Tong quat: a^3+1=(a+1)[a^2-a+1]=(a+1)[(a-0,5)^2+0,75]

                 a^3-1=(a-1)[a^2+a+1]=(a-1)[(a+0,5)^2+0,75]

Tu so cua A=(2+1).[(2-0,5)^2+0,75].(3+1).[(3-0,5)^2+0,75].(4+1).[(4-0,75)^2+0,75]....(10+1).[(10-0,5)^2+0,75]

                 =3.[1,5^2+0,75].4.[2,5^2+0,75].5.[3,5^2+0,75]....11.[9,5^2+0,75]

Mau so cua A= (2-1).[(2+0,5)^2+0,75].(3-1).[(3+0,5)^2+0,75].(4-1).[(4+0,75)^2+0,75]....(10-1).[(10+0,5)^2+0,75]

                 =[2,5^2+0,75].2.[3,5^2+0,75].3.[4,5^2+0,75]....9.[10,5^2+0,75]

Vay A=3.[1,5^2+0,75].4.[2,5^2+0,75].5.[3,5^2+0,75]....11.[9,5^2+0,75]/[2,5^2+0,75].2.[3,5^2+0,75].3.[4,5^2+0,75]....9.[10,5^2+0,75]

         =(3.4.5...11/1.2.3...9).[(1,5^2+0,75)(2,5^2+0,75)(3,5^2+0,75)...(9,5^2+0,75)/(2,5^2+0,75)(3,5^2+0,75)(4,5^2+0,75)...(10,5^2+0,75)]

         =11.10.(1,5^2+0,75)/2.(10,5^2+0,75)

         Con bao nhieu ban tu tinh tiep nha 

Tai vi may minh bi lag nen khong danh phan so duoc vi vay minh phai tach mau, tu ra. sorry

cảm ơn bạn nhiều

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em chiu a. chang hieu gi hihi

1)(5-x²).(x⁴+5x²+25)

2)15.(x-1)-(3x-1)

3)(x²-2)²

4)36x².(y-1)

5)(7-y).(z-x)

6)(x+3).(x+5)

7)(x-10).(x+2)

8)(x+5).(3y+1)

9)(-(y-x-3)).(y-x+3)

10)(11-x).(y+x)

11)(y-x+3)).(y+x-3)

12)(-(y+2x-5)).(y+2x+5)

13)4.(tz+y²+(-x).y-t²

14)(8-x).(y-x)

a, \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)

\(\Leftrightarrow72-20x-36x+84=30x-240-6x-84\)

\(\Leftrightarrow156-56x=24x-324\)

\(\Leftrightarrow-80x+480=0\Leftrightarrow x=-6\)

b, \(5\left(3x+5\right)-4\left(2x-3\right)=5x+3\left(2x-12\right)+1\)

\(\Leftrightarrow15x+25-8x+12=5x+6x-36+1\)

\(\Leftrightarrow7x+37=11x-35\)

\(\Leftrightarrow-4x+72=0\Leftrightarrow x=18\)

c, \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-2x-1=12x-5\)

\(\Leftrightarrow-14x+4=0\Leftrightarrow x=\frac{2}{7}\)

d, \(5x-3\left\{4x-2\left[4x-3\left(5x-2\right)\right]\right\}=182\)

\(\Leftrightarrow5x-3\left[4x-15x+6\right]=182\)

\(\Leftrightarrow5x-3\left(-11x+6\right)=182\)

\(\Leftrightarrow5x+33x-18-182=0\)

\(\Leftrightarrow38x-200=0\Leftrightarrow x=\frac{100}{19}\)

Tách : \(a-b=-\left(c-a\right)-\left(b-c\right)\)

Ta có : \(a^2b^2\left(a-b\right)+b^2c^2\left(b-c\right)+c^2a^2\left(c-a\right)\)

