Tính:
a,2x(x - 1) - 3(x2 + 4x) + x(x + 2)
b,(2x - 3) (3x + 5) - (x - 1) (6x + 2) + 3 - 5x
c,(x - y)(x2 + xy + y2) - (x + y)(x2- y2)
Tính:
a) 3.(2x-3)+5.(x+2)
b) 3x.(2x-8)+(6x-2).(5-x)
c) (x-3).(x+3)-(x-5)2
d) (x-y)3-(x-y).(x2+xy+y2)
a) \(3\left(2x-3\right)+5\left(x+2\right)=6x-9+5x+10=11x+1\)
b) \(3x\left(2x-8\right)+\left(6x+2\right)\left(5-x\right)=6x^2-24x+30x-6x^2+10-2x=4x+10\)
c) \(\left(x-3\right)\left(x+3\right)-\left(x-5\right)^2=x^2-9-x^2+10x-25=10x-34\)
d) \(\left(x-y\right)^3-\left(x-y\right)\left(x^2+xy+y^2\right)=x^3-3x^2y+3xy^2-y^3-x^3+y^3=3xy^2-3x^2y\)
a: Ta có: \(3\left(2x-3\right)+5\left(x+2\right)\)
\(=6x-9+5x+10\)
\(=11x+1\)
b: Ta có: \(3x\left(2x-8\right)-\left(6x-2\right)\left(x-5\right)\)
\(=6x^2-24x-6x^2+30x+2x-10\)
\(=8x-10\)
c: Ta có: \(\left(x-3\right)\left(x+3\right)-\left(x-5\right)^2\)
\(=x^2-9-x^2+10x-25\)
\(=10x-34\)
Thực hiện phép tính:
a)(2x-3)2+(2x+3).(5-2x)
b)3.(2x-3)+5.(x+2)
c)3x.(2x-8)+(6x-2).(5-x)
d)(x-3).(x+3)-(x-5)2
e)(x-y)3-(x-y).(x2+xy+y2)
Lời giải:
a.
$(2x-3)^2+(2x+3)(5-2x)=(4x^2-12x+9)-(-4x^2+4x+15)$
$=4x^2-12x+9+4x^2-4x-15$
$=24-8x$
b.
$3(2x-3)+5(x+2)=6x-9+5x+10=11x+1$
c.
$3x(2x-8)+(6x-2)(5-x)=(6x^2-24x)+(-6x^2+32x-10)$
$=6x^2-24x-6x^2-32x+10$
$=8x-10$
d.
$(x-3)(x+3)-(x-5)^2=(x^2-9)-(x^2-10x+25)$
$=x^2-9-x^2+10x-25=10x-34$
e.
$(x-y)^3-(x-y)(x^2+xy+y^2)=(x^3-3x^2y+3xy^2-y^3)-(x^3-y^3)$
$=-3x^2y+3xy^2=3xy(y-x)$
a: ta có: \(\left(2x-3\right)^2+\left(2x+3\right)\left(5-2x\right)\)
\(=4x^2-12x+9+2x-4x^2+15-6x\)
\(=-16x+24\)
b: Ta có: \(3\left(2x-3\right)+5\left(x+2\right)\)
\(=6x-9+5x+10\)
\(=11x+1\)
c: ta có: \(3x\left(2x-8\right)+\left(6x-2\right)\left(5-x\right)\)
\(=6x^2-24x+30x-6x^2-10+2x\)
\(=8x-10\)
Bài 1: Thực hiện phép tính:
a) x(3x2 – 2x + 5) b) 1/3 x2 y2 (6x + 2/3x2 – y)
c) ( 1/3x + 2)(3x – 6) d) ( 1/3x + 2)(3x – 6)
