tìm min,max:
H=1/4X^2 + X - 2
I= - X^2 - Y^2 + 2X - 2Y - 3
K=X^2 + X + 20
M= - X^2 + 3X + 2
N= - 2X^2 + X
P= -X^2 - 2Y^2 + 2XY - Y + 1
giúp mik vs nha
\(Tìm Min : B=2x²-4x-8 C=x²-2xy+2y²+2x-10y+17 D=x²-xy+y²-2x-2y E=(x²+x-6)(x²+x+2) F=(x+1)(x+2)(x+3)(x+4) Tìm Max G= 4x-x2 H=25-x-5x2 \)
1.Tìm Min
A=x^4-8xy-x^3y+x^2y^2-xy^3+y^4+1017
B=x^2+xy+y^2-3x-3y
2.Tìm Max
A=-x^2+2xy-4y^2+2x+10y+5
B= -x2 - 2y2 - 2xy + 2x - 2y -15
Tìm Min :
B=2x²-4x-8
C=x²-2xy+2y²+2x-10y+17
D=x²-xy+y²-2x-2y
E=(x²+x-6)(x²+x+2)
F=(x+1)(x+2)(x+3)(x+4)
Tìm Max
G= 4x-x2
H=25-x-5x2
\(B=2x^2-4x-8=2\left(x^2-2x-4\right)\)
\(=2\left(x^2-2x+1-5\right)\)
\(=2\left[\left(x-1\right)^2-5\right]\)
\(=2\left(x-1\right)^2-10\ge-10\)
Vậy \(B_{min}=-10\Leftrightarrow x-1=0\Leftrightarrow x=1\)
\(F=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)\)
Đặt \(x^2+5x+4=t\)
\(\RightarrowĐT=t\left(t+2\right)=t^2+2t+1-1\)
\(=\left(t+1\right)^2-1\ge-1\)
hay \(\left(x^2+5x+5\right)^2-1\ge-1\)
Vậy \(F_{min}=-1\Leftrightarrow x^2+5x+5=0\)
\(\Leftrightarrow x^2+5x+\frac{25}{4}-\frac{5}{4}=0\)
\(\Leftrightarrow\left(x+\frac{5}{2}\right)^2=\frac{5}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{5}{2}=\sqrt{\frac{5}{4}}\\x+\frac{5}{2}=-\sqrt{\frac{5}{4}}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{\frac{5}{4}}-\frac{5}{2}\\x=-\sqrt{\frac{5}{4}}-\frac{5}{2}\end{cases}}\)
\(G=4x-x^2=-\left(x^2-4x+4-4\right)\)
\(=-\left[\left(x-2\right)^2-4\right]=-\left(x-2\right)^2+4\le4\)
Vậy \(G_{max}=4\Leftrightarrow x-2=0\Leftrightarrow x=2\)
\(H=25-x-5x^2=-5\left(x^2+\frac{x}{5}-5\right)\)
\(=-5\left(x^2+2x.\frac{1}{10}+\frac{1}{100}-\frac{501}{100}\right)\)
\(=-5\left[\left(x+\frac{1}{10}\right)^2-\frac{501}{100}\right]\)
\(=-5\left(x+\frac{1}{10}\right)^2+\frac{101}{20}\le\frac{101}{2}\)
Vậy \(H_{max}=\frac{101}{2}\Leftrightarrow x+\frac{1}{10}=0\Leftrightarrow x=-\frac{1}{10}\)
bài 1: tìm đa thức M biết
a, \(M+x^2\)\(-3xy-y^2\)=\(2x^2\) \(-y^2+xy\)
b,\(x^2y^2-2x^2y^3+2x^2-y^3-P=x^2y^3-3x^2y^2-x^2\)
bài 2: tìm nghiệm của các đa thức sau
a, \(5\left(x-2\right)-2\left(x+3\right)\)
b, \(5x^2-125\)
c,\(2x^2-x-3\)
giúp mik vs ạ
2:
a: A(x)=0
=>5x-10-2x-6=0
=>3x-16=0
=>x=16/3
b: B(x)=0
=>5x^2-125=0
=>x^2-25=0
=>x=5 hoặc x=-5
c: C(x)=0
=>2x^2-x-3=0
=>2x^2-3x+2x-3=0
=>(2x-3)(x+1)=0
=>x=3/2 hoặc x=-1
a. 3x^2-3y^2-x-y
b. 2x^2+4xy-16+2y^2
c. -x^2-x+2
d. 3x^2-7x+4
e.-2x^2+3x-1
f. x^2+2xy+y^2-2x-2y
g.x^3-2x^2+1
h.4x^2-3x-1
k. 2x^2+5x+3
l. x^2-2x-y^2+1
a) \(3x^2-3y^2-x-y\)
\(\Leftrightarrow3\left(x^2-y^2\right)-x-y\)
\(\Leftrightarrow3\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(\Leftrightarrow3\left(x-y\right)\)
d) \(3x^2-7x+4\)
\(\Leftrightarrow3x^2-7x+7-3\)
\(\Leftrightarrow\left(3x^2-3\right)-\left(7x-7\right)\)
\(\Leftrightarrow3\left(x^2-1\right)-7\left(x-1\right)\)
\(\Leftrightarrow3\left(x-1\right)\left(x+1\right)-7\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(3\left(x+1\right)-7\right)\)
\(\Leftrightarrow\left(x+1\right)\left(3x-6\right)\)
e) \(-2x^2+3x-1\)
\(\Leftrightarrow\left(-2x^2-1^2\right)+3x\)
