1,\(f\left(x\right)=3x^2-2x-7\)

\(=3\left(x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)-\dfrac{22}{3}\)

\(=2\left(x-\dfrac{1}{3}\right)^2-\dfrac{22}{3}\ge-\dfrac{22}{3}\forall x\)

Vậy GTNN của biểu thức là \(-\dfrac{22}{3}\) khi \(x-\dfrac{1}{3}=0\Rightarrow x=\dfrac{1}{3}\)

\(b,f\left(x\right)=5x^2+7x=5\left(x^2+\dfrac{7}{5}x+\dfrac{49}{100}\right)-\dfrac{49}{20}\)\(=5\left(x+\dfrac{7}{10}\right)^2-\dfrac{49}{20}\ge-\dfrac{49}{20}\forall x\)

Vậy Giá trị nhỏ nhất của biểu thức là \(-\dfrac{49}{20}\) khi \(x+\dfrac{7}{10}=0\Rightarrow x=-\dfrac{7}{10}\)

\(c,f\left(x\right)=-5x^2+9x-2=-5\left(x^2-\dfrac{9}{5}x+\dfrac{81}{100}\right)+\dfrac{41}{20}\)\(=-5\left(x-\dfrac{9}{10}\right)^2+\dfrac{41}{20}\le\dfrac{41}{20}\forall x\)

Vậy GTLN của biểu thức là \(\dfrac{41}{20}\) khi \(x-\dfrac{9}{10}=0\Rightarrow x=\dfrac{9}{10}\)

\(d,f\left(x\right)=-7x^2+3x=-7\left(x^2-\dfrac{3}{7}x+\dfrac{9}{196}\right)+\dfrac{9}{28}\)\(=-7\left(x-\dfrac{3}{14}\right)^2+\dfrac{9}{28}\le\dfrac{9}{28}\forall x\)

Vậy GTLN của biểu thức là \(\dfrac{9}{28}\) khi \(x-\dfrac{3}{14}=0\Rightarrow x=\dfrac{3}{14}\)

1/ \(f\left(x\right)=3x^2-2x-7\)

\(=3\left(x^2-\dfrac{2}{3}x-7\right)\)

\(=3\left(x^2-\dfrac{2}{3}+\dfrac{1}{9}-\dfrac{64}{9}\right)\)

\(=3\left(x-\dfrac{1}{3}\right)^2-\dfrac{64}{3}\)

Ta có: \(3\left(x-\dfrac{1}{3}\right)^2\ge0\forall x\Rightarrow3\left(x-\dfrac{1}{3}\right)^2-\dfrac{64}{3}\ge-\dfrac{64}{3}\forall x\)

Dấu "=" xảy ra khi \(x-\dfrac{1}{3}=0\) hay \(x=\dfrac{1}{3}\)

Vậy MIN_f(x) = \(-\dfrac{64}{3}\) khi x = \(\dfrac{1}{3}\).

2/ \(f\left(x\right)=5x^2+7x\)

\(=5\left(x^2+\dfrac{7}{5}x\right)=5\left(x^2+\dfrac{7}{5}x+\dfrac{49}{100}-\dfrac{49}{100}\right)\)

\(=5\left(x+\dfrac{7}{10}\right)^2-\dfrac{49}{20}\)

Ta có: \(5\left(x+\dfrac{7}{10}\right)^2\ge0\forall x\Rightarrow5\left(x+\dfrac{7}{10}\right)^2-\dfrac{49}{20}\ge-\dfrac{49}{20}\forall x\)

Dấu "=" xảy ra khi \(x+\dfrac{7}{10}=0\) hay \(x=-\dfrac{7}{10}\)

Vậy MIN_f(x) = \(-\dfrac{49}{20}\) khi x = \(-\dfrac{7}{10}\).

