khó nhề

Đặt \(f\left(x\right)=x+x^2+x^3+x^4+...+x^{100}\)

\(\Rightarrow f'\left(x\right)=1+2x+3x^2+...+100x^{99}=P\) (1)

Mặt khác, ta có \(f\left(x\right)\) cũng là tổng của cấp số nhân với \(\left\{{}\begin{matrix}u_1=x\\q=x\\n=100\end{matrix}\right.\)

Do đó: \(f\left(x\right)=u_1.\dfrac{q^{100}-1}{q-1}=x.\dfrac{x^{100}-1}{x-1}=\dfrac{x^{101}-x}{x-1}\)

\(\Rightarrow f'\left(x\right)=\dfrac{\left(x^{101}-x\right)'.\left(x-1\right)-\left(x-1\right)'.\left(x^{101}-x\right)}{\left(x-1\right)^2}=\dfrac{100x^{101}-101x^{100}+1}{\left(x-1\right)^2}\) (2)

(1);(2) \(\Rightarrow P=\dfrac{100x^{101}-101x^{100}+1}{\left(x-1\right)^2}\)

Chia nhỏ ra ik ạ

\(\left(2x+3\right)\left(2x-3\right)-4x\left(x+5\right)=4x^2-9-4x^2-20x=-20x-9\)

\(5x\left(x-3\right)+\left(x-2\right)^2=5x^2-15x+x^2-4x+4=6x^2-19x+4\)

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

a)2x( 2x-1) -(2x-1)

=(2x-1)(2x-1)

=(2x-1)²

b)2x( 4x + 2x + 1) - ( 4x + 2x +1)

=(2x-1)(4x+2x+1)

=(2x-1)(6x+1)

a) \(2x\left(2x-1\right)-\left(2x-1\right)=\left(2x-1\right)\left(2x-1\right)\)

b) \(2x\left(4x+2x+4\right)-\left(4x+2x+4\right)=\left(2x-1\right)\left(4x+2x+4\right)\)

Mik làm cho vui thôi chứ chẳng ai T mik đâu

Naruto lục đạo:làm thiếu chả ko T :D

Bài 1: 

\(\left(2x-5\right)^2-4\left(2x-5\right)+4=0\)

\(\left(2x-5\right)^2-2\left(2x-5\right)\left(2\right)+2^2=0\)

\(\left(2x-5-2\right)^2=0\)

\(2x-5-2=0\)

\(2x-7=0\)

\(2x=0+7\)

\(2x=7\)

\(x=\frac{7}{2}\)

Bài 3: 

\(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=46\)

\(\left(4x\right)^2-3^2-16x^2+40x-25=46\)

\(4^2x^2-3^2-16x^2+40x-25=46\)

\(16x^2-9-16x^2+40x-25=46\)

\(-34+40x=46\)

\(40x-34=46\)

\(40x=46+34\)

\(40x=80\)

\(x=2\)

bài 2:

a) \(81^2=\left(80+1\right)^2=80^2+2.80+1=6400+160+1=6561\)

b) \(99^2=\left(100-1\right)^2=100^2-2.100+1=10000-200+1=8801\)

ĐKXĐ : \(x\ne\pm\frac{1}{2}\)

\(E=\left(\frac{\left(4x^2+2x\right)\left(1+4x^2\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}-\frac{\left(4x^2-2x\right)\left(1-4x^2\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}\right):\left(\frac{\left(1+2x\right)\left(1+2x\right)}{\left(1-2x\right)\left(1+2x\right)}-\frac{\left(1-2x\right)\left(1-2x\right)}{\left(1+2x\right)\left(1-2x\right)}\right)\)

\(E=\left(\frac{16x^4+8x^3+4x^2+2x+16x^4-8x^3-4x^2+2x}{1-16x^4}\right):\left(\frac{1+2x+x^2-1+2x-x^2}{1-4x^2}\right)\)

\(E=\frac{32x^4+4x}{1-16x^4}:\frac{4x}{1-4x^2}\)

\(E=\frac{4x\left(8x^3+1\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}.\frac{1-4x^2}{4x}\)

\(E=\frac{8x^3+1}{1+4x^2}\)

Study well 

E=\(\left(\frac{4x^2+2x}{1-4x^2}-\frac{4x^2-2x}{1+4x^2}\right):\left(\frac{1+2x}{1-2x}-\frac{1-2x}{1+2x}\right)\)

E=\(\left(\frac{\left(4x^2+2x\right)\left(1+4x^2\right)-\left(4x^2-2x\right)\left(1-4x^2\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}\right):\)\(\left(\frac{\left(1+2x\right)^2-\left(1-2x\right)^2}{\left(1-2x\right)\left(1+2x\right)}\right)\)

E=\(\frac{4x^2+16x^4+2x+8x^3-\left(4x^2-16x^4-2x+8x^3\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}:\)\(\left(\frac{\left(1+4x+4x^2\right)-\left(1-4x+4x^2\right)}{\left(1-2x\right)\left(1+2x\right)}\right)\)

E=\(\frac{4x^2+16x^4+2x+8x^3-4x^2+16x^4+2x-8x^3}{\left(1-4x^2\right)\left(1+4x^2\right)}:\)\(\left(\frac{1+4x+4x^2-1+4x-4x^2}{\left(1-2x\right)\left(1+2x\right)}\right)\)

E=\(\frac{16x^4+2x+16x^4+2x}{\left(1-4x^2\right)\left(1+4x^2\right)}:\)\(\left(\frac{8x}{\left(1-2x\right)\left(1+2x\right)}\right)\)

E=\(\frac{32x^4+8x}{\left(1-4x^2\right)\left(1+4x^2\right)}.\frac{1-4x^2}{8x}\)

E=\(\frac{8x\left(4x^3+1\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}.\frac{1-4x^2}{8x}\)

E=\(\frac{4x^3+1}{1+4x^2}\)

Góp ý : 

Cô giáo ơi

Cô tính nhầm rồi 2x + 2x = 4x cô ạ chứ không phải 8x đâu cô

mong cô xem lại ạ 

E=\(\frac{\left(4x^2+2x\right)\left(1+4x^2\right)-\left(4x^2-2x\right)\left(1-4x^2\right)}{\left(1-4x^2\right)\left(1+4x^2\right)}:\frac{\left(1+2x\right)^2-\left(1-2x\right)^2}{1-4x^2}\)

E=\(\frac{4x^2+16x^4+2x+8x^3-4x^2+16x^2+2x-8x^3}{\left(1-4x^2\right)\left(1+4x^2\right)}.\frac{1-4x^2}{1+4x+4x^2-1+4x-4x^2}\)

E=\(\frac{32x^4+4x}{8x\left(1+4x^2\right)}=\frac{8x^3+1}{2\left(1+4x^2\right)}\)