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1: \(y=x^2+2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{41}{4}\)
\(=\left(x+\dfrac{5}{2}\right)^2-\dfrac{41}{4}\ge-\dfrac{41}{4}\forall x\)
Dấu '=' xảy ra khi x=-5/2
2: \(y=2\left(x^2-2x+\dfrac{5}{2}\right)\)
\(=2\left(x^2-2x+1+\dfrac{3}{2}\right)\)
\(=2\left(x-1\right)^2+3\ge3\forall x\)
Dấu '=' xảy ra khi x=1
3: \(y=x^2-4x+4-3=\left(x-2\right)^2-3\ge-3\forall x\)
Dấu '=' xảy ra khi x=2
4: \(2x^2-8x+3\)
\(=2\left(x^2-4x+\dfrac{3}{2}\right)\)
\(=2\left(x^2-4x+4-\dfrac{5}{2}\right)\)
\(=2\left(x-2\right)^2-5\ge-5\forall x\)
Dấu '=' xảy ra khi x=2
Bài 1:
\(x^2+y^2-2x-4y+5=0\)
\(\Leftrightarrow (x^2-2x+1)+(y^2-4y+4)=0\)
\(\Leftrightarrow (x-1)^2+(y-2)^2=0\)
Vì $(x-1)^2; (y-2)^2\geq 0$ với mọi $x,y\in\mathbb{R}$ nên để tổng của chúng bằng $0$ thì $(x-1)^2=(y-2)^2=0$
$\Rightarrow x=1; y=2$
Vậy...........
Bài 2:
Ta có:
\(a(a-b)+b(b-c)+c(c-a)=0\)
\(\Leftrightarrow 2a(a-b)+2b(b-c)+2c(c-a)=0\)
\(\Leftrightarrow (a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)=0\)
\(\Leftrightarrow (a-b)^2+(b-c)^2+(c-a)^2=0\)
Lập luận tương tự bài 1, ta suy ra :
\((a-b)^2=(b-c)^2=(c-a)^2=0\Rightarrow a=b=c\)
Khi đó, thay $b=c=a$ ta có:
\(P=a^3+b^3+c^3-3abc+3ab-3c+5\)
\(=3a^3-3a^3+3a^2-3a+5=3a^2-3a+5\)
\(=3(a^2-a+\frac{1}{4})+\frac{17}{4}=3(a-\frac{1}{2})^2+\frac{17}{4}\geq \frac{17}{4}\)
Vậy $P_{\min}=\frac{17}{4}$
Giá trị này đạt được tại $b=c=a=\frac{1}{2}$
ừ thì mình sẽ giúp bạn mà câu a bạn viết sai đề nha
1/a)\(2x^2+3x-5=2x^2-2x+5x-5=2x\left(x-1\right)+5\left(x-1\right)=\left(2x+5\right)\left(x-1\right)\)
b)\(4x^2-3x-1=4x^2-4x+x-1=4x\left(x-1\right)+\left(x-1\right)=\left(4x+1\right)\left(x-1\right)\)
c)Sai đề: \(3x^2+6xy+3y^2-3z^2\)
\(=3\left(x^2+2xy+y^2-z^2\right)\)
\(=3\left[\left(x+y\right)^2-z^2\right]\)
\(=3\left(x+y+z\right)\left(x+y-z\right)\)
d)Sai đề:\(x^3-2x^2y+xy^2-9x=x\left(x-2xy+y^2-9\right)=x\left[\left(x-y\right)^2-9\right]=x\left(x-y+3\right)\left(x-y-3\right)\)
e)\(2x-2y-x^2+2xy-y^2=2\left(x-y\right)-\left(x-y\right)^2=\left(x-y\right)\left(2-x+y\right)\)
f)Hình như sai đề đúng không?
\(x^3-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y+1\right)\left(x+y-1\right)\)
2/a.\(7x-6x^2-2=0\)
\(\Leftrightarrow-\left(6x^2-3x-4x+2\right)=0\)
\(\Leftrightarrow3x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(x-1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}3x-2=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\x=1\end{matrix}\right.\)
b.\(16x-5x^2-3=0\)
\(\Leftrightarrow-\left(5x^2-15x-x+3\right)=0\)
\(\Leftrightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(5x-1\right)\left(x-3\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}5x-1=0\\x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\x=3\end{matrix}\right.\)
c.\(2x^2+3x-5=0\)
\(\Leftrightarrow2x^2-2x+5x-5=0\)
\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}2x+5=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}=-2,5\\x=1\end{matrix}\right.\)
a: \(=4x^4y+6x^2y^2z-2x^5y\)
b: \(=\dfrac{24x^5}{6x^2}-\dfrac{12x^4}{6x^2}+\dfrac{6x^2}{6x^2}=4x^3-2x^2+1\)
c: \(=\dfrac{\left(2x-1\right)^2}{2x-1}=2x-1\)
d: \(=\dfrac{\left(x+5\right)\left(x^2-1\right)}{x+5}=x^2-1\)
Bài 1
A= (5-x)(x3-2x2+x-1)
=5x3-10x2+5x-5-x4+2x3-x2+x
=-x4+7x3-12x2+6x-5
B= 2y-x - {2x-y-[y+3x-(5y-x)]}
B= 2y-x- [ 2x-y- ( y + 3x-5y+x)
B= 2y-x-( 2x-y-y-3x+5y-x)
B= 2y-x-2x+y+y+3x-5y+x
B= -y +x
a: \(\Leftrightarrow8x^3+12x^2+6x+1-8x^3-1-3\left(4x^2-4x+1\right)=15\)
=>\(12x^2+6x-12x^2+12x-3=15\)
=>18x=18
=>x=1
b: \(\Leftrightarrow y^3+6y^2+9y-y^3-8-6y^2+150=97\)
=>9y+142=97
=>9y=-45
=>y=-5