giải phương trình
x\(^3\) - 9x\(^2\) + 19x-11=0
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22 tháng 3 2021
a) Ta có: \(x^3-9x^2+19x-11=0\)
\(\Leftrightarrow x^3-x^2-8x^2+8x+11x-11=0\)
\(\Leftrightarrow x^2\left(x-1\right)-8x\left(x-1\right)+11\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-8x+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2-8x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\sqrt{5}+4\\x=-\sqrt{5}+4\end{matrix}\right.\)
Vậy: \(S=\left\{1;\sqrt{5}+4;-\sqrt{5}+4\right\}\)
TI
1
BT
19 tháng 3 2018
a) \(x^5-27+x^3-27x^2\) = 0
\(\Leftrightarrow x^3\left(x^2+1\right)-27\left(x^2+1\right)\)= 0
\(\Leftrightarrow\left(x^2+1\right)\left(x^3-27\right)=0\)
\(\Leftrightarrow x^3-27=0\) (Vì \(x^2+1>0\))
\(\Leftrightarrow\left(x-3\right)\left(x^2+3x+9\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+2\dfrac{3}{2}x+\dfrac{9}{4}+\dfrac{27}{4}\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[\left(x+\dfrac{3}{2}\right)^2+\dfrac{27}{4}\right]=0\)
\(\Leftrightarrow x-3=0\) (Vì \(\left(x+\dfrac{3}{2}\right)^2+\dfrac{27}{4}>0\))
\(\Leftrightarrow x=3\)
Vậy tập nghiệm của phương trình là S = {3}
b)\(x^3-9x^2+19x-11=0\)
\(\Leftrightarrow\left(x^3-x^2\right)-\left(8x^2-8x\right)+\left(11x-11\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)-8x\left(x-1\right)+11\left(x-1\right)=0\)
\(\Leftrightarrow\)\(\left(x-1\right)\left(x^2-8x+11\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2-\left(4+\sqrt{5}\right)x-\left(4-\sqrt{5}\right)x+11\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left\{x\left[x-\left(4+\sqrt{5}\right)\right]-\left(4-\sqrt{5}\right)\left[x-\left(4+\sqrt{5}\right)\right]\right\}=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-4-\sqrt{5}\right)\left(x-4+\sqrt{5}\right)=0\)
\(\Leftrightarrow x-1=0\) hoặc \(x-4-\sqrt{5}=0\) hoặc \(x-4+\sqrt{5}=0\)
\(\Leftrightarrow x=1\) hoặc \(x=4+\sqrt{5}\) hoặc \(x=4-\sqrt{5}\)
Vậy phương trình có tập nghiệm là \(S=\left\{1;4+\sqrt{5};4-\sqrt{5}\right\}\)
BB
6 tháng 3 2018
x^3 - 9X^2 +19x -11 =0
<=> (x^3 - x^2) - (8x^2 - 8x) +(11x-11)=0
<=> x^2(x-1) - 8x(x-1) + 11(x-1)=0
<=> (x-1)(x^2-8x+11) = 0
<=> x-1=0
<=> x=1
TQ
1
TM
16 tháng 4 2021
a, \(x^2\)≥1
\(\Leftrightarrow\) x>1
b, \(x^2\)<1
\(\Rightarrow\) x∈∅
c, \(x^2\)+3x ≥ 0
\(\Leftrightarrow\) \(x^2\)≥-3x
\(\Leftrightarrow\) x≥-3
d, \(x^2\)+3x+3≥0
\(\Leftrightarrow\) \(\left(x+\dfrac{3}{2}\right)^2\)+\(\dfrac{3}{4}\)≥0+\(\dfrac{3}{4}\)
\(\Leftrightarrow\) \(x^2\)+\(\dfrac{3}{2}^2\)≥0
\(\Leftrightarrow\)\(x^2\)≥\(\dfrac{9}{4}\)
\(\Leftrightarrow\)x≥\(\dfrac{3}{2}\)
TK
25 tháng 5 2021
\(x-4\sqrt{x-2}+1=0\)(Đk x>2)
⇔\(x-2-4\sqrt{x-2}+4-1=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-2\right)^2-1=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-3\right)\left(\sqrt{x-2}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}-3=0\\\sqrt{x-2}-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=3\\\sqrt{x-2}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=9\\x-2=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=11\\x=3\end{matrix}\right.\)(thảo đk)
Vậy\(\left[{}\begin{matrix}x=11\\x=3\end{matrix}\right.\)là nghiệm của pt
PH
1
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 5 2021
\(\left\{{}\begin{matrix}x^2+xy+y^2=1\\x-y-xy=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2+3xy=1\\x-y-xy=3\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x-y=u\\xy=v\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u^2+3v=1\\u-v=3\end{matrix}\right.\)
\(\Rightarrow u^2+3\left(u-3\right)=1\)
\(\Leftrightarrow u^2+3u-10=0\Rightarrow\left[{}\begin{matrix}u=2\Rightarrow v=-1\\u=-5\Rightarrow v=-8\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}u=2\\v=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x-y=2\\xy=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=x-2\\xy=-1\end{matrix}\right.\)
\(\Rightarrow x\left(x-2\right)=-1\Leftrightarrow\left(x-1\right)^2=0\Rightarrow x=1\Rightarrow y=-1\)
TH2: \(\left\{{}\begin{matrix}u=-5\\v=-8\end{matrix}\right.\) \(\Rightarrow...\) bạn tự làm tương tự
\(\Leftrightarrow x^3-x^2-8x^2+8x+11x-11=0\)
\(\Leftrightarrow x^2\left(x-1\right)-8x\left(x-1\right)+11\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-8x+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2-8x+11=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4-\sqrt{5}\\x=4+\sqrt{5}\end{matrix}\right.\)