\(=a^2b^2\left[-\left(c-a\right)-\left(b-c\right)\right]+b^2c^2\left(b-c\right)+c^2a^2\left(c-a\right)\)

\(=-a^2b^2\left(c-a\right)+c^2a^2\left(c-a\right)-a^2b^2\left(b-c\right)+b^2c^2\left(b-c\right)\)

\(=a^2\left(c-a\right)\left(c^2-b^2\right)+b^2\left(b-c\right)\left(c^2-a^2\right)\)

\(=a^2\left(c-a\right)\left(c-b\right)\left(c+b\right)+b^2\left(b-c\right)\left(c-a\right)\left(c+a\right)\)

\(=\left(c-a\right)\left(b-c\right)\left(b^2c+b^2a-a^2c-a^2b\right)\)

\(=\left(c-a\right)\left(b-c\right)\left[-\left(a-b\right)\left(ab+bc+ac\right)\right]\)

\(=-\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(ab+bc+ac\right)\)

a²b²(a-b)+ b²c²(b-c)+ c²a²(c-a)= a³b²-a²b³+b³c²-b²c³+c²a²(c-a)=b³(c²-a²)-b²(c³-a³)+c²a²(c-a)

Ta có ; \(\frac{MA'}{AA'}=\frac{S_{BMC}}{S_{ABC}}\) ; \(\frac{MB'}{BB'}=\frac{S_{AMC}}{S_{ABC}}\) ; \(\frac{MC'}{CC'}=\frac{S_{ABM}}{S_{ABC}}\)

\(\Rightarrow\frac{MA'}{AA'}+\frac{MB'}{BB'}+\frac{MC'}{CC'}=\frac{S_{BMC}+S_{AMC}+S_{AMB}}{S_{ABC}}=\frac{S_{ABC}}{S_{ABC}}=1\)

Áp dụng bất đằng thức Cauchy : \(\frac{MA'}{AA'}.\frac{MB'}{BB'}.\frac{MC'}{CC'}\le\left(\frac{MA'+MB'+MC'}{3}\right)^3=\left(\frac{1}{3}\right)^2\)

\(\Rightarrow\frac{MA'}{AA'}.\frac{MB'}{BB'}.\frac{MC'}{CC'}\le\frac{1}{27}\). Dấu "=" xảy ra khi và chỉ khi \(\hept{\begin{cases}\frac{MA'}{AA'}=\frac{MB'}{BB'}=\frac{MC'}{CC'}\\\frac{MA'}{AA'}+\frac{MB'}{BB'}+\frac{MC'}{CC'}=1\end{cases}}\)\(\Rightarrow\frac{MA'}{AA'}=\frac{MB'}{BB'}=\frac{MC'}{CC'}=\frac{1}{3}\)

Vậy dấu "=" xảy ra khi M là trọng tâm của tam giác ABC.

cam on

Ta có : \(A=x^4-2x^3+3x^2+ax+b\)

Vì A là bình phương của một đa thức nên giả sử: \(A=\left(x^2+cx+d\right)^2\)\(\Leftrightarrow x^4+c^2x^2+d^2+2\left(cx^3+cdx+dx^2\right)=x^4-2x^3+3x^2+ax+b\)

\(\Leftrightarrow x^3\left(2c+2\right)+x^2\left(c^2+2d-3\right)+x\left(2cd-a\right)+\left(d^2-b\right)=0\)

Suy ra được : (2c+2) = 0 ; c²+2d-3 = 0 ; 2cd-a = 0 ; d² - b = 0

\(\Rightarrow c=-1;d=1;a=-2;b=1\)

Vậy \(A=x^4-2x^3+3x^2-2x+1=\left(x^2-x+1\right)^2\)

ta đặt A=(x²`+cx+d)²=x⁴ +2cx³+(2d+c²)x²+2cdx+d²

đồng nhất hệ số ta được2c=-2;2d+c²=3;2cd=a;b=d²

giải ra ta được a=-2; b=1