2x^2- 3xy+ 9 với x=4 ; y=-2
mk nghĩ nó là 17 nhưng phần luyện tập cho rằng 65 cho hỏi ý kiến nào đúng. vì sao?
1, ( x+1/3)^3
2, ( 2x+y^2)^3
3, ( 1/2x^2+1/3y)^3
4, ( 3x^2-2y)^3
5, ( 2/3x^2-1/2y)^3
6, ( 2x+1/2)^3
7, ( x-3)^3
8, ( x+1).(X^2+3x+9)
9, ( x-3).( x^2+3x+9)
10, ( x-2).( x^2+2x+4)
11, ( x+4).( x^2-4x+16)
12, ( x-3y).( x^2+3xy+9y^2)
13, ( x^2-1/3). ( x^4+1/3x^2+1/9)
14, ( 1/3x+2y).( 1/9x^2-2/3xy+4y^2)
Đưa về HĐT
Tìm x thỏa mãn điều kiện
(2x+1)^3-(2x+1).(4x^2-2x+1)-3.(2x-1)^2=15
y.(y+3)^2-(y+2).(y^2-2y+4)-6.(y+5).(y-5)=97
(x-3)^3-(x-3).(x^2+3x+9)+9.(x+1)^2=18
x.(x-4).(x+4)-(x-5).(x^2+5x+25)=13
2.Rút gọn biểu thức rồi tính giá trị
3.(x-1).(x^2+x+1)+(x-1)^3-4x.(x+1).(x-1) tại x=-1
(3xy-2).(9x^2y^2+6xy+4)-3xy.(3xy+1)^2 tại x=-2010,y=-1/2010
Bài 2:
a: \(3\left(x-1\right)\left(x^2+x+1\right)+\left(x-1\right)^3-4x\left(x+1\right)\left(x-1\right)\)
\(=3\left(x^3-1\right)+x^3-3x^2+3x-1-4x\left(x^2-1\right)\)
\(=3x^3-3+x^3-3x^2+3x-1-4x^3+4x\)
\(=-3x^2+7x-4\)
\(=-3\cdot\left(-1\right)^2+7\cdot\left(-1\right)-4\)
=-3-4-7=-14
b: \(=27x^3y^3-8-3xy\left(9x^2y^2+6xy+1\right)\)
\(=27x^3y^3-8-27x^3y^3-18x^2y^2-3xy\)
\(=-18x^2y^2-3xy-8\)
\(=-18\cdot\left[\left(-2010\right)\cdot\left(-\dfrac{1}{2010}\right)\right]^2-3\cdot\left(-2010\right)\cdot\dfrac{-1}{2010}-8\)
\(=-18-3-8=-29\)
Tính 2x-3y+1 biết 2x-3y>0 và \(x^2-3xy+\frac{9}{4}y^2=9\).
\(x^2-3xy+\frac{9}{4}y^2=9\) \(\Rightarrow\left(x-\frac{3}{2}y\right)^2=9\)\(\Rightarrow\orbr{\begin{cases}x-\frac{3}{2}y=3\\x-\frac{3}{2}y=-3\end{cases}\Rightarrow}\orbr{\begin{cases}x=3+\frac{3}{2}y\\x=\frac{3}{2}y-3\end{cases}}\)
Th1: Thay \(x=3+\frac{3}{2}y\) vào 2x - 3y + 1
Ta có: \(2\left(3+\frac{3}{2}y\right)-3y+1=6+3y-3y+1=7\)
Th2: Thay \(x=\frac{3}{2}y-3\) vào 2x - 3y + 1
Ta có: \(2\left(\frac{3}{2}y-3\right)-3y+1=3y-6-3y+1=-5\)
BÀI 1: NHÂN ĐƠN THỨC VỚI ĐA THỨC
6) 5x +3 ( x2 -x - 1)
7) -\(\dfrac{2}{3}\)x ( -x4y2 -2x2 - 10y2)
8) \(\dfrac{2}{3}\)xy ( 3 x2y -3xy + y2)
9) (-2x).(3x2 - 2x +4)
10) 3x4 ( -2x3 + 5x2 - \(\dfrac{2}{3}\)x + \(\dfrac{1}{3}\))
9: \(\left(-2x\right)\left(3x^2-2x+4\right)=-6x^3+4x^2-8x\)
14 Tính :a) (x+2)(x^2+2x-9) ; b) (x^2y-6)(x^2-5) ; c) (x+y)(xy-4+y)
d) (x^2y^2-x+3/4)(x-1/2) ; e) (2x^n+1-3y^n).2xy(x^n+1-2y^n).3xy
Giups mình với mình đang vội lắm !Thank các bạn nha!
