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x(x+4)(x+6)(x+10)+128
=(x2+10x)(x2+10x+24)+128
=(x2+10x+12-12)(x2+10x+12+12)+128
=(x2+10x+12)2-122+128
=(x2+10x+12)2-16
=(x2+10x+12-16)(x2+10x+12+16)
=(x2+10x-4)(x2+10x+28)
![](https://rs.olm.vn/images/avt/0.png?1311)
x(x+4)(x+6)(x+10)+128
=x.(x+10)(x+4)(x+6)+128
=(x2+10x)(x2+10x+24)+128
=(x2+10x)[(x2+10x)+24]+128
=(x2+10x)2+24.(x2+10x)+128
=(x2+10x)2+8.(x2+10x)+16.(x2+10x)+128
=(x2+10x)(x2+10x+8)+16.(x2+10x+8)
=(x2+10x+8)(x2+10x+16)
=(x2+10x+8)(x2+2x+8x+16)
=(x2+10x+8)[x.(x+2)+8.(x+2)]
=(x2+10x+8)(x+2)(x+8)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128=0\)
\(\Leftrightarrow x\left(x+10\right)\left(x+4\right)\left(x+6\right)+128=0\)
\(\Leftrightarrow\left(x^2+10x\right)\left(x^2+10x+24\right)+128=0\)
Đặt \(x^2+10x+12=t\)
\(\Rightarrow\left(t-12\right)\left(t+12\right)+128=0\)
\(\Leftrightarrow t^2-144+128=0\)\(\Leftrightarrow t^2-16=0\)
\(\Leftrightarrow\left(t-4\right)\left(t+4\right)=0\)\(\Leftrightarrow\left(x^2+10x+12-4\right)\left(x^2+10x+12+4\right)=0\)
\(\Leftrightarrow\left(x^2+10x+8\right)\left(x^2+10x+16\right)=0\)
\(\Leftrightarrow\left(x^2+10x+8\right)\left(x+2\right)\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-8\end{cases}}\)
Vậy tập nghiệm của phương trình là \(S=\left\{-8;-2\right\}\)
Ta có : \(x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128=0\)
\(\Leftrightarrow\left(x^2+10x\right)\left(x^2+10x+24\right)+128=0\) (2)
Đặt \(x^2+10x=t\) Khi đó pt (2) có dạng :
\(t\cdot\left(t+24\right)+128=0\)
\(\Leftrightarrow t^2+24t+128=0\)
\(\Leftrightarrow\left(t+12\right)^2-16=0\)
\(\Leftrightarrow\left(t+12-4\right)\left(t+12+4\right)=0\)
\(\Leftrightarrow\left(t+8\right)\left(t+16\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}t+8=0\\t+16=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}t=-8\\t=-16\end{cases}}\)
+) Với \(t=-8\) thì \(x^2+10x=-8\)
\(\Leftrightarrow\left(x+5\right)^2=17\)
\(\Leftrightarrow\orbr{\begin{cases}x+5=\sqrt{17}\\x+5=-\sqrt{17}\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=-5+\sqrt{17}\\x=-5-\sqrt{17}\end{cases}}\) ( thỏa mãn )
+) Với \(t=-16\) thì \(x^2+10x=-16\)
\(\Leftrightarrow\left(x+5\right)^2-9=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+14\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x+14=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=4\\x=-14\end{cases}}\) ( thỏa mãn )
Vậy : phương trình đã cho có tập nghiệm \(S=\left\{-5\pm\sqrt{17},4,-14\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128\)
\(=x\left(x+10\right)\left(x+4\right)\left(x+6\right)+128\)
\(=\left(x^2+10x\right)\left(x^2+10x+24\right)+128\)
\(=\left(x^2+10x\right)^2+24\left(x^2+10x\right)+128\)
\(=\left(x^2+10x\right)^2+2.\left(x^2+10x\right).12+12^2-16\)
\(=\left(x^2+10x+12\right)^2-4^2\)
\(=\left(x^2+10x+12-4\right) \left(x^2+10x +12+4\right)\)
\(=\left(x^2+10x-8\right)\left(x^2+10x+16\right)\)
\(=\left(x^2+10x-8\right)\left(x^2+2x+8x+16\right)\)
\(=\left(x^2+10x-8\right)\left[x\left(x+2\right)+8\left(x+2\right)\right]\)
\(=\left(x^2+10x-8\right)\left(x+2\right)\left(x+8\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A= x(x+4)(x+6)(x+10) +128
=[(x(x+10)] [(x+4)(x+6)] +128
=(x^2+10)(x^2+10+24)+128
Đặt: x^2+10+12=y
Ta có: A=(y+12)(y-12)+128
=(y^2-12^2)+128
=y^2-12^2+128
=y^2-16
=y^2-4^2
=(y-4)(y+4)
Thay vào bt A ta có:A= ( x^2+10x+12-4)(x^2+10x+12+4)
=(x^2+10x+8)(x^2+10x+16)
=(x^2+10x+8)(x+8)(x+2)x
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ta có
\(x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128=\left(x^2+10x\right)\left(x^2+10x+24\right)+128\)
\(=\left[\left(x^2+10x+12\right)-12\right]\left[\left(x^2+10x+12\right)+12\right]+128\)
\(=\left(x^2+10x+12\right)^2-12^2+128=\left(x^2+10x+12\right)^2-16\)
\(=\left(x^2+10x+12-4\right)\left(x^2+10x+12+4\right)=\left(x^2+10x+8\right)\left(x^2+10x+16\right)\)
\(=\left(x+2\right)\left(x+8\right)\left(x^2+10x+8\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x . ( x + 4 ) . ( x + 6 ) . ( x + 10 ) + 128
= ( x2 + 10x ) . ( x2 + 10x + 24 ) + 128
đặt x2 + 10x + 12 = y, đa thức đã cho có dạng :
( y - 12 ) . ( y + 12 ) + 128 = y2 - 16 = ( y - 4 ) . ( y + 4 )
= ( x2 + 10x + 16 ) . ( x2 + 10x + 8 ) = ( x + 2 ) . ( x + 8 ) . ( x2 + 10x + 8 )