cho \(4x^2+2y^2+2z^2-4xy-4xz+2yz-6y-10z+34=0\)
tính \(M=\left(x-44\right)^{22}+\left(y-4\right)^6+\left(z-4\right)^{2013}\)
1) Cho \(4x^2+2y^2+2z^2-4xy-4xz+2yz-6y-10z+34=0\)
Tính M = \(\left(x-4\right)^{22}+\left(y-4\right)^6+\left(z-4\right)^{2013}\)
4x2 + 2y2 + 2z2-4xy - 4xz+2yz-6y-10z+34=0
<=>(-2x+y+z)2+(y-3)2+(z-5)2=0
<=>\(\left\{{}\begin{matrix}-2x+y+z=0\\y=3\\z=5\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x=4\\y=3\\z=5\end{matrix}\right.\)
Vậy A=\(\left(4-4\right)^{22}+\left(3-4\right)^6+\left(5-4\right)^{2013}=0^{22}+\left(-1\right)^6+1^{2013}=0+1+1=2\)
Cho x + 2y = -5. Tính giá trị biểu thức:
a/ \(x\left(x-3\right)+2y\left(2y-3\right)-4xy+19\)
b/\(x^2\left(x+1\right)+4y^2\left(2y+1\right)+2xy\left(3x+6y+2\right)+70\)
Sửa đề: \(x\left(x-3\right)+2y\left(2y-3\right)+4xy+19\)
a: \(x\left(x-3\right)+2y\left(2y-3\right)+4xy+19\)
\(=x^2-3x+4y^2-6y+4xy+19\)
\(=\left(x^2+4xy+4y^2\right)-3\left(x+2y\right)+19\)
\(=\left(x+2y\right)^2-3\left(x+2y\right)+19\)
\(=\left(-5\right)^2-3\cdot\left(-5\right)+19\)
=25+15+19=59
b: \(=x^3+x^2+8y^3+4y^2+2xy\left[3\left(x+2y\right)+2\right]+70\)
\(=x^3+8y^3+x^2+4y^2+2xy\cdot\left[3\cdot\left(-5\right)+2\right]+70\)
\(=\left(x+2y\right)^3-3\cdot x\cdot2y\left(x+2y\right)+\left(x+2y\right)^2-4xy+2xy\cdot\left(-13\right)+70\)
\(=\left(-5\right)^3+\left(-5\right)^2-6xy\cdot\left(-5\right)-4xy-26xy\)+70
\(=-125+25+70=-30\)
tìm x biết
1, 4x^2+4xy+4y^2-6y+3=0
2, \(\left(x+2\right)^2-x^4+4=0\)
3,\(4y^2+4xy+4y^2-6y+3=0\)
2,phân tích cac da tthuc thanh nhân tử
1,1,\(\left(x+3\right)\left(x+5\right)\left(x+7\right)\left(x+9\right)-125\)
2,\(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)
Tính tổng:
a/ \(\left(-5x^2y+3xy^2+7\right)+\left(-6x^2y+4xy^2-5\right)\)
b/ \(\left(2,4a^3-10a^2b\right)+\left(7a^2b-2,4a^3+3ab^2\right)\)
c/ \(\left(15x^2y-7xy^2-6y^3\right)+\left(2x^3-12x^2y+7xy^2\right)\)
B1 : a) làm tính nhân : \(\left(4x^3+3xy^2-2y^3\right)\left(3x^2-5xy-6y^2\right)\)
b) phân tích đa thức thành nhân tử :
\(8x^3+4x^2y-2xy^2-y^3\)
\(4x^2y^2-4x^2-4xy-y^2\)
c) tính GTBT :
\(\left(5xy-4y^2\right)\left(3x^2+4xy\right)-15xy\left(x+y\right)\left(x-y\right)\)tại \(x=2,y=16\)
d) thực hiện phép chia :
\(\left(9x^5-6x^3+18x^2-35x-42\right):\left(3x^3+5x+6\right)\)
a, \(=12x^5+9x^3y^2-6x^2y^3-20x^4y-15x^2y^3-10xy^4-24x^3y^2-18xy^4+12y^5\)
(tự rút gọn cái :P)
b, \(8x^3+4x^2y-2xy^2-y^3\)
\(=4x^2\left(2x+y\right)-y^2\left(2x+y\right)=\left(2x+y\right)^2\left(2x-y\right)\)
\(4x^2y^2-4x^2-4xy-y^2=4x^2y^2-\left(2x+y\right)^2\)
\(=\left(2x+y+2xy\right)\left(2xy-2x+y\right)\)
Mấy cái còn lại nhân tung ra là được mà :))))
Giải phương trình:
1. \(\left\{{}\begin{matrix}4x-2y=3\\6x-3y=5\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}2x-3y=5\\4x+6y=10\end{matrix}\right.\)
3. \(\left\{{}\begin{matrix}3x-4y+2=0\\5x+2y=14\end{matrix}\right.\)
4. \(\left\{{}\begin{matrix}2x+5y=3\\3x-2y=14\end{matrix}\right.\)
1) \(\left\{{}\begin{matrix}3x-2y=4\\4x+2y=10\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}3x-2y=4\\7x=14\end{matrix}\right.< =>\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
2)\(\left\{{}\begin{matrix}2x+3y=5\\4x+6y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+6y=10\\4x=6y=10\end{matrix}\right.\)
=> Hệ có vô số nghiệm.
