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a) \(\left(2x-3\right)\left(x^2-2x+1\right)+2\left(2-x\right)^3\)
\(=2x\left(x^2-2x+1\right)-3\left(x^2-2x+1\right)+2\left(2^3-3\cdot2^2\cdot x+3\cdot2\cdot x^2-x^3\right)\)
\(=2x^3-4x^2+2x-3x^2+6x-3+2\left(8-12x+6x^2-x^3\right)\)
\(=2x^3-4x^2+2x-3x^2+6x-3+16-24x+12x^2-2x^3\)
\(=\left(2x^3-2x^3\right)+\left(-4x^2-3x^2+12x^2\right)+\left(2x+6x-24x\right)+\left(-3+16\right)\)
\(=5x^2-16x+13\)
b)
2x^3 - 7x^2 + 2x + 3 x^2 - 4x + 3 2x^3 - 8x^2 + 6x x^2 - 4x + 3 2x + 1 - x^2 + 4x + 3 0
Vậy \(\left(2x^3-7x^2+2x+3\right):\left(x^2-4x+3\right)=2x+1\)
Câu b thêm dấu " - " ở chỗ 2x3 - 7x2 + 2x +3 và 2x3 - 8x2 + 6x nhé :)))
\(ĐKXĐ:x\ne3;x\ne-1\)
Nếu x=0 là nghiệm của phương trình
Nếu x khác 0 ta có:
\(\frac{1}{2\left(x-3\right)}+\frac{1}{2\left(x-1\right)}=\frac{2}{\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow\frac{x-1+x-3}{\left(x-1\right)\left(x-3\right)}=\frac{4}{\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow\frac{2x-4}{\left(x-1\right)\left(x-3\right)}=\frac{4}{\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow2x-4=4\)
\(\Leftrightarrow x=4\)
\(\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\left(x\ne-1;x\ne3\right)\)
<=> \(\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}-\frac{2x}{\left(x+1\right)\left(x-3\right)}=0\)
<=> \(\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}-\frac{2x\cdot2}{2\left(x+1\right)\left(x-3\right)}=0\)
<=> \(\frac{x^2+x+x^2-3x-4x}{2\left(x-3\right)\left(x+1\right)}=0\)
=> 2x2-6x=0
<=> 2x(x-3)=0
<=> \(\orbr{\begin{cases}2x=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
ĐCĐK x khác -1 và x khác 3 => x=0
Vậy x=0 là nghiệm của phương trình
\(\left(\dfrac{1}{2}x-1\right)\left(2x-3\right)=x^2-\dfrac{3}{2}x-2x+3=x^2-\dfrac{1}{2}x+3\)\(b,\left(x-7\right)\left(x-5\right)=x^2-5x-7x+35=x^2-12x+35\)\(c,\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)\left(4x-1\right)=\left(x^2-\dfrac{1}{4}\right)\left(4x-1\right)=4x^3-x^2-x+\dfrac{1}{4}\)
a)\(\left(2x^2-3x\right)\left(5x^2-2x+1\right)\)
\(=2x^2\left(5x^2-2x+1\right)-3x\left(5x^2-2x+1\right)\)
\(=10x^4-4x^3+2x^2-15x^3+6x^2-3x\)
\(=10x^4-19x^3+8x^2-3x\)
a. \(\left(2x^2-3x\right)\left(5x^2-2x+1\right)\)
\(=10x^4-4x^3+2x^2-15x^3+6x^2-3x\)
\(=10x^4-19x^3+8x^2-3x\)
b. \(\left(2x^4-x^3+3x^2\right):\left(\frac{1}{3}x^2\right)\)
\(=\left(2x^4-x^3+3x^2\right).\frac{3}{x^2}\)
\(=0,6x^2-3x+0,9\)
a) (x2 – 2x + 3) (1212x – 5)
= 1212x3 - 5x2 - x2 +10x + 3232x – 15
= 1212x3 – 6x2 + 232232x -15
b) (x2 – 2xy + y2)(x – y)
= x3 - x2 y - 2x2 y + 2xy2 +xy2- y3
= x3 - 3x2 y + 3xy2 - y3
a) (x2 – 2x + 3) ( 1/2x – 5) = \(\dfrac{1}{2}\)x3 – 5x2 – x2 + 10x +\(\dfrac{3}{2}\)x - 15
= \(\dfrac{1}{2}\)x3 – 6x2 + \(\dfrac{23}{2}\) x – 15.
b) (x2 – 2xy + y2)( x – y) = x3 – x2y – 2x2y + xy2 – y3 = x3 – 3x2y + 3xy2 – y3
Bài làm:
đk: \(x\ne-3;x\ne1\)
Ta có: \(\frac{x^2+6x+9}{1-x}\cdot\frac{\left(x-1\right)^3}{2\left(x+3\right)^3}\)
\(=\frac{\left(x+3\right)^2}{-\left(x-1\right)}\cdot\frac{\left(x-1\right)^3}{2\left(x+3\right)^3}\)
\(=\frac{-\left(x-1\right)^2}{2\left(x+3\right)}\)
\(=-\frac{x^2-2x+1}{2x+6}\)
\(ĐKXĐ:\hept{\begin{cases}x\ne-3\\x\ne1\end{cases}}\)
\(\frac{x^2+6x+9}{1-x}.\frac{\left(x-1\right)^3}{2\left(x+3\right)^3}=\frac{-\left(x+3\right)^2}{x-1}.\frac{\left(x-1\right)^3}{2\left(x+3\right)^3}=\frac{-\left(x-1\right)^2}{2\left(x+3\right)}\)
2x2 + 3(x - 1)(x + 1) - 5x(x + 1)
= 2x2 + 3.(x2 - 1) - 5x(x + 1)
= 2x2 + 3x2 - 3 - 5x2 - 5x
= -3 - 5x
= 2x2 + 3 (x2 - 1) - 5x2 - 5x ( hằng đẳng thức số 3 )
= 2x2 + 3x2 - 3 - 5x2 - 5x (nhân đơn với đa )
= -5x - 3 (thực hiện phép tính )
(không chắc nữa !!! pn tự kiểm tra lại nhé )
\(\left(x+2\right)^2-\left(x+3\right)\left(x-3\right)+1=x^2+4x+4-\left(x^2-9\right)+1\)
\(=x^2+4x+5-x^2+9=4x+14\)