1. Thực hiện phép tính
a, (x-1) . (x+1) . (x+2)
b,1/2x2y2 . (2x+y) . (2x-y)
c, (x-1/2). (x+1/2) . (4x-1)
Bài 1: Thực hiện phép tính
a, (5x-2y)(x\(^2\)-xy+1)
b, (x-1)(x+1)(x+2)
c, \(\dfrac{1}{2}\)x\(^2\)y\(^2\)(2x+y)(2x-y)
d, (x-\(\dfrac{1}{2}\))(x+\(\dfrac{1}{2}\))(4x-1)
e, (x-7)(x+5)-(2x+1)(3-x)
a: =5x^3-5x^2y+5x-2x^2y+2xy^2-2y
=5x^3-7x^2y+2xy^2+5x-2y
b: =(x^2-1)(x+2)
=x^3+2x^2-x-2
c: =1/2x^2y^2(4x^2-y^2)
=2x^4y^2-1/2x^2y^4
d: =(x^2-1/4)(4x-1)
=4x^3-x^2-x+1/4
e: =x^2-2x-35+(2x+1)(x-3)
=x^2-2x-35+2x^2-6x+x-3
=3x^2-7x-38
Bài 1: Thực hiện phép tính
a) 2x2y (2x2y2 - xy2) b) (x - 1)(2x + 3)
c) (20x3y4 + 10x2y3 - 5xy) : 5xy d) (y - 3x)2 - (y2 - 6xy)
Bài 2: Phân tích đa thức thành nhân tử
a) 4xy + 4xz b) x2 - y2 + 9 - 6x
Bài 3: Thực hiện phép tính
a) 3xy/y+z + 3xz/y+z b) x/x+2 - x/x-2
Bài 4: Một túi quà có dạng hình chóp tứ giác đều có độ dài cạnh đáy là 12cm, đường cao mặt bên xuất phát từ đỉnh 10cm.
a) Tính thể tích túi quà.
b) Tính số tiền để mua giấy màu làm túi quà (không tính mép dán). Biết rằng giá 1m2 giấy màu là 200.000đ.
Bài 1:
\(a,2x^2y\left(2x^2y^2-xy^2\right)\\ =2x^2x^2y^2y-2x^2x.y^2.y=2x^4y^3-2x^3y^3\\ b,\left(x-1\right)\left(2x+3\right)\\ =x.2x+x.3-1.2x-1.3=2x^2+3x-2x-3\\ =2x^2+x-3\\ c,\left(20x^3y^4+10x^2y^3-5xy\right):5xy\\ =20x^3y^4:5xy+10x^2y^3:5xy-5xy:5xy\\ =\left(20:5\right).\left(x^3:x\right).\left(y^4:y\right)+\left(10:5\right).\left(x^2:x\right).\left(y^3:y\right)-\left(5:5\right).\left(x:x\right).\left(y:y\right)\\ =4x^2y^3+2xy^2-1\\ d,\left(y-3x\right)^2-\left(y^2-6xy\right)\\ =\left[y^2-2.y.3x+\left(3x\right)^2\right]-\left(y^2-6xy\right)\\ =y^2-6xy+9x^2-y^2+6xy =9x^2\)
Bài 2:
\(a,4xy+4xz=4x\left(y+z\right)\\ b,x^2-y^2+9-6x\\ =\left(x^2-6x+9\right)-y^2\\ =\left(x-3\right)^2-y^2\\ =\left(x-3-y\right)\left(x-3+y\right)\)
Bài 3:
\(a,\dfrac{3xy}{y+z}+\dfrac{3xz}{y+z}\\=\dfrac{3xy+3xz}{y+z}\\ =\dfrac{3x\left(y+z\right)}{\left(y+z\right)}=3x\left(Với:y\ne-z\right)\\ b,\dfrac{x}{x+2}-\dfrac{x}{x-2}\\ =\dfrac{x\left(x-2\right)-x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}\\ =\dfrac{x^2-2x-x^2-2x}{\left(x+2\right)\left(x-2\right)}=0\)
bài 1 :thực hiện phép tính
a/ 2x (x – 5) + (x – 2)(x + 3) b/ (x + 2)2 + (x – 5)(3 – x) c/ (2x – 3)2 – x (4x – 1)
c: \(=4x^2-12x+9-4x^2+x=-11x+9\)
