Đề bài bn ghi thek thì ai làm nổi cho bn :V ?

mng giúp em với tối em nộp bài rồi a

cức + điên= lan ngọc cức điên

lớp 8 sao giống lớp 6 bọn em vậy

ko có biết

2. Ta có: A = x² - 6x + 5 = (x² - 6x + 9) - 4 = (x - 3)² - 4 

Ta luôn có: (x - 3)² \(\ge\)0 \(\forall\)x

=> (x - 3)² - 4 \(\ge\)-4 \(\forall\)x

Dấu "=" xảy ra <=> x - 3 = 0 <=> x = 3

Vậy MinA = -4 tại  x = 3

Ta có: B = 4x² - 8x + 7 = 4(x² - 2x + 1) + 3 = 4(x - 1)² + 3

Ta luôn có: 4(x - 1)² \(\ge\)0 \(\forall\)x

=> 4(x - 1)² + 3 \(\ge\)3 \(\forall\)x

Dấu "=" xảy ra <=> x - 1 = 0 <=> x = 1

vậy MinB = 3 tại x = 1

Ta có: C = 2x² + 4x - 6 = 2(x² + 2x + 1) - 8 = 2(x + 1)² - 8

Ta luôn có: 2(x + 1)² \(\ge\)0 \(\forall\)x

=> 2(x + 1)² - 8 \(\ge\)-8 \(\forall\)x

Dấu "=" xảy ra <=> x + 1 = 0 <=> x = -1

Vậy MinC = -8 tại x = -1

1/

\(A=x^2-6x+5\)

\(A=x^2-2\cdot3x+3^2-3^2+5\)

\(A=\left(x-3\right)^2-3^2+5\)

\(A=\left(x-3\right)^2-9+5\)

\(A=\left(x-3\right)^2-4\)

mà \(\left(x-3\right)^2\ge0\Rightarrow\left(x-3\right)^2-4\ge-4\)

\(\Rightarrow GTNNA\left(x^2-6x+5\right)=-4\)

với \(\left(x-3\right)^2=0;x=3\)

\(B=4x^2-8x+7\)

\(B=4\left(x^2-2x+\frac{7}{4}\right)\)

\(B=4\left(x^2-2\cdot1x+1-1+\frac{7}{4}\right)\)

\(B=4\left(x-1\right)^2+3\)

\(\left(x-1\right)^2\ge0\Rightarrow4\left(x^2-1\right)^2+3\ge3\)

\(\Rightarrow GTNNB=3\)

với \(\left(x-1\right)^2=0;x=1\)

\(C=2x^2+4x-6\)

\(C=2\left(x^2+2x-3\right)\)

\(C=2\left(x^2+2\cdot1x+1-1-3\right)\)

\(C=\left(x+1\right)^2-8\)

có\(\left(x+1\right)^2\ge0\Rightarrow\left(x+1\right)^2-8\ge-8\)

\(\Rightarrow GTNNC=-8\)

với \(\left(x+1\right)^2=0;x=-1\)

3

\(A=-x^2+2x-3.\)

\(A=-\left(x^2-2x+3\right)\)

\(A=-\left(x^2-2\cdot1x+1-1+3\right)\)

\(A=-\left(x-1\right)^2-2\)

có \(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2-2\ge-2\)

\(\Rightarrow GTLNA=-2\) với \(\left(x-1\right)^2=0;x=1\)

\(B=-9x^2+6x-4\)

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

\(B=-9\left(x^2-2\cdot\frac{1}{3}x+\frac{1}{9}-\frac{1}{9}+\frac{4}{9}\right)\)

\(B=-9\left(x^2-2\cdot\frac{1}{3}x+\frac{1}{9}-\frac{3}{9}\right)\)

\(B=-9\left(x-\frac{1}{3}\right)^2+3\)

có \(\left(x-\frac{1}{3}\right)^2\ge0\Rightarrow-9\left(x-\frac{1}{3}\right)^2+3\ge3\)

\(\Rightarrow GTLNB=3\)với \(\left(x-\frac{1}{3}^2=0;\right)x=\frac{1}{3}\)

2.

c) \(C=2x^2+4x-6=2\left(x^2+2x+1\right)-8\)

\(=2\left(x+1\right)^2-8\ge-8\forall x\)

Dấu"=" xảy ra<=> \(2\left(x+1\right)^2=0\Leftrightarrow x=-1\)

3.

c) \(C=-3x^2-6x+9=-3\left(x^2+2x+1\right)+12\)

\(=-3\left(x+1\right)^2+12\le12\forall x\)

Dấu "=" xảy ra<=> \(-3\left(x+1\right)^2=0\Leftrightarrow x=-1\)

\(2,GTNN\)

\(A=x^2-6x+5=x^2+6x+9-4\)

\(=\left(x+3\right)^2-4\ge-4\)

\(A_{min}=-4\Leftrightarrow\left(x+3\right)^2=0\Rightarrow x=-3\)

\(B=4x^2-8x+7=4\left(x^2-2x+\frac{7}{4}\right)\)

\(=4\left(x^2-2x+1+\frac{3}{4}\right)=4\left(x-1\right)^2+3\ge3\)

\(\Rightarrow B_{min}=3\Leftrightarrow\left(x-1\right)^2=0\Rightarrow x=1\)

\(C=2x^2+4x-6=2\left(x^2+2x-3\right)\)

\(=2\left(x^2+2x+1-4\right)=2\left(x+1\right)^2-8\ge-8\)

\(\Rightarrow C_{min}=-8\Leftrightarrow\left(x+1\right)^2=0\Rightarrow x=-1\)

\(3,GTLN\)

\(A=-x^2+2x-3=-\left(x^2-2x+3\right)\)

\(=-\left(x^2-2x+1-4\right)=-\left(x-1\right)^2+4\le4\)

\(A_{max}=4\Leftrightarrow-\left(x-1\right)^2=0\Rightarrow x=1\)

\(B=-9x^2+6x-4=-\left[9x^2-6x+4\right]\)

\(=-\left[\left(3x\right)^2-6x+1+3\right]=-\left(3x-1\right)^2-3\)

\(B_{max}=-3\Leftrightarrow-\left(3x-1\right)^2=0\Rightarrow x=\frac{1}{3}\)

\(C=-3x^2-6x+9=-3\left(x^2+2x-3\right)\)

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

\(C_{max}=12\Leftrightarrow-3\left(x+1\right)^2=0\Rightarrow x=-1\)

2/

A= X²-6X+5

  = X²-2.3.X+3²-4

 = (X-3)²-4 ( giá trị nhỏ nhất là -4)

B=4X²-8X+7

 =(2X)²-2.2.2X+2²+3

 =(2X-2)²+3 ( giá trị nhỏ nhất là 3 )

C= 2X²+ 4X-6

  = 2(X²+2X+1²-7)

  = 2((X+1)²-7)

  = 2(X+1)²-14 ( giá trị nhỏ nhất là -14 )