Tìm min
D=2x2+9y2-6xy-6x+12y+2012
E=x2-2xy+4y2-2x+10+29
F=\(\frac{3}{2x-x^2-4}\)
G=\(\frac{2}{6x-5-9x^2}\)
Tìm GTNN:
1. G=2x2+9y2-6xy-6x-12y+2021
2. H=2x2+4y2+4xy+4y+9
3. I= x2-4xy+5y2+10x-22y+28
4. K=x2+5y2-4xy+6x-14y+15
1) Tìm x, y, z
a) 9x2 +y2 + 2z2 – 18x +4z – 6y +20 = 0
b) 5x2 +5y2 +8xy+2y – 2x+2 = 0
c) 5x2 +2y2 + 4xy – 2x + 4y +5 = 0
d) x2 + 4y2 + z2 =2x + 12y – 4z – 14
e) x2 +y2 – 6x + 4y +2= 0
2) Phân tích đa thức thành nhân tử
a) 3xy2 – 3x3 – 6xy +3x
b) 3x2 + 11x + 6
c) –x3 – 4xy2 + 4x2y +16x
d) xz – x2 – yz +2xy – y2
e) 4x2 – y2 – 6x + 3y
f) X4 – x3 – 10x2 + 2x +4
g) (x3 – x2 + x)(121 – 25y2 – 10y) – (x3 – x2 + x) – (121 – 25y2 – 10y) +1
h) X4 – 14x3 + 71x2 – 154x + 120
Giúp mik vs cần gấp!!!
\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11
e: Ta có: \(x^2-6x+y^2+4y+2=0\)
\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Dấu '=' xảy ra khi x=3 và y=-2
1) Tìm x, y, z
a) 9x2 +y2 + 2z2 – 18x +4z – 6y +20 = 0
b) 5x2 +5y2 +8xy+2y – 2x+2 = 0
c) 5x2 +2y2 + 4xy – 2x + 4y +5 = 0
d) x2 + 4y2 + z2 =2x + 12y – 4z – 14
e) x2 +y2 – 6x + 4y +2= 0
2) Phân tích đa thức thành nhân tử
a) 3xy2 – 3x3 – 6xy +3x
b) 3x2 + 11x + 6
c) –x3 – 4xy2 + 4x2y +16x
d) xz – x2 – yz +2xy – y2
e) 4x2 – y2 – 6x + 3y
f) X4 – x3 – 10x2 + 2x +4
g) (x3 – x2 + x)(121 – 25y2 – 10y) – (x3 – x2 + x) – (121 – 25y2 – 10y) +1
h) X4 – 14x3 + 71x2 – 154x + 120
Giúp mik với mik đang cần rất gấp ạ!!!
bài 2: viết cá đa thức sau dưới dạng hằng đẳng thức đáng nhớ sau :
a,x2+2x+1=
b,y2+4y+4=
c,9-6x+x2=
d,a2-14a+49=
e,m2-4m+4=
f,4x2-4x+1=
g,a2+10a+25=
h,100-20z+z2=
i,x2+6xy+9y2=
j,4x2-12xz+25b2=
k,a2+10ab+25b2=
l,x4+2x2+1=
m,y6-2y3+1=
n,c10-10c5+25=
o,9x4+12x2y+4y2=
p,25m4n6-10m2n3=
em đang cần gấp ,giúp em với
\(a,=\left(x+1\right)^2\\ b,=\left(y-2\right)^2\\ c,=\left(x-3\right)^2\\ d,=\left(a-7\right)^2\\ e,=\left(m-2\right)^2\\ f,=\left(2x-1\right)^2\\ g,=\left(a+5\right)^2\\ h,=\left(z-10^2\right)\\ i,=\left(x+3y\right)^2\\ j,=\left(2x-5b\right)^2\\ k,=\left(a+5\right)^2\\ l,=\left(x^2+1\right)^2\\ m,=\left(y^3-1\right)^2=\left(y-1\right)^2\left(y^2+y+1\right)^2\\ n,=\left(c^5-5\right)^2\\ o,=\left(3x^2+2y\right)^2\\ p,=5m^2n^3\left(5m^2n^3-2\right)\)
phân tích đa thức thành nhân tử
a) xy+y2-x-y
b) 25-x2+4xy-4y2
c) xy+yz-2y-2z
d) x2-6xy+9y2-25z2
e) 3x2-3y2-12x+12y
f) 4x3+4xy2+8x2y-16x
g) x2-5x+4
h) x4+5x2+4
i) 2x2+3x-5
k) x3-2x2+6x-5
l) x2-4x+3
Mong mọi người giúp đỡ em cảm ơn ạ
Bài làm
a) xy + y2 - x - y
= ( xy + y2 ) - ( x + y )
= y( x + y ) - ( x + y )
= ( x + y )( y - 1 )
b) 25 - x2 + 4xy - 4y2
= 25 - ( x2 - 4xy + 4y2 )
