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(2x - 1)2 - 2(4x2 - 1) + (2x + 1)2
= (2x - 1)2 - 2(2x - 1)(2x + 1) + (2x + 1)2
= (2x - 1 - 2x - 1)2
= (-2)2 = 4
\(\left(2x-1\right)^2-2\left(4x^2-1\right)+\left(2x+1\right)^2\)
\(=\left(2x-1\right)^2-2\left(2x-1\right)\left(2x+1\right)+\left(2x+1\right)^2\)
\(=\left[\left(2x-1\right)-\left(2x+1\right)\right]^2\)
\(=\left(2x-1-2x-1\right)^2=\left(-2\right)^2=4\)
Ta có: \(2x^2+2y^2+z^2+25-6y-2xy-8x+2z\left(y-x\right)=0\)
\(\Leftrightarrow\left(x^2-8x+16\right)+\left(y^2-6y+9\right)+\left(x^2-2xy+y^2\right)-2\left(x-y\right)z+z^2=0\)
\(\Leftrightarrow\left(x-4\right)^2+\left(y-3\right)^2+\left[\left(x-y\right)^2-2\left(x-y\right)z+z^2\right]=0\)
\(\Leftrightarrow\left(x-4\right)^2+\left(y-3\right)^2+\left(x-y-z\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(x-4\right)^2=0\\\left(y-3\right)^2=0\\\left(x-y-z\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\y=3\\z=1\end{cases}}\)
Chỗ (x²-8x+16)
16 là ở đâu ra vậy bạn
Chỗ (y²-6y+9 )
9 là ở đâu ra nx v
Ta có:
\(G=x^2+3y^2+2xy-6y+3\)
\(G=\left(x^2+2xy+y^2\right)+\left(2y^2-6y+\frac{18}{4}\right)-\frac{3}{2}\)
\(G=\left(x+y\right)^2+2\left(y-\frac{3}{2}\right)^2-\frac{3}{2}\ge-\frac{3}{2}\left(\forall x,y\right)\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x+y\right)^2=0\\2\left(y-\frac{3}{2}\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{3}{2}\\y=\frac{3}{2}\end{cases}}\)
Vậy Min(G) = -3/2 khi \(\hept{\begin{cases}x=-\frac{3}{2}\\y=\frac{3}{2}\end{cases}}\)
G = x2 + 3xy2 + 2xy - 6y + 3
<=> G = ( x2 + 2xy + y2 ) + ( y2 - 6y + 9 ) - 6
<=> G = ( x + y )2 + ( y - 3 )2 - 6
Vì ( x + y )2\(\ge\)0 ; ( y - 3 )2\(\ge\)0\(\forall\)x ; y
=> G = ( x + y )2 + ( y - 3 )2 - 6\(\ge\)- 6
Dấu "=" xảy ra <=>\(\orbr{\begin{cases}\left(x+y\right)^2=0\\\left(y-3\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-y\\y=3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\y=3\end{cases}}\)
Vậy minG = - 6 <=> x = - 3 ; y = 3
\(axz^2-ax-ayz^2+ax+ay+az^3\)
= \(axz^2-ayz^2+ay+az^3\)
\(=a\left(xz^2-yz^2+y+z^3\right)\)
Bạn vẫn nên kiểm tra đề bài lại nhé
4x ( x - 5 ) - ( x - 1 ) ( 4x - 3 ) = 5
<=> 4x2 - 20x - ( 4x2 - 7x + 3 ) = 5
<=> 4x2 - 20x - 4x2 + 7x - 3 = 5
<=> - 13x = 8
<=> x = - 8/13
\(\left(x-3\right)\left(x+3\right)-\left(x+2\right)\left(x-1\right)\)
\(=x^2-3^2-\left(x^2+x-2\right)\)
\(=x^2-9-x^2-x+2\)
\(=-x-7\)
x = 1/3
\(=-\frac{1}{3}-7\)
\(=-\frac{22}{3}\)
a) = -15x4y7 - (-10x5y6) - (-5x3y5)
= -15x4y7 + 10x5y6 + 5x3y5
Bài 2:
a) \(x^2-y^2+3x-3y=\left(x^2-y^2\right)+\left(3x-3y\right)\)
\(=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\)
b) \(5x-5y+x^2-2xy+y^2=\left(5x-5y\right)+\left(x^2-2xy+y^2\right)\)
\(=5\left(x-y\right)+\left(x-y\right)^2=\left(x-y\right)\left(x-y+5\right)\)
c) \(x^2-5x+4=x^2-x-4x+4=\left(x^2-x\right)-\left(4x-4\right)\)
\(=x\left(x-1\right)-4\left(x-1\right)=\left(x-1\right)\left(x-4\right)\)