1:

a: =x^2-7x+49/4-5/4

=(x-7/2)^2-5/4>=-5/4

Dấu = xảy ra khi x=7/2

b: =x^2+x+1/4-13/4

=(x+1/2)^2-13/4>=-13/4

Dấu = xảy ra khi x=-1/2

e: =x^2-x+1/4+3/4=(x-1/2)^2+3/4>=3/4

Dấu = xảy ra khi x=1/2

f: x^2-4x+7

=x^2-4x+4+3

=(x-2)^2+3>=3

Dấu = xảy ra khi x=2

2:

a: A=2x^2+4x+9

=2x^2+4x+2+7

=2(x^2+2x+1)+7

=2(x+1)^2+7>=7

Dấu = xảy ra khi x=-1

b: x^2+2x+4

=x^2+2x+1+3

=(x+1)^2+3>=3

Dấu = xảy ra khi x=-1

\(A=x^2-4x+20=x^2-4x+4+16=\left(x-2\right)^2+16\)

Do \(\left(x-2\right)^2\ge0\)

\(\Rightarrow\left(x-2\right)^2+16\ge16\)

\(\Rightarrow Min\left(A\right)=16\)

\(B=x^2-3x+7=x^2-3x+\dfrac{9}{4}-\dfrac{9}{4}+7=\left(x-\dfrac{3}{2}\right)^2+\dfrac{19}{4}\)

Do \(\left(x-\dfrac{3}{2}\right)^2\ge0\)

\(\Rightarrow\left(x-\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)

\(\Rightarrow Min\left(B\right)=\dfrac{19}{4}\)

\(C=-x^2-10x+70=-\left(x^2+10x+25\right)+25+70=-\left(x-5\right)^2+95\)

Do \(-\left(x-5\right)^2\le0\)

\(\Rightarrow-\left(x-5\right)^2+95\le95\)

\(\Rightarrow Max\left(C\right)=95\)

\(D=-4x^2+12x+1=-\left(4x^2-12x+9\right)+9+1=-\left(2x-3\right)^2+10\)

Do \(-\left(2x-3\right)^2\le0\)

\(\Rightarrow-\left(2x-3\right)^2+10\le10\)

\(\Rightarrow Max\left(D\right)=10\)

Bài 1: 

a: \(\Leftrightarrow x^2-5x+6< =0\)

=>(x-2)(x-3)<=0

=>2<=x<=3

b: \(\Leftrightarrow\left(x-6\right)^2< =0\)

=>x=6

c: \(\Leftrightarrow x^2-2x+1>=0\)

\(\Leftrightarrow\left(x-1\right)^2>=0\)

hay \(x\in R\)

Bài 3: 

a) Ta có: \(A=25x^2-20x+7\)

\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)

\(=\left(5x-2\right)^2+3>0\forall x\)(đpcm)

d) Ta có: \(D=x^2-2x+2\)

\(=x^2-2x+1+1\)

\(=\left(x-1\right)^2+1>0\forall x\)(đpcm)

Bài 1: 

a) Ta có: \(A=x^2-2x+5\)

\(=x^2-2x+1+4\)

\(=\left(x-1\right)^2+4\ge4\forall x\)

Dấu '=' xảy ra khi x=1

b) Ta có: \(B=x^2-x+1\)

\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

MK KO BT MK MỚI HO C LỚP 6

AI HỌC LỚP 6 CHO MK XIN

dài thế ai trả lời đc hả ?

tu lam di luoi vua thoi

\(a,=\left(x+1\right)\left(x+3\right)\\ b,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ c,=2x^2+2x+5x+5=\left(2x+5\right)\left(x+1\right)\\ d,=2x^2-2x+5x-5=\left(x-1\right)\left(2x+5\right)\\ e,=x^3+x^2-4x^2-4x+x+1=\left(x+1\right)\left(x^2-4x+1\right)\\ f,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)

a) `3x+5 =0`

`3x=-5`

`x=-5/3`

`b) -4x+8=0`

`-4x =-8`

`x=2`

`c) 3x -6=0`

`3x=6`

`x=2`

`d)x^2 +x =0`

`x(x+1) =0`

`=>[(x=0),(x=-1):}`

`e) x^2 -4 =0`

`x^2 =4`

`=> x = +-2`

`f) x^3 -27 =0`

`x^3 =27`

`=> x=3`

`g) 3x^2 +4 =0`

`3x^2 =-4`

`x^2 =-4/3(vô-lí)`

=> Đa thức ko có nghiệm

h) `x^3 -4x =0`

`x(x^2 -4) =0`

`=>[(x=0),(x^2=4 => x=+-2):}`

i) `2x^3 -32x =0`

`2x(x^2 -16)=0`

`=>[(2x=0),(x^2=16):}`

`=>[(x=0),(x=+-4):}`

a)-(x-y)(x²+xy-1)=-(x³+x²y-x-x²y-xy²+y)

                          =-(x³-xy²-x+y)

                          =-x³+xy²+x-y

b)x²(x-1)-(x³+1)(x-y)=x³-x²-x³+x²y-x+y

                                =-x²+x²y-x+y

c)(3x-2)(2x-1)+(-5x-1)(3x+2)=6x²-3x-4x+2-15x²-10x-3x-2

                                             =-9x²-20x

d) hình như bạn ghi lỗi

Bài 2: C=x(x²-y)-x²(x+y)+y(x²-x)

             =x³-xy-x³-x²y+x²y-xy

             =-2xy

Thay x=1/2,y=-1 vào C, ta có:

        C=-2.1/2.(-1)=1

Vậy C=1 khi x=1/2 và y=-1.