Sai từ bước 3 bởi vì  

f ' 0 - = lim x → 0 - f x - f 0 x - 0 = lim x → 0 - - x - 0 x - 0 = 1

Do f ' 0 + ≠ f ' 0 -  nên f '(0) không tồn tại

Đáp án C

ta có: \(S=1+1.2+2.3+3.4+...+38.39+39.40+40\)

\(\Rightarrow3S=3+1.2.3+2.3.3+3.4.3+...+38.39.3+39.40.3+40.3\)

\(3S=3+1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+38.39.\left(40-37\right)\)

           \(+39.40.\left(41-38\right)+120\)

\(3S=3+1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+38.39.40-37.38.39\)

             \(+39.40.41-38.39.40+120\)

\(3S=\left(3+1.2.3+2.3.4+3.4.5+...+39.40.41+120\right)-\left(1.2.3+2.3.4+...+38.39.40\right)\)

\(3S=3+39.40.41+120\)

\(\Rightarrow S=\frac{3+39.40.41+120}{3}\)

\(S=21361\)

Mình tìm ra kết quả là 21443 . Sai thì các bn cứ k sai cho mình nhé . Mình sẽ rút kinh nhiệm . Còn bn nào thông minh có thể giúp mình được không . Mình sẽ k .

S=1+1*2+2*3+3*4+4*5.....*39+39*40+40=21361

Đặt biểu thức là A, ta có:

\(A=\frac{x^{40}+x^{30}+x^{20}+x^{10}+1}{x^{45}+x^{40}+x^{35}+...+x^{10}+x^5+1}\)

\(\Rightarrow A.x^5=\frac{x^{45}+x^{35}+x^{25}+x^{15}+x^5}{x^{45}+x^{40}+x^{35}+...+x^{10}+x^5+1}\)

\(\Rightarrow A.x^5+A=\frac{x^{45}+x^{40}+x^{35}+x^{25}+x^{15}+x^5+x^{40}+x^{30}+x^{20}+x^{10}+1}{x^{45}+x^{40}+x^{35}+...+x^{10}+x^5+1}\)

\(\Rightarrow A.x^5+1=1\)

\(\Rightarrow A=\frac{1}{x^5+1}\)

(x⁸⁰+x⁴⁰+1)/(x²⁰+x¹⁰+1)

=x⁶⁰+x³⁰

=x³⁰(x²+1)

HIHI ÁNH BÉO TỚ LÀM BỪA

\(\dfrac{x+1}{2}+\dfrac{x+1}{3}+\dfrac{x+1}{4}=\dfrac{x+1}{5}+\dfrac{x+1}{6}\)

\(\Leftrightarrow\dfrac{x+1}{2}+\dfrac{x+1}{3}+\dfrac{x+1}{4}-\dfrac{x+1}{5}-\dfrac{x+1}{6}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}\right)=0\)

Mà \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}\ne0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

Vậy ..

\(\dfrac{x+1}{2}+\dfrac{x+1}{3}+\dfrac{x+1}{4}=\dfrac{x+1}{5}+\dfrac{x+1}{6}\)

=> \(\dfrac{x+1}{2}+\dfrac{x+1}{3}+\dfrac{x+1}{4}-\dfrac{x+1}{5}-\dfrac{x+1}{6}\)= 0

(x + 1).(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}\)) = 0

Ta thấy \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}\) > 0

=> x + 1 = 0

x = 0 - 1

x = -1

sai rui ahihi

\(A=\left(x+1\right)^3-\left(x+3\right)^2\left(x+1\right)+4x^2+8\)

\(A=x^3+3x^2+3x+1-\left(x^2+6x+9\right)\left(x+1\right)+4x^2+8\)

\(A=x^3+3x^2+3x+1-\left(x^3+6x^2+9x+x^2+6x+9\right)+4x^2+8\)

\(A=x^3+3x^2+3x+1-x^3-6x^2-9x-x^2-6x-9+4x^2+8\)

\(A=\left(x^3-x^3\right)+\left(3x^2-6x^2-x^2+4x^2\right)+\left(3x-9x-6x\right)+\left(1-9+8\right)\)

\(A=-12x\)

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

\(B=x^3+2x^2+4x-2x^2-4x-8-\left(x^3+3x^2+3x+1\right)+3\left(x^2-1\right)\)

\(B=x^3+2x^2+4x-2x^2-4x-8-x^3-3x^2-3x-1+3x^2-3\)

\(B=\left(x^3-x^3\right)+\left(2x^2-2x^2-3x^2+3x^2\right)+\left(4x-4x-3x\right)+\left(-8-3-1\right)\)

\(B=-3x-12\)

Câu C tương tự.

Chúc bạn học tốt!!!

A = \(\left(x+1\right)^3-\left(x+3\right)^2.\left(x+1\right)+4x^2+8\)

A = \(\left(x+1\right)\left(x+1-x-3\right)\left(x+1+x+3\right)+4x^2+8\)

A = \(\left(x+1\right).\left(-2\right).\left(2x+4\right)+4x^2+8\)

A = \(\left(-2\right)\left(2x^2+4x+2x+4\right)+4x^2+8\)

A = \(\left(-2\right)\left(2x^2+6x+4\right)+4x^2+8\)

A = \(-4x^2-12x-8+4x^2+8=-12x\)

b) B = \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)

B = \(x^3-8-\left(x+1\right)\left(x^2+2x+1+3x-3\right)\)

B = \(x^3-8-\left(x+1\right)\left(x^2+5x-2\right)\)

B = \(x^3-8-x^3-5x^2+2x-x^2-5x+2\)

B = \(-6x^2-3x-6\)