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HM
0
21 tháng 1
Bài 1:
a: ĐKXĐ: \(x+4\ne0\)
=>\(x\ne-4\)
b: ĐKXĐ: \(2x-1\ne0\)
=>\(2x\ne1\)
=>\(x\ne\dfrac{1}{2}\)
c: ĐKXĐ: \(x\left(y-3\right)\ne0\)
=>\(\left\{{}\begin{matrix}x\ne0\\y-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\y\ne3\end{matrix}\right.\)
d: ĐKXĐ: \(x^2-4y^2\ne0\)
=>\(\left(x-2y\right)\left(x+2y\right)\ne0\)
=>\(x\ne\pm2y\)
e: ĐKXĐ: \(\left(5-x\right)\left(y+2\right)\ne0\)
=>\(\left\{{}\begin{matrix}x\ne5\\y\ne-2\end{matrix}\right.\)
Bài 2:
a: \(\dfrac{-12x^3y^2}{-20x^2y^2}=\dfrac{12x^3y^2}{20x^2y^2}=\dfrac{12x^3y^2:4x^2y^2}{20x^2y^2:4x^2y^2}=\dfrac{3x}{5}\)
b: \(\dfrac{x^2+xy-x-y}{x^2-xy-x+y}\)
\(=\dfrac{\left(x^2+xy\right)-\left(x+y\right)}{\left(x^2-xy\right)-\left(x-y\right)}\)
\(=\dfrac{x\left(x+y\right)-\left(x+y\right)}{x\left(x-y\right)-\left(x-y\right)}=\dfrac{\left(x+y\right)\left(x-1\right)}{\left(x-y\right)\left(x-1\right)}\)
\(=\dfrac{x+y}{x-y}\)
c: \(\dfrac{7x^2-7xy}{y^2-x^2}\)
\(=\dfrac{7x\left(x-y\right)}{\left(y-x\right)\left(y+x\right)}\)
\(=\dfrac{-7x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{-7x}{x+y}\)
d: \(\dfrac{7x^2+14x+7}{3x^2+3x}\)
\(=\dfrac{7\left(x^2+2x+1\right)}{3x\left(x+1\right)}\)
\(=\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)}{3x}\)
e: \(\dfrac{3y-2-3xy+2x}{1-3x-x^3+3x^2}\)
\(=\dfrac{3y-2-x\left(3y-2\right)}{1-3x+3x^2-x^3}\)
\(=\dfrac{\left(3y-2\right)\left(1-x\right)}{\left(1-x\right)^3}=\dfrac{3y-2}{\left(1-x\right)^2}\)
g: \(\dfrac{x^2+7x+12}{x^2+5x+6}\)
\(=\dfrac{\left(x+3\right)\left(x+4\right)}{\left(x+3\right)\left(x+2\right)}\)
\(=\dfrac{x+4}{x+2}\)
12 tháng 10 2021
Bài 2:
a: \(3x^2-3xy=3x\left(x-y\right)\)
b: \(x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\)
c: \(3x-3y+xy-y^2=\left(x-y\right)\left(3+y\right)\)
d: \(x^2-y^2+2y-1=\left(x-y+1\right)\left(x+y-1\right)\)
25 tháng 10 2023
a: \(a^2+6ab+9b^2-1\)
\(=\left(a+3b\right)^2-1^2\)
\(=\left(a+3b+1\right)\left(a+3b-1\right)\)
b: \(4x^2-25+\left(2x+7\right)\left(5-2x\right)\)
\(=\left(2x-5\right)\left(2x+5\right)-\left(2x+7\right)\left(2x-5\right)\)
\(=\left(2x-5\right)\left(2x+5-2x-7\right)\)
\(=-2\left(2x-5\right)\)
c: \(5\left(x+3y\right)-15x\left(x+3y\right)\)
\(=\left(x+3y\right)\left(-15x+5\right)\)
\(=-5\left(3x-1\right)\left(x+3y\right)\)
