5x^2-8x^2+24=-11x+24\)
c) \(\left(7x-3\right)\left(2x+1\right)-\left(5x-2\right)\left(x+4\right)-9x^2+17x=\left(7x-3\right).2x+\left(7x-3\right)-\left[\left(5x-2\right).x+4\left(5x-2\right)\right]-9x^2+17x=14x^2-6x+7x-3-\left(5x^2-2x+20x-8\right)-9x^2+17x=5x^2+18x-3-\left(5x^2+18x-8\right)=5x^2+18x-3-5x^2-18x+8=5\)
d) \(\left(6x-5\right)\left(x+8\right)-\left(3x-1\right)\left(2x+3\right)-9\left(4x-3\right)=\left(6x-5\right).x+8\left(6x-5\right)-\left[\left(3x-1\right).2x+3\left(3x-1\right)\right]-36x+27=6x^2-5x+48x-40-\left(6x^2-2x+9x-3\right)-36x+27=6x^2+7x-13-\left(6x^2+7x-3\right)=6x^2+7x-13-6x^2-7x+3=-10\)
Bài 6
\(a,ĐK:x\ne\pm5\\ b,P=\dfrac{x-5+2x+10-2x-10}{\left(x-5\right)\left(x+5\right)}=\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}=\dfrac{1}{x+5}\\ c,P=-3\Leftrightarrow\dfrac{1}{x+5}=-3\Leftrightarrow-3\left(x+5\right)=1\Leftrightarrow x=-\dfrac{16}{3}\\ \Leftrightarrow Q=\left(3x-7\right)^2=\left[3\cdot\left(-\dfrac{16}{3}\right)-7\right]^2=529\)
Bài 7:
\(a,ĐK:x\ne\pm3\\ b,P=\dfrac{3x-9+x+3+18}{\left(x-3\right)\left(x+3\right)}=\dfrac{4\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{4}{x-3}\\ b,P=4\Leftrightarrow4\left(x-3\right)=4\Leftrightarrow x=4\)