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PT
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ND
2 tháng 8 2019
Bài 1:
a, \(\left(x-y\right)^2=x^2+y^2+2xy-4xy=\left(x+y\right)^2-4xy\)
Thay \(x+y=3,xy=-4\), ta có:
\(\left(x-y\right)^2=3^2-4.\left(-4\right)=25\)
b, \(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)\)
Thay \(x+y=3,xy=-4\),ta có:
\(x^3+y^3=3^3-3.\left(-4\right).3=63\)
c, Giải \(\left\{{}\begin{matrix}x+y=3\\xy=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=4\\y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=4\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^3-y^3=65\\x^3-y^3=-65\end{matrix}\right.\)
2 tháng 8 2019
Bài 1:
\(a,\left(x-y\right)^2=\left(x+y\right)^2-4xy=3^2-4.\left(-4\right)=25\)
\(b,x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=3\left[\left(x+y\right)^2-3xy\right]\)
\(=3\left(3^2-3.\left(-4\right)\right)=63\)
\(c,\)\(x+y=3\Rightarrow x=3-y\)
Thay vào xy = -4 ,có :
\(\left(3-y\right)y=-4\Leftrightarrow-y^2+3y+4=0\Leftrightarrow\left[{}\begin{matrix}y=4\\y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3-4=-1\\x=3-\left(-1\right)=4\end{matrix}\right.\)
\(TH1:x^3-y^3=\left(4^3\right)-\left(-1\right)^3=65\)
\(TH2:x^3-y^3=\left(-1\right)^3-4^3=-65\)
Bài 2:
\(A=x^2-3x=\left(x^2-3x+\frac{9}{4}\right)-\frac{9}{4}\)\(=\left(x+\frac{3}{2}\right)^2-\frac{9}{4}\ge-\frac{9}{4}\)
Dấu = xảy ra \(\Leftrightarrow x=-\frac{3}{2}\)
Vậy \(Min_A=-\frac{9}{4}\Leftrightarrow x=-\frac{3}{2}\)
\(B=2x^2+x=2\left(x^2+\frac{1}{2}x+\frac{1}{16}\right)-\frac{1}{8}\)
\(=2\left(x+\frac{1}{4}\right)^2-\frac{1}{8}\ge-\frac{1}{8}\)
Dấu = xảy ra \(\Leftrightarrow x=-\frac{1}{4}\)
\(Min_B=-\frac{1}{8}\Leftrightarrow x=-\frac{1}{4}\)
TN
7 tháng 6 2016
1)M=3x(2x-5y)+(3x-y)(-2x)-1/2(2-26xy)
=-1
2)
a)7x(x-2)-5(x-1)=21x^2-14x^2+3
<=>7x2-19x+5=7x2+3
<=>-19x=-2
<=>x=\(\frac{2}{19}\)
DT
19 tháng 8 2020
a, -x - y2 + x2 - y = (x2 - y2) - (x + y)
= (x - y)(x + y) - (x + y)
= (x + y)(x - y - 1)
b, x( x + y ) - 5x - 5y = x(x + y) - 5(x + y)
= (x - 5)(x + y)
c, x2 - 5x + 5y - y2 = (x - y)(x + y) - 5(x - y)
= (x - y)(x + y - 5)
d, 5x3 - 5x2y - 10x2 + 10xy = 5x2(x - y) - 10x(x - y)
= 5x(x - y)(x - 2)
e, 27x3 - 8y3 = (3x - 2y)(9x2 + 6xy + 4y2)
f, x2 - y2 - x - y = (x - y)(x + y) - (x + y)
= (x + y)(x - y - 1)
g, x2 - y2 - 2xy + y2 = (x2 - 2xy + y2) - y2
= (x - y)2 - y2
= (x - y - y)(x - y + y) = x(x - 2y)
h, x2 - y2 + 4 - 4x = (x2 - 4x + 4) - y2
= (x - 2)2 - y2
