(x^3-4x)^2-3x^2.|y-3|=0. tim x,y
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10 tháng 11 2016
Bài 2:
\(A=-2x^2+3x-5\)
\(=-2\left(x^2+\frac{3x}{2}-\frac{5}{2}\right)\)
\(=-2\left(x^2-\frac{3x}{2}+\frac{9}{16}\right)-\frac{31}{8}\)
\(=-2\left(x-\frac{3}{4}\right)^2-\frac{31}{8}\le-\frac{31}{8}\)
Dấu = khi \(-2\left(x-\frac{3}{4}\right)^2=0\Leftrightarrow x-\frac{3}{4}=0\Leftrightarrow x=\frac{3}{4}\)
Vậy \(Max_A=-\frac{31}{8}\Leftrightarrow x=\frac{3}{4}\)
Y
1 tháng 8 2018
Bài 1:
a) \(8\left(x-2\right)-2\left(3x-4\right)=2\)
\(\Rightarrow2\left[4\left(x-2\right)-\left(3x-4\right)\right]=2\)
\(\Rightarrow4\left(x-2\right)-3x+4=0\)
\(\Rightarrow4x-8-3x+4=0\)
\(\Rightarrow x-4=0\)
\(\Rightarrow x=4\)
b) \(10\left(3x-2\right)-3\left(5x+2\right)+5\left(11-4x\right)=25\)
\(\Rightarrow5\left[2\left(3x-2\right)+11-4x\right]-3\left(5x+2\right)=25\)
\(\Rightarrow5\left(6x-4+11-4x\right)-3\left(5x+2\right)=25\)
\(\Rightarrow5\left(2x+7\right)-3\left(5x+2\right)=25\)
\(\Rightarrow10x+35-15x-6=25\)
\(\Rightarrow-5x+29=25\)
\(\Rightarrow-5x=25-29\)
\(\Rightarrow-5x=-4\)
\(\Rightarrow x=\dfrac{4}{5}\)
c) \(2x\left(x+1\right)-x^2\left(x+2\right)+x^3-x+4=0\)
\(\Rightarrow2x^2+2x-x^3-2x^2+x^3-x+4=0\)
\(\Rightarrow x+4=0\)
\(\Rightarrow x=-4\)
d) \(4x\left(3x+2\right)-6x\left(2x+5\right)+21\left(x-1\right)=0\)
\(\Rightarrow12x^2+8x-12x^2-30x+21x-21=0\)
\(\Rightarrow-x-21=0\)
\(\Rightarrow x=-21\)
Bài 2:
a) \(P=\left(4x^2-3y\right)2y-\left(3x^2-4y\right)3y\)
\(P=8x^2y-6y^2-9x^2y+12y^2\)
\(P=-x^2y+6y^2\)
Thay x = -1 ; y = 2 vào P ta được
\(P=-\left(-1\right)^2.2+6.2^2\)
\(P=-2+24=22\)
b) \(Q=4x^2\left(5x-3y\right)-x^2\left(4x+y\right)\)
\(Q=20x^3-12x^2y-4x^3-x^2y\)
\(Q=16x^3-13x^2y\)
Thay x = -1 ; y = 2 vào Q ta được
\(Q=16\left(-1\right)^3-13\left(-1\right)^2.2\)
\(Q=-16-26\)
\(Q=-42\)
c) \(H=x\left(x^3-y\right)+x^2\left(y-x^2\right)-y\left(x^2-3x\right)\)
\(H=x^4-xy+x^2y-x^4-x^2y+3xy\)
\(H=2xy\)
Thay x = 1/4 ; y = 2012 vào H ta được
\(H=2.\dfrac{1}{4}.2012\)
\(H=1006\)
T
1 tháng 8 2018
1.a)\(8\left(x-2\right)-2\left(3x-4\right)=2\)
\(\Leftrightarrow8x-16-6x+8=2\)
\(\Leftrightarrow2x-8=2\Leftrightarrow2x=10\Leftrightarrow x=5\)
b)\(10\left(3x-2\right)-3\left(5x+2\right)+5\left(11-4x\right)=25\)
\(\Leftrightarrow30x-20-15x-6+55-20x=25\)
\(\Leftrightarrow-5x+29=25\Leftrightarrow-5x=-4\Leftrightarrow x=\dfrac{4}{5}=0,8\)
\(c)2x\left(x+1\right)-x^2\left(x+2\right)+x^3-x+4=0\)
\(\Leftrightarrow2x^2+2x-x^3-2x^2+x^3-x+4=0\)
\(\Leftrightarrow x+4=0\Leftrightarrow x=-4\)
\(d)4x\left(3x+2\right)-6x\left(2x+5\right)+21\left(x-1\right)=0\)
\(\Leftrightarrow12x^2+8x-12x^2-30x+21x-21=0\)
\(\Leftrightarrow-x-21=0\Leftrightarrow-x=21\Leftrightarrow x=-21\)
2.
