Thực hiện phép tính và rút gọn:
2x(x-3)+(x-1)(x+1)
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a) \(2x\left(x+5\right)-2x^2=2x^2+10x-2x^2=10x\)
b) \(\left(x+3\right)^2+\left(x-1\right)\left(3+2x\right)=x^2+6x+9+3x+2x^2-3-2x\)
\(=3x^2+7x+6\)
a: \(2x\left(x+5\right)-2x^2=2x^2+10x-2x^2=10x\)
b: \(\left(x+3\right)^2+\left(2x+3\right)\left(x-1\right)\)
\(=x^2+6x+9+2x^2-2x+3x-3\)
\(=3x^2+7x+6\)
Bài 3:
\(\Leftrightarrow x^3+64-x^3+25x=264\)
hay x=8
\(1,C=6x^2+23x-55-6x^2-23x-21=-76\\ 2,=\left(2x^4-x^2+2x^3-x-6x^2+6-3\right):\left(2x^2-1\right)\\ =\left[\left(2x^2-1\right)\left(x^2+x-6\right)-3\right]:\left(2x^2-1\right)\\ =x^2+x-6\left(dư.-3\right)\\ 3,\Leftrightarrow x^3+64-x^3+25x=264\\ \Leftrightarrow25x=200\Leftrightarrow x=8\)
2x ( x - 5 ) x . ( 3 - 2x ) = 26
2x\(^2\)- 10x . 3x - 2x\(^2\)= 26
2x\(^2\). ( 10x - 3x ) = 26
2x\(^2\). 7x = 26
14x\(^3\) = 26
x\(^3\) = 26 : 14
x\(^3\) = \(\frac{13}{7}\)
→ X = 1.229.... \(\approx\)1,3
\(2x\left(x-5\right)\cdot x\left(3-2x\right)=26\)
\(\Leftrightarrow\left(2x^2-10x\right)\left(3x-6x\right)=26\)
\(\Leftrightarrow6x^3-30x^2-12x^3+60x^2=26\)
\(\Leftrightarrow-12x^3+30x^2=26\)
\(\Leftrightarrow2\left(-6x^3+15x^2\right)=26\)
\(\Leftrightarrow-6x^3+15x^2=13\)
\(\Leftrightarrow-6x^3+15x^2-13=0\)
...............mình chỉ làm được đến đây thôi!
a: \(=25x^4-10x^3+5x^2\)
c: \(=2x^3-3x-5x^3-x^2+x^2=-3x^3-3x\)
a: \(=\dfrac{x^3}{\left(x-2\right)\left(x+2\right)}-\dfrac{x}{x-2}-\dfrac{2}{x+2}\)
\(=\dfrac{x^3-x^2-2x-2x+4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^3-x^2-4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x-2\right)\left(x+2\right)\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}=x-1\)
Bài 1:
\(3a.\left(2a^2-ab\right)=6a^3-3a^2b\)
\(\left(4-7b^2\right).\left(2a+5b\right)=8a+20b-14ab^2-35b^3\)
Bài 2:
\(2x^2-6x+xy-3y=2x.\left(x-3\right)+y.\left(x-3\right)=\left(x-3\right).\left(2x+y\right)\)
Bài 3: Tại x = 3/2, y =1/3 thì Q = 67/9
Bài 4:
\(\left(\frac{1}{x+1}+\frac{2x}{1-x^2}\right).\left(\frac{1}{x-1}\right)\) \(\frac{1}{\left(x+1\right).\left(x-1\right)}+\frac{2x}{\left(1-x^2\right).\left(x-1\right)}=\frac{x-1}{\left(x+1\right).\left(x-1\right)^2}+\frac{-2x}{\left(x-1\right)^2.\left(x+1\right)}\)
= \(\frac{x-1-2x}{\left(x+1\right).\left(x-1\right)^2}=\frac{-\left(x+1\right)}{\left(x+1\right).\left(x-1\right)^2}=\frac{-1}{\left(x-1\right)^2}\)
a: \(-2y\left(5xy+3x^2\right)=-10xy^2-6x^2y\)
b: \(\left(2x+y\right)\left(4x-y\right)\)
\(=8x^2-2xy+4xy-y^2\)
\(=8x^2+2xy-y^2\)
\(=2x^2-6x+x^2-1=x^2-6x-1\)
\(2x\left(x-3\right)+\left(x-1\right)\left(x+1\right)\)
\(=2x^2-6x+x^2-1\)
\(=3x^2-6x+1\)