Giải pt:
\(x^2\left(x-1\right)^2=\left(2x-1\right)^2+2\)
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12 tháng 3 2023
\(a.x^2+\dfrac{1}{x^2}=x+\dfrac{1}{x}\) ( ĐKXĐ : \(x\ne0\) )
\(\Leftrightarrow x^2+\dfrac{1}{x^2}-x-\dfrac{1}{x}=0\Leftrightarrow\left(x^2-\dfrac{1}{x}\right)+\left(\dfrac{1}{x^2}-x\right)=0\)
\(\Leftrightarrow-x\left(\dfrac{1}{x^2}-x\right)+\left(\dfrac{1}{x^2}-x\right)=0\Leftrightarrow\left(\dfrac{1}{x^2}-x\right)\left(1-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}1-x=0\\\dfrac{1}{x^2}-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\1-x^3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(1-x\right)\left(1+x+x^2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=1\end{matrix}\right.\Leftrightarrow x=1\) ( x2 + x + 1 loại nhé nếu phân tích ra thì ta được \(x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\in R\) )
Vậy \(S=\left\{1\right\}\)
b, \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=24\)
\(\Leftrightarrow x\left(x+3\right).\left(x+1\right)\left(x+2\right)-24=0\)
\(\Leftrightarrow\left(x^2+3x\right)\left(x^2+3x+2\right)-24=0\)
\(\Leftrightarrow\left(x^2+3x+1-1\right)\left(x^2+3x+1+1\right)-24=0\)
\(\Leftrightarrow\left(x^2+3x+1\right)-1-24=0\Leftrightarrow\left(x^2+3x+1\right)-25=0\)
\(\Leftrightarrow\left(x^2+3x+1-5\right)\left(x^2+3x+1+5\right)=0\Leftrightarrow\left(x^2+3x-4\right)\left(x^2+3x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+3x-4=0\\x^2+3x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x+4\right)=0\\\left(x+\dfrac{3}{2}\right)^2+\dfrac{15}{4}\ge\dfrac{15}{4}\forall x\in R\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
Vậy \(S=\left\{-4;1\right\}\)
e, \(\left(x^2+x+1\right)-2x^2-2x=5\Leftrightarrow\left(x^2+x+1\right)-2x^2-2x-2-3=0\)
\(\Leftrightarrow\left(x^2+x+1\right)-2\left(x^2+x+1\right)-3=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left(x^2+x-1\right)-3=0< =>\left(x^2+x\right)^2-4=0\)
\(\Leftrightarrow\left(x^2+x-2\right)\left(x^2+x+2\right)=0\)
\(\Leftrightarrow x^2+x-2=0\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\) ( x^2 + x + 2 loại nhé y như mấy câu trên luôn khác 0 ! )
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy \(S=\left\{-2;1\right\}\)
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 11 2021
Chú ý:
\(\left(x^2+2x\right)^2+4\left(x+1\right)^2=\left(x^2+2x\right)^2+4\left(x^2+2x+1\right)=\left(x^2+2x\right)^2+4\left(x^2+2x\right)+4\)
\(=\left(x^2+2x+2\right)^2\)
\(x^2+\left(x+1\right)^2+\left(x^2+x\right)^2\)
\(=\left(x^2+x\right)+x^2+x^2+2x+1\)
\(=\left(x^2+x\right)^2+2x^2+2x+1\)
\(=\left(x^2+x\right)^2+2\left(x^2+x\right)+1\)
\(=\left(x^2+x+1\right)^2\)
HH
1
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 8 2021
\(\Leftrightarrow x^4=\left(1-x\right)\left(x^2+2x-2-4x+4\right)\)
\(\Leftrightarrow x^4=\left(1-x\right)\left(x^2+2x-2\right)+\left(2x-2\right)^2\)
\(\Leftrightarrow x^4-\left(2x-2\right)^2+\left(x-1\right)\left(x^2+2x-2\right)=0\)