e) (x2 – 3x + 1)(2x – 5) f) ( 1/2x + 3)(2x2 – 4x + 6)
Bài 2: Tìm x, biết:
a) 3(2x – 3) + 2(2 – x) = –3 b) x(5 – 2x) + 2x(x – 1) = 13
c) 5x(x – 1) – (x + 2)(5x – 7) = 6 d) 3x(2x + 3) – (2x + 5)(3x – 2) = 8
Bài 3: Chứng tỏ rằng giá trị của biểu thức sau không phụ thuộc vào giá trị của biến: a) A = x(2x + 1) – x2 (x + 2) + x3 – x + 3
b) B = (2x + 11)(3x – 5) – (2x + 3)(3x + 7) + 5
Bài 4: Tính giá trị của biểu thức
a) A = 2x( 1/2x2 + y) – x(x2 + y) + xy(x3 – 1) tại x = 10; y = – 1 10
b) B = 3x2 (x2 – 5) + x(–3x3 + 4x) + 6x2 tại x = –5
\(1,\\ a,=3x^3-2x^2+5x\\ b,=2x^3y^2+\dfrac{2}{9}x^4y^2-\dfrac{1}{3}x^2y^3\\ c,=x^2-2x+6x-12=x^2+4x-12\\ 2,\\ a,\Rightarrow6x-9+4-2x=-3\\ \Rightarrow4x=2\Rightarrow x=\dfrac{1}{2}\\ b,\Rightarrow5x-2x^2+2x^2-2x=13\\ \Rightarrow3x=13\Rightarrow x=\dfrac{13}{3}\\ c,\Rightarrow5x^2-5x-5x^2+7x-10x+14=6\\ \Rightarrow-8x=-8\Rightarrow x=1\\ d,\Rightarrow6x^2+9x-6x^2+4x-15x+10=8\\ \Rightarrow-2x=-2\Rightarrow x=1\)
\(3,\\ A=2x^2+x-x^3-2x^2+x^3-x+3=3\\ B=6x^2-10x+33x-55-6x^2-14x-9x-21=-76\)
Bài 4:
b: Ta có: \(B=3x^2\left(x^2-5\right)+x\left(-3x^3+4x\right)+6x^2\)
\(=3x^4-15x^2-3x^3+4x^2+6x^2\)
\(=-5x^2\)
\(=-5\cdot25=-125\)
a) 3x(x+1)-x(3x+2)
b) 2x(x2-5x+6)+(x-1)(x+3)
c) (x2-xy+y2)-(x2+2xy+y2)
d) (2/5xy+x-y)-(3x+4y)-2/5xy
e) 2xy(x2-4xy+4y2)
f) (x+y)(xy+5)
g) (x3-2x2-x+2):(x-1)
h) (2x2+3x-2):(2x-1)
c) C = x(y2 +z2)+y(z2 +x2)+z(x2 +y2)+2xyz.
d) D = x3(y−z)+y3(z−x)+z3(x−y).
e) E = (x+y)(x2 −y2)+(y+z)(y2 −z2)+(z+x)(z2 −x2).
b) x2 +2x−24 = 0.
d) 3x(x+4)−x2 −4x = 0.
f) (x−1)(x−3)(x+5)(x+7)−297 = 0.
(2x−1)2 −(x+3)2 = 0.
c) x3 −x2 +x+3 = 0.
e) (x2 +x+1)(x2 +x)−2 = 0.
a) A = x2(y−2z)+y2(z−x)+2z2(x−y)+xyz.
b) B = x(y3 +z3)+y(z3 +x3)+z(x3 +y3)+xyz(x+y+z). c) C = x(y2 −z2)−y(z2 −x2)+z(x2 −y2).
Đề bài yêu cầu gì vậy em.
