\(6,\\ a,\\ 1,A=x^2+3x+7=\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)

Dấu \("="\Leftrightarrow x=-\dfrac{3}{2}\)

\(2,B=\left(x-2\right)\left(x-5\right)\left(x^2-7x+10\right)=\left(x-2\right)^2\left(x-5\right)^2\ge0\)

Dấu \("="\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

\(b,\\ 1,A=11-10x-x^2=-\left(x+5\right)^2+36\le36\)

Dấu \("="\Leftrightarrow x=-5\)

a) Đặt \(A=10+2x-5x^2\)

\(-A=5x^2-2x-10\)

\(-5A=25x^2-10x-50\)

\(-5A=\left(25x^2-10x+1\right)-51\)

\(-5A=\left(5x-1\right)^2-51\)

Do \(\left(5x-1\right)^2\ge0\forall x\)

\(\Rightarrow-5A\ge-51\)

\(A\le\frac{51}{5}\)

Dấu "=" xảy ra khi : \(5x-1=0\Leftrightarrow x=\frac{1}{5}\)

Vậy Max A = \(\frac{51}{5}\Leftrightarrow x=\frac{1}{5}\)

b) Đặt \(B=x^2-6x+10\)

\(B=\left(x^2-6x+9\right)+1\)

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

Mà \(\left(x-3\right)^2\ge0\forall x\)

\(B\ge1\)

Dấu "=" xảy ra khi :

\(x-3=0\Leftrightarrow x=3\)

Vậy Min B \(=1\Leftrightarrow x=3\)

a, \(A=\left(-5x+4\right)\left(3x-2\right)+\left(-2x+3\right)\left(x-2\right)\)

\(=-15x^2+10x+12x-8=-15x^2+22x-8\)

Thay x = -2 vào biểu thức ta có : \(-15\left(-2\right)^2+22\left(-2\right)-8\)

\(=-15.4-44-8=-112\)

b, \(B=\left(x-9\right)\left(2x+3\right)-2\left(x+7\right)\left(x-5\right)\)

\(=2x^2+3x-18x-27=2x^2-15x-27\)

Thay x = -1/2 vào biểu thức ta có : \(2\left(-\frac{1}{2}\right)^2-15\left(-\frac{1}{2}\right)-27\)

\(=2.\frac{1}{4}+\frac{15}{2}-27=\frac{11}{2}+\frac{15}{2}+27=40\)

Bài làm:

a) \(A=\left(-5x+4\right)\left(3x-2\right)+\left(-2x+3\right)\left(x-2\right)\)

\(A=-15x^2+22x-8-2x^2+7x-6\)

\(A=-17x^2+29x-14\)

Thay x = -2 vào ta được:

\(A=-17.\left(-2\right)^2+29.\left(-2\right)-14\)

\(A=-68-58-14\)

\(A=-140\)

b) \(B=\left(x-9\right)\left(2x+3\right)-2\left(x+7\right)\left(x-5\right)\)

\(B=2x^2-15x-27-2\left(x^2+2x-35\right)\)

\(B=2x^2-15x-27-2x^2-4x+70\)

\(B=-19x+43\)

Thay x = -1/2 vào B ta được:

\(B=-19.\left(-\frac{1}{2}\right)+43=\frac{19}{2}+43=\frac{105}{2}\)

A = ( -5x + 4 )( 3x - 2 ) + ( -2x + 3 )( x - 2 )

A = -15x² + 22x - 8 - 2x² + 7x - 6

A = -17x² + 29x - 14

Thế x = -2 ta được :

A = -17(-2)² + 29(-2) - 14 = -68 - 58 - 14 = -140

b) B = ( x - 9 )( 2x + 3 ) - 2( x + 7 )( x - 5 )

B = 2x² - 15x - 27 - 2( x² + 2x - 35 )

B = 2x² - 15x - 27 - 2x² - 4x + 70

B = -19x + 43

Thế x = -1/2 ta được :

B = -19.(-1/2) + 43 = 19/2 + 43 = 105/2

a:Ta có: \(A=-4x^2+x-1\)

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

\(=-4\left(x^2-2\cdot x\cdot\dfrac{1}{8}+\dfrac{1}{64}+\dfrac{63}{64}\right)\)

\(=-4\left(x-\dfrac{1}{8}\right)^2-\dfrac{63}{16}\le-\dfrac{63}{16}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{8}\)

b: Ta có: \(B=-3x^2+5x+6\)

\(=-3\left(x^2-\dfrac{5}{3}x-2\right)\)

\(=-3\left(x^2-2\cdot x\cdot\dfrac{5}{6}+\dfrac{25}{36}-\dfrac{97}{36}\right)\)

\(=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{97}{12}\le\dfrac{97}{12}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{5}{6}\)

c: Ta có: \(C=-x^2+3x+4\)

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

\(=-\left(x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{25}{4}\right)\)

\(=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)

Toán lớp 1 cái gì,xạo.Toán trung học thì có.

Lớp 1 mà làm được cái này thì...THIÊN TÀI

khó hiểu quá 

bn giải giúp mình đi

1)   P = \(3+15x-5x^2\)\(=-5x^2+15x+3=-5\left(x^2-3x-\frac{3}{5}\right)\)  \(=-5\left(x^2-2.\frac{3}{2}.x+\frac{9}{4}-\frac{9}{4}-\frac{3}{5}\right)\)= \(-5\left[\left(x-\frac{3}{2}\right)^2-\frac{57}{20}\right]=-5.\left(x-\frac{3}{2}\right)^2+\frac{57}{4}\)

vì \(\left(x-\frac{3}{2}\right)^2>=0\) => \(-5.\left(x-\frac{3}{2}\right)^2+\frac{57}{4}>=0\)  =>\(-5.\left(x-\frac{3}{2}\right)^2+\frac{57}{4}>=\frac{57}{4}\)

 => GTLN  của P là \(\frac{57}{4}\)tại x =\(\frac{3}{2}\)

2) GTNN của B là -36

a: =-x^2+6x-4

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

=-(x^2-6x+9-5)

=-(x-3)^2+5<=5

Dấu = xảy ra khi x=3

b: =3(x^2-5/3x+7/3)

=3(x^2-2*x*5/6+25/36+59/36)

=3(x-5/6)^2+59/12>=59/12

Dấu = xảy ra khi x=5/6

c: \(=-\left(x-3\right)^2+2\left|x-3\right|\)

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

\(=-\left(\left|x-3\right|-1\right)^2+1< =1\)

Dấu = xảy ra khi x=4 hoặc x=2

a) Ta có: \(\left|x-2\right|\ge0\forall x\)

\(\Leftrightarrow\left|x-2\right|+15\ge15\forall x\)

Dấu '=' xảy ra khi x=2

b) Ta có: \(\left|x-5\right|\ge0\forall x\)

\(\Leftrightarrow2\left|x-5\right|\ge0\forall x\)

\(\Leftrightarrow2\left|x-5\right|-4\ge-4\forall x\)

Dấu '=' xảy ra khi x=5