\(\Leftrightarrow\left(x^2-2x+2\right)\left(x^2+2x-2\right)+\left(x-1\right)\left(x^2+2x-2\right)=0\)
\(\Leftrightarrow\left(x^2+2x-2\right)\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left(x^2+2x-2\right)\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]=0\)
\(\Leftrightarrow x^2+2x-2=0\) (bấm máy)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 7 2021
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-1\right)\left(x+y-2\right)=6\\\left(x-1\right)^2+\left(y-1\right)^2=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-1\right)\left(x+y-2\right)=6\\\left(x+y-2\right)^2-2\left(x-1\right)\left(y-1\right)=5\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\left(x-1\right)\left(y-1\right)=v\\x+y-2=u\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}uv=6\\u^2-2v=5\end{matrix}\right.\) \(\Rightarrow u^2-\dfrac{12}{u}=5\)
\(\Rightarrow u^3-5u-12=0\)
\(\Leftrightarrow\left(u-3\right)\left(u^2+3u+4\right)=0\)
\(\Leftrightarrow u=3\Rightarrow v=2\)
\(\Rightarrow\left\{{}\begin{matrix}x+y-2=3\\\left(x-1\right)\left(y-1\right)=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=5-x\\\left(x-1\right)\left(y-1\right)=2\end{matrix}\right.\)
\(\Rightarrow\left(x-1\right)\left(5-x-1\right)=2\)
\(\Leftrightarrow...\) em tự hoàn thành bài toán
NM
1
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 4 2021
ĐKXĐ: ...
\(\left(\dfrac{x-1}{x+2}\right)^2-4\left(\dfrac{x+2}{x-3}\right)^2+3\left(\dfrac{x-1}{x-3}\right)=0\)
Đặt \(\left\{{}\begin{matrix}\dfrac{x-1}{x+2}=a\\\dfrac{x+2}{x-3}=b\end{matrix}\right.\)
\(\Rightarrow a^2-4b^2+3ab=0\Leftrightarrow\left(a-b\right)\left(a+4b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a-b=0\\a+4b=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x-1}{x+2}-\dfrac{x+2}{x-3}=0\\\dfrac{x-1}{x+2}+\dfrac{4x+8}{x-3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x-3\right)-\left(x+2\right)^2=0\\\left(x-\right)\left(x-3\right)+4\left(x+2\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
\(x^2\left(x-1\right)^2=\left(2x-1\right)^2+2\)
\(\Leftrightarrow x^2\left(x-1\right)^2=\left[x+\left(x-1\right)\right]^2+2\)
\(\Leftrightarrow x^2\left(x-1\right)^2=4x^2-4x+1+2\)
\(\Leftrightarrow x^2\left(x-1\right)^2-4x\left(x-1\right)-3=0\) (1)
Đặt \(a=x\left(x-1\right)\)
\(x\left(x-1\right)=x^2-x=\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
\(ĐK:a^2\ge2\Leftrightarrow\left|a\right|\ge\sqrt{2}\)
\(\left(1\right)\Leftrightarrow a^2-4a-3=0\)
\(\Delta=\left(-4\right)^2-4.\left(-3\right)=16+12=28>0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{4+\sqrt{28}}{2}=2+\sqrt{7}\left(tm\right)\\a=\dfrac{4-\sqrt{28}}{2}=2-\sqrt{7}\left(ktm\right)\end{matrix}\right.\)
\(\rightarrow x\left(x-1\right)=2+\sqrt{7}\)
\(\Leftrightarrow x^2-x-\left(2+\sqrt{7}\right)=0\)
\(\Leftrightarrow x=\dfrac{1\pm\sqrt{9+4\sqrt{7}}}{2}\)
Vậy \(S=\left\{\dfrac{1\pm\sqrt{9+4\sqrt{7}}}{2}\right\}\)
=)) khai triển HĐT sai gòi:v \(\left(2x-1\right)^2=4x^2-4x+1\) nha bạn