a) x2(x-2)2-(x-2)^2 - x2 +1 b) x3-4x2+8x-8
c)1+6x-6x2-x3
d)x3-y3-3x2+3x-1
e)(x+y+z)^3-x3-y3-z3
rút gọn A,B,C
A=(3x+7)(2x+3)-(3x-5)(2x+11)
B=(x2-2)(x2+x-1)-x(x3+x2-3x-2)
C=x(x3+x2-3x-2)-(x2-2)(x2+x-1)
\(A=6x^2+23x+21-\left(6x^2+23x-55\right)=76\\ B=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x\\ =2\\ C=x^4+x^3-3x^2-2x-\left(x^4+x^3-x^2-2x^2-2x+2\right)\\ =-2\)
a)A=3x(2/3x2-3x4)+(3x2)(x3-1)+(-2+9).x2-12
b)B=x(2x3+x+2)-2x2(x2+1)+x2-2x+1
c)C=x.(2x+1)-x2(x+2)+x3-x+3
a, \(A=2x^3-9x^5+3x^5-3x^2+7x^2-12=-6x^5+2x^3+4x^2-12\)
b, \(B=2x^4+x^2+2x-2x^3-2x^2+x^2-2x+1=2x^4-2x^3+1\)
c, \(C=2x^2+x-x^3-2x^2+x^3-x+3=3\)
Chứng tỏ rằng mỗi biểu thức sau không phụ thuộc vào giá trị của biến x :
A=(x2-2)(x2+x-1)-x(x3+x2-3x-2)
B=2(2x+x2)-x2(x+2)+(x3-4x+3).
Thực hiện phép tính g) (x + 2)(1 + x - x2 + x3 - x4) - (1 - x)(1 + x +x2 + x3 + x4); a) (x + 1)(1 + x - x2 + x3 - x4) - (x - 1)(1 + x + x2 + x3 + x4); b) ( 2b2 - 2 - 5b + 6b3)(3 + 3b2 - b); c) (4a - 4a4 + 2a7)(6a2 - 12 - 3a3); d) (2ab + 2a2 + b2)(2ab2 + 4a3 - 4a2b) e) (2a3 - 0,02a + 0,4a5)(0,5a6 - 0,1a2 + 0,03a4).
\(a,=x+x^2-x^3+x^4-x^5+1+x-x^2+x^3-x^4-x-x^2+x^3-x^4+x^5+1+x-x^2+x^3-x^4\\ =2x-2x^2+2x^3-2x^4\)
Thực hiện phép chia:
a) ( x 3 - 3x - 2) : (x - 2);
b) ( x 3 + 6 x 2 + 8x - 3): ( x 2 + 3x -1);
c) (2 x 4 – 7 x 3 + 9 x 2 - 7x + 2): (2 x 2 - 5x + 2).
a) x 2 + 2x + 1. b) x + 3. c) x 2 – x + 1.
giải các phương trình sau:
a)(√x+1+1)3+2√x−1=2−x(x+1+1)3+2x−1=2−x
b)x3=x4+x3+x2+x+2x3=x4+x3+x2+x+2
c)2(x2+x+1)2−7(x−1)2=13(x3−1)2(x2+x+1)2−7(x−1)2=13(x3−1)
d)8x2+√1x=52
Phân tích đa thức thành nhân tử:
a) x 4 + 1 - 2 x 2 ; b) x 2 - y 2 - 5y + 5x;
c) y 2 - 4 x 2 +4x - 1; d) x3 ( 2 + x ) 2 - ( x + 2 ) 2 + 1 - x 3 .
Bài 3: Giải phương trình:
a) x3+ 2x2 + x +2 = 0
b) x3 – x2 – 21x + 45 = 0
c) x3 + 3x2+4x + 2 = 0
d) x4+ x2 +6x – 8 = 0
e) (x2 + 1)2 = 4 ( 2x – 1 )
Bài 4: Giải phương trình:
a) ( x2-5x)2 + 10( x2 – 5x) + 24 = 0
b) ( x2 + 5x)2 - 2( x2 + 5x) = 24
c) ( x2 + x – 2)(x2 + x – 3) = 12
d) x ( x+1) (x2 + x + 1) = 42
Bài 1
a/ \(x\left(x^2+1\right)+2\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+1\right)=0\Rightarrow x=-2\)
b/
\(\Leftrightarrow x^3-6x^2+9x+5x^2-30x+45=0\)
\(\Leftrightarrow x\left(x-3\right)^2+5\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-3\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)
1.
