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bạn tự làm đk nhé
pt <=> \(2\left(x^2-2x-2\right)=3\sqrt{\left(x+3\right)\left(x^2-x+1\right)}\\ \)
Đặt a=x^2-x+1
b=x+3
pt<=> \(2\left(a-b\right)=3\sqrt{ab}\)
\(2a-2b-3\sqrt{ab}=0\)
\(\left(2a-4\sqrt{ab}\right)+\left(\sqrt{ab}-2b\right)=0\)
\(2\sqrt{a}\left(\sqrt{a}-2\sqrt{b}\right)+\sqrt{b}\left(\sqrt{a}-2\sqrt{b}\right)=0\)
\(\left(a-2\sqrt{b}\right)\left(2\sqrt{a}+\sqrt{b}\right)=0\)
tới đây bạn tự giải nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x^2-3x+3\right)\left(x^2-2x+3\right)=2x^2\)
TH1 : \(x^2-3x+3=2x^2\Leftrightarrow-x^2-3x+3=0\)
\(\Delta=\left(-3\right)^2-4.\left(-1\right).3=9+15=21>0\)
Nên phương trình có 2 nghiệm phân biệt
\(x_1=\frac{3-\sqrt{21}}{2.\left(-1\right)}=\frac{3-\sqrt{21}}{-2}=\frac{-3-\sqrt{21}}{2}\)
\(x_2=\frac{3+\sqrt{21}}{2.\left(-1\right)}=\frac{3+\sqrt{21}}{-2}=\frac{-3+\sqrt{21}}{2}\)
TH2 : \(x^2-2x+3=2x^2\Leftrightarrow-x^2-2x+3=0\)
\(\Delta=\left(-2\right)^2-4.\left(-1\right).3=4+12=16>0\)
Nên phương trình có 2 nghiệm phân biệt
\(x_1=\frac{2-\sqrt{16}}{2.1}=\frac{2-4}{2}=-\frac{2}{2}=-1\)
\(x_2=\frac{2+\sqrt{16}}{2.1}=\frac{2+4}{2}=\frac{6}{2}=3\)
Thực hiện tiếp nha cj, cách này khá dài ...
Cách này nha.
\(\left(x^2-3x+3\right)\left(x^2-2x+3\right)=2x^2\)
\(x^4-5x^3+12x^2-15x+9=2x^3\)
\(x^4-5x^3+10x^2-15x+9=0\)
\(\left(x-1\right)\left(x^3-4x^2+6x-9\right)=0\)
TH1 : \(x-1=0\Leftrightarrow x=1\)
\(x^3-4x^2+6x-9=0\Leftrightarrow\left(x^2-x+3\right)\left(x-3\right)=0\)
TH2 : \(x-3=0\Leftrightarrow x=3\)
TH3 : \(x^2-x+3=0\)
\(\Delta=\left(-1\right)^2-4.1.3=1-12=-11< 0\)
Nên phuwong trình vô nghiệm
Vậy \(S=\left\{1;3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left\{{}\begin{matrix}xy\left(x+y\right)=2\\\left(x+y\right)^3-3xy\left(x+y\right)+\left(xy\right)^3+7\left(xy+x+y+1\right)=31\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)=2\\\left(x+y\right)^3+\left(xy\right)^3+7\left(xy+x+y\right)=30\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\) với \(u^2\ge4v\)
\(\Rightarrow\left\{{}\begin{matrix}uv=2\\u^3+v^3+7\left(u+v\right)=30\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\\left(u+v\right)^3-3uv\left(u+v\right)+7\left(u+v\right)=30\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\\left(u+v\right)^3+\left(u+v\right)-30=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\u+v=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u=2\\v=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=2\\xy=1\end{matrix}\right.\) \(\Leftrightarrow\left(x;y\right)=\left(1;1\right)\)
2.
ĐKXĐ: \(0\le x\le\dfrac{3}{2}\)
\(\Leftrightarrow9x\left(3-2x\right)+81+54\sqrt{x\left(3-2x\right)}=49x+25\left(3-2x\right)+70\sqrt{x\left(3-2x\right)}\)
\(\Leftrightarrow9x^2-14x-3+8\sqrt{x\left(3-2x\right)}=0\)
\(\Leftrightarrow9\left(x^2-2x+1\right)-4\left(3-x-2\sqrt{x\left(3-2x\right)}\right)=0\)
\(\Leftrightarrow9\left(x-1\right)^2-\dfrac{36\left(x-1\right)^2}{3-x+2\sqrt{x\left(3-2x\right)}}=0\)
\(\Leftrightarrow9\left(x-1\right)^2\left(1-\dfrac{4}{3-x+2\sqrt{x\left(3-2x\right)}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\3-x+2\sqrt{x\left(3-2x\right)}=4\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2\sqrt{x\left(3-2x\right)}=x+1\)
\(\Leftrightarrow4x\left(3-2x\right)=x^2+2x+1\)
\(\Leftrightarrow9x^2-10x+1=0\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{9}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(2x^2\left(2x^2+3\right)=2-x^2\)
Đặt \(x^2=a;a\ge0\)
`->` pt trở thành:
`<=>2a(2a+3)=2-a`
`<=>4a^2+7a-2=0`
\(\Leftrightarrow\left[{}\begin{matrix}a=\dfrac{1}{4}\left(tm\right)\\a=-2\left(ktm\right)\end{matrix}\right.\)
`=>`\(x=\pm\sqrt{\dfrac{1}{4}}=\pm\dfrac{1}{2}\)
Vậy \(S=\left\{\pm\dfrac{1}{2}\right\}\)
\(\Leftrightarrow4x^4+6x^2-2+x^2=0\)
\(\Leftrightarrow4x^4+5x^2-2=0\)(1)
Đặt \(x^2=a\left(a>=0\right)\)
(1) trở thành \(4a^2+5a-2=0\)(2)
\(\text{Δ}=5^2-4\cdot4\cdot\left(-2\right)=25+32=57>0\)
Do đó: Phương trình (2) có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}a_1=\dfrac{-5-\sqrt{57}}{8}\left(loại\right)\\a_2=\dfrac{-5+\sqrt{57}}{8}\left(nhận\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\pm\sqrt{\dfrac{\sqrt{57}-5}{8}}\)
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b: \(\Leftrightarrow\left(x^2-2x+1-1\right)^2-2\left(x-1\right)^2-1=0\)
\(\Leftrightarrow\left[\left(x-1\right)^2-1\right]^2-2\left(x-1\right)^2-1=0\)
\(\Leftrightarrow\left(x-1\right)^4-2\left(x-1\right)^2+1-2\left(x-1\right)^2-1=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left(x-3\right)\left(x+1\right)=0\)
hay \(x\in\left\{1;3;-1\right\}\)
a: \(\Leftrightarrow2x^3-3x-10=-2\left(8-12x+6x^2-x^3\right)\)
\(\Leftrightarrow2x^3-3x-10=-16+24x-12x^2+2x^3\)
\(\Leftrightarrow-3x-10+16-24x+12x^2=0\)
=>\(12x^2-27x+6=0\)
hay \(x\in\left\{2;\dfrac{1}{4}\right\}\)
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