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\(P=\dfrac{15x^5y^3-10x^3y^2+20x^4y^4}{5x^2y^2}\)
\(=\dfrac{15x^5y^3}{5x^2y^2}-\dfrac{10x^3y^2}{5x^2y^2}+\dfrac{20x^4y^4}{5x^2y^2}\)
\(=3x^3y-2x+4x^2y^2\)
Khi x=-1 và y=2 thì \(P=3\cdot\left(-1\right)^3\cdot2-2\cdot\left(-1\right)+4\cdot\left(-1\right)^2\cdot2^2\)
\(=-6+2+16=4+16=20\)
a: Khi x=2 và y=-3 thì \(x^2+2y=2^2+2\cdot\left(-3\right)=4-6=-2\)
b: \(A=x^2+2xy+y^2=\left(x+y\right)^2\)
Khi x=4 và y=6 thì \(A=\left(4+6\right)^2=10^2=100\)
c: \(P=x^2-4xy+4y^2=\left(x-2y\right)^2\)
Khi x=1 và y=1/2 thì \(P=\left(1-2\cdot\dfrac{1}{2}\right)^2=\left(1-1\right)^2=0\)
a: M=x^3+27-(27-8x^3)
=x^3+27-27+8x^3
=9x^3
=9*20^3=72000
b: \(M=x^3-\left(2y\right)^3+16y^3=x^3+8y^3\)
=(x+2y)(x^2-2xy+4y^2)
=0
1/
\(M=3x^2-4x+3=3\left(x^2-\frac{4}{3}x+1\right)=3\left(x^2-2x\cdot\frac{2}{3}+\frac{4}{9}\right)+\frac{5}{3}=3\left(x-\frac{2}{3}\right)^2+\frac{5}{3}\ge\frac{5}{3}>0\)
\(N=5x^2-10x+2018=5\left(x^2-2x+1\right)+2013=5\left(x-1\right)^2+2013\ge2013>0\)
\(P=x^2+2y^2-2xy+4y+7=\left(x^2-2xy+y^2\right)+\left(y^2+4y+4\right)+3=\left(x-y\right)^2+\left(y+2\right)^2+3\ge3>0\)
2/
\(A=10x-6x^2+7=-6x^2+10x+7=-6\left(x^2-\frac{10}{6}x+\frac{25}{36}\right)-\frac{11}{6}=-6\left(x-\frac{5}{6}\right)^2-\frac{11}{6}\le-\frac{11}{6}< 0\)
\(B=-3x^2+7x+10=-3\left(x^2-\frac{7}{3}x+\frac{49}{36}\right)-\frac{311}{12}=-3\left(x-\frac{7}{6}\right)^2-\frac{311}{12}\le-\frac{311}{12}< 0\)
\(C=2x-2x^2-y^2+2xy-5=\left(2x-x^2-1\right)-\left(x^2-2xy+y^2\right)-4=-\left(x^2-2x+1\right)-\left(x-y\right)^2-4=-\left(x-1\right)^2-\left(x-y\right)^2-4\)\(\le-4< 0\)
Ta có :
\(\left(x+2y\right)^2+\left(y-1\right)^2+\left(x-z\right)^2=0\)
=> \(\hept{\begin{cases}\left(x+2y\right)=0\\\left(y-1\right)=0\\\left(x-z\right)=0\end{cases}}\)=> \(\hept{\begin{cases}x=-2y\\y=1\\x=z\end{cases}}\)
=> \(\hept{\begin{cases}x=-2\\y=1\\z=-2\end{cases}}\)
M = x + 2y + 3z = -2 + 2 - 6 = (-6)
Chọn C
Bài 1:
a: \(\left(\dfrac{1}{3}x+2\right)\left(3x-6\right)\)
\(=x^2-3x+6x-12\)
\(=x^2+3x-12\)
b: \(\left(x+3\right)\left(x^2-3x+9\right)=x^3+27\)
c: \(\left(-2xy+3\right)\left(xy+1\right)\)
\(=-2x^2y^2-2xy+3xy+3\)
\(=-2x^2y^2+xy+3\)
d: \(x\left(xy-1\right)\left(xy+1\right)\)
\(=x\left(x^2y^2-1\right)\)
\(=x^3y^2-x\)
Bài 2:
a: Ta có: \(M=\left(3x+2\right)\left(9x^2-6x+4\right)\)
\(=27x^3+8\)
\(=27\cdot\dfrac{1}{27}+8=9\)
b: Ta có: \(N=\left(5x-2y\right)\left(25x^2+10xy+4y^2\right)\)
\(=125x^3-8y^3\)
\(=125\cdot\dfrac{1}{125}-8\cdot\dfrac{1}{8}\)
=0
\(5x^2+2y^2+6xy-8x-4y+4=0\)
\(\Leftrightarrow4x^2+x^2+y^2+y^2+2xy+4xy-8x-4y+4=0\)
\(\Leftrightarrow\left(4x^2+y^2+4+4xy-8x-4y\right)+\left(x^2+2xy+y^2\right)=0\)
\(\Leftrightarrow\left[\left(2x\right)^2+4xy+y^2-4\left(2x+y\right)+2^2\right]+\left(x+y\right)^2=0\)
\(\Leftrightarrow\left[\left(2x+y\right)^2-2\cdot\left(2x+y\right)\cdot2+2^2\right]+\left(x+y\right)^2=0\)
\(\Leftrightarrow\left(2x+y-2\right)^2+\left(x+y\right)^2=0\)
Ta có: \(\left\{{}\begin{matrix}\left(2x+y-2\right)^2\ge0\forall x,y\\\left(x+y\right)^2\ge0\forall x,y\end{matrix}\right.\)
\(\Rightarrow\left(2x+y-2\right)^2+\left(x+y\right)^2\ge0\forall x,y\)
Mặt khác: \(\left(2x+y-2\right)^2+\left(x+y\right)^2=0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}2x+y-2=0\\x+y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\cdot\left(-y\right)+y-2=0\\x=-y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2y+y-2=0\\x=-y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-y=2\\x=-y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-2\\x=2\end{matrix}\right.\)
Thay x,y vào P ta có:
\(P=2^{2023}+\left(-2\right)^{2023}=2^{2023}-2^{2023}=0\)
Vậy: ...
