Cho f(x)=x^7-90x^6+90x^5-90x^4+...+90x+1928
Tính f(89)
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x=89 nên x+1=90
\(f\left(x\right)=x^7-x^6\left(x+1\right)+x^5\left(x+1\right)-x^4\left(x+1\right)+...+x\left(x+1\right)+1928\)
\(=x^7-x^7-x^6+x^6-...+x^2+x+1928\)
=x+1928=89+1928=2017
\(90x-6750=75x-x^2\)
\(\Leftrightarrow180x-6750=75x-x^2\)
\(\Leftrightarrow x^2+105x-6750=0\)
\(\Leftrightarrow x^2-45x+150x-6750=0\)
\(\Leftrightarrow\left(x-45\right)\left(x+150\right)=0
\)
\(\Leftrightarrow\left[{}\begin{matrix}x-45=0\\x+150=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=45\\x=-150\end{matrix}\right.\)
Vậy x=45 hoặc x=-150
a, \(\left(x+2\right)^3-x\left(x^2+6x-3\right)=0\Leftrightarrow x^3+4x^2+4x+2x^2+8x+8-x^3-6x^2+3x=0\)
\(\Leftrightarrow15x+8=0\Leftrightarrow x=-\frac{8}{15}\)
b, \(\left(x+4\right)^3-x\left(x+6\right)^2=7\Leftrightarrow12x+64=0\Leftrightarrow x=-\frac{19}{4}\)làm tắt:P
Tự làm nốt nhé
a, \(\left(x-y+4\right)^2-\left(2x+3y-1\right)^2=\left(x-y+4+2x+3y-1\right)\left(x-y+4-2x-3y+1\right)\)
\(=\left(3x+2y+3\right)\left(-x-4y+5\right)\)
b, \(x^6+y^6=\left(x^2\right)^3+\left(y^2\right)^3=\left(x^2+y^2\right)\left(x^4-x^2y^2+y^4\right)\)
c, \(x^{16}-1=\left(x^2\right)^8-1=\left[\left(x^2\right)^4\right]^2-1=\left(x^8-1\right)\left(x^8+1\right)\)
\(=\left(3x+15\right)^2-\left(x-7\right)^2=\left(4x+8\right)\left(2x+22\right)=8\left(x+2\right)\left(x+11\right)\)
(x-7)^2 = x^2-14x+49
<=> 9x^2+90x+225 -x^2+14x-49
= 8x^2+104x+176
= 8(x^2+13+22)
<=> 8(x+2)(x+11)
9x2+90x+225-(x-7)2
=(3x+15)2-(x-7)2
=(3x+15-x+7)(3x+15+x-7)
=(2x+22)(4x+8)
=2 (x+11)4 (x+2)
=8 (x+11)(x+2)
\(=x^4-9x^3+9x^3-81x^2-9x^2+81x+10x-90\)
\(=\left(x-9\right)\left(x^3+9x^2-9x+10\right)\)
\(=\left(x-9\right)\left(x^3+10x^2-x^2-10x+x+10\right)\)
\(=\left(x-9\right)\left(x+10\right)\left(x^2-x+1\right)\)
\(f\left(x\right)=x^7-90x^6+90x^5-90x^4+...+90x+1928\)\(\Rightarrow f\left(89\right)=x^7-\left(x+1\right)x^6+\left(x+1\right)x^5\)\(-\left(x+1\right)x^4+...+\left(x+1\right)x+1928\)
\(\Rightarrow f\left(89\right)=x^7-x^7-x^6+x^6+x^5-x^5\)\(-x^4+...+\)\(x^2+x+1928\)\(=89+1928=2017\)
2017