\(\left(x+2\right)\left(x^2+2x-9\right)\)
\(=x^3+2x^2-9x+2x^2+4x-18\)
\(=x^3+4x^2-5x-18\)
\(\left(x^{2y}-6\right)\left(x^2-5\right)\)
\(=x^{4y}-5x^{2y}-6x^2+30\)
\(\left(x+y\right)\left(xy-4+y\right)\)
\(=x^2y-4x+xy+xy^2-4y+y^2\)
câu còn lại tương tự nha
Phân tích đa thức sau thành nhân tử
a ) 9(x+y-1)^2 - 4 (2x+3y+1)^2
b ) 3x^4y^2 +3x^3y^2 +3xy^2 +3y^2
c ) ( x+y )^3 - 1 -3xy( x + y -1)
d ) x^3 + 3x^2 + 3x +1 - 27z^3
Giúp với ạ ! Cảm ơn
a) \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)
\(=\left(3x+3y-3\right)^2-\left(4x+6y+2\right)^2\)
\(=\left(3x+3y-3-4x-6y-2\right)\left(3x+3y-3+4x+6y+2\right)\)
\(=\left(-x-3y-5\right)\left(7x+9y-1\right)\)
b) \(3x^4y^2+3x^3y^2+3xy^2+3y^2\)
\(=\left(3x^4y^2+3xy^2\right)+\left(3x^3y^2+3y^2\right)\)
\(=3xy^2\left(x^3+1\right)+3y^2\left(x^3+1\right)\)
\(=\left(3xy^2+3y^2\right)\left(x^3+1\right)\)
\(=3y^2\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
\(=3y^2\left(x+1\right)^2\left(x^2-x+1\right)\)
c) \(\left(x+y\right)^3-1-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left[\left(x+y\right)^2+x+y+1\right]-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2+2xy+y^2+x+y+1-3xy\right)\)
\(=\left(x+y-1\right)\left(x^2+x+y^2+y+1-xy\right)\)
Ai giải đúng chỗ mình mình sẽ đánh giá 5 sao và đúng mình cần gấp lắm a)(x+2)(x^2-24+4)(x^3+8) b)(2x-1/2)(4x^2+x+1/4) c)(x^2+y)(x^2-y)+y^2+x^4 d)(x+3)(x^2-3x+9)-x^3 e)(3x+y)(9x^2-3xy+y^2)-26x^3 g)(x+3y)(x^2-3xy+9y^2)+(3x-y)(9x^2+3xy+y^2)
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) \(\left(x+3y\right)\left(x^2-3xy+9y^2\right)+\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)
\(=\left(x+3y\right)\left[x^2-x\cdot3y+\left(3y\right)^2\right]+\left(3x-y\right)\left[\left(3x\right)^2+3x\cdot y+y^2\right]\)
\(=\left[x^3+\left(3y\right)^3\right]+\left[\left(3x\right)^3-y^3\right]\)
\(=x^3+27y^3+27x^3-y^3\)
\(=28x^3+26y^3\)
a) Sửa đề:
(x + 2)(x² - 2x + 4)(x³ + 8)
= (x³ + 8)(x³ + 8)
= (x³ + 8)²
b) (2x - 1/2)(4x² + x + 1/4)
= (2x)³ - (1/2)³
= 8x³ - 1/8
c) (x² + y)(x² - y) + y² + x⁴
= (x²)² - y² + y² + x⁴
= 2x⁴
d) (x + 3)(x² - 3x + 9) - x³
= x³ + 3³ - x³
= 27
e) (3x + y)(9x² - 3xy + y²) - 26x³
= (3x)³ + y³ - 26x³
= 27x³ + y³ - 26x³
= x³ + y³
g) (x + 3y)(x² - 3xy + 9y²) + (3x - y)(9x² + 3xy + y²)
= x³ + (3y)³ + (3x)³ - y³
= x³ + 27y³ + 27x³ - y³
= 28x³ + 26y³
14 Tính a) (x+2)(x^2+2x-9) ; b) (x^2y-6)(x^2-5) ; c) (x+y)(xy-4+y)
d) ( x^2y^2-x+3/4)(x-1/2) ; e) (2x^n+1-3y^n).2xy-(x^n+1-2y^n).3xy
Giups mình với mình sắp phải đi học rùi
a: \(=x^3+2x^2-9x+2x^2+4x-18\)
\(=x^3+4x^2-5x-18\)
b: \(=x^4y-5x^2y-6x^2+30\)
c: \(=x^2y-4x+xy+xy^2-4y+y^2\)
d: \(=x^3y^2-\dfrac{1}{2}x^2y^2-x^2+\dfrac{1}{2}x+\dfrac{3}{4}x-\dfrac{3}{8}\)
\(=x^3y^2-\dfrac{1}{2}x^2y^2-x^2+\dfrac{5}{4}x-\dfrac{3}{8}\)
Bai 1 TINH
a) x ^2 . x -2x^3
b) 6 x^2y . 3xy - 2y^2.x +y
C) 4x^2 +5x -1 .2x^3 -3x
d)-8x^3y + 2y^4 . 3xy^3 - 2x^4 +7y^4
GIUP MINH NHÀ