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1 ) \(B=x^4-2x^3+3x^2-2x+1\)
\(B=x^2\left(x^2-2x+3-\frac{2}{x}+\frac{1}{x^2}\right)\)
\(B=x^2\left[\left(x^2+2+\frac{1}{x^2}\right)-2\left(x+\frac{1}{x}\right)+1\right]\)
\(B=x^2\left[\left(x+\frac{1}{x}\right)^2-2\left(x+\frac{1}{x}\right)+1\right]\)
\(B=x^2\left(x+\frac{1}{x}-1\right)^2\)
\(B=\left[x\left(x+\frac{1}{x}-1\right)\right]^2\)
\(B=\left(x^2-x+1\right)^2\)
Xét \(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)
\(\Rightarrow B=\left(x^2-x+1\right)^2\ge\left(\frac{3}{4}\right)^2=\frac{9}{16}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x=\frac{1}{2}\)
2 ) \(A=ax^2+bx+c\)
\(A=a\left(x^2+\frac{bx}{a}+\frac{c}{a}\right)\)
\(A=a\left(x^2+2.x.\frac{b}{2a}+\frac{b^2}{4a^2}+\frac{c}{a}-\frac{b^2}{4a^2}\right)\)
\(A=a\left[\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a^2}\right]\)
\(A=a\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a}\ge\frac{4ac-b^2}{4a}\forall x;a;b;c\)
Dấu : = " xảy ra \(\Leftrightarrow x=-\frac{b}{2a}\)
Chúc bạn học tốt !!!
\(A=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)\\ A=\left(x^2-5x+4\right)\left(x^2-5x+6\right)\\ A=\left(x^2-5x+5-1\right)\left(x^2-5x+5+1\right)\\ A=\left(x^2-5x+5\right)^2-1\ge-1\)
đẳng thức xảy ra khi :
\(x^2-5x+5=0\\ x^2-2.\dfrac{5}{2}x+\dfrac{25}{4}=\dfrac{25}{4}-5\\ \left(x-\dfrac{5}{2}\right)^2=\dfrac{5}{4}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{5}{2}=\sqrt{\dfrac{5}{4}}\\x-\dfrac{5}{2}=-\sqrt{\dfrac{5}{4}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{5}{4}}+\dfrac{5}{2}=\dfrac{\sqrt{5}+5}{2}\\x=-\sqrt{\dfrac{5}{4}}+\dfrac{5}{2}=\dfrac{5-\sqrt{5}}{2}\end{matrix}\right.\)
vậy GTNN của A =-1 tại \(\left[{}\begin{matrix}x=\sqrt{\dfrac{5}{4}}+\dfrac{5}{2}=\dfrac{\sqrt{5}+5}{2}\\x=-\sqrt{\dfrac{5}{4}}+\dfrac{5}{2}=\dfrac{5-\sqrt{5}}{2}\end{matrix}\right.\)
a: \(=\dfrac{27a^6b^3\cdot a^2b^6}{a^8b^8}=27b\)
b: \(=3y^2-5x^2y^3-2y^2+3x^2y^3\)
\(=y^2-2x^2y^3\)
c: \(=6x-y+2x^2+3y-2x^2+x\)
\(=7x+2y\)
d: \(=x-y+2y^2-6xy+\dfrac{10x^2}{y}\)
1,
a,\(2x\left(3x^2-5x+3\right)\)
\(=6x^3-10x^2+6x\)
b,\(-2x\left(x^2+5x-3\right)\)
\(=-2x^3-10x^2+6x\)
c,\(-\dfrac{1}{2}x\left(2x^3-4x+3\right)\)
\(=-x^4+2x^2-\dfrac{3}{2}x\)
Bài 2:
a) \(\left(2x-1\right)\left(x^2-5-4\right)\)
\(=\left(2x-1\right)\left(x^2-9\right)\)
\(=2x^3-18x-x^2+9\)
b) \(-\left(5x-4\right)\left(2x+3\right)\)
\(=-\left(10x^2+15x-8x-12\right)\)
\(=-10x^2-7x+12\)
c) \(\left(2x-y\right)\left(4x^2-2xy+y^2\right)\)
\(=8x^3-y^3\)
a: \(=3y^2-5x^2y^3-2y^2+3x^2y^3=y^2-2x^2y^3\)
b: \(=6x-y+2x^2+3y^2-2x^2+x=7x-y+3y^2\)
c: \(=x-y+4y^2-6xy+\dfrac{10x^2}{y}\)
\(a.\left(9x^2y^3-15x^4y^4\right):3x^2y-\left(2-3x^2y\right)y^2\)
\(=3y^2-5x^2y^3-2y^2+3x^2y^3\)
\(=y^2-2x^2y^3\)
\(b.\left(6x^2-xy\right):x+\left(2x^3y+3xy^2\right):xy-\left(2x-1\right)x\)
\(=6x-y+2x^2+3y-2+x\)
\(=2x^2+7x+2y-2\)
\(c.\left(x^2-xy\right):x+\left(6x^2y^5-9x^3y^4+15x^4y^3\right):\dfrac{3}{2}x^2y^3\)
\(=x-y+4y^2-6xy+10x^2\)
1) \(D=\left|x^2+x+3\right|+\left|x^2+x-6\right|\)
\(D=\left|x^2+x+3\right|+\left|6-x^2-x\right|\)
Áp dụng bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có :
\(D\ge\left|x^2+x+3+6-x^2-x\right|=\left|9\right|=9\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x^2+x+3\right)\left(6-x^2-x\right)\ge0\Leftrightarrow-3\le x\le2\)
2) \(C=x^2+xy+y^2-3x-3y\)
\(C=\left(x^2-2x+1\right)+\left(y^2-2y+1\right)+\left(xy-x-y+1\right)-3\)
\(C=\left(x-1\right)^2+\left(y-1\right)^2+\left(x-1\right)\left(y-1\right)-3\)
\(C=\left(x-1\right)^2+2\cdot\left(x-1\right)\cdot\frac{\left(y-1\right)}{2}+\frac{\left(y-1\right)^2}{4}+\frac{3\left(y-1\right)^2}{4}-3\)
\(C=\left(x-1-\frac{y-1}{2}\right)^2+\frac{3\left(y-1\right)^2}{4}-3\ge-3\forall x;y\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-1-\frac{y-1}{2}=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
3) \(B=x^4-2x^3+3x^2-2x+1\)
\(B=x^2\left(x^2-2x+3-\frac{2}{x}+\frac{1}{x^2}\right)\)
\(B=x^2\left[\left(x^2+2+\frac{1}{x^2}\right)-2\left(x+\frac{1}{x}\right)+1\right]\)
\(B=x^2\left[\left(x+\frac{1}{x}\right)^2-2\left(x+\frac{1}{x}\right)+1\right]\)
\(B=x^2\left(x+\frac{1}{x}-1\right)^2\)
\(B=\left[x\left(x+\frac{1}{x}-1\right)\right]^2\)
\(B=\left(x^2-x+1\right)^2\)
Xét \(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)
\(\Rightarrow B=\left(x^2-x+1\right)^2\ge\left(\frac{3}{4}\right)^2=\frac{9}{16}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=\frac{1}{2}\)