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\(4x^4+4x^3+5x^2+8x-6\)
\(=4x^4-2x^3+6x^3-3x^2+8x^2-4x+12x-6\)
\(=2x^3\left(2x-1\right)+3x^2\left(2x-1\right)+4x\left(2x-1\right)+6\left(2x-1\right)\)
\(=\left(2x^3+3x^2+4x+6\right)\left(2x-1\right)\)
\(=\left[x^2\left(2x+3\right)+2\left(2x+3\right)\right]\left(2x-1\right)\)
\(=\left(x^2+2\right)\left(2x+3\right)\left(2x-1\right)\)
\(4x^4+6x^3-4x^2+9x-15\)
\(=4x^4-4x^3+10x^3-10x^2+6x^2-6x+15x-15\)
\(=4x^3\left(x-1\right)+10x^2\left(x-1\right)+6x\left(x-1\right)+15\left(x-1\right)\)
\(=\left(4x^3+10x^2+6x+15\right)\left(x-1\right)\)
\(=\left[2x^2\left(2x+5\right)+3\left(2x+5\right)\right]\left(x-1\right)\)
\(=\left(2x^2+3\right)\left(2x+5\right)\left(x-1\right)\)
- 6x2+13x+6
=6x2+9x+4x+6
=3x(2x+3)+2(2x+3)
=(2x+3)(3x+2)
2. 6x2-15x+6
=6x2-12x-3x+6
=6x(x-2)-3(x-2)
=(x-2).3(2x-1)
3. 8x2 -2x-3
= 8x2-6x+4x-3
=2x(4x-3)+(4x-3)
=(4x-3)(2x+1)
4. 8x2-10x-3
=8x2+12x-2x-3
=4x(2x+3)-(2x+3)
=(2x+3)(4x-1)
5. -10x2+4x+6
=-10x2+10x-6x+6
=-10x(x-1)-6(x-1)
=(x-1).(-2)(5x+3)
6. 10x2-28x-6
=10x2-30x+2x-6
=10x(x-3)+2(x-3)
=(x-3).2(5x+1)
6x2 + 13x + 6 = 6x2 + 9x + 4x + 6 = 3x( 2x + 3 ) + 2( 2x + 3 ) = ( 2x + 3 )( 3x + 2 )
6x2 - 15x + 6 = 6x2 - 12x - 3x + 6 = 6x( x - 2 ) - 3( x - 2 ) = 3( x - 2 )( 2x - 1 )
\(a,A=6x^2-6x+1\)
\(=6\left(x^2-x+\frac{1}{4}\right)-\frac{1}{2}\)
\(=6\left(x-\frac{1}{2}\right)^2-\frac{1}{2}\ge-\frac{1}{2}\)
Dấu = xảy ra \(\Leftrightarrow x=\frac{1}{2}\)
Vậy \(Min_A=-\frac{1}{2}\Leftrightarrow x=\frac{1}{2}\)
\(b,B=3+2x+3x^2\)
\(=3\left(x^2+\frac{2}{3}x+\frac{1}{9}\right)+\frac{8}{3}\)
\(=3\left(x+\frac{1}{3}\right)^2+\frac{8}{3}\ge\frac{8}{3}\)
Dấu = xảy ra \(\Leftrightarrow x=-\frac{1}{3}\)
Vậy \(Min_B=\frac{8}{3}\Leftrightarrow x=-\frac{1}{3}\)
\(c,C=4x+2x^2-3\)
\(=2\left(x^2+2x+1\right)-5\)
\(=2\left(x+1\right)^2-5\ge-5\)
Dấu = xảy ra \(\Leftrightarrow x=-1\)
Vậy \(Min_C=-5\Leftrightarrow x=-1\)
\(d,D=10x+6+x^2\)
\(=\left(x^2+10x+25\right)-19\)
\(=\left(x+5\right)^2-19\ge-19\)
Dấu = xảy ra \(\Leftrightarrow x=-5\)
Vậy \(Min_D=-19\Leftrightarrow x=-5\)
\(e,E=8x^2-6x+3\)
\(=8\left(x^2-\frac{3}{4}x+\frac{9}{64}\right)+\frac{15}{8}\)
\(=8\left(x-\frac{3}{8}\right)^2+\frac{15}{8}\ge\frac{15}{8}\)
Dấu = xảy ra \(\Leftrightarrow x=\frac{3}{8}\)
Vậy \(Min_E=\frac{15}{8}\Leftrightarrow x=\frac{3}{8}\)
a) Ta có:A = 6x2 - 6x + 1 = 6(x2 - x + 1/4) - 1/2 = 6(x - 1/2)2 - 1/2
Ta luôn có : (x - 1/2)2 \(\ge\)0 \(\forall\)x --> 6(x - 1/2)2 \(\ge\) 0 \(\)x
=> 6(x - 1/2)2 - 1/2 \(\ge\)-1/2 \(\forall\)x
hay A \(\ge\)-1/2 \(\forall\)x
Dấu "=" xảy ra khi : (x - 1/2)2 = 0 <=> x - 1/2 = 0 <=> x = 1/2
Vậy Amin = -1/2 tại x = 1/2
\(a,A=6x^2-6x+1\)
\(=6\left(x^2-x+\frac{1}{6}\right)\)
\(=6\left[\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}\right)-\frac{1}{4}+\frac{1}{6}\right]\)
\(=6\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{12}\right]\)
\(=6\left(x-\frac{1}{2}\right)^2-\frac{1}{2}\)
\(A_{min}=-\frac{1}{12}\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\)
\(\Rightarrow x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)
a) ta có 3x^2 -11x+6= 3x^2 - 9x -2x + 6 = 3x (x-3) - 2(x-3) =(x-3) (3x-2)
b) ta có 8x^2 + 10x - 3 = 2x (4x-1) - 3(4x- 1) = (2x-3)(4x-1)
c) ta có 8x^2 -2x - 1 = 8x^2 - 4x +2x-1 = 4x(2x-1) + 2x- 1 = (4x+1)(2x-1) k nha bạn
a, 3x2 - 11x+6 = 3x2 - 9x - 2 x + 6 = 3x(x-3) - 2(x-3) = (3x-2)(x-3)
3x-2 = 0 và x - 3 =0
x = 2/3 x = 3
Mik ghi nhầm số 30 thành 20. Xin lỗi nhé!