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KN
17 tháng 8 2021
1. ( x2 - x + 2 )4 - 3x2 ( x2 - x + 2 )2 + 2x4
Đặt t = x2 - x + 2 , ta có :
t4 - 3x2t2 + 2x4
= t4 - 2x2t2 - x2t2 + 2x4
= t2 ( t2 - 2x2 ) - x2 ( t2 - 2x2 )
= ( t2 - x2 ) ( t2 - 2x2 )
= ( t - x ) ( t + x ) ( t2 - 2x2 )
= ( x2 - x + 2 - x ) ( x2 - x + 2 + x ) [ ( x2 - x + 2 )2 - 2x2 ]
= ( x2 - 2x + 2 ) ( x2 + 2x ) ( x2 - 3x + 2 ) ( x2 + x + 2 )
2. 3 ( - x2 + 2x + 3 )4 - 26x2 ( - x2 + 2x + 3 )2 - 9x4
Đặt y = - x2 + 2x + 3 , ta có :
3y4 - 26x2y2 - 9x4
= x2y2 + 3y4 - 9x4 - 27x2y2
= y2 ( x2 + 3y2 ) - 9x2 ( x2 + 3y2 )
= ( y2 - 9x2 ) ( x2 + 3y2 )
= ( y - 3x ) ( y + 3x ) ( x2 + 3y2 )
= ( - x2 + 2x + 3 - 3x ) ( - x2 + 2x + 3 + 3x ) [ x2 + 3 ( - x2 + 2x + 3 )2 ]
= ( - x2 - x + 3 ) ( - x2 + 5x + 3 ) ( 3x4 - 12x3 - 5x2 + 36x + 27 )
NL
Nguyễn Lê Phước Thịnh
CTVHS
3 tháng 6 2023
1: Sửa đề: 3x-5
\(=\dfrac{-x^2\left(3x-5\right)-3\left(3x-5\right)}{3x-5}=-x^2-3\)
2: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)
=5x^2+14x^2+12x+8
3: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)
4: \(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}=x^2+1-2x\)
5: \(=\dfrac{x^2\left(5-3x\right)+3\left(5-3x\right)}{5-3x}=x^2+3\)
P
11 tháng 7 2023
a) \(\left(x+2\right)\left(x^2-24+4\right)\left(x^3+8\right)\)
\(=\left(x+2\right)\left(x^2-20\right)\left(x^3+8\right)\)
\(=\left(x^3-20x+2x^2-40\right)\left(x^3+8\right)\)
\(=x^6+8x^3-20x^4+160x+2x^5+16x^2-40x^3-120\)
\(=x^6+2x^5-20x^4-32x^3+16x^2+160x-120\)
b) \(\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
\(=8x^3+2x^2+\dfrac{1}{2}x-2x^2-\dfrac{1}{2}x-\dfrac{1}{8}\)
\(=8x^3-\dfrac{1}{8}\)
c) \(\left(x^2+y\right)\left(x^2-y\right)+y^2+x^4\)
\(=\left(x^2\right)^2-y^2+y^2+x^4\)
\(=x^4-y^2+y^2+x^4\)
\(=2x^4\)
d) \(\left(x+3\right)\left(x^2-3x+9\right)-x^3\)
\(=\left(x+3\right)\left(x^2-3\cdot x+3^2\right)-x^3\)
\(=x^3+3^3-x^3\)
\(=27\)
e) \(\left(3x+y\right)\left(9x^2-3xy+y^2\right)-26x^3\)
\(=\left(3x+y\right)\left[\left(3x\right)^2-3x\cdot y+y^2\right]-26x^3\)
\(=\left(3x\right)^3+y^3-26x^3\)
\(=27x^3+y^3-26x^3\)
\(=x^3+y^3\)
g) \(\left(x+3y\right)\left(x^2-3xy+9y^2\right)+\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)
\(=\left(x+3y\right)\left[x^2-x\cdot3y+\left(3y\right)^2\right]+\left(3x-y\right)\left[\left(3x\right)^2+3x\cdot y+y^2\right]\)
\(=\left[x^3+\left(3y\right)^3\right]+\left[\left(3x\right)^3-y^3\right]\)
\(=x^3+27y^3+27x^3-y^3\)
\(=28x^3+26y^3\)
NX
Tim x,
a,2x^4-6x^3+x^2+6x-3=0
b,x^3-9x^2+26x+24=0
c, P= 2x^4 - 4x^3 + 6x^2 - 4x + 5 biet rang x^2 - x=7
1
28 tháng 11 2016
a)\(2x^4-6x^3+x^2+6x-3=0\)