\(\Leftrightarrow\left(-2x-1\right)\left(-2x+1\right)+3x\)
f) \(x^2+2xy+y^2-2x-2y\)
\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)
\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)
k) \(2x^2+5x+3\)
\(\Leftrightarrow2x^2+2x+3x+3\)
\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)\)
\(\Leftrightarrow\left(2x+3\right)\left(x+1\right)\)
l) \(x^2-2x-y^2+1\)
\(\Leftrightarrow\left(x^2-2x+1\right)-y^2\)
\(\Leftrightarrow\left(x-1\right)^2-y^2\)
\(\Leftrightarrow\left(x-1-y\right)\left(x-1+y\right)\)
a) \(3x^2-3y^2-x-y\)
\(\Leftrightarrow3\left(x^2-y^2\right)-x-y\)
\(\Leftrightarrow3\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(\Leftrightarrow3\left(x-y\right)\)
d) \(3x^2-7x+4\)
\(\Leftrightarrow3x^2-7x+7-3\)
\(\Leftrightarrow\left(3x^2-3\right)-\left(7x-7\right)\)
\(\Leftrightarrow3\left(x^2-1\right)-7\left(x-1\right)\)
\(\Leftrightarrow3\left(x-1\right)\left(x+1\right)-7\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(3\left(x+1\right)-7\right)\)
\(\Leftrightarrow\left(x+1\right)\left(3x-6\right)\)
e) \(-2x^2+3x-1\)
\(\Leftrightarrow\left(-2x^2-1^2\right)+3x\)
\(\Leftrightarrow\left(-2x-1\right)\left(-2x+1\right)+3x\)
f) \(x^2+2xy+y^2-2x-2y\)
\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)
\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)
k) \(2x^2+5x+3\)
\(\Leftrightarrow2x^2+2x+3x+3\)
\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)\)
\(\Leftrightarrow\left(2x+3\right)\left(x+1\right)\)
l) \(x^2-2x-y^2+1\)
\(\Leftrightarrow\left(x^2-2x+1\right)-y^2\)
\(\Leftrightarrow\left(x-1\right)^2-y^2\)
\(\Leftrightarrow\left(x-1-y\right)\left(x-1+y\right)\)
Triển khai giúp mình
a) ( 5x - x^2 ) . ( 5 + x^2 )
b) ( 2x - y ) . ( 4x^2 + 2xy + y^2)
c) ( x + 3) . ( x^2 - 3x + 9)
d) -x3 + 3x^2 - 3x + 1
e) x^2 - 2x + 9
g) ( x + 1) . ( x-1)
h) ( x -2y) . ( x + 2y)
i) 25a^2 + 4b^2 - 20ab
It's khai triển :)
a) \(\left(5x-x^2\right)\left(5x+x^2\right)=25x^2-x^4\)
b) \(\left(2x-y\right)\left(4x^2+2xy+y^2\right)=8x^3-y^3\)
c) \(\left(x+3\right)\left(x^2-3x+9\right)=x^3-27\)
d) \(-x^3+3x^2-3x+1=\left(1-x\right)^3\)
e) \(x^2-2x+9=\left(x-1\right)^2+8??\) ko ra gì cả-.-
g) \(\left(x+1\right)\left(x-1\right)=x^2-1\)
h) \(\left(x-2y\right)\left(x+2y\right)=x^2-4y^2\)
i) \(25a^2+4b^2-20ab=\left(5a-2b\right)^2\)
tìm max y-2y^2+x^2-5x và
7xy-3x^2-4y^2+2x-3y+5
tìm min
3y^2-2xy+6x^2 -x +2y-1
Tìm min: a, A=9x^2 - 6x +5 b, B= 2x^2 + 2xy + y^2 -2x +2y+2
Tìm max: a, M= -2x^2 +3x +1 b, N =-x^2 + 2xy - 4y^2 + 2x+ 10y +5
1. Tìm min:
a, x2-x+1
b, 3x2+5x-2
c, x2+2y2-2xy-4y+5
d, x2+2y2+2xy-4x+2y+2017
e, 2x2+4y2-4xy-4x-4y+2003
2. Tìm max:
a, -x2+3x
b, -2x2+x-1
c, -x2-y2+xy+2x+2y
d, -5x2-2xy-2y2+14x+10y
e, -8x2-3y2-26x+6y+100
a ) \(x^2-x+1\)
\(\Leftrightarrow\left(x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right)+\dfrac{3}{4}\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có : \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Vậy GTNN là \(\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}.\)