= 25 - ( x - 2y )2
= ( 5 - x + 2y )( 5 + x - 2y )
c) xy + xz - 2y - 2z
= ( xy + xz ) - ( 2y + 2z )
= x( y + z ) - 2( y + z )
= ( y + z )( x - 2 )
d) x2 - 6xy + 9y2 - 25z2
= ( x2 - 6xy + 9y2 ) - 25z2
= ( x - 3y )2 - 25z2
= ( x - 3y - 5z )( z - 3y + 5z )
e) 3x2 - 3y2 - 12x + 12y
= 3( x - y )( x + y ) - 12( x - y )
= ( x - y )[ 3( x + y ) - 12 ]
f) 4x3 + 4xy2 + 8x2y - 16x
= 4x( x2 + y2 + 2xy - 4 )
= 4x[ ( x + y)2 - 4 ]
= 4x( x + y - 2 )( x + y + 2 )
g) x2 - 5x + 4
= x2 - x - 4x + 4
= x( x - 1 ) - 4( x - 1 )
= ( x - 1 )( x - 4 )
h) x4 + 5x2 + 4
= x4 + x2 + 4x2 + 4
= x2( x2 + 1 ) + 4( x2 + 1 )
= ( x2 + 1 )( x2 + 4 )
i) 2x2 + 3x - 5
= 2x2 - 5x + 2x - 5
= 2x( x + 1 ) - 5( x + 1 )
= ( x + 1 )( 2x - 5 )
k) x3 - 2x2 + 6x - 5 ( không biết làm )
l) x2 - 4x + 3
= ( x2 - 4x + 4 ) - 1
= ( x - 2 )2 - 1
= ( x - 3 )( x - 1 )
# Học tốt #
1) x3-x2+2x-2 4) ax-2x-a2+2a 7) x2-6xy-25z2+9y2
2) x2-y2+2x+2y 5) 2xy +3z+6y+xz 8) x3-2x2+x
3) x2/4+2xy+4y2-25 6) x2y2+yz+y3+zx2 9) x4+4
Bài 2: Rút gọn biểu thức
A=(x-2)(x2+2x+4)-(128+x3)
B=(2x+3y)(4x2-6xy+9y2)-(3x-2y)(9x2+6xy+4y2)
\(A=x^3-8-128-x^3=-136\\ B=8x^3+27y^3-27x^3+8y^3=-19x^3+35y^3\)
\(A=\left(x-2\right)\left(x^2+2x+4\right)-\left(128+x^3\right)=x^3-8-128-x^3=-136\)
\(B=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)-\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)=8x^3+27y^3-27x^3+8y^3=-19x^3+35y^3\)
\(A=x^3+2x^2+4x-2x^2-4x-8-128-x^3\)
\(A=-136\)
\(B=\left(2x+3y\right)\left(2x-3y\right)^2-\left(3x-2y\right)\left(3x+2y\right)^2\)
\(B=\left(2x+3y\right)\left(2x-3y\right)\left(2x-3y\right)-\left(3x-2y\right)\left(3x+2y\right)\left(3x+2y\right)\)
\(B=\left(4x^2-9y^2\right)\left(2x-3y\right)-\left(9x^2-4y^2\right)\left(3x+2y\right)\)
\(B=8x^3-12x^2y-18xy^2-27y^3-27x^3-18x^2y+12xy^2+8y^3\)
\(B=-19x^3-30x^2y-6xy^2-19y^3\)
\(B=-19\left(x^3-y^3\right)-6xy\left(5x+y\right)\)
a, -x2 + 2x + 3
b, x2 - 2x + 4y2 - 4y + 8 c, -x2 - y2 + xy + 2x + 2y + 4 d, x2 + 5y2 - 4xy - 2y + 2015 e, 2x2 + y2 + 6x + 2y + 2xy + 2018A= -x2+2x+3
=>A= -(x2-2x+3)
=>A= -(x2-2.x.1+1+3-1)
=>A=-[(x-1)2+2]
=>A= -(x+1)2-2
Vì -(x+1)2 ≤0=> A≤-2
Dấu "=" xảy ra khi
-(x+1)2=0 => x=-1
Vây A lớn nhất= -2 khi x= -1
B=x2-2x+4y2-4y+8
=> B= (x2-2x+1)+(4y2-4y+1)+6
=> B=(x-1)2+(2y+1)2+6
=> B lớn nhất=6 khi x=1 và y=-1/2
Tìm GTLN hoặc GTNN của các bt sau:
a)A=1+6x-x2
b) B=-2x2+ 6x+ 8
c) C=x2+ 3y2- 2xy- 2y
d) D=2x2+ y2+ 2xy- 2x+ 2y+ 2
e) E= x2+ 2y2+ 9z2- 2x+ 12y+ 6z+ 24
f) F=\(\frac{7}{10x-x^2-30}\)
g) G=\(\frac{6x+17}{x^2+2}\)
h) H=\(\frac{\left(2+x^2\right)\left(8+x^2\right)}{x^2}\)
i) I=\(\frac{3x^2+6x+10}{x^2+2x+3}\)