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\(2x\left(x-1\right)-x^2+6=0\)
\(2x^2-2x-x^2+6=0\)
\(x^2-2x+6=0\)
\(x^2-2x+1+5=0\)
\(\left(x-1\right)^2+5=0\)
Ta có: \(\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-1\right)^2+5\ge5>0\forall x\)
Mà: \(\left(x-1\right)^2+5=0\) => vô lí
Vậy : ko có giá trị của c thỏa mãn
=.= hok tốt!!
Ta có \(2x.\left(x-1\right)-x^2+6=0\)
\(\Rightarrow2x^2-2x-x^2+6=0\)
\(\Rightarrow x^2-2x+6=0\)
\(\Rightarrow\left(x^2-2x+1\right)+5=0\)
\(\Rightarrow\left(x-1\right)^2=-5\)
Vì \(\left(x-1\right)^2\ge0\)với mọi x nên không tìm được x
Vậy...
Bài 2:
1) \(7x^2+2x=0\)
\(\Leftrightarrow x\left(7x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\7x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{2}{7}\end{matrix}\right.\)
2) \(2x\left(x-9\right)+5\left(x-9\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-9=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-\dfrac{5}{2}\end{matrix}\right.\)
3) \(x^2+8x+16=0\)
\(\Leftrightarrow\left(x+4\right)^2=0\)
\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Bài 1:
2) \(24x-18y+30=6\left(4x-3y+5\right)\)
5) \(x^2+14x+49=\left(x+7\right)^2\)
6) \(27x^3+y^3=\left(3x+y\right)\left(9x^2-3xy+y^2\right)\)
\(2x-1\left(x+2\right)-3\left(x+2\right)=0\)
\(\Rightarrow\left(2x-4\right)\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-4=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}}\)
(2x-1)(x+2)-3(x+2)=0
<=>2x2+3x-2-3x-6=0
<=>2x2-8=0
<=>2(x2-4)=0
<=>x2-4=0
<=>(x+2)(x-2)=0
=>\(\orbr{\begin{cases}x+2=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=2\end{cases}}}\)
Vậy...
1) đặt 2x+1 = a => \(a^4-3a^2+2=\left(a^2-1\right)\left(a^2-2\right)=\)\(\left(a-1\right)\left(a+1\right)\left(a-\sqrt{2}\right)\left(a+\sqrt{2}\right)\)
=(2x+1-1)(2x+1+1)(2x+1-\(\sqrt{2}\))(2x+1+\(\sqrt{2}\)) = 4x(x+1)(2x+1-\(\sqrt{2}\))(2x+1+\(\sqrt{2}\))
2) =(x2-x)(x2-x-2)-3
đặt x2-x = b => b(b-2)-3 = b2-2b-3 = (b+1)(b-3) = (x2-x+1)(x2-x-3)
3) đặt x2+2x-1 = c => c2-3xc+2x2 = (c-x)(c-2x) = (x2+2x-1-x)(x2+2x-1-2x) = (x2+x-1)(x2-1) = (x2+x-1)(x-1)(x+1)
tìm x
x3-8 +(x-2)(x+1)=0 <=> (x-2)(x2+2x+4)+(x-2)(x+1)=0 <=>(x-2)(x2+2x+4+x+1)=0 <=> x=2 (vì x2+3x+5= (x+\(\frac{3}{2}\))2 +\(\frac{11}{4}\)>0)
vậy x=2
2) \(x\left(x-1\right)\left(x+1\right)\left(x-2\right)-3\)
\(=\left(x^2-x\right)\left(x^2-x-2\right)-3\)(1)
Đặt \(x^2-x=t\)
\(\Rightarrow\left(1\right)=t\left(t-2\right)-3=t^2-2t+1-4\)
\(=\left(t-1\right)^2-4\)
\(=\left(t+3\right)\left(t-5\right)\)
Thay \(x^2-x=t\), ta được:
\(BTDNT=\left(x^2-x+3\right)\left(x^2-x-5\right)\)
1. Ta có:
\(x^3-9x^2+27x-26=x^3-2x^2-7x^2+14x+13x-26\)
\(=x^2\left(x-2\right)-7x\left(x-2\right)+13\left(x-2\right)=\left(x-2\right)\left(x^2-7x+13\right)\)
Thay x = 23, ta có: \(C=\left(23-2\right)\left(23^2-7.23+13\right)=8001\)
2.
a) \(x^2+4y^2+6x-12y+18=0\)
\(\Leftrightarrow\left(x^2-6x+9\right)+\left(4y^2-12y+9\right)=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(2y-3\right)^2=0\)
Mà \(\left(x-3\right)^2\ge0\) với mọi x, \(\left(2y-3\right)^2\ge0\) với mọi y
\(\Rightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)và \(\left(2y-3\right)^2=0\Leftrightarrow2y-3=0\Leftrightarrow y=\frac{3}{2}\)
Vậy \(\left(x,y\right)=\left(3;\frac{3}{2}\right)\)
b) \(2x^2+2y^2+2xy-10x-8y+41=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2-10x+25\right)+\left(y^2-8y+16\right)=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x-5\right)^2+\left(y-4\right)^2=0\)
.....................................
Rồi giải tương tự như trên
\(2x^2+10xy+14y^2+2x+2y+2=0\)
\(\Leftrightarrow\left(x^2+4y^2+1+2x+4xy+4y\right)+\left(x^2+6xy+9y^2\right)+\left(y^2-2y+1\right)=0\)
\(\Leftrightarrow\left(x+2y+1\right)^2+\left(x+3y\right)^2+\left(y-1\right)^2=0\)
Vì \(\hept{\begin{cases}\left(x+2y+1\right)^2\ge0;\forall x,y\\\left(x+3y\right)^2\ge0;\forall x,y\\\left(y-1\right)^2\ge0;\forall x,y\end{cases}}\)
\(\Rightarrow\left(x+2y+1\right)^2+\left(x+3y\right)^2+\left(y-1\right)^2\ge0;\forall x,y\)
Do đó :\(\left(x+2y+1\right)^2+\left(x+3y\right)^2+\left(y-1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(x+2y+1\right)^2=0\\\left(x+3y\right)^2=0\\\left(y-1\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}y=1\\x=-3\\y=1\end{cases}}\)
Vậy x=-3 và y=1
Kiến thức bổ sung
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ca\)
\(\Leftrightarrow4x^2+20xy+28y^2+4x+4y+4=0\)
\(\Leftrightarrow\left(4x^2+4x+20xy+25y^2+10y+1\right)+\left(3y^2-6y+3\right)=0\)
\(\Leftrightarrow\left(2x+5y+1\right)^2+3\left(y-1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}2x+5y+1=0\\y-1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-3\\y=1\end{cases}}\)