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Bài 3:
a) Ta có: \(A=25x^2-20x+7\)
\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)
\(=\left(5x-2\right)^2+3>0\forall x\)(đpcm)
d) Ta có: \(D=x^2-2x+2\)
\(=x^2-2x+1+1\)
\(=\left(x-1\right)^2+1>0\forall x\)(đpcm)
Bài 1:
a) Ta có: \(A=x^2-2x+5\)
\(=x^2-2x+1+4\)
\(=\left(x-1\right)^2+4\ge4\forall x\)
Dấu '=' xảy ra khi x=1
b) Ta có: \(B=x^2-x+1\)
\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
a) Ta có: \(x^2\left(x+1\right)+x+1=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow x+1=0\)
hay x=-1
b) Ta có: \(x^2-x=-2x^2+2x\)
\(\Leftrightarrow3x^2-3x=0\)
\(\Leftrightarrow3x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
c) Ta có: \(2x^2\left(x-1\right)+x^2=x\)
\(\Leftrightarrow2x^2\left(x-1\right)+x^2-x=0\)
\(\Leftrightarrow2x^2\left(x-1\right)+x\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\cdot\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{-1}{2}\end{matrix}\right.\)
d) Ta có: \(\left(x-2\right)\left(x^2+4\right)=x^2-2x\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+4\right)-x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-x+4\right)=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
1. Đề bài sai, các biểu thức này chỉ có giá trị lớn nhất, không có giá trị nhỏ nhất
2.
\(A=\left(2x\right)^3-3^3-\left(8x^3+2\right)\)
\(=8x^3-27-8x^3-2\)
\(=-29\)
\(B=x^3+9x^2+27x+27-\left(x^3+9x^2+27x+243\right)\)
\(=27-243=-216\)
sửa đề lại thành tìm Max nhé1, vì mấy ý này ko có min
\(1,=>D=-\left(x^2-4x-3\right)=-\left(x^2-2.2x+4-7\right)\)
\(=-[\left(x-2\right)^2-7]=-\left(x-2\right)^2+7\le7\)
dấu"=" xảy ra<=>x=2
2, \(E=-2\left(x^2-x+\dfrac{5}{2}\right)=-2[x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{9}{4}]\)
\(=-2[\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}]\le-\dfrac{9}{2}\) dấu"=" xảy ra<=>x=1/2
3, \(F=-\left(x^2+4x-20\right)=-\left(x^2+2.2x+4-24\right)\)
\(=-[\left(x+2\right)^2-24]\le24\) dấu"=" xảy ra<=>x=-2
a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
a) \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=\left(x^2+3x+1\right)^2+x\)
\(\Leftrightarrow\left(x^2+3x\right)\left(x^2+3x+2\right)=\left(x^2+3x+1\right)^2+x\)
\(\Leftrightarrow\left(t-1\right)\left(t+1\right)=t^2+x\) (với \(t=x^2+3x+1\))
\(\Leftrightarrow t^2-1=t^2+x\)
\(\Leftrightarrow x=-1\).
b) \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)=\left(x^2+8x+11\right)^2+2x\)
\(\Leftrightarrow\left(x^2+8x+7\right)\left(x^2+8x+15\right)=\left(x^2+8x+11\right)^2+2x\)
\(\Leftrightarrow\left(t-4\right)\left(t+4\right)=t^2+2x\) (với \(t=x^2+8x+11\))
\(\Leftrightarrow t^2-16=t^2+2x\)
\(\Leftrightarrow x=-8\)
c) \(\left(x^2-x+1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x+1\right)=63\)
\(\Leftrightarrow\left(x^3-1\right)\left(x^3+1\right)=63\)
\(\Leftrightarrow x^6-1=63\)
\(\Leftrightarrow x^6=64\)
\(\Leftrightarrow x=\pm2\)
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\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)
\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)
\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Ta có :
\(x^2-4x+5=\left(x^2-2.2x+2^2\right)+1=\left(x-2\right)^2+1\ge1>0\)
Vậy đa thức \(x^2-4x+5\) vô nghiệm với mọi giá trị của x
Chúc bạn học tốt ~