1/ \(f\left(x\right)=-5x^2+9x-2\)

\(=-5\left(x^2-\dfrac{9}{5}x+\dfrac{2}{5}\right)\)

\(=-5\left(x^2-\dfrac{9}{5}x+\dfrac{81}{100}-\dfrac{41}{100}\right)\)

\(=-5\left(x-\dfrac{9}{10}\right)^2+\dfrac{41}{20}\)

Ta có: \(-5\left(x-\dfrac{9}{10}\right)^2\le0\forall x\Rightarrow-5\left(x-\dfrac{9}{10}\right)^2+\dfrac{41}{20}\le\dfrac{41}{20}\forall x\)

Dấu "=" xảy ra khi \(x-\dfrac{9}{10}=0\) hay \(x=\dfrac{9}{10}\)

Vậy MAX_f(x) = \(\dfrac{41}{20}\) khi x = \(\dfrac{9}{10}\)

2/ \(f\left(x\right)=-7x^2+3x=-7\left(x^2-\dfrac{3}{7}x+\dfrac{9}{196}\right)+\dfrac{9}{28}\)

\(=-7\left(x-\dfrac{3}{14}\right)^2+\dfrac{9}{28}\)

Ta có: \(-7\left(x-\dfrac{3}{14}\right)^2\le0\forall x\Rightarrow-7\left(x-\dfrac{3}{14}\right)^2+\dfrac{9}{28}\le\dfrac{9}{28}\forall x\)

Dấu "=" xảy ra khi \(x-\dfrac{3}{14}=0\) hay x = \(\dfrac{3}{14}\)

Vậy MAX_f(x) = \(\dfrac{9}{28}\) khi x = \(\dfrac{3}{14}\).

a) 5x - 14 = x - 34

5x - x = -34 + 14

4x = -20

x = -20 : 4

x = -5

b) x - 3 - (3x + 2) = -15

x - 3 - 3x - 2 = -15

x - 3x = -15 + 3 + 2

-2x = -10

x = (-10) : (-2)

x = 5

c) 2(x  - 12) + 19 = x + (-34)

2x - 24 + 19 = x + (-34)

2x - x = -34 + 24 - 19

x = -29

d) 2(x - 33) - 3(x - 43) = 96 - 115

2x - 66 - 3x + 129 = 96 - 115

2x - 3x = 96 - 115 + 66 - 129

-x = -82

x = 82

e) 2x + 3 - 9x = -11

2x - 9x = -11 - 3

-7x = -14

x = (-14) : (-7)

x = 2

f) -(x + 3 - 84) = (x + 70 - 71) - 6

-x - 3 + 84 = x + 70 - 71 - 6

-x - x = 70 - 71 - 6 - 84

-2x = -91

x = (-91) : (-2)

x = 45,5

F(x)=6²+5x+8+3x-3x²+3x³

      =(36+8)+(5x+3x)-3x²+3x³

      =3x³-3x²+8x+44

G(x)=12x²-6-9x²+3x³

       =3x³+(12x²-9x²)-6

       =3x³+3x²-6

F(x)+G(x)=3x³-3x²+8x+44+3x³+3x²-6

                =(3x³+3x³)+(-3x²+3x²)+8x+(44-6)

                =6x³+8x+38

\(F\left(x\right)=G\left(x\right)\\ \Rightarrow6^2-5x+8+3x-3x^2+3x^3=12x^2-6-9x^2+3x^3\\ \Leftrightarrow-3x^2-2x+44=3x^2-6\\ \Leftrightarrow6x^2+2x-50=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1+\sqrt{301}}{6}\\x=\dfrac{-1-\sqrt{301}}{6}\end{matrix}\right.\)

Ta có : F(x)=G(x) suy ra: 3x^3 + 3x^2 - 2x + 8 = 3x^3+3x^2 -6

                                        3x^3+ 3x^2 -2x +8 -3X^3- 3x^2+6=0

                                         (3x^3-3x^3)+(3x^2-3x^2)-2x+(8+6)=0

                                         -2x +14 =0

                                          2x         =14

                                            x          = 7