d: \(x\left(x+y\right)^2-y\left(x+y\right)^2+xy-x^2\)
\(=\left(x+y\right)^2\cdot\left(x-y\right)-x\left(x-y\right)\)
\(=\left(x-y\right)\left[\left(x+y\right)^2-x\right]\)
e: \(a^2-6a+9-b^2\)
\(=\left(a-3\right)^2-b^2\)
\(=\left(a-3-b\right)\left(a-3+b\right)\)
f: \(x^3-y^3-3x^2+3x-1\)
\(=\left(x^3-3x^2+3x-1\right)-y^3\)
\(=\left(x-1\right)^3-y^3\)
\(=\left(x-1-y\right)\left[\left(x-1\right)^2+y\left(x-1\right)+y^2\right]\)
11 tháng 10 2021
a: \(x\left(2x-y\right)-y\left(2x-y\right)=\left(2x-y\right)\left(x-y\right)\)
c: \(x^2-3x+3y-y^2\)
\(=\left(x-y\right)\left(x+y\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-3\right)\)
b: \(x^2-6x-7=\left(x-7\right)\left(x+1\right)\)
11 tháng 10 2021
a) \(x\left(2x-y\right)-y\left(2x-y\right)=\left(2x-y\right)\left(x-y\right)\)
b) \(x^2-6x-7=x\left(x-7\right)+\left(x-7\right)=\left(x-7\right)\left(x+1\right)\)
c) \(x^2-3x+3y-y^2=\left(x-y\right)\left(x+y\right)-3\left(x-y\right)=\left(x-y\right)\left(x+y-3\right)\)
d) \(x^3-xy+2y-8=\left(x-2\right)\left(x^2+2x+4\right)-y\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x+4-y\right)\)
25 tháng 7 2021
a, \(=\left(xy+1+x-y\right)\left(xy+1-x+y\right)\)
b, \(\left(x+y-x+y\right)[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2]\)
\(=2y[x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2]\)
\(=2y\left(3x^2+y^2\right)\)
c,\(=3\left(x+1\right)^2\left(x^2-x+1\right)y^2\)
25 tháng 7 2021
câu a, b áp dụng hằng đẳng thức rồi làm nha
c) 3x4y2 + 3x3y2 + 3xy2 + 3y2
= ( 3x4y2 + 3x3y2 ) + ( 3xy2 + 3y2 )
= 3x3y2 ( x + 1) + 3y2 ( x + 1 )
= ( 3x3y2 + 3y2 ) ( x + 1 )
= 3y2 ( x3 + 1 ) ( x + 1 )
= 3y2 ( x + 1 ) ( x2 - x + 1 ) ( x + 1 )
= 3y2 ( x + 1 )2 ( x2 - x + 1 )
DT
1
6 tháng 11 2019
\(A=x\left(x-3\right)\left(x-4\right)\left(x-7\right)\)
\(=\left[x\left(x-7\right)\right]\left[\left(x-3\right)\left(x-4\right)\right]\)
\(=\left[x^2-7x\right]\left[x^2-7x+12\right]\)
Đặt: \(t=x^2-7x\)
=> \(A=t\left(t+12\right)=t^2+12t+36-36\)
\(=\left(t+6\right)^2-36\ge-36\)
Dấu "=" xảy ra <=> \(t=-6\)
khi đó: \(x^2-7x=-6\Leftrightarrow x^2-x-6x+6=0\)
<=> \(x\left(x-1\right)-6\left(x-1\right)=0\)
<=> (x - 6 ) ( x - 1) =0
<=> x = 6 hoặc x =1
Vậy GTNN của A là -36 đạt tại x =6 hoặc x =1 .
b) \(B=x^2+xy-y^2-3x-3y\)
Xem lại đề nhé \(y^2\)hay \(-y^2\)?
PT
0