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\(a,2x^2-2xt-5x+5y\)
\(=\left(2x^2-5x\right)-\left(2xy-5y\right)\)
\(=x\left(2x-5\right)-y\left(2x-5\right)\)
\(=\left(2x-5\right)\left(x-y\right)\)
\(b,8x^2+4xy-2ax-ay\)
\(=\left(8x^2-2ax\right)+\left(4xy-ay\right)\)
\(=2x\left(4x-a\right)+y\left(4x-a\right)\)
\(=\left(4x-a\right)\left(2x+y\right)\)
\(c,x^3-4x^2+4x\)
\(=x^3-2x^2-2x^2+4x\)
\(=\left(x^3-2x^2\right)-\left(2x^2-4x\right)\)
\(=x^2\left(x-2\right)-2x\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-2x\right)=x\left(x-2\right)\left(x-2\right)\)
\(=x\left(x-2\right)^2\)
\(d,2xy-x^2-y^2+16\)
\(=-\left(x^2-2xy+y^2-16\right)\)
\(=-\left[\left(x-y\right)^2-4^2\right]\)
\(=-\left(x-y-4\right)\left(x-y+4\right)\)
\(e,x^2-y^2-2yz-z^2\)
\(=x^2-\left(y^2+2yz+z^2\right)\)
\(=x^2-\left(y+z\right)^2=\left(x-y-z\right)\left(x+y+z\right)\)
a )x2+2y2-2xy+2x-4y+2=0
<=>x2-2x(y-1)+y2-2y+1+y2-2y+1=0
<=>x2-2x(y-1)+(y-1)2+(y-1)2=0
<=>(x-y+1)2+(y-1)2=0
<=>x-y+1=0 va y-1=0
<=>x=y-1 y=1
<=>x=1-1=0 y=1
d, (x2 + 4x + 8)2 + 3x(x2 + 4x + 8) + 2x2 = 0
Đặt x2 + 4x + 8 = t ta được:
t2 + 3xt + 2x2 = 0
\(\Leftrightarrow\) t2 + xt + 2xt + 2x2 = 0
\(\Leftrightarrow\) t(t + x) + 2x(t + x) = 0
\(\Leftrightarrow\) (t + x)(t + 2x) = 0
Thay t = x2 + 4x + 8 ta được:
(x2 + 4x + 8 + x)(x2 + 4x + 8 + 2x) = 0
\(\Leftrightarrow\) (x2 + 5x + 8)[x(x + 4) + 2(x + 4)] = 0
\(\Leftrightarrow\) (x2 + 5x + \(\frac{25}{4}\) + \(\frac{7}{4}\))(x + 4)(x + 2) = 0
\(\Leftrightarrow\) [(x + \(\frac{5}{2}\))2 + \(\frac{7}{4}\)](x + 4)(x + 2) = 0
Vì (x + \(\frac{5}{2}\))2 + \(\frac{7}{4}\) > 0 với mọi x
\(\Rightarrow\left[{}\begin{matrix}x+4=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-2\end{matrix}\right.\)
Vậy S = {-4; -2}
Mình giúp bn phần khó thôi!
Chúc bn học tốt!!
c) \(\frac{1}{x-1}\)+\(\frac{2x^2-5}{x^3-1}\)=\(\frac{4}{x^2+x+1}\) (ĐKXĐ:x≠1)
⇔\(\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)+\(\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}\)=\(\frac{4\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
⇒x2+x+1+2x2-5=4x-4
⇔3x2-3x=0
⇔3x(x-1)=0
⇔x=0 (TMĐK) hoặc x=1 (loại)
Vậy tập nghiệm của phương trình đã cho là:S={0}
\(A=\left(x+1\right)^2+\left(x+2\right)^2=\left(x+1\right)^2+\left(-2-x\right)^2\ge\frac{1}{2}\left(x+1-2-x\right)^2=\frac{1}{2}.1^2=\frac{1}{2}\Rightarrow A_{min}=\frac{1}{2}\Leftrightarrow x=\frac{3}{2}\)
\(B=-2x^2-4\le0-4=-4\Rightarrow B_{max}=-4\Leftrightarrow x=0\)
\(C=-5x^2+10x-7=-5x^2+10x-5-2=-5\left(x-1\right)^2-2\le0-2=-2\Rightarrow C_{min}=-2\Leftrightarrow x-1=0\Leftrightarrow x=1\)
a) \(\left(4x-1\right)^2-\left(3x+2\right)\left(3x-2\right)=\left(7x-1\right)\left(x+2\right)+\left(2x+1\right)^2-\left(4x^2+7\right)\)(1)
\(\Leftrightarrow\left(16x^2-8x+1\right)-\left(9x^2-4\right)=\left(7x^2+14x-x-2\right)+\left(4x^2+4x+1\right)-\left(4x^2+7\right)\)
\(\Leftrightarrow16x^2-8x+1-9x^2+4=7x^2+13x-2+4x^2+4x+1-4x^2-7\)
\(\Leftrightarrow7x^2-8x+5=7x^2+17x-8\)
\(\Leftrightarrow7x^2-8x-7x^2-17x=-8-5\)
\(\Leftrightarrow-25x=-13\)
\(\Leftrightarrow x=\dfrac{13}{25}\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{13}{25}\right\}\)
a) \(x^2-2x=24\)
\(\Rightarrow x^2-2x-24=0\)
\(\Rightarrow x^2-6x+4x-24=0\)
\(\Rightarrow x\left(x-6\right)+4\left(x-6\right)=0\)
\(\Rightarrow\left(x-6\right)\left(x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-6=0\\x+4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
b) \(\left(5-2x\right)^2-16=0\)
\(\Rightarrow\left(5-2x\right)^2-4^2=0\)
\(\Rightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\)
\(\Rightarrow\left(1-2x\right)\left(9-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}1-2x=0\\9-2x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=1\\2x=9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\)
c)Sửa đề
\(x^2-4x+4-9x^2+6x-1=0\)
\(\Rightarrow\left(x^2-4x+4\right)-\left(9x^2-6x+1\right)=0\)
\(\Rightarrow\left(x-2\right)^2-\left(3x-1\right)^2=0\)
\(\Rightarrow\left(x-2-3x+1\right)\left(x-2+3x-1\right)=0\)
\(\Rightarrow\left(-2x-1\right)\left(4x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-2x-1=0\\4x-3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-2x=1\\4x=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\)
d) \(2x^2+y^2+2xy-4x+4=0\)
\(\Rightarrow\left(x^2+2xy+y^2\right)+\left(x^2-4x+4\right)=0\)
\(\Rightarrow\left(x+y\right)^2+\left(x-2\right)^2=0\)
Vì \(\left(x+y\right)^2\ge0\) với mọi x và y
\(\left(x-2\right)^2\ge0\) với mọi x
\(\Rightarrow\left(x+y\right)^2+\left(x-2\right)^2\ge0\) với mọi x và y
Mà \(\left(x+y\right)^2+\left(x-2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x+y=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}y=-x\\x=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}y=-2\\x=2\end{matrix}\right.\)