a)
\(\left(x^3-7x^2+14.x\right)-\left(x^2-7x+14\right)\)
\(x\left(x^2-7x+14\right)-\left(x^2-7x+14\right)=\left(x^2-7x+14\right)\left(x-1\right)\)\(\left[\left(x-\dfrac{7}{2}\right)^2+\dfrac{7}{4}\right]\left(x-1\right)=0\)
\(\left[{}\begin{matrix}\left(x-\dfrac{7}{2}\right)^2+\dfrac{7}{4}>0vnghiem\\x-1=0=>x=1\end{matrix}\right.\) Kết luận x=1
a,x3 - 8x2 + 21x -14 = 0
\(\Leftrightarrow\)x3-x2-7x2+7x+14x-14=0
\(\Leftrightarrow\left(x^3-x^2\right)-\left(7x^2-7x\right)+\left(14x-14\right)=\)0
\(\Leftrightarrow x^2\left(x-1\right)-7x\left(x-1\right)+14\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-7x+14\right)\)=0
\(\Leftrightarrow\)x-1=0\(\Leftrightarrow\)x=1
vì x2-7x+14=(x2-2.\(\dfrac{7}{2}\)x+\(\dfrac{49}{4}\))+\(\dfrac{7}{4}\)=(x-\(\dfrac{7}{2}\))2+\(\dfrac{7}{4}\)\(\ge\dfrac{7}{4}\forall x\)
vậy pt có một nghiệm duy nhất là x=1
a) x3-8x2+21x-14=0
<=>(x3-x2)-(7x2-7x)+(14x-14)=0
<=>x2(x-1)-7x(x-1)+14(x-1)=0
<=>(x-1)(x2-7x+14)=0(1)
Vì \(x^2-7x+14=\left(x-\dfrac{7}{2}\right)^2+\dfrac{7}{4}>0\) với mọi x
nên (1)<=>x-1=0
<=>x=1.
Vậy S={1}
b, ( x+4 ) (x + 5) (x + 7) (x + 8) = 4
\(\Leftrightarrow\)(x+4)(x+8)(x+5)(x+7)=4
\(\Leftrightarrow\)(x2+12x+32)(x2+12x+35)=4
đặt x2+12x+32=a ta có
a(a+3)-4=0\(\Leftrightarrow\)a2+3a-4=0\(\Leftrightarrow\)(a2-a)+(4a-4)=0\(\Leftrightarrow\)(a-1)(a+4)=0\(\Leftrightarrow\)\(\left[{}\begin{matrix}a-1=0\\a+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=1\\a=-4\end{matrix}\right.\)
với a=1 ta có x2+12x+32=1\(\Leftrightarrow\)x2+12x+31=0\(\Leftrightarrow\)(x2+12x+36)-5=0\(\Leftrightarrow\)(x+6)2-5=0\(\Leftrightarrow\)(x+6-\(\sqrt{5}\))(x+6+\(\sqrt{5}\))=0\(\Leftrightarrow\left[{}\begin{matrix}x+6-\sqrt{5}=0\\x+6+\sqrt{5}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6+\sqrt{5}\\x=-6-\sqrt{5}\end{matrix}\right.\)
với x=-4 ta có x2+12x+32=-4\(\Leftrightarrow\)x2+2.6x+36=0\(\Leftrightarrow\)(x+6)2=0\(\Leftrightarrow\)x+6=0\(\Leftrightarrow\)x=-6
vậy pt có tập nghiệm là S=\(\left\{-6+\sqrt{5};-6-\sqrt{5};-6\right\}\)
