(x-1)(x-3)(x-5)(x-7)-30
(x+1)(x+2)(x+3)(x+4)-24
phan tich thanh nhan tu
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(9x^2+6x-8\)
\(=9x^2+12x-6x-8\)
\(=3x\left(3x+4\right)-2\left(3x+4\right)\)
\(=\left(3x+4\right)\left(3x-2\right)\)
b) \(x^2-7xy+10y^2\)
\(=x^2-2xy-5xy+10y^2\)
\(=x\left(x-2y\right)-5y\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x-5y\right)\)
c) \(x^8+x^7+1\)
\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)
a, \(A=x^2-6x+11\)
\(=x^2-2.3.x+9+2\)
\(=\left(x-3\right)^2+2\)
Ta có: \(\left(x-3\right)^2\ge0\Leftrightarrow\left(x-3\right)^2+2\ge2\)
Dấu "=" xảy ra \(\Leftrightarrow x-3=0\)\(\Leftrightarrow x=3\)
Vậy \(MinA=3\Leftrightarrow x=3\)
b, \(B=2x^2+10x-1\)
\(=2\left(x^2+5x\right)-1\)
\(=2\left(x^2+2.\frac{5}{2}x+\frac{25}{4}\right)-\frac{21}{4}\)
\(=2\left(x+\frac{5}{2}\right)^2-\frac{21}{4}\)
Ta có: \(\left(x+\frac{5}{2}\right)^2\ge0\Leftrightarrow\left(x+\frac{5}{2}\right)^2-\frac{21}{4}\ge-\frac{21}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x+\frac{5}{2}\right)^2=0\Leftrightarrow x+\frac{5}{2}=0\Leftrightarrow x=-\frac{5}{2}\)
Vậy \(MinB=-\frac{21}{4}\Leftrightarrow x=-\frac{5}{2}\)
c, \(C=5x-x^2\)
\(=-x^2+5x\)
\(=-\left(x^2+2.\frac{5}{2}x+\frac{25}{4}\right)+\frac{25}{4}\)
\(=-\left(x+\frac{5}{2}\right)^2+\frac{25}{4}\)
Ta có: \(-\left(x+\frac{5}{2}\right)^2\le0\Leftrightarrow-\left(x+\frac{5}{2}\right)^2+\frac{25}{4}\le\frac{25}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x+\frac{5}{2}\right)^2=0\Leftrightarrow x=-\frac{5}{2}\)
Vậy \(MaxB=\frac{25}{4}\Leftrightarrow x=-\frac{5}{2}\)
đk: \(x\ne0\); \(x\ne\pm3y\)
\(\frac{x+9y}{x^2-9y^2}-\frac{3y}{x^2+3xy}\)
\(=\frac{x+9y}{\left(x-3y\right)\left(x+3y\right)}-\frac{3y}{x\left(x+3y\right)}\)
\(=\frac{x\left(x+9y\right)}{x\left(x-3y\right)\left(x+3y\right)}-\frac{3y\left(x-3y\right)}{x\left(x-3y\right)\left(x+3y\right)}\)
\(=\frac{x^2+9xy-3xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}\)
\(=\frac{\left(x+3y\right)^2}{x\left(x-3y\right)\left(x+3y\right)}\)
\(=\frac{x+3y}{x\left(x-3y\right)}\)
\(\frac{x^2+25y^2-10xy}{x^2-25y^2}\)
\(\frac{\left(x-5y\right)^2}{\left(x-5y\right)\cdot\left(x+5y\right)}\)
\(\frac{x-5y}{x+5y}\)
\(x^2-3x-4x+12+x^2+10x+25\)
\(2x^2-7x+10x+27\)
\(2x^2+3x+27\)
\(\left(a^2+9\right)^2-36a^2\)
\(=\left(a^2+9+6a\right)\left(a^2+9-6a\right)\)
\(=\left(a+3\right)^2.\left(a-3\right)^2\)
p/s:chúc bạn hk tốt
\(\left(a^2+9\right)^2-36a^2\)
\(=\left(a^2+9+36a^2\right)\cdot\left(a^2+9-36a^2\right)\)
\(=\left(37a^2+9\right)\cdot\left(-35a^2+9\right)\)
\(\left(a^2+4b^2\right)^2-16a^2\)
\(=\left(a^2+4b^2-4a\right)\left(a^2+4b^2+4a\right)\)
p/s: chúc bạn học tốt
\((4x-y)(a+b)(4x-y)(c-1)\)
\(=\left(4x-y\right)\left(4x-y\right)=\left(4x-y\right)^{1+1}=\left(4y-2\right)^2\)
\(=\left(a+b\right)\left(4x-y\right)^2\left(c-1\right)\)
(4x-y)(a+b)(4x-y)(c-1)
= ( 4x - y ) ( 4x - y ) = ( 4x - y ) 1 + 1 = ( 4y - 2 ) 2
= (a + b ) ( 4x - y )2 ( c - 1 )
a, k ph đc
b,Đặt \(A=...=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-24=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)
Đặt x^2+5x+4=t,ta có:
\(A=t\left(t+2\right)-24=t^2+2t-24=t^2-4t+6t-24=t\left(t-4\right)+6\left(t-4\right)=\left(t-4\right)\left(t+6\right)\)
\(=\left(x^2+5x+4-4\right)\left(x^2+5x+4+6\right)=x\left(x+5\right)\left(x^2+5x+10\right)\)