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ĐKXĐ: bla bla bla
\(3x\left(x-2\right)\sqrt{3x-1}=2\left(x^3-5x^2+7x-2\right)\)
\(\Leftrightarrow3x\left(x-2\right)\sqrt{3x-1}=2\left(x-2\right)\left(x^2-3x+1\right)\)
TH1: \(x=2\)
TH2: \(3x\sqrt{3x-1}=2\left(x^2-3x+1\right)\)
Đặt \(\sqrt{3x-1}=t\ge0\)
\(\Rightarrow3tx=2\left(x^2-t^2\right)\)
\(\Leftrightarrow2x^2-3tx-2t^2=0\)
\(\Leftrightarrow\left(2x+t\right)\left(x-2t\right)=0\)
\(\Rightarrow x=2t\)
\(\Leftrightarrow x=2\sqrt{3x-1}\)
\(\Leftrightarrow x^2=4\left(3x-1\right)\)
\(\Leftrightarrow x^2-12x+4=0\)
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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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![](https://rs.olm.vn/images/avt/0.png?1311)
vào câu hỏi tương tự nhé bạn, với lại mình chưa học lớp 9
\(1\text{) }a=\sqrt{2x^2-4x+3}\Rightarrow x^2-2x=\frac{a^2-3}{2}\)
Pt trở thành \(\frac{a^2-3}{2}+3=2a\)
\(3\text{) }pt\Leftrightarrow2\left(x^2-2x+4\right)+\left(x+2\right)=3\sqrt{x+2}\sqrt{x^2-2x+4}\)
\(\Leftrightarrow\left(2\sqrt{x^2-2x+4}+\sqrt{x+2}\right)\left(\sqrt{x^2-2x+1}-\sqrt{x+2}\right)=0\)
![](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)
Bài 1:
\(\left\{{}\begin{matrix}x+2y=1\\2x^2-5xy=48\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=1-2y\left(1\right)\\2x^2-5xy=48\left(2\right)\end{matrix}\right.\)
Thay (1) vào (2)\(\Leftrightarrow2\left(1-2y\right)^2-5\left(1-2y\right)y=48\Leftrightarrow2\left(1-4y+4y^2\right)-5y+10y^2=48\Leftrightarrow2-8y+8y^2-5y+10y^2=48\Leftrightarrow18y^2-13y-46=0\Leftrightarrow\left(y-2\right)\left(18y+23\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}y=2\\y=-\frac{23}{18}\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-3\\x=\frac{32}{9}\end{matrix}\right.\)
Vậy (x;y)={(\(-3;2\));(\(\frac{32}{9};-\frac{23}{18}\))}
Bài 2:
a) Đặt a=x2-1(a\(\ge-1\))
Vậy pt\(\Leftrightarrow a^2-4a=5\Leftrightarrow a^2-4a-5=0\Leftrightarrow\left(a-5\right)\left(a+1\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}a=5\\a=-1\end{matrix}\right.\)(tm)
TH1: a=5\(\Leftrightarrow x^2-1=5\Leftrightarrow x^2=6\Leftrightarrow x=\pm\sqrt{6}\)
TH2: a=-1\(\Leftrightarrow x^2-1=-1\Leftrightarrow x^2=0\Leftrightarrow x=0\)
Vậy S={\(-\sqrt{6};0;\sqrt{6}\)}
b) \(\left(x+2\right)^2-3x-5=\left(1-x\right)\left(1+x\right)\Leftrightarrow x^2+4x+4-3x-5=1-x^2\Leftrightarrow2x^2+x-2=0\Leftrightarrow\)\(\left[{}\begin{matrix}x=\frac{-1+\sqrt{17}}{4}\\x=\frac{-1-\sqrt{17}}{4}\end{matrix}\right.\)
Vậy S={\(\frac{-1+\sqrt{17}}{4};\frac{-1-\sqrt{17}}{4}\)}
c) Đặt a=\(x^2-3x+2\)
Vậy pt\(\Leftrightarrow\left(a+2\right)a=3\Leftrightarrow a^2+2a-3=0\Leftrightarrow\left(a-1\right)\left(a+3\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}a=1\\a=-3\end{matrix}\right.\)(tm)
TH1:\(a=1\Leftrightarrow x^2-3x+2=1\Leftrightarrow x^2-3x+1=0\Leftrightarrow\)\(\left[{}\begin{matrix}x=\frac{3+\sqrt{5}}{2}\\x=\frac{3-\sqrt{5}}{2}\end{matrix}\right.\)
TH2: a=-3\(\Leftrightarrow x^2-3x+2=-3\Leftrightarrow x^2-3x+5=0\)(vô nghiệm)
Vậy S=\(\left\{\frac{3+\sqrt{5}}{2};\frac{3-\sqrt{5}}{2}\right\}\)
\(\left(x^2-3x+3\right)\left(x^2-2x+3\right)=2x^2\)
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 \(S=\left\{1;3\right\}\)