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1a/ z2 - 6z + 5 - t2 - 4t = z2 - 2 . 3z + 32 - 4 - t2 - 4t = (z2 - 2 . 3z + 32) - (22 + 2 . 2t + t2) = (z - 3)2 - (2 + t)2
b/ x2 - 2xy + 2y2 + 2y2 + 1 = x2 - 2xy + y2 + y2 + 2y + 1 = (x2 - 2xy + y2) + (y2 + 2y + 1) = (x - y)2 + (y + 1)2
c/ 4x2 - 12x - y2 + 2y + 8 = (2x)2 - 12x - y2 + 2y + 32 - 1 = [ (2x)2 - 2 . 3 . 2x + 32 ] - (y2 - 2y + 1) = (2x - 3)2 - (y - 1)2
2a/ (x + y + 4)(x + y - 4) = x2 + xy - 4x + xy + y2 - 4y + 4x + 4y + 16 = x2 + (xy + xy) + (-4x + 4x) + (-4y + 4y) + y2 + 16
= x2 + 2xy + y2 + 42 = (x + y)2 + 42
b/ (x - y + 6)(x + y - 6) = x2 + xy - 6x - xy - y2 + 6y + 6x + 6y - 36 = x2 + (xy - xy) + (-6x + 6x) + (6y + 6y) - y2 - 36
= x2 - y2 + 12y - 62 = x2 - (y2 - 12y + 62) = x2 - (y2 - 2 . 6y + 62) = x2 - (y - 6)2
c/ (y + 2z - 3)(y - 2z - 3) = y2 -2yz - 3y + 2yz - 4z2 - 6z - 3y + 6z + 9 = y2 + (-2yz + 2yz) + (-3y - 3y) + (-6z + 6z) - 4z2 + 9
= y2 - 6y - 4z2 + 9 = (y2 - 6y + 9) - 4z2 = (y - 3)2 - (2z)2
d/ (x + 2y + 3z)(2y + 3z - x) = 2xy + 3xz - x2 + 4y2 + 6yz - 2xy + 6yz + 9z2 - 3xz = (2xy - 2xy) + (3xz - 3xz) - x2 + (6yz + 6yz) + 9z2 + 4y2
= -x2 + 4y2 + 12yz + 9z2 = (4y2 + 12yz + 9z2) - x2 = [ (2y)2 + 2 . 2 . 3yz + (3z)2 ] - x2 = (2y + 3z)2 - x2
2a) \(4x^2-1=\left(2x\right)^2-1^2=\left(2x+1\right)\left(2x-1\right)\)
b) \(x^2+16x+64=\left(x+8\right)^2\)
c) \(x^3-8y^3=x^3-\left(2y\right)^3\)
\(=\left(x-2y\right)\left(x^2+2xy+4y^2\right)\)
d) \(9x^2-12xy+4y^2=\left(3x-2y\right)^2\)
x + y + z = 0 \(\Rightarrow\) x = - ( y + z )
\(\Rightarrow\) \(x^2\) = \((y+z)^2\) = \(y^2\) + \(z^2 \) + 2yz
\(\Rightarrow\) \(x^2\) - \(y^2\) - \(z^2 \) = 2xy
\(\Rightarrow\) (\(x^2-y^2-z^2\) )\(^2 \) = \((2xy)^2\)= \(4x^2y^2\)
\(\Rightarrow\) \(x^4 + y^4 + z^4\) - \(2x^2y^2\) - \(2x^2z^2\) = \(4x^2y^2\)
\(\Rightarrow\) \(x^4+y^4+z^4\) = \(4y^2z^2\) - \(2y^2z^2\) + \(2x^2y^2\) = \(2x^2y^2 + 2y^2z^2+ 2x^2z^2\)
\(\Rightarrow\) 2 (\(x^4+y^4+z^4\) ) = \((x^2+y^2+z^2)^2\) (đpcm)
\(x+y+z=0\Rightarrow x=-\left(y+z\right)\)
\(\Rightarrow x^2=\left(y+z\right)^2=y^2+z^2+2yz\)
\(\Rightarrow x^2-y^2-z^2=2xy\)
\(\Rightarrow\left(x^2-y^2-z^2\right)^2=\left(2xy\right)^2=4x^2y^2\)
\(\Rightarrow x^4+y^4+z^4-2x^2y^2-2x^2z^2+2y^2z^2=4x^2y^2\)
\(\Rightarrow x^4+y^4+x^4=4y^2z^2-2y^2z^2+2x^2z^2+2x^2y^2=2x^2y^2+2y^2z^2+2x^2z^2\)
\(\Rightarrow2\left(x^4+y^4+z^4\right)=\left(x^2+y^2+z^2\right)^2\) (đpcm)
Bài 1
a)\(=x^2+2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2+2\)
\(=\left(x+\frac{3}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)
MIN = \(-\frac{1}{4}\)khi \(x+\frac{3}{2}=0\Rightarrow x=-\frac{3}{2}\)
a)Đặt A= \(x^2+2x+11=\left(x+1\right)^2+10\)
vì \(\left(x+1\right)^2\ge0;\forall x\)
\(\Rightarrow\left(x+1\right)^2+11\ge11;\forall x\)
Hay \(A\ge11>0;\forall x\)
phần b và c mình sẽ giải ra hằng đẳng thức lập luận tương tự phần a
b)\(4x^2+8x+5\)
\(\left(2x\right)^2+2.2x.2+2^2+1\)
\(=\left(2x+2\right)^2+1\)
c) \(x^2+x+2=x^2+2.x.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+2\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{7}{4}\)
Ta có :
\(\left(x+2y\right)^2+\left(y-1\right)^2+\left(x-z\right)^2=0\)
=> \(\hept{\begin{cases}\left(x+2y\right)=0\\\left(y-1\right)=0\\\left(x-z\right)=0\end{cases}}\)=> \(\hept{\begin{cases}x=-2y\\y=1\\x=z\end{cases}}\)
=> \(\hept{\begin{cases}x=-2\\y=1\\z=-2\end{cases}}\)
M = x + 2y + 3z = -2 + 2 - 6 = (-6)
Chọn C