1.(x+1)^2 -y-2x-2
2.3x^3-3xy-5x+5y
3.(x+2)^2-y^2
4.(x+5)^2-y^2
5.16x^2-(x+1)^2
A = \(\dfrac{5xy^2-3z}{3xy}+\dfrac{4x^2y+3z}{3xy}\)
B = \(\dfrac{3y+5}{y-1}+\dfrac{-y^2-4y}{1-y}+\dfrac{y^2+y+7}{y-1}\)
C = \(\dfrac{6x}{x^2-9}+\dfrac{5x}{x-3}+\dfrac{x}{x+3}\)
D = \(\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}+\dfrac{3x-2}{2x-4x^2}\)
E = \(\dfrac{x^3+2x}{x^3+1}+\dfrac{2x}{x^2-x+1}+\dfrac{1}{x+1}\)
b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)
\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)
\(=\dfrac{2y^2+8y+12}{y-1}\)
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: \(=27x^3y^3-8-3xy\left(9x^2y^2+6xy+1\right)\)
\(=27x^3y^3-8-27x^3y^3-18x^2y^2-3xy\)
\(=-18x^2y^2-3xy-8\)
\(=-18\cdot\left[\left(-2010\right)\cdot\left(-\dfrac{1}{2010}\right)\right]^2-3\cdot\left(-2010\right)\cdot\dfrac{-1}{2010}-8\)
\(=-18-3-8=-29\)
Rút gọn rồi tính giá trị của biểu thức :
A = 5x(x-y) - y(5x-y) với x =-1 ; y =3
B = 4y(x2 - 3xy + 3y2) - 2xy(2x-6y-3) với x = 5; y = -1
C = 5x2(x-y2) + 3x(xy2-y) -5x3 với x=-2 ; y=-5
D = 6x2(y2-xy+2x2y) -3xy(2xy-x2+4x3) với x=11; y=-1
A = 5x(x - y) - y(5x - y)
A = 5x2 - 5xy - 5xy + y2
A = 5x2 - 10xy + y2 (1)
Thay x = -1; y = 3 vào (1), ta có:
5.(-1)2 - 10.(-1).3 + 32 = 44
B = 4y(x2 - 3xy + 3y2) - 2xy(2x - 6y - 3)
B = 4x2y - 12x2 + 12y3 - 4x2y + 12xy2 + 6xy
B = 12y3 + 6xy (1)
Thay x = 5; y = -1 vào (1), ta có:
12.(-1)3 + 6.5.(-1) = -42
C = 5x2(x - y2) + 3x(xy2 - y) - 5x3
C = 5x3 - 5x2y2 + 3x2y2 - 3xy - 5x3
C = -2x2y2 - 3xy (1)
Thay x = -2; y = -5 vào (1), ta có:
-2.(-2)2.(-5)2 - 3.(-2).(-5) = -230
D = 6x2(y2 - xy + 2x2y) - 3xy(2xy - x2 + 4x3)
D = 6x2y2 - 6x3y + 12x4y - 6x2y2 + 3x3y - 12x4y
D = -3x3y (1)
Thay x = 11; y = -1 vào (1), ta có:
-3.113.(-1) = 3993
Thực hiện các phép tính sau: a) (x²/x²+3x)+(3/x+3)+(3/x) b) (2/x+2)+(-4/2-x)+(5x+2/4-x²) c) (1/x-y)+(3xy/y³-x³)+(x-y/x²+xy+y²) d) (3-3x/2x)+(3x-1/2x-1)+(11x-5/2x-4x²)
Rút gọn biểu thức:
a) P= (5x-1)+2x(1-5x)x(4+5x)+(5x+4)^2
b) Q= (x-y)^3+(y+x)^3+(y-x)^3-3xy(x+y)
c) 12(5^2+1)(5^4+1)(5^8+1)(5^16+1)
Rút gọn biểu thức:
a) P= (5x-1)+2x(1-5x)x(4+5x)+(5x+4)^2
b) Q= (x-y)^3+(y+x)^3+(y-x)^3-3xy(x+y)
c) 12(5^2+1)(5^4+1)(5^8+1)(5^16+1)
a: \(P=\left(5x-1-5x-4\right)^2=\left(-3\right)^2=9\)
b: \(Q=\left(x+y\right)^3-3xy\left(x+y\right)=x^3+y^3\)
c: \(=\dfrac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(=\dfrac{\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(=\dfrac{\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(=\dfrac{5^{32}-1}{2}\)
giải các hệ phương trình sau
a.{ x + 3y = -2
{ 5x - 4y = 11
b.{ 3xy = 5
{ 5x + 2y = 23
c.{ 3x +5y = 1
{ 2x - y = -8
d.{ x - 2y + 6 = 0
{ 5x - 3y - 5 = 0
e.{ 2(x + y) + 3(x - y) = 4
{ (x + y) + 2(x - y) = 5
\(a,\Leftrightarrow\left\{{}\begin{matrix}5x+15y=-10\\5x-4y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19y=-21\\5x-4y=11\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{21}{19}\\5x-4\left(-\dfrac{21}{19}\right)=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{25}{19}\\y=-\dfrac{21}{19}\end{matrix}\right.\)
\(c,\Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\10x-5y=-40\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\13x=-39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=2\end{matrix}\right.\\ d,\Leftrightarrow\left\{{}\begin{matrix}5x-10y=-30\\5x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-3y=5\\-7y=-35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=5\end{matrix}\right.\\ e,\Leftrightarrow\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\2\left(x+y\right)+4\left(x-y\right)=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=6\\2\left(x+y\right)+3\cdot6=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=6\\x+y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-\dfrac{13}{2}\end{matrix}\right.\)
Phân tích đa thức thành nhân tử
1, x\(^2\)-xy+x-y
2, xz+yz-5x-5y
3, 3x\(^2\)-3xy-5x+5y
4, x\(^2\)-2xy+y\(^2\)-xz+yz
5, 3x\(^2\)+6xy+3y\(^2\)-3z\(^2\)
6, 45+x\(^{^{ }3}\)-5x\(^2\)-9x
7, x\(^2\)-6x+5
8,x\(^2\)+7x+12
9, 2x\(^2\)-7x+3
10, 3x\(^2\)-12y\(^2\)
11, 5xy\(^2\)-10xyz+5xz\(^2\)
12,x\(^2\)-y\(^2\)-x+y
13, a\(^3\)x-ab+b-x
14, (1+x\(^2\))-4x(1-x\(^2\))
CÁC CẬU GIẢI CHI TIẾT GIÚP MÌNH VỚI Ạ
13: =x(a^3-1)-b(a-1)
=x(a-1)(a^2+a+1)-b(a-1)
=(a-1)(a^2x+a*x+x-b)
12: =(x-y)(x+y)-(x-y)
=(x-y)(x+y-1)
10: =3(x^2-4y^2)
=3(x-2y)*(x+2y)
7: =x^2-x-5x+5=(x-1)(x-5)
8: =x^2+3x+4x+12=(x+3)(x+4)
9: =2x^2-6x-x+3=(x-3)(2x-1)
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