a)\(P=\left(4x^2-3y\right)2y-\left(3x^2-4y\right)3y\)
\(\Leftrightarrow8x^2y-6y^2-9x^2y-12y^2\)
\(\Leftrightarrow x^2y-18y^2\)
tại x=-1 , y=2
ta có:\(x^2y-18y^2=\left(-1\right)^2.2-18.2^2=2-72=-70\)
vậy \(P=\left(4x^2-3y\right)2y-\left(3x^2-4y\right)3y=-70\) tại x=-1,y=2
b)\(Q=4x^2\left(5x-3y\right)-x^2\left(4x+y\right)\)
\(\Leftrightarrow20x^3-12x^2y-4x^3-x^2y\)
\(\Leftrightarrow17x^3-13x^2y\)
tại x=-1,y=2
ta có:\(17x^3-13x^2y=17\left(-1\right)^3-13\left(-1\right)^2.2=-17-26=-43\)
vậy \(Q=4x^2\left(5x-3y\right)-x^2\left(4x+y\right)=-43\)
c)\(H=x\left(x^3-y\right)+x^2\left(y-x^2\right)-y\left(x^2-3x\right)\)
\(\Leftrightarrow x^4-xy+x^2y-x^3-x^2y+3xy\)
\(\Leftrightarrow x^4+2xy-x^3\)
tại x=1/4 và y=2012
ta có:\(x^4+2xy-x^3=\left(\dfrac{1}{4}\right)^4+2.\dfrac{1}{4}.2012-\left(\dfrac{1}{4}\right)^3\approx1006\)
TN
0
14 tháng 10 2017
a)
\(A=3\left(x-y\right)^2-2\left(x+y\right)^2-\left(x-y\right)\left(x+y\right)\)\(2A=\left[\left(x-y\right)-\left(x+y\right)\right]^2+5\left(x-y\right)^2-5\left(x+y\right)^2\)
\(2A=4y^2+5\left[\left(x-y\right)-\left(x+y\right)\right]\left[\left(x-y\right)+\left(x+y\right)\right]\)\(2A=4y^2+5\left[-2y\right]\left[2x\right]=4y^2-20xy=4y\left(y-5x\right)\\ \)\(A=2y\left(y-5x\right)\)
PH
2
8 tháng 10 2020
a) \(A=-4x^2-8x+3=-4\left(x^2+2x+1\right)+7=-4\left(x+1\right)^2+7\le7\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(x+1\right)^2=0\Rightarrow x=-1\)
Vậy Max(A) = 7 khi x = -1
b) \(B=6x-x^2+2=-\left(x^2-6x+9\right)+11=-\left(x-3\right)^2+11\le11\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(x-3\right)^2=0\Rightarrow x=3\)
Vậy Max(B) = 11 khi x = 3
c) \(C=x\left(2-3x\right)=-3\left(x^2-\frac{2}{3}x+\frac{1}{9}\right)+\frac{1}{3}=-3\left(x-\frac{1}{3}\right)^2+\frac{1}{3}\le\frac{1}{3}\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(x-\frac{1}{3}\right)^2=0\Rightarrow x=\frac{1}{3}\)
Vậy Max(C) = 1/3 khi x = 1/3
8 tháng 10 2020
d) \(D=3x-x^2+2=-\left(x^2-3x+\frac{9}{4}\right)+\frac{17}{4}=-\left(x-\frac{3}{2}\right)^2+\frac{17}{4}\le\frac{17}{4}\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(x-\frac{3}{2}\right)^2=0\Rightarrow x=\frac{3}{2}\)
Vậy Max(D) = 17/4 khi x = 3/2
e) \(E=3-2x^2+2xy-y^2-2x\)
\(E=-\left(x^2-2xy+y^2\right)-\left(x^2+2x+1\right)+4\)
\(E=-\left(x-y\right)^2-\left(x+1\right)^2+4\le4\left(\forall x,y\right)\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(x+1\right)^2=0\end{cases}}\Rightarrow x=y=-1\)
Vậy Max(E) = 4 khi x = y = -1
x = 0, y = 3