= (x - y - 2)(x + y - 2)
i, x3 + 3x2 + 3x + 1 - 27z3 = (x + 1)3 - 27z3
= (x+1-3z)(x2+2x+1+3xz+3z+9z2)
k, 4x2 + 4x - 9y2 + 1 = (2x + 1)2 - 9y2
= (2x - 3y + 1)(2x + 3y + 1)
m, x2 - 3x + xy - 3y = x(x - 3) + y(x - 3)
= (x - 3)(x + y)
TG
19 tháng 8 2020
a) \(-x-y^2+x^2-y\)
\(=\left(x^2-y^2\right)-\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right).1\)
\(=\left(x+y\right)\left(x-y-1\right)\)
b) \(x\left(x+y\right)-5x-5y\)
\(=x\left(x+y\right)-5\left(x+y\right)\)
\(=\left(x+y\right)\left(x-5\right)\)
c) \(x^2-5x+5y-y^2\)
\(=\left(x^2-y^2\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-5\right)\)
d) \(5x^3-5x^2y-10x^2+10xy\)
\(=5x\left(x^2-xy-2x+2y\right)\)
\(=5x\left[x\left(x-y\right)-2\left(x-y\right)\right]\)
\(=5x\left(x-y\right)\left(x-2\right)\)
e) \(27x^3-8y^3\)
\(=\left(3x\right)^3-\left(2y\right)^3\)
\(=\left(3x-2y\right)\left[\left(3x\right)^2+3x2y+\left(2y\right)^2\right]\)
\(=\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)\)
f) \(x^2-y^2-x-y\)
\(=\left(x^2-y^2\right)-\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)
g) \(x^2-y^2-2xy+y^2\)
\(=\left(x^2-2xy+y^2\right)-y^2\)
\(=\left(x-y\right)^2-y^2\)
\(=\left(x-y-y\right)\left(x-y+y\right)\)
\(=\left(x-y^2\right)x\)
h) \(x^2-y^2+4-4x\)
\(=\left(x^2-4x+4\right)-y^2\)
\(=\left(x^2-2.2x+2^2\right)-y^2\)
\(=\left(x-2\right)^2-y^2\)
\(=\left(x-2-y\right)\left(x-2+y\right)\)
i) \(x^6-y^6\)
\(=\left(x^3\right)^2-\left(y^3\right)^2\)
\(=\left(x^3-y^3\right)\left(x^3+y^3\right)\)
\(=\left[\left(x-y\right)\left(x^2+xy+y^2\right)\right]\left[\left(x+y\right)\left(x^2-xy+y^2\right)\right]\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)
5 tháng 8 2018
Bài 2:
\(A=-x^2-4x-2=-\left(x^2+4x+4\right)+2=-\left(x+2\right)^2+2\le2\)
Vậy GTLN của A là 2 khi x = -2
\(B=-2x^2-3x+5=-2\left(x^2+\dfrac{3}{2}x+\dfrac{9}{16}\right)+\dfrac{49}{8}=-2\left(x+\dfrac{3}{4}\right)^2+\dfrac{49}{8}\le\dfrac{49}{8}\)
Vậy GTLN của B là \(\dfrac{49}{8}\) khi x = \(-\dfrac{3}{4}\)
\(M=x^2+xy+y^2-3x-3\)
\(=\dfrac{1}{4}x^2+xy+y^2+\dfrac{3}{4}x^2-3x-3\)
\(=\left(\dfrac{1}{2}x+y\right)^2+3\left(\dfrac{1}{4}x^2-x-1\right)\)
\(=\left(\dfrac{1}{2}x+y\right)^2+3\left(\dfrac{1}{4}x^2-x+1-2\right)\)
\(=\left(\dfrac{1}{2}x+y\right)^2+3\left(\dfrac{1}{2}x-1\right)^2-6>=-6\forall x,y\)
Dấu = xảy ra khi \(\left\{{}\begin{matrix}\dfrac{1}{2}x-1=0\\\dfrac{1}{2}x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-\dfrac{1}{2}x=-\dfrac{1}{2}\cdot2=-1\end{matrix}\right.\)