\(a,=2x^2-10x+x^2+x-6=3x^2-9x-6\\ b,=x^2+4x+4-x^2+8x-15=12x-11\\ c,=4x^2-12x+9-4x^2+x=-11x+9\)
Bài1:thực hiện phép tính
a) (4x-1).(2x^2-x-1)
b) (4x^3+8x^2-2x):2x
c) (6x^3-7x^2-16x+12):(2x+3)
Bài2: phân tích da thức thành nhân tử
a) 2x^3-8x^2+8x
b) 2xy+2x+yz+z
c) x^2+2x+1-y^2
\(Bài1:\\ a,\left(4x-1\right)\left(2x^2-x-1\right)=4x\left(2x^2-x-1\right)-\left(2x^2-x-1\right)=8x^3-4x^2-4x-2x^2+x+1=8x^3-6x^2-3x+1\\ b,\left(4x^3+8x^2-2x\right):2x\\ =2x\left(2x^2+4x-1\right):2x\\ =2x^2+4x-1\)
\(Bài2:\\ a,2x^3-8x^2+8x=2x\left(x^2-4x+4\right)=2x\left(x-2\right)^2\\ b,2xy+2x+yz+z=2x\left(y+1\right)+z\left(y+1\right)=\left(y+1\right)\left(2x+z\right)\\ c,x^2+2x+1-y^2=\left(x+1\right)^2-y^2=\left(x-y+1\right)\left(x+y+1\right)\)
Thực hiện phép tính
a) \(\dfrac{3-x}{x-5}+\dfrac{2x-8}{x-5}\)
b) \(\dfrac{1}{x-y}+\dfrac{1}{x+y}+\dfrac{2x}{x^2-y^2}\)
a,\(\dfrac{3-x}{x-5}+\dfrac{2x-8}{x-5}=\dfrac{3-x+2x-8}{x-5}=\dfrac{x-5}{x-5}=1\)
b, \(\dfrac{1}{x-y}+\dfrac{1}{x+y}+\dfrac{2x}{x^2-y^2}=\dfrac{x+y}{\left(x-y\right)\left(x+y\right)}+\dfrac{x-y}{\left(x-y\right)\left(x+y\right)}+\dfrac{2x}{\left(x-y\right)\left(x+y\right)}=\dfrac{x+y+x-y+2x}{\left(x-y\right)\left(x+y\right)}=\dfrac{4x}{\left(x-y\right)\left(x+y\right)}\)
Thực hiên phép tính
a)\(\dfrac{x^2+2}{x^3+1}\)-\(\dfrac{1}{x+1}\)
b)\(\dfrac{x}{x^2-2x}\)-\(\dfrac{x^2+4x}{x^3-4x}\)-\(\dfrac{2}{x^2+2x}\)
c)\(\dfrac{1}{2-2x}\)-\(\dfrac{3}{2+2x}\)+\(\dfrac{2x}{x^2-1}\)
\(a,\dfrac{x^2+2}{x^3+1}-\dfrac{1}{x+1}\left(ĐKXĐ:x\ne-1\right)\\ =\dfrac{x^2+2-\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\\ c,\dfrac{1}{2-2x}-\dfrac{3}{2+2x}+\dfrac{2x}{x^2-1}\\ =\dfrac{-1}{2\left(x-1\right)}-\dfrac{3}{2\left(x+1\right)}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}\left(ĐKXĐ:x\ne\pm1\right)\\ =\dfrac{-1\left(x+1\right)-3\left(x-1\right)+2x.2}{2\left(x+1\right)\left(x-1\right)}\\ =\dfrac{-x-1-3x+3+4x}{2\left(x+1\right)\left(x-1\right)}=\dfrac{2}{2\left(x+1\right)\left(x-1\right)}=\dfrac{1}{\left(x-1\right)\left(x+1\right)}\)
\(\dfrac{x}{x^2-2x}-\dfrac{x^2+4x}{x^3-4x}-\dfrac{2}{x^2+2x}\) (ĐK: \(x\ne0;x\ne\pm2\) )
\(=\dfrac{x}{x\left(x-2\right)}-\dfrac{x\left(x+4\right)}{x\left(x^2-4\right)}-\dfrac{2}{x\left(x+2\right)}\)
\(=\dfrac{1}{x-2}-\dfrac{x+4}{\left(x+2\right)\left(x-2\right)}-\dfrac{2}{x\left(x+2\right)}\)
