Bài 6:
\(A=x^2-6x+5\\ =\left(x^2-6x+9\right)-4\\ =\left(x-3\right)^2-4\ge-4\forall x\in R\\ Vậy:min_A=-4.khi.x=3\\ B=x^2+4x-1\\ =\left(x^2+4x+4\right)-5\\ =\left(x+2\right)^2-5\ge-5\forall x\in R\\ Vậy:min_B=-5.khi.x=-2\\ C=x^2-3x+5\\=x^2-2.x.\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2+\dfrac{11}{4}\\ =\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\forall x\in R\\ Vậy:min_C=\dfrac{11}{4}khi.x=\dfrac{3}{2}\\ D=x^2+x-1\\ =x^2+2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{5}{4}\\ =\left(x+\dfrac{1}{2}\right)^2-\dfrac{5}{4}\ge\dfrac{5}{4}\forall x\in R\\ Vậy:min_D=\dfrac{5}{4}.khi.x=-\dfrac{1}{2}\\ E=x^2-\dfrac{1}{3}x+5\\ =x^2-2.x.\dfrac{1}{6}+\left(\dfrac{1}{6}\right)^2+\dfrac{179}{36}\\ =\left(x-\dfrac{1}{6}\right)^2+\dfrac{179}{36}\ge\dfrac{179}{36}\forall x\in R\)
\(Vậy:min_E=\dfrac{179}{36}khi.x=\dfrac{1}{6}\)
4:
a: x^3-2x^2-2x+a chia hết cho x+2
=>x^3+2x^2-4x^2-8x+6x+12+a-12 chia hết cho x+2
=>a-12=0
=>a=12
b: 2x^3+5x^2-2x+a chia hết cho 2x^2-x+1
=>2x^3-x^2+x+6x^2-3x+3+a-3 chia hết cho 2x^2-x+1
=>a-3=0
=>a=3
5:
a: 2x^2-5xy
=x*2x-x*5y
=x(2x-5y)
b: x^2-xy+2x-2y
=x(x-y)+2(x-y)
=(x-y)(x+2)
c: x^2+4x+4-y^2
=(x+2)^2-y^2
=(x+2+y)(x+2-y)
d: =a^2(a+3)+4(a+3)
=(a+3)(a^2+4)
e: =3(x^2+2xy+y^2-z^2)
=3[(x+y)^2-z^2]
=3(x+y+z)(x+y-z)
f: =(x-y)^2-(z-t)^2
=(x-y-z+t)(x-y+z-t)
\(Bài.6:\\ F=2x^2+4x-1\\ =2.\left(x^2+2x-\dfrac{1}{2}\right)\\ =2.\left(x^2+2x+1\right)-3\\ =2.\left(x+1\right)^2-3\ge3\forall x\in R\\ Vậy:min_F=3.khi.x=-1\\ G=3x^2-2x+1\\ =3.\left(x^2+\dfrac{2}{3}x+\dfrac{1}{3}\right)\\ =3.\left(x^2+2.x.\dfrac{1}{3}+\dfrac{1}{9}\right)+\dfrac{2}{3}\\ =3.\left(x+\dfrac{1}{3}\right)^2+\dfrac{2}{3}\ge\dfrac{2}{3}\forall x\in R\\ Vậy:min_G=\dfrac{2}{3}.khi.x=-\dfrac{1}{3}\\ H=x^2+y^2+2x-6y+20\\ =\left(x^2+2x+1\right)+\left(y^2-6y+9\right)+10\\ =\left(x+1\right)^2+\left(y-3\right)^2+10\ge10\forall x,y\in R\\ Vậy:min_H=10.khi.x=-1.và.y=3\)
\(Bài.6:\\ I=x^2+2y^2+2xy-2y+10\\ =\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)+9\\ =\left(x+y\right)^2+\left(y-1\right)^2+9\ge9\forall x,Y\in R\\ Vậy:min_I=9.khi.x=y=1\\ K=x^2+5y^2+4x+2y+2024\\ =\left(x^2+4x+4\right)+5.\left(y^2+2.\dfrac{1}{5}y+\dfrac{1}{25}\right)+\dfrac{10099}{5}\\ =\left(x+2\right)^2+5.\left(y+\dfrac{1}{5}\right)^2+\dfrac{10099}{5}\ge\dfrac{10099}{5}\forall x,y\in R\\ Vậy:min_K=\dfrac{10099}{5}.khi.x=-2.và.y=-\dfrac{1}{5}\)
\(N=x^2+3y^2+4x+4y+2024\\ =\left(x^2+4x+4\right)+3.\left(y^2+2.\dfrac{2}{3}y+\dfrac{4}{9}\right)+\dfrac{6056}{3}\\ =\left(x+2\right)^2+3.\left(y+\dfrac{2}{3}\right)^2+\dfrac{6056}{3}\ge\dfrac{6056}{3}\forall x,y\in R\\ Vậy:min_N=\dfrac{6056}{3}.khi.x=-2.và.y=-\dfrac{2}{3}\)
Câu 1:
\(1.\\ 4x\left(2x-3\right)-\left(2x-3\right)\left(x+5\right)-\left(x+1\right)^2\\ =\left(2x-3\right)\left(4x-x-5\right)-\left(x^2+2x+1\right)\\ =\left(2x-3\right)\left(3x-5\right)-\left(x^2+2x+1\right)\\ =6x^2-9x-10x+15-\left(x^2+2x+1\right)\\ =6x^2-x^2-9x-10x-2x+15-1\\ =5x^2-21x+14\\ 2.\\ x^3-\left(x+3\right)\left(x-3\right)-\left(x-2\right)^3\\ =x^3-\left(x^2-9\right)-\left(x^3-6x^2+6x-8\right)\\ =x^3-x^3-x^2+6x^2-6x+9+8\\ =5x^2-6x+17\)
Bài 1:
\(3.\\ \left(2x+1\right)^2-2\left(4x^2-1\right)+\left(2x-1\right)^2\\ =\left(4x^2+4x+1\right)-8x^2+2+\left(4x^2+4x-1\right)\\ =4x^2+4x^2-8x^2+4x+4x+1+2-1\\ =8x+2\\ ---\\ 4.\\ \left(x^2-1\right)\left(x+2\right)-\left(x-2\right)\left(x^2+2x+4\right)\\ =x^3-x+2x^2-2-\left(x^3-2^3\right)\\ =x^3-x^3+2x^2-x-2+8\\ =2x^2-x+6\)
