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a.\(-x^4+2x^3-2x^2+2x-1=-\left(x^4+2x^2+1\right)+\left(2x^3+2x\right)\)
=\(-\left(x^2+1\right)^2+2x\left(x^2+1\right)=\left(x^2+1\right)\left(-1+2x\right)\)
b.\(-2x^2-y^2+2xy+4x-10=-\left(x^2-4x+4\right)-\left(x^2-2xy+y^2\right)-6\)
=\(-\left(x-2\right)^2-\left(x-y\right)^2-6\)
a.=\(-\left(x^4+2x^2+\text{1}\right)+\left(2x^3+2x\right)\)=\(-\left(x^2+1\right)^2+2x\left(x^2+1\right)=\left(x^2+1\right)\left(-x^2+2x-1\right)\)
=\(-\left(x^2+1\right)\left(x-1\right)^2\)
.
b.=\(-\left(x^2-4x+4\right)-\left(x^2-2xy+y^2\right)-36\)=\(-\left(x-2\right)^2-\left(x-y\right)^2-36\)
a) \(2\left(x^2-x\right)-2x^2=3\)
\(\Leftrightarrow2x^2-2x-2x^2=3\)
\(\Leftrightarrow-2x=3\Leftrightarrow x=-\frac{3}{2}\)
b) \(2\left(x^2+x\right)-2x=8\)
\(\Leftrightarrow2x^2+2x-2x=8\)
\(\Leftrightarrow2x^2=8\Leftrightarrow x^2=4\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-2\end{array}\right.\)
a) 2 . ( x^2 - x ) - 2x^2 = 3
2x2-2x-2x2=0
-2x=0
x=0
Vậy x=0
b) 2 . ( x^2 + x ) - 2x = 8
2x2+2x-2x=8
2x2=8
x2=4=22=(-2)2
Vậy x=2;-2
a.\(2x\left(7x^2-5x-1\right)=14x^3-10x^2-2x\)
b.\(-2x^3\left(2x^2-3y+5yz\right)=-4x^5+6x^3y-10x^3yz\)
c.\(\left(2x-y\right)\left(4x^2-2xy+y^2\right)=2x\left(4x^2-2xy+y^2\right)-y\left(4x^2-2xy+y^2\right)\)
\(=8x^2-4x^2y+4xy^2-4x^2y+2xy^2-y^3\)
a.2x(7x2−5x−1)=14x3−10x2−2x
b.−2x3(2x2−3y+5yz)=−4x5+6x3y−10x3yz
c.(2x−y)(4x2−2xy+y2)=2x(4x2−2xy+y2)−y(4x2−2xy+y2)
=8x2−4x2y+4xy2−4x2y+2xy2−y3
a,\(\frac{3}{2x^2+2x}+\frac{2x-1}{x^2-1}-\frac{2}{x}\)
\(=\frac{3}{2x\left(x+1\right)}+\frac{2x-1}{\left(x-1\right)\left(x+1\right)}-\frac{2}{x}\)
\(=\frac{3\left(x-1\right)}{2x\left(x+1\right)\left(x-1\right)}+\frac{\left(2x-1\right).2x}{2x\left(x-1\right)\left(x+1\right)}-\frac{2.2\left(x+1\right)\left(x-1\right)}{2x\left(x+1\right)\left(x-1\right)}\)
\(=\frac{3x-3}{2x\left(x+1\right)\left(x-1\right)}+\frac{4x^2-2x}{2x\left(x-1\right)\left(x+1\right)}-\frac{4x^2-4}{2x\left(x+1\right)\left(x-1\right)}\)
\(=\frac{3x-3+4x^2-2x-4x^2+4}{2x\left(x+1\right)\left(x-1\right)}\)
\(=\frac{x+1}{2x\left(x+1\right)\left(x-1\right)}=\frac{1}{2x\left(x-1\right)}\)
\(b,\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
\(=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x-y\right)}\)
\(=\frac{3x.2\left(x-y\right)}{10\left(x+y\right).\left(x-y\right)}-\frac{x.\left(x+y\right)}{10\left(x-y\right).\left(x+y\right)}\)
\(=\frac{6x^2-6xy}{10\left(x+y\right)\left(x-y\right)}-\frac{x^2+xy}{10\left(x-y\right)\left(x+y\right)}\)
\(=\frac{6x^2-6xy-x^2+xy}{10\left(x+y\right)\left(x-y\right)}\)
\(=\frac{5x^2-5xy}{10\left(x+y\right)\left(x+y\right)}\)
\(=\frac{5x\left(x-y\right)}{10\left(x-y\right)\left(x+y\right)}=\frac{x}{2\left(x+y\right)}\)
1.
$2x^3-21x^2+67x-60=2x^2(x-5)-11x(x-5)+12(x-5)$
$=(x-5)(2x^2-11x+12)$
$\Rightarrow (2x^3-21x^2+67x-60):(x-5)=2x^2-11x+12$
2.
$x^4+2x^3+x-25=x^2(x^2+5)+2x(x^2+5)-5x^2-9x-25$
$=x^2(x^2+5)+2x(x^2+5)-5(x^2+5)-9x=(x^2+5)(x^2+2x-5)-9x$
$\Rightarrow (x^4+2x^3+x-25):(x^2+5)=x^2+2x-5$ và dư $-9x$
cac bạn cho Anh sửa đề a , kết quả = 44 chứ k phải 41 đâu nhé
c: \(=\dfrac{1}{3x-2}-\dfrac{4}{3x+2}+\dfrac{3x-6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+2-12x+8+3x-6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{-6x+4}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{-2}{3x+2}\)
d: \(=\dfrac{x^2-4-x^2+10}{x+2}=\dfrac{6}{x+2}\)
e: \(=\dfrac{1}{2\left(x-y\right)}-\dfrac{1}{2\left(x+y\right)}-\dfrac{y}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{x+y-x+y-2y}{2\left(x-y\right)\left(x+y\right)}=\dfrac{0}{2\left(x-y\right)\left(x+y\right)}=0\)
2400 - (60 - 2x)(40 - 2x)
= 2400 - 2400 + 120x + 80x - 4x²
= -4x² + 200x