c/ \(\Leftrightarrow x^3+2x^2+2x+x^2+2x+2=0\)
\(\Leftrightarrow x\left(x^2+2x+2\right)+x^2+2x+2=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+2x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x^2+2x+2=0\left(vn\right)\end{matrix}\right.\)
d/
\(\Leftrightarrow x^4+x^3-2x^2-x^3-x^2+2x+4x^2+4x-8=0\)
\(\Leftrightarrow x^2\left(x^2+x-2\right)-x\left(x^2+x-2\right)+4\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x^2-x+4\right)\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+4=0\left(vn\right)\\x^2+x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Bài 1:
e/ \(\Leftrightarrow x^4+2x^2-8x+5=0\)
\(\Leftrightarrow x^4-2x^3+x^2+2x^3-4x^2+2x+5x^2-10x+5=0\)
\(\Leftrightarrow x^2\left(x-1\right)^2+2x\left(x-1\right)^2+5\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x^2+2x+5\right)\left(x-1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x+5=0\left(vn\right)\\x=1\end{matrix}\right.\)
Bài 2:
a/ Đặt \(x^2-5x=t\)
\(t^2+10t+24=0\Rightarrow\left[{}\begin{matrix}t=-4\\t=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x=-4\\x^2-5x=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x+4=0\\x^2-5x+6=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=4\\x=2\\x=3\end{matrix}\right.\)
Tìm x biết:
a) x 6 + 2 x 3 +1 = 0; b) x(x - 5) = 4x - 20;
c) x 4 -2 x 2 =8-4 x 2 ; d) ( x 3 - x 2 ) - 4 x 2 + 8x-4 = 0.
a) x = -1. b) x = 4 hoặc x = 5.
c) x = ± 2 . d) x = 1 hoặc x = 2.
Bài 1:Thực hiện các phép tính
a. (x5 +4x3 - 6x2):4x2
b. (x3 +x2-12) : (x-2)
c. (-2x5+3x2-4x3):2x2
d. (x3 - 64):(x2 + 4x + 16)
Bài 2:Rút gọn biểu thức
a. 3x (x - 2)- 5x (1 - x) - 8(x2 - 3)
b.(x - y) (x2 + xy + y2)+2y3
c. (x - y)2 + (x+y)2 - 2(x-y) (x+y)
a) \(\left(x^5+4x^3-6x^2\right):4x^2\)
\(=\left(x^5:4x^2\right)+\left(4x^3:4x^2\right)+\left(-6x^2:4x^2\right)\)
\(=\dfrac{1}{4}x^3+x-\dfrac{3}{2}\)
b)
Vậy \(\left(x^3+x^2-12\right):\left(x-2\right)=x^2+3x+6\)
c) (-2x5 : 2x2) + (3x2 : 2x2) + (-4x^3 : 2x^2)
= \(-x^3+\dfrac{3}{2}-2x\)
d) \(\left(x^3-64\right):\left(x^2+4x+16\right)\)
\(=\left(x-4\right)\left(x^2+4x+16\right):\left(x^2+4x+16\right)\)
\(=x-4\)
(dùng hẳng đẳng thức thứ 7)
Bài 2 :
a) 3x(x - 2) - 5x(1 - x) - 8(x2 - 3)
= 3x2 - 6x - 5x + 5x2 - 8x2 + 24
= (3x2 + 5x2 - 8x2) + (-6x - 5x) + 24
= -11x + 24
b) (x - y)(x2 + xy + y2) + 2y3
= x3 - y3 + 2y3
= x3 + y3
c) (x - y)2 + (x + y)2 - 2(x - y)(x + y)
= (x - y)2 - 2(x - y)(x + y) + (x + y)2
= [(x - y) + x + y)2 = [x - y + x + y] = (2x)2 = 4x2
Bài 1 :
a]= \(\frac{1}{4}\)x3 + x - \(\frac{3}{2}\).
b] => [x3 + x2 -12 ] = [ x2 +3 ][x-2] + [-6]
c]= -x3 -2x +\(\frac{3}{2}\).
d] = [ x3 - 64 ] = [ x2 + 4x + 16][ x- 4].