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Trả lời:
M = ( x - 3 )3 - ( x + 1 )3 - 12x ( x - 1 )
= x3 - 3.x2.3 + 3.x.32 - 33 - ( x3 + 3.x2.1 + 3.x.12 + 13 ) - 12x2 - 12x
= x3 - 9x2 + 27x - 27 - x3 - 3x2 - 3x - 1 - 12x2 - 12x
= - 24x2 + 12x - 28
\(\left(\frac{9}{x.x^2-9.x}+\frac{1}{x+_{ }3}\right):\left(\frac{x-3}{x.3+x^2}-\frac{x}{3.x+9}\right)\) đk (x\(\ne\)o; công trừ 3)
<=>\(9+\frac{x.\left(x-3\right)}{x.\left(x^2-9\right)}\):\(\frac{3.\left(x-3\right)-x^2}{3x.\left(x+3\right)}\)
<=>\(-\frac{3}{x-3}=\frac{3}{3-x}\)
Bạn ơi mk k hiểu sao lại ra bước 2 ... bạn giải chi tiết giùm mk nha
dù sao cx cảm ơn bạn đã giúp mk
Suy ra:4^3-12x+24x-12= 64 +12x-12
= 12x+52
mk ko bik co dung ko sai thi thoi nha!
a: \(=\dfrac{\left(3x-1\right)^3}{3x-1}\cdot\dfrac{8xy}{12x^3}=\dfrac{2y\left(3x-1\right)^2}{3x^2}\)
b: \(=\dfrac{\left(x-5\right)\left(x+5\right)}{x\left(x-5\right)}=\dfrac{x+5}{x}\)
a)\(P=\left(5x-1\right)+2\left(1-5x\right)\left(4+5x\right)+\left(5x+4\right)\)
\(P=5x-1+5x+4+\left(2-10x\right)\left(4+5x\right)\)
\(P=10x+3+8+10x-40x-50x^2\)
\(P=-20x+11-50x^2\)
b)\(Q=\left(x-y\right)^3+\left(y+x\right)^3+\left(y-x\right)^3-3xy\left(x+y\right)\)
\(Q=x^3-3x^2y+3xy^2-y^3+y^3+3y^2x+3yx^2+x^3+y^3-3y^2x+3yx^2-x^3-3x^2y+3xy^2\)
\(Q=x^3+y^3\)
a) P = (5x - 1) + [(-2).(-1).(1-5x).(4+5x)] + (5x+4)^2
= (5x - 1).[(-1)-2.(-1).(4+5x)] + (25x^2+40x+16)
= [(5x - 1).(10x + 7)] + (25x^2+40x+16)
= 50x^2 +35x - 10x - 7 + 25x^2+40x+16
= 75x^2 + 65x -9
b) Q = (x-y)^3 + (x+y)^3 + (y-x)^3 - [3xy(x+y)]
Xét (x-y)^3 = x^3 - 3x^2y + 3xy^2 - y^3 = x^3 + 3x^2y + 3xy^2 + y^3 = y^3 - 3xy^2 + 3x^2y - x^3
Do đó Q = (2x^3 + 6xy^2) + (y^3 - 3xy^2 + 3x^2y - x^3) - [3xy(x+y)]
= (x^3 + 3xy^2 + 3x^2y + y^3) - 3x^2y - 3xy^2
= x^3 + y^3
a, P=(5x-1)+2(1-5x)(4+5x)+(5x+4)
=5x-1+2.(4-15x-5x2)+5x+4
=5x-1+8-30x-10x2+5x+4
=-10x2-20x+11
b,Q=(x-y)^3 +(y+x)^3+ (y-x)^3 -3xy(x+y)
=x3-3x2y+3xy2-y3+y3+3x2y+3xy2+x3+y3-3y2x+3yx2-x3-3x2y-3xy2
=x3+y3
a)
P=(5x−1)+2(1−5x)(4+5x)+(5x+4)2
P=(5x−1)2+2(1−5x)(4+5x)+(5x+4)2−(5x−1)2
P=(5x−1+5x+4)2−(5x−1)2
P=(5x−1+5x+4−5x+1)(5x−1+5x+4+5x−1
b)
Q= (x-y)^3 + (x+y)^3 + (y-x)^3 - [3xy(x+y)]
(x-y)^3 = x^3 - 3x^2y + 3xy^2 - y^3
(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3
(y-x)^3 = y^3 - 3xy^2 + 3x^2y - x^3
Q= (2x^3 + 6xy^2) + (y^3 - 3xy^2 + 3x^2y - x^3) - [3xy(x+y)]
= (x^3 + 3xy^2 + 3x^2y + y^3) - 3x^2y - 3xy^2
Q= x^3 + y^3
(x - 3)3 - (x + 1)3 + 12x (x - 1)
= x3 - 3x2 . 3 + 3x . 32 - 27 - (x3 + 3x2 . 1 + 3x . 12 + 13) + 12x . x + 12x . (-1)
= x3 - 9x2 + 27x - 27 - x3 - 3x2 - 3x - 1 + 12x2 - 12x
= (x3 - x3) + (12x2 - 9x2 - 3x2) + (27x - 3x - 12x) - (27 + 1)
= 12x - 28
\(\left(x-3\right)^3-\left(x+1\right)^3+12x\left(x-1\right)\)
\(\Leftrightarrow\left(x^3-3x^23+3x3^2-3^3\right)-\left(x^3+3x^21+3x1^2+1^3\right)+12x^2-12x\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3-3x^2-3x-1+12x^2-12x\)
\(\Leftrightarrow12x-28=0\)
\(\Leftrightarrow12x=28\)
\(\Leftrightarrow x=\frac{7}{3}\)
Vậy S={\(\frac{7}{3}\)} là nghiệm pt