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![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\) Đặt \(A=\dfrac{3x^2-2x+3}{x^2+1}\Leftrightarrow Ax^2+A=3x^2-2x+3\)
\(\Leftrightarrow x^2\left(A-3\right)-2x+A-3=0\)
Coi đây là PT bậc 2 ẩn x, PT có nghiệm
\(\Leftrightarrow\Delta=4-4\left(A-3\right)^2\ge0\\ \Leftrightarrow\left(A-3\right)^2\le1\Leftrightarrow2\le A\le4\)
Vậy \(A_{min}=4\Leftrightarrow\dfrac{3x^2-2x+3}{x^2+1}=4\Leftrightarrow x=...\)
\(b,\) Đặt \(B=\dfrac{3x^2-4x+4}{x^2+2}\Leftrightarrow Bx^2+2B=3x^2-4x+4\)
\(\Leftrightarrow x^2\left(B-3\right)+4x+2B-4=0\)
Coi đây là PT bậc 2 ẩn x, PT có nghiệm
\(\Leftrightarrow\Delta=16-8\left(B-2\right)\left(B-3\right)\ge0\\ \Leftrightarrow\left(B-2\right)\left(B-3\right)\le2\\ \Leftrightarrow B^2-5B+4\le0\\ \Leftrightarrow\left(B-1\right)\left(B-4\right)\le0\\ \Leftrightarrow1\le B\le4\)
Vậy\(B_{min}=4\Leftrightarrow\dfrac{3x^2-4x+4}{x^2+2}=4\Leftrightarrow x=...\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\Leftrightarrow2x^3-x^2+ax+b=\left(x-1\right)\left(x+1\right)\cdot a\left(x\right)\)
Thay \(x=1\Leftrightarrow2-1+a+b=0\Leftrightarrow a+b=-1\)
Thay \(x=-1\Leftrightarrow-2-1-a+b=0\Leftrightarrow b-a=3\)
Từ đó ta được \(\left\{{}\begin{matrix}a+b=-1\\-a+b=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-2\\b=1\end{matrix}\right.\)
\(b,\Leftrightarrow ax^3+bx^2+2x-1=\left(x-1\right)\left(x+6\right)\cdot b\left(x\right)\)
Thay \(x=1\Leftrightarrow a+b+2-1=0\Leftrightarrow a+b=-1\)
Thay \(x=-6\Leftrightarrow-216a+36b+12-1=0\Leftrightarrow216a-36b=11\)
Từ đó ta được \(\left\{{}\begin{matrix}a+b=-1\\216a-36b=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{25}{252}\\b=-\dfrac{227}{252}\end{matrix}\right.\)
\(c,\Leftrightarrow ax^4+bx^3+1=\left(x+1\right)^2\cdot c\left(x\right)\)
Thay \(x=-1\Leftrightarrow a-b+1=0\Leftrightarrow b=a+1\)
\(\Leftrightarrow ax^4+\left(a+1\right)x^3+1⋮\left(x+1\right)\\ \Leftrightarrow ax^4+ax^3+x^3+1⋮\left(x+1\right)\\ \Leftrightarrow ax^3\left(x+1\right)+\left(x+1\right)\left(x^2-x+1\right)⋮\left(x+1\right)\\ \Leftrightarrow\left(x+1\right)\left(ax^3+x^2-x+1\right)⋮\left(x+1\right)\\ \Leftrightarrow ax^3+x^2-x+1⋮\left(x+1\right)\)
Thay \(x=-1\Leftrightarrow-a+1+1+1=0\Leftrightarrow a=3\Leftrightarrow b=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b) \(2x^2+4y^2+z^2-4xy-2x-2z+5=0\)
\(\Leftrightarrow\left(x^2-4xy+4y^2\right)+\left(x^2-2x+1\right)+\left(z^2-2z+1\right)+3=0\)
....
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^2+5y^2-4xy+6y+9=0\)
\(\Leftrightarrow\left(x^2-4xy+4y^2\right)+\left(y^2+6y+9\right)=0\)
\(\Leftrightarrow\left(x-2y\right)^2+\left(y+3\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=0\\y+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2y=2.\left(-3\right)=-6\\y=-3\end{matrix}\right.\)
Vậy : \(\left(x,y\right)=\left(-6,-3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) Nhìn cái pt hết ham, nhưng bấm nghiệm đẹp v~`~
\(\left(\sqrt{2}+2\right)\left(x\sqrt{2}-1\right)=2x\sqrt{2}-\sqrt{2}\)
\(\Leftrightarrow\left(\sqrt{2}+2\right)\left(x\sqrt{2}-1\right)-2x\sqrt{2}+\sqrt{2}=0\)
\(\Leftrightarrow2x-\sqrt{2}+2x\sqrt{2}-2-2x\sqrt{2}+\sqrt{2}=0\)
\(\Leftrightarrow2x-2=0\Leftrightarrow2x=2\Rightarrow x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
đề bài là tìm x à bạn? đề có cho điều kiện ko vậy ạ? (ví dụ như x nguyên?)
\(\left(x-1\right)^3+\left(x^3-8\right).3x.\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right).\left[\left(x-1\right)^2+\left(x^3-8\right).3x\right]=0\)
TH1: \(x-1=0\Leftrightarrow x=1\)
TH2: \(\left(x-1\right)^2+\left(x^3-8\right).3x=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(x^3-8\right).3x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left\{{}\begin{matrix}x^3-8=0\\3x=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\\left\{{}\begin{matrix}x=2\\x=0\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x\in\left\{0;1;2\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=x^3+3x^2+3x^2y+3xy^2+y^3+3y^2+6xy+3x+3y+2019\)
\(=\left(x+y\right)^3-3\left(x+y\right)^2+3\left(x+y\right)+2019\)
\(=\left[\left(x+y\right)^3-3\left(x+y\right)^2+3\left(x+y\right)+1\right]+2018\)
\(=\left(x+y-1\right)^3+2018\)
Mà \(x+y=101\)
\(B=\left(101-1\right)^3+2018=1002018\)
Đang 3x2+3y2 sao lại ra -3(x+y)2 ?? Phải là +3(x2+y2) chứ :v Không nhớ hằng đẳng thức 1 và 3 à :v với cả 6xy đâu?