Giải PT sau:
a) 9x2+29y2+30xy=6(x+5y−4)−29x2+29y2+30xy=6(x+5y−4)−2
b)5x2+5y2+8xy+2y−2x+2=05x2+5y2+8xy+2y−2x+2=0
c)y2−2y+3=6x2+2x+4y2−2y+3=6x2+2x+4
d)−9x2+18x−17x2−2x+3=y(y+4)
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
Giúp mik vs cần gấp!!!
\(a,\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,\Leftrightarrow\left(4x^2+4xy+y^2\right)+\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
a,9x^2+y^2+2z^2−18x+4z−6y+20=0
⇔9(x−1)^2+(y−3)^2+2(z+1)^2=0
⇔x=1;y=3;z=−1
b,5x^2+5y^2+8xy+2y−2x+2=0
⇔4(x+y)2+(x−1)2+(y+1)2=0
⇔x=−y;x=1y=−1⇔x=1y=−1
c,5x^2+2y^2+4xy−2x+4y+5=0
⇔(2x+y)^2+(x−1)^2+(y+2)^2=0
⇔2x=−y;x=1;y=−2
⇔x=1;y=−2
⇔(x−1)^2+(2y−3)^2+(z+2)^2=0
\(d,\Leftrightarrow\left(x^2-2x+1\right)+\left(4y^2-12y+9\right)+\left(z^2+4z+4\right)=0\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
\(\Rightarrow\)PT vô nghiệm vì 11 không phải là tổng 2 số chính phương
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!!!
Tìm GTNN
A= x2 + y2 – 6x + 4y + 20
B= 9x2 + y2 + 2z2 – 18x + 4z – 6y +30
C= x2 +y2 + z2 – xy – yz – zx + 3
D= 5x2 + 2y2 + 4xy – 2x + 4y + 2021
E= x2 – 2x+ 4y2 + 4y + 2014
F= 5x2 + 5y2 + 8xy + 2y – 2x + 30
K= x2 + 4y2 + z2 – 2x + 12y – 4z +44
Giúp mik vs cần gấp!!!!
$A=x^2+y^2-6x+4y+20=(x^2-6x+9)+(y^2+4y+4)+7$
$=(x-3)^2+(y+2)^2+7\geq 0+0+7=7$
Vậy $A_{\min}=7$. Giá trị này đạt tại $(x-3)^2=(y+2)^2=0$
$\Leftrightarrow x=3; y=-2$
---------------------
$B=9x^2+y^2+2z^2-18x+4z-6y+30$
$=(9x^2-18x+9)+(y^2-6y+9)+(2z^2+4z+2)+10$
$=9(x^2-2x+1)+(y^2-6y+9)+2(z^2+2z+1)+10$
$=9(x-1)^2+(y-3)^2+2(z+1)^2+10\geq 10$
Vậy $B_{\min}=10$. Giá trị này đạt tại $(x-1)^2=(y-3)^2=(z+1)^2$
$\Leftrightarrow x=1; y=3; z=-1$
$C=x^2+y^2+z^2-xy-yz-xz+3$
$2C=2x^2+2y^2+2z^2-2xy-2yz-2xz+6$
$=(x^2-2xy+y^2)+(y^2-2yz+z^2)+(x^2-2xz+z^2)+6$
$=(x-y)^2+(y-z)^2+(z-x)^2+6\geq 6$
$\Rightarrow C\geq 3$
Vậy $C_{\min}=3$. Giá trị này đạt tại $x-y=y-z=z-x=0$
$\Leftrihgtarrow x=y=z$
--------------------------------------
$D=5x^2+2y^2+4xy-2x+4y+2021$
$=2(y^2+2xy+x^2)+3x^2-2x+4y+2021$
$=2(x+y)^2+4(x+y)+3x^2-6x+2021$
$=2(x+y)^2+4(x+y)+2+3(x^2-2x+1)+2016$
$=2[(x+y)^2+2(x+y)+1]+3(x^2-2x+1)+2016$
$=2(x+y+1)^2+3(x-1)^2+2016\geq 2016$
Vậy $D_{\min}=2016$ khi $x+y+1=x-1=0$