1.
\(\frac{25x^4y^3-15x^3y^5+20x^2y^4}{5x^2y^3}\)
\(=\frac{5x^2y^3\left(5x^2-3xy^2+4y\right)}{5x^2y^3}\)
\(=5x^2-3xy^2+4y\)
2.
a) \(27x^4-8x=x\left(27x^3-8\right)\)
\(=x\left(3x-2\right)\left(9x^2+6x+4\right)\)
b) \(16x^2y-4xy^2-4x^3+x^2y\)
\(=4xy\left(4x-y\right)-x^2\left(4x-y\right)\)
\(=x\left(4x-y\right)\left(4y-x\right)\)
c) \(x^2-2x-5+2\sqrt{5}\)
\(=\left(x-1\right)^2-6+2\sqrt{5}\)
\(=\left(x-1\right)^2-\left(6-2\sqrt{5}\right)=\left(x-1\right)^2-\left(\sqrt{5}-1\right)^2\)
\(=\left(x-\sqrt{5}\right)\left(x-2+\sqrt{5}\right)\)
Bài 1:
\(\left(25x^4y^3-15x^3y^5+20x^2y^4\right):\left(5x^2y^3\right)\)
\(=\frac{25x^4y^3-15x^3y^5+20x^2y^4}{5x^2y^3}\)
\(=\frac{5x^2y^3\left(5x^2-3xy^2+4y\right)}{5x^2y^3}\)
\(=5x^2-3xy^2+4y\)
Bài 2:
a) \(27x^4-8x\)
\(=x\left(3x-2\right)\left(3^2x^2+2.3x+2^2\right)\)
\(=x\left(3x-2\right)\left(9x^2+6x+4\right)\)
b) \(16x^2y-4xy^2-4x^3+x^2y\)
\(=4y^2+x^2-\left(4x^2\right)^2\)
\(=x\left(-4x^2+xy+4y^2\right)\)
Bài 1:
$2xy=(x+y)^2-(x^2+y^2)=4^2-10=6\Rightarrow xy=3$
$M=x^6+y^6=(x^3+y^3)^2-2x^3y^3$
$=[(x+y)^3-3xy(x+y)]^2-2(xy)^3=(4^3-3.3.4)^2-2.3^3=730$
Bài 2:
$8x^3-32y-32x^2y+8x=0$
$\Leftrightarrow (8x^3+8x)-(32y+32x^2y)=0$
$\Leftrightarrow 8x(x^2+1)-32y(1+x^2)=0$
$\Leftrightarrow (8x-32y)(x^2+1)=0$
$\Rightarrow 8x-32y=0$ (do $x^2+1>0$ với mọi $x$)
$\Leftrightarrow x=4y$
Khi đó:
$M=\frac{3.4y+2y}{3.4y-2y}=\frac{14y}{10y}=\frac{14}{10}=\frac{7}{5}$
Ta có:
\(15x^4y^4-M=10x^2y^4+6x^2y^4\)
\(\Leftrightarrow M=15x^4y^4-\left(10x^2y^4+6x^2y^2\right)\)
\(\Leftrightarrow M=15x^4y^4-16x^2y^4\)
Thay \(x=-\dfrac{1}{2};x=2\) vào M ta có:
\(M=15\cdot\left(-\dfrac{1}{2}\right)^4\cdot2^4-16\cdot\left(-\dfrac{1}{2}\right)^2\cdot2^4=-49\)