Rút gọn
B=(x3x4−4x2+1212−6x+1x+2):(10−x2x+2+x−2)B=\left(\dfrac{x^3}{x^4-4x^2}+\dfrac{12}{12-6x}+\dfrac{1}{x+2}\right):\left(\dfrac{10-x^2}{x+2}+x-2\right)B=(x4−4x2x3+12−6x12+x+21):(x+210−x2+x−2)
A=(1+2x4+2x−x3x−6+2x212−3x2).24−12x6+13xA=\left(\dfrac{1+2x}{4+2x}-\dfrac{x}{3x-6}+\dfrac{2x^2}{12-3x^2}\right).\dfrac{24-12x}{6+13x}A=(4+2x1+2x−3x−6x+12−3x22x2).6+13x24−12x=(1+2x2(x+2)−x3(x−2)−2x23(x−2)(x+2)).−12(x−2)13x+6=\left(\dfrac{1+2x}{2\left(x+2\right)}-\dfrac{x}{3\left(x-2\right)}-\dfrac{2x^2}{3\left(x-2\right)\left(x+2\right)}\right).\dfrac{-12\left(x-2\right)}{13x+6}=(2(x+2)1+2x−3(x−2)x−3(x−2)(x+2)2x2).13x+6−12(x−2)=(3(1+2x)(x−2)6(x+2)(x−2)−2x(x+2)6(x+2)(x−2)−4x26x(x+2)(x−2)).−2(x−2)13x+6=\left(\dfrac{3\left(1+2x\right)\left(x-2\right)}{6\left(x+2\right)\left(x-2\right)}-\dfrac{2x\left(x+2\right)}{6\left(x+2\right)\left(x-2\right)}-\dfrac{4x^2}{6x\left(x+2\right)\left(x-2\right)}\right).\dfrac{-2\left(x-2\right)}{13x+6}=(6(x+2)(x−2)3(1+2x)(x−2)−6(x+2)(x−2)2x(x+2)−6x(x+2)(x−2)4x2).13x+6−2(x−2)=6x2−9x−6−2x2−4x−4x26(x+2)(x−2).−12(x−2)13x+6=\dfrac{6x^2-9x-6-2x^2-4x-4x^2}{6\left(x+2\right)\left(x-2\right)}.\dfrac{-12\left(x-2\right)}{13x+6}=6(x+2)(x−2)6x2−9x−6−2x2−4x−4x2.13x+6−12(x−2)=−(13x+6)6(x+2)(x−2).−12(x−2)13x+6=\dfrac{-\left(13x+6\right)}{6\left(x+2\right)\left(x-2\right)}.\dfrac{-12\left(x-2\right)}{13x+6}=6(x+2)(x−2)−(13x+6).13x+6−12(x−2)
=2x+2=\dfrac{2}{x+2}=x+22
Rút gọn:
A=(1+2x4+2x−x3x−6+2x212−3x2)⋅24−12x6+13xA=\left(\dfrac{1+2x}{4+2x}-\dfrac{x}{3x-6}+\dfrac{2x^2}{12-3x^2}\right)\cdot\dfrac{24-12x}{6+13x}A=(4+2x1+2x−3x−6x+12−3x22x2)⋅6+13x24−12x
=[x3x2(x2−4)+126(2−x)+1x+2]:(10−x2x+2+x−2)=\left[\dfrac{x^3}{x^2\left(x^2-4\right)}+\dfrac{12}{6\left(2-x\right)}+\dfrac{1}{x+2}\right]:\left(\dfrac{10-x^2}{x+2}+x-2\right)=[x2(x2−4)x3+6(2−x)12+x+21]:(x+210−x2+x−2)
=[x(x−2)(x+2)−2x−2+1x+2]:(10−x2x+2+x−2)=\left[\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right]:\left(\dfrac{10-x^2}{x+2}+x-2\right)=[(x−2)(x+2)x−x−22+x+21]:(x+210−x2+x−2)
=x−2x−4+x−2(x−2)(x+2):(10−x2x+2+x−2)=\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\left(\dfrac{10-x^2}{x+2}+x-2\right)=(x−2)(x+2)x−2x−4+x−2:(x+210−x2+x−2)
=−6(x−2)(x+2):10−x2+x2+2x−2x−4x+2=-\dfrac{6}{\left(x-2\right)\left(x+2\right)}:\dfrac{10-x^2+x^2+2x-2x-4}{x+2}=−(x−2)(x+2)6:x+210−x2+x2+2x−2x−4
=−6(x−2)(x+2).x+26=-\dfrac{6}{\left(x-2\right)\left(x+2\right)}.\dfrac{x+2}{6}=−(x−2)(x+2)6.6x+2
=−1x−2=\dfrac{-1}{x-2}=x−2−1
16x2-9(x+1)2=0
giúp với ạk
Tìm x biết :
x−12013+x−22012+x−32011+...\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}+..._{ }2013x−1+2012x−2+2011x−3+...+x−20122=2012+\dfrac{x-2012}{2}=2012+2x−2012=2012
1. Tìm hệ số a,b,c biết: 3x2(ax2 - 2bx - 3c) = 3x4 - 12x3 + 27x2
A. a= -1 ; b= 2 ; c= 3
B. a= -1 ; b= -2 ; c= -3
C. a= 1 ; b= 2 ; c= -3
D. a= -3 ; b= 2; c= 1
1)chứng minh rằng 3x2+2y2+5 luôn dương với x,y
Cho xeyxe yxey . Chứng minh x3+xy2−x2y−y3x−y≥y2\dfrac{x^3+xy^2-x^2y-y^3}{x-y}\ge y^2x−yx3+xy2−x2y−y3≥y2
5
56.a+65.b=12.1 và a+b=0.2 tìm giá trị a và b chỉ em cách làm dạng bài này vs tks mn.
Tìm x,y biết:
a) −2x(10x−3)+5x(4x+1)=25-2x\left(10x-3\right)+5x\left(4x+1\right)=25−2x(10x−3)+5x(4x+1)=25
b) y(5−2y)+2y(y−1)=15y\left(5-2y\right)+2y\left(y-1\right)=15y(5−2y)+2y(y−1)=15
c) x(x+1)−(x+1)=35x\left(x+1\right)-\left(x+1\right)=35x(x+1)−(x+1)=35
d) x(x2+x+1)−x2(x+1)=0x\left(x^2+x+1\right)-x^2\left(x+1\right)=0x(x2+x+1)−x2(x+1)=0
Chứng minh rằng với mọi giá trị nguyên của n ta luôn có:
a) (n2+3n−1)(n+2)−n3+2⋮5\left(n^2+3n-1\right)\left(n+2\right)-n^3+2⋮5(n2+3n−1)(n+2)−n3+2⋮5
b) (6n+1)(n+5)−(3n+5)(2n−1)⋮2\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)⋮2(6n+1)(n+5)−(3n+5)(2n−1)⋮2
Cho P = 1 + x+x2 +x3 +...+x10 .Chứng minh rằng: x.P - P = x11 - 1 Giusp mik với,mik cảm ơn
tính giá trị của biểu thức ;
A = 8x15 - 8x14 + 8x13 - 8x12 + -... - 8x2 +8x - 5 tại x =7
giúp giùm mình sớm nhất có thể nhaaaa