\(=\dfrac{x\left(x+2\right)}{x\left(x+2\right)\left(x-2\right)}-\dfrac{x\left(x+4\right)}{x\left(x+2\right)\left(x-2\right)}-\dfrac{2\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2+2x-x^2-4x-2x+4}{x\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{-4x+4}{x\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{4-4x}{x^3-4x}\)
\(b,\dfrac{x}{x^2-2x}-\dfrac{x^2+4x}{x^3-4x}-\dfrac{2}{x^2+2x}\\ =\dfrac{x}{x\left(x-2\right)}-\dfrac{x^2+4x}{x\left(x^2-4\right)}-\dfrac{2}{x\left(x+2\right)}\left(ĐKXĐ:x\ne0;x\ne\pm2\right)\\ =\dfrac{x\left(x+2\right)-\left(x^2+4x\right)-2\left(x-2\right)}{x\left(x+2\right)\left(x-2\right)}\\ =\dfrac{x^2-x^2+2x-4x-2x+4}{x\left(x+2\right)\left(x-2\right)}\\ =\dfrac{-4x+4}{x\left(x+2\right)\left(x-2\right)}\)
Bài1:Thực hiện phép tính
a) (4x-1).(2x^2-x-1)
b) (4x^3+8x^2-2x):2x
c) (6x^3-7x^2-16x+12):(2x+3)
Bài2:phân tích đa thức thành nhân tử
a) 2x^3-8x^2+8x
b) 2xy+2x+yz+z
c) x^2+2x+1-y^2
Câu3: tìm m để đa thức A(x)=3x^2+5x+m chia hết cho đa thức B(x)=x-2.
giúp mk với mk đang cần gấp UwU
Thực hiện phép tính a, ½x²y . (2x³-x²+4x-1) b,(-2x²y+½xy²-3) . (-2x²y) c,(x+2). (3x²-4x) d, (5x-2y) . (x²-xy+1)
c: \(=3x^3-4x^2+6x^2-8x=3x^2+2x^2-8x\)
Thực hiện phép tính :
Thực hiện phép tính :
5.x^2(x-y+1)+(x^2-1)(x+y)
Bài 2:
1: \(A=\left(x+2\right)\left(x^2-2x+4\right)+2\left(x+1\right)\left(1-x\right)\)
\(=\left(x+2\right)\left(x^2-x\cdot2+2^2\right)-2\left(x+1\right)\left(x-1\right)\)
\(=x^3+2^3-2\left(x^2-1\right)\)
\(=x^3+8-2x^2+2=x^3-2x^2+10\)
\(B=\left(2x-y\right)^2-2\left(4x^2-y^2\right)+\left(2x+y\right)^2+4\left(y+2\right)\)
\(=\left(2x-y\right)^2-2\cdot\left(2x-y\right)\left(2x+y\right)+\left(2x+y\right)^2+4\left(y+2\right)\)
\(=\left(2x-y-2x-y\right)^2+4\left(y+2\right)\)
\(=\left(-2y\right)^2+4\left(y+2\right)\)
\(=4y^2+4y+8\)
2: Khi x=2 thì \(A=2^3-2\cdot2^2+10=8-8+10=10\)
3: \(B=4y^2+4y+8\)
\(=4y^2+4y+1+7\)
\(=\left(2y+1\right)^2+7>=7>0\forall y\)
=>B luôn dương với mọi y
Bài 1:
5: \(x^2\left(x-y+1\right)+\left(x^2-1\right)\left(x+y\right)\)
\(=x^3-x^2y+x^2+x^3+x^2y-x-y\)
\(=2x^3-x+x^2-y\)
6: \(\left(3x-5\right)\left(2x+11\right)-6\left(x+7\right)^2\)
\(=6x^2+33x-10x-55-6\left(x^2+14x+49\right)\)
\(=6x^2+23x-55-6x^2-84x-294\)
=-61x-349