Bài 3: Rút gọn các biểu thức sau:
A = 3x(x2 – 2x + 3) – x2(3x – 2) + 5(x2 – x)
B = x(x2 + xy + y2) – y(x2 + xy + y2)
\(A=3x^3-6x^2+9x-3x^3+2x^2+5x^2-5x=x^2+4x\\ B=\left(x^2+xy+y^2\right)\left(x-y\right)=x^3-y^3\)
Bài 3: Rút gọn các biểu thức sau:
1) ( x+ 3)(x2 -3x + 9) - (x3 + 54)
2) (2x + y)(4x2 + 2xy + y2 ) - (2x – y)(4x2 + 2xy + y2 )
3) (x – 1)3 – (x + 2)(x2 -2x +4) +3(x +4)(x – 4)
4) x(x + 1)(x - 1) – (x + 1)(x2 – x +1)
5) 8x3 - 5 (2x + 1)(4x2 – 4x + 1)
6) 27 + (x – 3)(x2 +3x + 9)
7) (x – 1)3 – (x +2)(x2 -2x + 4) +3(x +4)(x -4)
8) (x – 2)3 +6( x – 1)2 –(x +1)(x2 -x +1) +3x
1: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54\right)\)
\(=x^3+27-x^3-54\)
=-27
2: Ta có: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3\)
\(=2y^3\)
\(1,=x^3+270-x^3-54=-27\\ 2,=8x^3+y^3-8x^3+y^3=2y^3\\ 3,=x^3-3x^2+3x-1-x^3-8+3x^2-48=3x-57\\ 4,=x^3-x-x^3-1=-x-1\\ 5,=8x^3-5\left(8x^3+1\right)=-32x^3-5\\ 6,=27+x^3-27=x^3\\ 7,làm.ở.câu.3\\ 8,=x^3-6x^2+12x-8+6x^2-12x+6-x^3-1+3x\\ =3x-3\)
Thu gọn biểu thức
a, A= x (x2-x+1)+1/2x2(2-2x)
b, B= 3x (x-2)-x (1+3x)
c, C = x (x2+xy+y2)-y (x2+xy+y2)
d, D=3x (x2-2x-3)-x2(3x-2)+5(x2-x)
GIÚP MK VỚI MK LIKE NHA haha
quy đồng các mẫu thức sau
a 1 / x3-8 và 3 / 4-2x
b x / x2-1 và 1 / x2+2x+1
c 1 / x+2 ; x+1 / x2-4x-4 và 5 / 2-x
d 1 / 3x+3y;2x / x2-y2 và x2-xy+y2 / x2-2xy+y2
a) \(\dfrac{1}{x^3-8}=\dfrac{1}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{2}{2\left(x-2\right)\left(x^2+2x+4\right)}\)
\(\dfrac{3}{4-2x}=\dfrac{-3}{2\left(x-2\right)}=\dfrac{-3\left(x^2+2x+4\right)}{2\left(x-2\right)\left(x^2+2x+4\right)}\)
b) \(\dfrac{x}{x^2-1}=\dfrac{x}{\left(x+1\right)\left(x-1\right)}=\dfrac{x\left(x+1\right)}{\left(x+1\right)^2\left(x-1\right)}\)
\(\dfrac{1}{x^2+2x+1}=\dfrac{1}{\left(x+1\right)^2}=\dfrac{x-1}{\left(x+1\right)^2\left(x-1\right)}\)
c) \(\dfrac{1}{x+2}=\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)^2}\)
\(\dfrac{1}{x^2-4x+4}=\dfrac{1}{\left(x-2\right)^2}=\dfrac{x+2}{\left(x+2\right)\left(x-2\right)^2}\)
\(\dfrac{5}{2-x}=\dfrac{-5}{x-2}=\dfrac{-5\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)^2}\)
d) \(\dfrac{1}{3x+3y}=\dfrac{1}{3\left(x+y\right)}=\dfrac{\left(x-y\right)^2}{3\left(x+y\right)\left(x-y\right)^2}\)
\(\dfrac{2x}{x^2-y^2}=\dfrac{2x}{\left(x+y\right)\left(x-y\right)}=\dfrac{6x\left(x-y\right)}{3\left(x+y\right)\left(x-y\right)^2}\)
\(\dfrac{x^2-xy+y^2}{x^2-2xy+y^2}=\dfrac{x^2-xy+y^2}{\left(x-y\right)^2}=\dfrac{3\left(x^2-xy+y^2\right)\left(x+y\right)}{3\left(x+y\right)\left(x-y\right)^2}=\dfrac{3\left(x^3+y^3\right)}{3\left(x+y\right)\left(x-y\right)^2}\)