\(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15=0\)

\(\Leftrightarrow\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15=0\)

Đặt \(x^2+8x+11=y\Rightarrow x^2+8x+7=y-4;x^2+8x+15=y+4\)

Khi đó:

\(pt\Leftrightarrow\left(y-4\right)\left(y+4\right)+15=0\)

\(\Leftrightarrow y^2-1=0\)

\(\Leftrightarrow y=1;y=-1\)

Nếu \(y=1\Rightarrow x^2+8x+11=1\)

\(\Rightarrow x^2+8x+10=0\)

\(\Rightarrow-\left(6-x^2-8x-16\right)=0\)

\(\Rightarrow-\left[6-\left(x+4\right)^2\right]=0\)

\(\Rightarrow-\left(\sqrt{6}-x-4\right)\left(\sqrt{6}+x+4\right)=0\)

\(\Rightarrow x=-4-\sqrt{6};x=\sqrt{6}-4\)

Nếu \(y=-1\),ta có:

\(x^2+8x+11=-1\)

\(\Rightarrow x^2+8x+12=0\)

\(\Rightarrow x^2+2x+6x+12=0\)

\(\Rightarrow x\left(x+2\right)+6\left(x+2\right)=0\)

\(\Rightarrow\left(x+2\right)\left(x+6\right)=0\)

\(\Rightarrow x=-2;x=-6\)

Vậy \(x=-2;x=-6;x=-4-\sqrt{6};x=\sqrt{6}-4\)

b) (x+1)(x+7)(x+3)(x+5)+15=0

=> (x^2+7x+x+7)(x^2+5x+3x+15)+15=0

=> (x^2+8x+7)(x^2+8x+15)+15=0

\(\left|x-7\right|^{15}+\left|x-8\right|^{16}=1\)

Ta có: \(\hept{\begin{cases}\left|x-7\right|^{15}\ge0\\\left|x-8\right|^{16}\ge0\end{cases}}\)mà \(\left|x-7\right|^{15}+\left|x-8\right|^{16}=1\)

\(\Rightarrow\orbr{\begin{cases}\left|x-7\right|^{15}=1;\left|x-8\right|^{16}=0\\\left|x-7\right|^{15}=0;\left|x-8\right|^{16}=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=8\\x=0\end{cases}}\)

a: \(x+5\sqrt{x}-6=0\)

\(\Leftrightarrow\sqrt{x}-1=0\)

hay x=1

b: \(x-\sqrt{x}+\dfrac{1}{4}=0\)

\(\Leftrightarrow\sqrt{x}-\dfrac{1}{2}=0\)

hay \(x=\dfrac{1}{4}\)

a: \(x+5\sqrt{x}-6=0\)

\(\Leftrightarrow\sqrt{x}-1=0\)

hay x=1

b: \(x-\sqrt{x}+\dfrac{1}{4}=0\)

\(\Leftrightarrow\sqrt{x}-\dfrac{1}{2}=0\)

hay \(x=\dfrac{1}{4}\)

a: \(x+5\sqrt{x}-6=0\)

\(\Leftrightarrow\sqrt{x}-1=0\)

hay x=1

b: \(x-\sqrt{x}+\dfrac{1}{4}=0\)

\(\Leftrightarrow\sqrt{x}-\dfrac{1}{2}=0\)

hay \(x=\dfrac{1}{4}\)