=>24-6x-4x+4=2x+40

=>-10x+28=2x+40

=>-12x=12

=>x=-1

\(6\left(4-x\right)-4\left(x-1\right)=2x+40\)

\(\Rightarrow24-6x-4x+4=2x+40\)

\(\Rightarrow-6x-4x-2x=40-24-4\)

\(\Rightarrow-12=12\)

\(\Rightarrow x=\dfrac{12}{-12}\)

\(\Rightarrow x=-1\)

1: 

\(\Leftrightarrow\left(x^2+5x+6\right)\left(x^2+5x+4\right)=24\)

\(\Leftrightarrow\left(x^2+5x\right)^2+10\left(x^2+5x\right)=0\)

\(\Leftrightarrow x^2+5x=0\)

=>x=0 hoặc x=-5

3: \(\Leftrightarrow\left(x^2+x+6\right)\left(x^2+x-2\right)=0\)

=>(x+2)(x-1)=0

=>x=-2 hoặc x=1

a)

\(P=\dfrac{x^{10}-x^8+x^6-x^4+x^2-1}{x^4-1}\)

\(=\dfrac{x^8\left(x^2-1\right)+x^4\left(x^2-1\right)+\left(x^2-1\right)}{\left(x^2-1\right)\left(x^2+1\right)}\)

\(=\dfrac{\left(x^2-1\right)\left(x^8+x^4+1\right)}{\left(x^2-1\right)\left(x^2+1\right)}\)

\(=\dfrac{x^8+x^4+1}{x^2+1}\)

b)

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

\(=\dfrac{x^{40}+x^{30}+x^{20}+x^{10}+1}{\left(x^{45}+x^{35}+...+x^5\right)+\left(x^{40}+x^{30}+...+1\right)}\)

\(=\dfrac{x^{40}+x^{30}+x^{20}+x^{10}+1}{x^5\left(x^{40}+x^{30}+...+1\right)+\left(x^{40}+x^{30}+...+1\right)}\)

\(=\dfrac{x^{40}+x^{30}+x^{20}+x^{10}+1}{\left(x^{40}+x^{30}+...+1\right)\left(x^5+1\right)}\)

\(=\dfrac{1}{\left(x^5+1\right)}\)

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

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

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

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

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

\(A=-12x\)

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

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

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

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

\(B=-3x-12\)

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

\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=40\\ \Leftrightarrow\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]=40\\ \Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=40\\ \Leftrightarrow\left(x^2+5x+4\right)\left[\left(x^2+5x+4\right)+2\right]=40\\ \Leftrightarrow\left(x^2+5x+4\right)^2+2\left(x^2+5x+4\right)-40=0\)

Mình thấy nghiệm xấu lắm, bạn xem có đúng đề ko

Đặt \(x=y-3\).

\(\Rightarrow\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=\left(y-2\right)\left(y-1\right)\left(y+2\right)\left(y+1\right)=\left(y^2-1\right)\left(y^2-4\right)=40\)

\(\Rightarrow y^2=9\)

\(\Rightarrow x=\hept{\begin{cases}0\\-6\end{cases}}\)

https://diendantoanhoc.net/topic/170566-gi%E1%BA%A3i-pt-a-x1x2x4x540-b-x4-3x32x2-9x90/

tham khảo ở đó nha

\(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=40\)

\(\Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x\right)=40\)

Đặt \(x^2+6x=y\)

ta có \(y\left(y+5\right)=40\)

Đến đó bạn tự giải nhé

(x + 1)(x + 2)(x + 4)(x + 5) = 40

<=> (x + 1)(x + 5)(x + 2)(x + 4) - 40 = 0

<=> (x² + 6x + 5)(x² + 6x + 8) - 40 = 0

Đặt x² + 6x + 5 = a <=> a(a + 3) - 40 = 0

<=> a² + 3a - 40 = 0

<=> a² + 8a - 5a - 40 = 0

<=> (a + 8)(a - 5) = 0

<=> \(\orbr{\begin{cases}a+8=0\\a-5=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x^2+6x+5+8=0\\x^2+6x+5-5=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x^2+6x+9+4=0\\x^2+6x=0\end{cases}}\)

<=> \(\orbr{\begin{cases}\left(x+3\right)^2+4=0\left(vn\right)\\x\left(x+6\right)=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=0\\x+6=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=0\\x=-6\end{cases}}\) Vậy S  = {0; -6}