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

\(f\left(x\right)=x^7-90x^6+90x^5-90x^4+...+90x+1928\)\(\Rightarrow f\left(89\right)=x^7-\left(x+1\right)x^6+\left(x+1\right)x^5\)\(-\left(x+1\right)x^4+...+\left(x+1\right)x+1928\)

\(\Rightarrow f\left(89\right)=x^7-x^7-x^6+x^6+x^5-x^5\)\(-x^4+...+\)\(x^2+x+1928\)\(=89+1928=2017\)

2017

giúp mifnnh với ạ !!

CHữ số hay tích

dễ cậu kêu ai trả lời được cho 2 k

Ta có : \(x^3-27x^2+90x-81=0\)

=> \(x^3-3.x^2.9+3.x.9^2-729+648=0\)

=> \(\left(x-9\right)\left(x^2+9x+81\right)+648=0\)

=> \(\left(x-9\right)\left(x^2+9x+81\right)=-648\)

Ta thấy : \(x^2+9x+81=x^2+\frac{2.x.9}{2}+\frac{81}{4}+\frac{243}{4}\)

\(=\left(x+\frac{9}{2}\right)^2+\frac{243}{4}\ge\frac{243}{4}>0\)

=> \(x-9< 0\)

=> \(\left(x-9\right)\left(\left(x+\frac{9}{2}\right)^2+\frac{243}{4}\right)\le\frac{243\left(x-9\right)}{4}\)

- Dấu " = " xảy ra <=> \(\frac{243\left(x-9\right)}{4}=0\)

=> \(x-9=0\) ( KTM )

Vậy phương trình vô nghiệm .

Hic hỏi ngu quá hay sao mà bạn kia xoá bài giải luôn r

\(=\left(3x+15\right)^2-\left(x-7\right)^2=\left(4x+8\right)\left(2x+22\right)=8\left(x+2\right)\left(x+11\right)\)

(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)

\(=x^4-9x^3+9x^3-81x^2-9x^2+81x+10x-90\)

\(=\left(x-9\right)\left(x^3+9x^2-9x+10\right)\)

\(=\left(x-9\right)\left(x^3+10x^2-x^2-10x+x+10\right)\)

\(=\left(x-9\right)\left(x+10\right)\left(x^2-x+1\right)\)

a) \(x^2-90x+1400=0\)

\(\Leftrightarrow x^2-20x-70x+1400=0\)

\(\Leftrightarrow x\left(x-20\right)-70\left(x-20\right)=0\)

\(\Leftrightarrow\left(x-20\right)\left(x-70\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-20=0\\x-70=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=20\\x=70\end{cases}}\)

S={20;70} là tập nghiệm của  phương trình 

\(25x^2+90x+81=\left(5x+9\right)^2\)

\(64x^2-48x+9=\left(8x-3\right)^2\)