3)\(\left\{{}\begin{matrix}3x-4y=-2\\10x+4y=28\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}3x-4y=-2\\13x=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
4)\(\left\{{}\begin{matrix}6x+15y=9\\6x-4y=28\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}6x+15y=9\\19y=19\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-1\end{matrix}\right.\)
cho \(4x^2+2y^2+2z^2-4xy-4xz+2yz-6y-10z+34=0\)
tính \(S=\left(x-4\right)^{2014}+\left(y-4\right)^{2015}+\left(z-4\right)^{2016}\)
Lời giải:
\(4x^2+2y^2+2z^2-4xy-4xz+2yz-6y-10z+34=0\)
\(\Leftrightarrow (4x^2-4xy+y^2)+y^2+2z^2-2z(2x-y)-6y-10z+34=0\)
\(\Leftrightarrow (2x-y)^2-2z(2x-y)+z^2+y^2+z^2-6y-10z+34=0\)
\(\Leftrightarrow (2x-y-z)^2+(y^2-6y+9)+(z^2-10z+25)=0\)
\(\Leftrightarrow (2x-y-z)^2+(y-3)^2+(z-5)^2=0\)
Do \((2x-y-z)^2; (y-3)^2; (z-5)^2\geq 0, \forall x,y,z\), nên để tổng của chúng bẳng $0$ thì:
\((2x-y-z)^2=(y-3)^2=(z-5)^2=0\Rightarrow \left\{\begin{matrix}
y=3\\
z=5\\
x=4\end{matrix}\right.\)
\(\Rightarrow S=(x-4)^{2014}+(y-4)^{2015}+(z-4)^{2016}=0+(-1)^{2015}+1^{2016}=-1+1=0\)
Tìm số nguyên dương x,y biết:
a) \(x^2+5y^2+2x-4xy-10y-9=0\)
b) \(5x^2+5y^2+8xy+2+2y-2x=0\)
c) \(x^2+5y^2-4xy+10x-22y+\left|x+y+z\right|+26=0\)
d) \(2x^2+2y^2+z^2+2xy+2xz+2yz+10x+6y+34=0\)
a/
\(\Leftrightarrow\left(x^2+4y^2+1-4xy+2x-4y\right)+\left(y^2-6y+9\right)-19=0\)
\(\Leftrightarrow\left(x-2y+1\right)^2+\left(y-3\right)^2=19\)
Do 19 không thể phân tích thành tổng của 2 số chính phương nên pt vô nghiệm
b/
\(\left(4x^2+4y^2+8xy\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)
\(\Leftrightarrow\left(2x+2y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
Do x; y nguyên dương nên \(\left(2x+2y\right)^2>0\Rightarrow VT>0\)
Pt vô nghiệm
c/
\(\Leftrightarrow\left(x^2+4y^2+25-4xy+10x-20y+25\right)+\left(y^2-2y+1\right)+\left|x+y+z\right|=0\)
\(\Leftrightarrow\left(x-2y+5\right)^2+\left(y-1\right)^2+\left|x+y+z\right|=0\)
Do x;y;z nguyên dương nên \(\left|x+y+z\right|>0\Rightarrow VT>0\)
Vậy pt vô nghiệm
d/
\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2=0\)
Do x;y;z nguyên dương nên vế phái luôn dương
Pt vô nghiệm
