(x+1)(x+3)(x+5)(x+8)+15

=[(x+1)(x+7)][(x+3)(x+5)]+15

=(x²+8x+7)(x²+8x+15)+15

Đặt t=x²+8x+7

=>x²+8x+15=t+8

=>(x² +8x+7)(x²+8x+15)+15

=t(t+8)+15

=t²+8t+15

=t²+3t+5t+15

=t(t+3)+5(t+3)

=(t+3)(t+5)

=(x²+8x+10)(x²+8x+12)

Đặt \(A=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)

\(\Rightarrow A=\left(x+1\right)\left(x+7\right)\left(x+3\right)\left(x+5\right)+15\)

        \(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

Đặt \(x^2+8x+11=t\)

\(\Rightarrow A=\left(t-4\right)\left(t+4\right)+15=t^2-16+15=t^2-1=\left(t+1\right)\left(t-1\right)\)

\(=\left(x^2+8x+11+1\right)\left(x^2+8x+11-1\right)=\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)

\(=\left(x^2+2x+6x+12\right)\left(x^2+8x+10\right)\)\(=\left[x\left(x+2\right)+6\left(x+2\right)\right]\left(x^2+8x+10\right)\)

\(=\left(x+2\right)\left(x+6\right)\left(x^2+8x+10\right)\)

a, \(x^2+3x+3y+xy=\left(x^2+xy\right)+\left(3x+3y\right)=x\left(x+y\right)+3\left(x+y\right)=\left(x+y\right)\left(x+3\right)\)

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

\(\frac{2}{6}x-4=\frac{1}{3}x-4\ne\frac{1}{3}x-2\)

\(\left(x+5\right)^4+\left(x+3\right)^4=16\)

Đặt t = x+ 4 pt ban đầu trở thành

\(\left(t+1\right)^4+\left(t-1\right)^4=16\Leftrightarrow t^4+6t^2-7=0\)

PT \(t^2=7\left(vn\right)\)

\(PTt^2=1\) cho ta nghiệm \(t=1;t=-1\)

\(\Rightarrow PT\) ban đầu \(\Leftrightarrow\orbr{\begin{cases}x+4=-1\\x+4=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-5\\x=-3\end{cases}}\)

\(2y^2+12y+16=0\Rightarrow2\left(y^2+6y+8\right)=0\Rightarrow2\left(y^2+6y+9-1\right)=0\)

\(\Rightarrow2\left(\left(y+3\right)^2-1\right)=2\left(y+3-1\right)\left(y+3+1\right)=0\Rightarrow....\)(tự làm tiếp nha bạn)

\(2y^2+12y+16=0\)\(\Leftrightarrow2\left(y^2+6y+8\right)=0\)

\(\Leftrightarrow y^2+6y+8=0\)\(\Leftrightarrow y^2+2y+4y+8=0\)

\(\Leftrightarrow y\left(y+2\right)+4\left(y+2\right)=0\)\(\Leftrightarrow\left(y+2\right)\left(y+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y+2=0\\y+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}y=-2\\y=-4\end{cases}}\)

Vậy \(x=-2\)hoặc \(x=-4\)

bạn kiếm kiểu gì cx ko có ai giải đâu, đề này sai r, nãy mình sửa mới đúng