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NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 11 2021
Chú ý:
\(\left(x^2+2x\right)^2+4\left(x+1\right)^2=\left(x^2+2x\right)^2+4\left(x^2+2x+1\right)=\left(x^2+2x\right)^2+4\left(x^2+2x\right)+4\)
\(=\left(x^2+2x+2\right)^2\)
\(x^2+\left(x+1\right)^2+\left(x^2+x\right)^2\)
\(=\left(x^2+x\right)+x^2+x^2+2x+1\)
\(=\left(x^2+x\right)^2+2x^2+2x+1\)
\(=\left(x^2+x\right)^2+2\left(x^2+x\right)+1\)
\(=\left(x^2+x+1\right)^2\)
14 tháng 5 2022
\(x^2\left(x-1\right)^2=\left(2x-1\right)^2+2\)
\(\Leftrightarrow x^2\left(x-1\right)^2=\left[x+\left(x-1\right)\right]^2+2\)
\(\Leftrightarrow x^2\left(x-1\right)^2=4x^2-4x+1+2\)
\(\Leftrightarrow x^2\left(x-1\right)^2-4x\left(x-1\right)-3=0\) (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 \(S=\left\{\dfrac{1\pm\sqrt{9+4\sqrt{7}}}{2}\right\}\)
14 tháng 5 2022
=)) khai triển HĐT sai gòi:v \(\left(2x-1\right)^2=4x^2-4x+1\) nha bạn
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)
5 tháng 10 2021
\(ĐK:-1\le x\le1\\ PT\Leftrightarrow13\left(1-2x^2\right)\sqrt{\left(1-x^2\right)\left(1+x^2\right)}+9\left(1+2x^2\right)\sqrt{\left(1+x^2\right)\left(1-x^2\right)}=0\\ \Leftrightarrow\sqrt{1-x^4}\left(13-26x^2+9+18x^2\right)=0\\ \Leftrightarrow\sqrt{1-x^4}\left(22-8x^2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}1-x^4=0\\22-8x^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(1+x^2\right)\left(1-x\right)\left(1+x\right)=0\\x^2=\dfrac{22}{8}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\\\left[{}\begin{matrix}x=\dfrac{\sqrt{11}}{2}\left(ktm\right)\\x=-\dfrac{\sqrt{11}}{2}\left(ktm\right)\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
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
TA
0
P
1
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 3 2022
\(\Leftrightarrow\left\{{}\begin{matrix}4x^2-2y^2=2\\xy+x^2=2\end{matrix}\right.\)
Trừ vế cho vế:
\(\Rightarrow3x^2-xy-2y^2=0\)
\(\Leftrightarrow\left(x-y\right)\left(3x+2y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=x\\y=-\dfrac{3}{2}x\end{matrix}\right.\)
Thế vào pt đầu: \(\left[{}\begin{matrix}2x^2-x^2=1\\2x^2-\left(-\dfrac{3}{2}\right)x^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2=1\\-\dfrac{1}{4}x^2=1\left(vn\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=y=1\\x=y=-1\end{matrix}\right.\)
T
1
Đặt: \(x^2-2x+1=t\left(t\ge0\right)\)
pt <=> \(\left(t+1\right)\left(t-2\right)=2\)
\(\Leftrightarrow t^2-t-4=0\)
\(\Leftrightarrow\left(t^2-2\cdot\dfrac{1}{2}t+\dfrac{1}{4}\right)-\dfrac{17}{4}=0\)
\(\Leftrightarrow\left(t-\dfrac{1}{2}\right)^2=\dfrac{17}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}t-\dfrac{1}{2}=\dfrac{\sqrt{17}}{2}\\t-\dfrac{1}{2}=-\dfrac{\sqrt{17}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{1+\sqrt{17}}{2}\left(tm\right)\\t=\dfrac{1-\sqrt{17}}{2}\left(ktm\right)\end{matrix}\right.\)
Ta có: \(t=\dfrac{1+\sqrt{17}}{2}\)
\(\Rightarrow x^2-2x+1=\dfrac{1+\sqrt{17}}{2}\)
\(\Leftrightarrow\left(x-1\right)^2=\dfrac{1+\sqrt{17}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=\sqrt{\dfrac{1+\sqrt{17}}{2}}\\x-1=-\sqrt{\dfrac{1+\sqrt{17}}{2}}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1+\sqrt{\dfrac{1+\sqrt{17}}{2}}\\x=1-\sqrt{\dfrac{1+\sqrt{17}}{2}}\end{matrix}\right.\)
Vậy pt có 2 nghiệm........................