$\Leftrightarrow x=1; y=-2$
$E=x^2-2x+4y^2+4y+2014$
$=(x^2-2x+1)+(4y^2+4y+1)+2012$
$=(x-1)^2+(2y+1)^2+2012$
$\geq 2012$
Vậy $E_{\min}=2012$. Giá trị này đạt tại $x-1=2y+1=0$
$\Leftrightarrow x=1; y=\frac{-1}{2}$
----------------------
$F=5x^2+5y^2+8xy+2y-2x+30$
$=4(x^2+2xy+y^2)+x^2+y^2+2y-2x+30$
$=4(x+y)^2+(x^2-2x+1)+(y^2+2y+1)+28$
$=4(x+y)^2+(x-1)^2+(y+1)^2+28\geq 28$
Vậy $F_{\min}=28$. Giá trị này đạt tại $x+y=x-1=y+1=0$
$\Leftrightarrow x=1; y=-1$
Giải PT sau:
a) \(9x^2+29y^2+30xy=6\left(x+5y-4\right)-2\)
b)\(5x^2+5y^2+8xy+2y-2x+2=0\)
c)\(y^2-2y+3=\frac{6}{x^2+2x+4}\)
d)\(\frac{-9x^2+18x-17}{x^2-2x+3}=y\left(y+4\right)\)
a/
\(9x^2+25y^2+1+30xy-6x-10y+4y^2-20y+25=0\)
\(\Leftrightarrow\left(3x+5y-1\right)^2+\left(2y-5\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+5y-1=0\\2y-5=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\frac{23}{6}\\y=\frac{5}{2}\end{matrix}\right.\)
b/
\(4x^2+4y^2+8xy+x^2-2x+1+y^2+2y+1=0\)
\(\Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x-1=0\\y+1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
c/
\(y^2-2y+1+2=\frac{6}{x^2+2x+1+3}\)
\(\Leftrightarrow\left(y-1\right)^2+2=\frac{6}{\left(x+1\right)^2+3}\)
Ta có \(VT=\left(y-1\right)^2+2\ge2\)
\(\left(x+1\right)^2+3\ge3\Rightarrow VP=\frac{6}{\left(x+1\right)^2+3}\le\frac{6}{3}=2\)
\(\Rightarrow VT\ge VP\)
Dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}y-1=0\\x+1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
d/
\(\frac{-9x^2+18x-9-8}{x^2-2x+1+2}=y^2+4y+4-4\)
\(\Leftrightarrow\frac{-9\left(x-1\right)^2-8}{\left(x-1\right)^2+2}=\left(y+2\right)^2-4\)
\(\Leftrightarrow\frac{-9\left(x-1\right)^2-18+10}{\left(x-1\right)^2+2}=\left(y+2\right)^2-4\)
\(\Leftrightarrow-9+\frac{10}{\left(x-1\right)^2+2}=\left(y+2\right)^2-4\)
\(\Leftrightarrow\frac{10}{\left(x-1\right)^2+2}=\left(y+2\right)^2+5\)
Ta có \(\left(x-1\right)^2+2\ge2\Rightarrow\frac{10}{\left(x-1\right)^2+2}\le\frac{10}{2}=5\Rightarrow VT\le5\)
\(\left(y+2\right)^2+5\ge5\Rightarrow VP\ge5\)
\(\Rightarrow VT\le VP\)
Dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Bài 4:Cho các số x,y thoả mãn đẳng thức 5x2+5y2+8xy-2x+2y+2=0
Tính giá trị của biểu thức:M=(x+y)2019 + (x-2)2020 + (y+1)2021
