Ccá bạn giải dùm mình phương trình cuối
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QD
quynh do
9 tháng 2 2018
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Những câu hỏi liên quan
NT
9 tháng 2 2018
\(x^3-6x^2+11x-6=0\\ \Leftrightarrow\left(x^3-x^2\right)-\left(5x^2-5x\right)+\left(6x-6\right)=0\\ \Leftrightarrow x^2\left(x-1\right)-5x\left(x-1\right)+6\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-5x+6\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=3\end{matrix}\right.\)
8 tháng 2 2018
x3-3x2+4=0
⇔x3+x2-4x2-4x+4x+4=0
⇔(x3+x2)-(4x2+4x)+(4x+4)=0
⇔x2(x+1)-4x(x+1)+4(x+1)=0
⇔(x+1)(x2-4x+4)=0
⇔(x+1)(x-2)2=0
=>\(\left\{{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
vậy S={-1;2}
QT
8 tháng 2 2018
\(x^3+4x^2+x+6=0\)
\(\Leftrightarrow\text{ (x + 3).(x + 2).(x - 1) = 0 }\)
<=>
Tự làm nhé mk nhẩm ko nhầm là dậy :D
8 tháng 2 2018
\(x^3+4x^2+x-6=x^3-x^2+5x^2-5x+6x-6=x^2\left(x-1\right)+5x\left(x-1\right)+6\left(x-1\right)=\left(x-1\right)\left(x^2+5x+6\right)=\left(x-1\right)\left(x^2+2x+3x+6\right)=\left(x-1\right)\left(x+2\right)\left(x+3\right)\)
chúc bạn học tốt
QD
3
GH
8 tháng 2 2018
\(x^3-6x^2-x+6=0\)
\(\Leftrightarrow x^2\left(x-6\right)-\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x-1=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=1\\x=-1\end{matrix}\right.\)
Vậy ................................
8 tháng 2 2018
x3-6x2+11x-6=0
⇔x3-x2-5x2+5x+6x-6=0
⇔(x3-x2)-(5x2-5x)+(6x-6)=0
⇔x2(x-1)-5x(x-1)+6(x-1)=0
⇔(x-1)(x2-5x+6)=0
⇔(x-1)(x2-2x-3x+6)=0
⇔(x-1)[(x2-2x)-(3x-6)]=0
⇔(x-1)[x(x-2)-3(x-2)]=0
⇔(x-1)(x-2)(x-3)=0
=>\(\left\{{}\begin{matrix}x-1=0\\x-2=0\\x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=2\\x=3\end{matrix}\right.\)
Vậy S={1;2;3}
KK
8 tháng 2 2018
x3-4x2+x+6=0
⇔x3+x2-5x2-5x+6x+6=0
⇔(x3+x2)-(5x2-5x)+(6x+6)=0
⇔x2(x+1)-5x(x+1)+6(x+1)=0
⇔(x+1)(x2-5x+6)=0
⇔(x+1)(x2-2x-3x+6)=0
⇔(x+1)[(x2-2x)-(3x-6)]=0
⇔(x+1)[x(x-2)-3(x-2)]=0
⇔(x+1)(x-2)(x-3)=0
⇔\(\left[{}\begin{matrix}x+1=0\\x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\\x=3\end{matrix}\right.\)
vậy S={-1;2;3}
TA
8 tháng 2 2018
Dat x2+2x+2=a (a>0)
pt<=> \(\dfrac{a-1}{a}+\dfrac{a}{a+1}=\dfrac{7}{6}\)
=> \(\dfrac{\left(a-1\right)\left(a+1\right)}{a\left(a+1\right)}+\dfrac{a.a}{a\left(a+1\right)}=\dfrac{7}{6}\)
=> \(\dfrac{a^2-1}{a\left(a+1\right)}+\dfrac{a^2}{a\left(a+1\right)}=\dfrac{7}{6}\)
=> (2a2-1).6=7a(a+1)
=> 12a2-6=7a2+7a
=> 5a2-7a-6=0
KN
8 tháng 2 2018
\(\dfrac{x^2+2x+1}{x^2+2x+2}+\dfrac{x^2+2x+2}{x^2+2x+3}=\dfrac{7}{6}\)
Đặt x2 + 2x + 1 = t, ta có:
\(\dfrac{t}{t+1}+\dfrac{t+1}{t+2}=\dfrac{7}{6}\)
\(\Leftrightarrow\)\(\dfrac{t\left(t+2\right)}{\left(t+1\right)\left(t+2\right)}+\dfrac{\left(t+1\right)^2}{\left(t+2\right)\left(t+1\right)}=\dfrac{7}{6}\)
\(\Leftrightarrow\) \(\dfrac{t^2+2t}{t^2+3t+2}+\dfrac{t^2+2t+1}{t^2+3t+2}=\dfrac{7}{6}\)
\(\Leftrightarrow\)\(\dfrac{t^2+2t+t^2+2t+1}{t^2+3t+2}=\dfrac{7}{6}\)
\(\Leftrightarrow\)\(\dfrac{2t^2+4t+1}{t^2+3t+2}=\dfrac{7}{6}\)
\(\Leftrightarrow\)6(2t2+4t+1) = 7(t2 + 3t + 2)
\(\Leftrightarrow\) 12t2 + 24t + 6 = 7t2 + 21t + 14
\(\Leftrightarrow\) 12t2 + 24t + 6 - 7t2 - 21t - 14 = 0
\(\Leftrightarrow\) 5t2 + 3t - 8 = 0
\(\Leftrightarrow\) 5t2 - 5t + 8t - 8 = 0
\(\Leftrightarrow\) 5t(t - 1) + 8(t - 1) = 0
\(\Leftrightarrow\) (5t + 8)(t - 1) = 0
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5t+8=0\\t-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}t=-\dfrac{8}{5}\\t=1\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x^2+2x+1=-\dfrac{8}{5}\left(vôlívì:x^2+2x+1=\left(x+1\right)^2\ge0>-\dfrac{8}{5}\right)\\x^2+2x+1=1\end{matrix}\right.\)\(\Leftrightarrow\)x2 + 2x + 1 = 1
\(\Leftrightarrow\) x2 + 2x = 0
\(\Leftrightarrow\)x(x + 2) = 0
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
Vậy phương trình có n0 là S={-2;0}