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\(B=1+5y-y^2=-\left(y^2-5y-1\right)\)
\(=-\left(y^2-2.\frac{5}{2}x+\frac{25}{4}-\frac{29}{4}\right)\)
\(=-\left[\left(y-\frac{5}{2}\right)^2-\frac{29}{4}\right]\)
\(=-\left(y-\frac{5}{2}\right)^2+\frac{29}{4}\le\frac{29}{4}\)
\(C=4x-x^2+1=-\left(x^2-4x-1\right)\)
\(=-\left(x^2-4x+4-5\right)\)
\(=-\left[\left(x-2\right)^2-5\right]\)
\(=-\left(x-2\right)^2+5\le5\)
mk lm mẫu cho bạn 1 phần nhé
a) \(A=3x^2+y^2+10x-2xy+26\)
\(=\left(x^2-2xy+y^2\right)+2\left(x^2+5x+6,25\right)+13,5\)
\(=\left(x-y\right)^2+2\left(x+2,5\right)^2+13,5\ge13,5\)
Dấu "=" xảy ra <=> \(x=y=-2,5\)
Vậy MIN A = 13,5 khi x = y = - 2,5
\(=x^3+x^2-\left(4x+4\right)=x^2\left(x+1\right)-4\left(x+1\right)=\left(x^2-4\right)\left(x+1\right)\)
\(=\left(x-2\right)\left(x+1\right)\left(x+2\right)\)
\(x^4+x^3+x^2-1=x^3\left(x+1\right)+\left(x-1\right)\left(x+1\right)=\left(x+1\right)\left(x^3+x-1\right)\)
\(c,=\left(x+y\right)^2-2\left(x+y\right)+1=\left(x+y-1\right)^2\)
\(d,=x^2y^2-y^2-x^2+1=\left(x^2-1\right)\left(y^2-1\right)=\left(x-1\right)\left(y-1\right)\left(x+1\right)\left(y+1\right)\)
\(e,4x^2+4x-15=\left(4x^2+4x+1\right)-16=\left(2x+1\right)^2-4^2=\left(2x+5\right)\left(2x-3\right)\)
\(3x^2-7x+2=\left(3x^2-6x\right)-\left(x-2\right)=3x\left(x-2\right)-\left(x-2\right)=\left(3x-1\right)\left(x-2\right)\)
\(4x^2-5x+1=\left(4x^2-4x\right)-\left(x-1\right)=4x\left(x-1\right)-\left(x-1\right)=\left(4x-1\right)\left(x-1\right)\)
Phân tích à :v
a) x3 + x2 - 4x - 4 = x2( x + 1 ) - 4( x + 1 ) = ( x + 1 )( x2 - 4 ) = ( x + 1 )( x - 2 )( x + 2 )
b) x4 + x3 + x2 - 1 = x3( x + 1 ) + ( x - 1 )( x + 1 ) = ( x + 1 )( x3 + x - 1 )
c) x2 + 2xy + y2 - 2x - 2y + 1 = ( x2 + 2xy + y2 ) - ( 2x + 2y ) + 1 = ( x + y )2 - 2( x + y ) + 12 = ( x + y - 1 )2
d) x2y2 + 1 - x2 - y2 = ( x2y2 - x2 ) - ( y2 - 1 ) = x2( y2 - 1 ) - ( y2 - 1 ) = ( y2 - 1 )( x2 - 1 ) = ( y - 1 )( y + 1 )( x - 1 )( x + 1 )
e) 4x2 + 4x - 15 = ( 4x2 + 4x + 1 ) - 16 = ( 2x + 1 )2 - 42 = ( 2x + 1 - 4 )( 2x + 1 + 4 ) = ( 2x - 3 )( 2x + 5 )
g) 3x2 - 7x + 2 = 3x2 - 6x - x + 2 = 3x( x - 2 ) - ( x - 2 ) = ( x - 2 )( 3x - 1 )
h) 4x2 - 5x + 1 = 4x2 - 4x - x + 1 = 4x( x - 1 ) - ( x - 1 ) = ( x - 1 )( 4x - 1 )
a,\(2x^2-8x+y^2+2y+9=0\)
\(\Rightarrow2\left(x^2-4x+4\right)+\left(y^2+2y+1\right)=0\)
\(\Rightarrow2\left(x-2\right)^2+\left(y+1\right)^2=0\)
Mà \(2\left(x-2\right)^2\ge0\forall x\); \(\left(y+1\right)^2\ge0\forall y\)
\(\Rightarrow2\left(x-2\right)^2+\left(y+1\right)^2\ge0\forall x;y\)
Dấu "=" xảy ra<=> \(\hept{\begin{cases}2\left(x-2\right)^2=0\\\left(y+1\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\y=-1\end{cases}}}\)
Vậy x=2;y=-1
Ik mk nha, hôm nay ngày mai, ngày kia mk ik 3 lần lại cho bạn (thành 9 lần)
Nhớ kb với mìn lun nha!! Mk rất vui đc làm quen vs bạn, cảm ơn mn nhìu lắm
a) \(A=x^2-8x+17=\left(x-4\right)^2+1\ge1\)
Vậy MIN A = 1 khi x = 4
b) \(T=x^2-4x+7=\left(x-2\right)^2+3\ge3\)
Vậy MIN T = 3 khi x = 2
c) \(H=3x^2+6x-1=3\left(x+1\right)^2-4\ge-4\)
Vậy MIN H = -4 khi x = -1
d) \(E=x^2+y^2-4\left(x+y\right)+16=\left(x-2\right)^2+\left(y-2\right)^2+8\ge8\)
Vậy MIN E = 8 khi x = y = 2
e) \(K=4x^2+y^2-4x-2y+3=\left(2x-1\right)^2+\left(y-1\right)^2+1\ge1\)
Vậy MIN K = 1 khi x = 1/2; y = 1
f) \(M=\frac{3}{2}x^2+x+1=\frac{3}{2}\left(x+\frac{1}{3}\right)^2+\frac{5}{6}\ge\frac{5}{6}\)
Vậy MIN M = 5/6 khi x = -1/3