Bài 11 : rút gọn các biểu thức
a. ( 7x + 4 )2 - ( 7x + 4 ) ( 7x - 4 )
b. ( x + 2y)2 - 6xy ( x + 2y )
Bài 12 : Tính
a. (1/2x + 4)2
b. ( 7x - 5y )2
c. ( 6x2 + y2 ) ( y2 - 6x2 )
d . ( x + 2y )2
e. ( x - 3y ) ( x + 3y )
f. ( 5 - x )2
Bài 12:
a) \(\left(\dfrac{1}{2}x+4\right)^2\)
\(=\left(\dfrac{1}{2}x\right)^2+2\cdot\dfrac{1}{2}x\cdot4+4^2\)
\(=\dfrac{1}{4}x^2+4x+16\)
b) \(\left(7x-5y\right)^2\)
\(=\left(7x\right)^2-2\cdot7x\cdot5y+\left(5y\right)^2\)
\(=49x^2-70xy+25y^2\)
c) \(\left(6x^2+y^2\right)\left(y^2-6x^2\right)\)
\(=\left(y^2+6x^2\right)\left(y^2-6x^2\right)\)
\(=y^4-36x^4\)
d) \(\left(x+2y\right)^2\)
\(=x^2+2\cdot x\cdot2y+\left(2y\right)^2\)
\(=x^2+4xy+4y^2\)
e) \(\left(x-3y\right)\left(x+3y\right)\)
\(=x^2-\left(3y\right)^2\)
\(=x^2-9y^2\)
f) \(\left(5-x\right)^2\)
\(=5^2-2\cdot5\cdot x+x^2\)
\(=25-10x+x^2\)
\(11,\)
\(a,\left(7x+4\right)^2-\left(7x+4\right)\left(7x-4\right)\)
\(=\left(7x+4\right)\left(7x+4-7x+4\right)\)
\(=\left(7x+4\right).8=56x+32\)
\(b,\left(x+2y\right)^2-6xy\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x+2y-6xy\right)\)
Bài `12`
`(1/2x+4)^2`
`=(1/2x)^2 + 2 . 1/2x.4 + 4^2`
`= 1/4 x^2 +4x + 16`
__
`(7x-5y)^2`
`=(7x)^2-2.7x.5y+(5y)^2`
`= 49x^2 - 70xy + 25y^2`
__
`(6x^2+y^2)(y^2-6x^2)`
`=(y^2+6x^2)(y^2-6x^2)`
`=(y^2)^2 - (6x^2)^2`
`=y^4-36x^4`
__
`(x+2y)^2`
`=x^2+ 2.x.2y+(2y)^2`
`= x^2 + 4xy +4y^2`
__
`(x-3y)(x+3y)`
`=x^2 - (3y)^2`
`=x^2 - 9y^2`
__
`(5-x)^2`
`=5^2 -2.5.x+x^2`
`=25 - 10x+x^2`
Bài `11`
`(7x+4)^2 -(7x+4)(7x-4)`
`= (7x+4)(7x+4) -(7x+4)(7x-4)`
`=(7x+4)(7x+4-7x+4)`
`=8(7x+4)`
`= 56x+32`
__
`(x+2y)^2-6xy (x+2y)`
`= (x+2y) (x+2y-6xy)`
Rút gọn :
a. ( x + 2 ) ( x2 - 2x + 4 ) - ( 1 - 3x ) ( 1 + 3x + 9x2)
b . ( x + y ) ( y2 - 2y + 4 ) + ( 5 - y ) ( 25 + 5y + y2)
a: =x^3+8-1+27x^3=28x^3+7
b: Sửa đề: (2+y)(y^2-2y+4)+(5-y)(25+5y+y^2)
=8+y^3+125-y^3
=133
Hãy giải các phương trình sau đây :
1, x2 - 4x + 4 = 0
2, 2x - y = 5
3, x + 5y = - 3
4, x2 - 2x - 8 = 0
5, 6x2 - 5x - 6 = 0
6,( x2 - 2x )2 - 6 (x2 - 2x ) + 5 = 0
7, x2 - 20x + 96 = 0
8, 2x - y = 3
9, 3x + 2y = 8
10, 2x2 + 5x - 3 = 0
11, 3x - 6 = 0
1) Ta có: \(x^2-4x+4=0\)
\(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
Vậy: S={2}