\(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=40\\ \Leftrightarrow\left(x+1\right)\left(x+5\right)\left(x+2\right)\left(x+4\right)=40\\ \Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)=40\)

Đặt \(x^2+6x+5=t\)

\(\Rightarrow t\left(t+3\right)=40\\ \Rightarrow t^2+3t=40\\ \Rightarrow t^2+2.t.\dfrac{3}{2}+\dfrac{9}{4}=\dfrac{169}{4}\\ \Rightarrow\left(t+\dfrac{3}{2}\right)^2=\left(\dfrac{13}{2}\right)^2\\ \Leftrightarrow\left(t+\dfrac{3}{2}\right)^2-\left(\dfrac{13}{2}\right)^2=0\\ \Leftrightarrow\left(t+\dfrac{3}{2}-\dfrac{13}{2}\right)\left(t+\dfrac{3}{2}+\dfrac{13}{2}\right)=0\\ \Leftrightarrow\left(t-5\right)\left(t+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=5\\t=-8\end{matrix}\right.\)

TH1: t=5

\(\Rightarrow x^2+6x+5=5\\ \Rightarrow x\left(x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)

TH2 : t=-8

\(\Rightarrow x^2+6x+5=-8\\ \Rightarrow\left(x+3\right)^2=-4\left(voly\right)\)

=> x rông

Vậy x=0 hoặc x=-6

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

\(\Rightarrow\left(x+1\right)\left(x+5\right)\left(x+2\right)\left(x+4\right)-40=0\)

\(\Rightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)-40=0\)

Đặt \(a=x^2+6x+5\)

\(\Rightarrow a\left(a+3\right)-40=0\)

\(\Rightarrow a^2+3a-40=0\)

\(\Rightarrow a^2+8a-5a-40=0\)

\(\Rightarrow a\left(a+8\right)-5\left(a+8\right)=0\)

\(\Rightarrow\left(a+8\right)\left(a-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a+8=0\\a-5=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}a=-8\\a=5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2+6x+5+8=0\\x^2+6x+5-5=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2+6x+13=0\\x^2+6x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2+6x+13\ne0\\x\left(x+6\right)=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)

Kết luận:\(S=\left\{-6;0\right\}\)

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

<=>x^4+12*x^3+49*x^2+78*x=0

<=>x*(x+6)*(x^2+6*x+13)=0

suy ra

x=0

x=-6

x^2+6*x+13=0(mà phương trình này không thể phân tích nếu phân tích thì sẽ liên quan tới số vô tỉ lên lớp 9 mới học)

Vậy tập nghiệm của phương trình S=-6;0

Nhớ tich nha bạn

a: \(\dfrac{x+1}{5}+\dfrac{x+1}{6}=\dfrac{x+1}{7}+\dfrac{x+1}{8}\)

\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}-\dfrac{1}{8}\right)=0\)

=>x+1=0

hay x=-1

b: \(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)

=>x-2010=0

hay x=2010

c: \(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Leftrightarrow\dfrac{x}{\left(x+2\right)\left(x+17\right)}=\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}\)

=>x=15

A= \(\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{2}{x+3}-...+\frac{8}{x+5}-\frac{8}{x+6}\)

A=\(\frac{1}{x+1}+\frac{1}{x+3}+\frac{2}{x+4}+\frac{4}{x+5}-\frac{8}{x+6}\)

Rồi tiếp tục làm nhé bạn.

\(1,2\left(x-3\right)+1=2\left(x+1\right)-9\\ \Rightarrow2x-6+1=2x+2-9\\ \Rightarrow2x-5=2x-7\\ \Rightarrow-2=0\left(vô.lí\right)\)

\(2,\dfrac{5-x}{2}=\dfrac{3x-4}{6}\\ \Rightarrow30-6x=6x-8\\ \Rightarrow12x=38\\ \Rightarrow x=\dfrac{19}{6}\)

\(3,\left(x-1\right)^2+\left(x+2\right)\left(x-2\right)=\left(2x+1\right)\left(x-3\right)\\ \Rightarrow x^2-2x+1+x^2-4=2x^2-6x+x-3\\ \Rightarrow2x^2-2x-3=2x^2-5x-3\\ \Rightarrow3x=0\\ \Rightarrow x=0\)

\(4,\left(x+5\right)\left(x-1\right)-\left(x+1\right)\left(x+2\right)=1\\ \Rightarrow x^2+5x-x-5-x^2-2x-x-2=1\\ \\ \Rightarrow x-7=1\\ \Rightarrow x=8\)

\(5,\dfrac{6x-1}{15}-\dfrac{x}{5}=\dfrac{2x}{3}\\ \Rightarrow\dfrac{6x-1}{15}-\dfrac{3x}{15}=\dfrac{10x}{15}\\ \Rightarrow6x-1-3x=10x\\ \Rightarrow3x-1=10x\\ \Rightarrow7x=-1\\ \Rightarrow x=\dfrac{-1}{7}\)

\(6,\dfrac{5\left(x-2\right)}{2}-\dfrac{x+5}{3}=1-\dfrac{4\left(x-3\right)}{5}\\ \Rightarrow\dfrac{75\left(x-2\right)}{30}-\dfrac{10\left(x+5\right)}{30}=\dfrac{30}{30}-\dfrac{24\left(x-3\right)}{30}\\ \Rightarrow75\left(x-2\right)-10\left(x+5\right)=30-24\left(x-3\right)\\ \Rightarrow75x-150-10x-50=30-24x+72\\ \Rightarrow65x-200=102-24x\\ \Rightarrow89x=302\\ \Rightarrow x=\dfrac{320}{89}\)

Đặt \(x+1=t\)

PT\(\Leftrightarrow\left(t-2\right)^4+\left(t+2\right)^4=40\)

\(\Leftrightarrow\left[\left(t-2\right)^2\right]^2+\left[\left(t+2\right)^2\right]^2=40\)

\(\Leftrightarrow\left[\left(t-2\right)^2+\left(t+2\right)^2\right]^2-2\left(t-2\right)^2\left(t-2\right)^2=40\)

\(\Leftrightarrow\left(t^2-4t+4+t^2+4t+4\right)^2-2\left(t^2-4\right)^2=40\)

​\(\Leftrightarrow\left(2t^2+8\right)^2-2\left(t^2-4\right)^2=40\)​

\(\Leftrightarrow...\)

a: Ta có: \(\left(7x+4\right)^2-\left(7x-4\right)\left(7x+4\right)\)

\(=\left(7x+4\right)\left(7x+4-7x+4\right)\)

\(=8\left(7x+4\right)\)

=56x+32

b: Ta có: \(8\left(x-2\right)^2-3\left(x^2-4x-5\right)-5x^2\)

\(=8x^2-32x+32-3x^2+12x+15-5x^2\)

\(=-20x+47\)

c: Ta có: \(\left(x+1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(x+1\right)\)

\(=x^3+3x^2+3x+1-x^3+1-3x^2-3x\)

=2