\(\Leftrightarrow2x^4-6x^3+3x^2-2x^2+6x-3=0\)
\(\Leftrightarrow x^2\left(2x^2-6x+3\right)-\left(2x^2-6x+3\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(2x^2-6x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(2x^2-6x+3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x+1=0\\2x^2-6x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\\\Delta_{2x^2-6x+3}=\left(-6\right)^2-4\left(2.3\right)=12\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\\x_{1,2}=\frac{6\pm\sqrt{12}}{4}\end{array}\right.\)
b)\(x^3+9x^2+26x+24=0\)
\(\Leftrightarrow x^3+5x^2+6x+4x^2+20x+24=0\)
\(\Leftrightarrow x\left(x^2+5x+6\right)+4\left(x^2+5x+6\right)=0\)
\(\Leftrightarrow\left(x^2+5x+6\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+2=0\\x+3=0\\x+4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\\x=-3\\x=-4\end{array}\right.\)
8 tháng 4 2020
\(\left(x-1\right)^2-\left(x+1\right)^2=2\left(x+3\right)\)
\(\Leftrightarrow\left(x-1+x+1\right)\left(x-1-x-1\right)=2\left(x+3\right)\)
\(\Leftrightarrow2x\left(-2\right)=2\left(x+3\right)\)
\(\Leftrightarrow-4x=2x+6\)
\(\Leftrightarrow-6x=6\)
\(\Leftrightarrow x=-1\)
2) \(\left(2x-1\right)^2-\left(2x+1\right)^2=4\left(x-3\right)\)
\(\Leftrightarrow\left(2x-1+2x+1\right)\left(2x-1-2x-1\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow4x\left(-2\right)-4x+12=0\)
\(\Leftrightarrow-12x=-12\)
\(\Leftrightarrow x=1\)
3)\(\left(2x+3\right)^2-\left(2x+3\right)\left(2x-4\right)+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(2x+3\right)\left(2x+3-2x+4\right)+\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow7\left(2x+3\right)+x^2-4x+4=0\)
\(\Leftrightarrow x^2+10x+25=0\)
\(\Leftrightarrow\left(x+5\right)^2=0\)
\(\Leftrightarrow x=-5\)
4) \(8x^3-\left(x+1\right)^3=3x-3\)
\(\Leftrightarrow8x^3-\left(x^3+3x+3x^2+1\right)-3x+3=0\)
\(\Leftrightarrow7x^3-3x^2-6x+2=0\)
\(\Leftrightarrow\left(x-1\right)\left(7x^2+4x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{-2+3\sqrt{2}}{7}\\x=\frac{-2-3\sqrt{2}}{7}\end{matrix}\right.\)
5)\(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)
\(\Leftrightarrow\left(3x\right)^3-2^3-\left(\left(3x\right)^3-1^3\right)=x-4\)
\(\Leftrightarrow27x^3-8-\left(27x^3-1\right)=x-4\)
\(\Leftrightarrow-7=x-4\)
\(\Leftrightarrow x=-3\)
TK
0
IL
17 tháng 8 2016
Katherine Lilly Filbert nói rất đúng câu hỏi nhiều như vậy ai mà trả lời đc hết cơ chứ
Lời giải:
$3x(-x^2+2x+3)-26x^2(-x^2+2x+3)-9x^4$
$=(-x^2+2x+3)(3x-26x^2)-9x^4$
$=(-x^2+2x+3).x(3-26x)-9x^4$
$=x[(-x^2+2x+3)(3-26x)-9x^3]$
$=x(17x^3-55x^2-72x+9)$
bn đăng từng bài riêng lẻ thôi nhé!