Đặt \(A=x+\dfrac{1}{x}\)

\(A=\left(\dfrac{x}{25}+\dfrac{1}{x}\right)+\dfrac{24}{25}x\ge2\sqrt{\dfrac{x}{25x}}+\dfrac{24}{25}.5=\dfrac{26}{5}\)

\(A_{min}=\dfrac{26}{5}\) khi \(x=5\)

a) \(A=x^2-6x+10=\left(x^2-6x+9\right)+1=\left(x-3\right)^2+1\ge1\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=3\). \(min_A=1\)

b) \(B=3x^2+x-2=3\left(x^2+\dfrac{1}{3}x-\dfrac{2}{3}\right)=3\left(x^2+\dfrac{1}{3}x+\dfrac{1}{36}-\dfrac{25}{36}\right)=3\left(x+\dfrac{1}{6}\right)^2-\dfrac{25}{12}\ge\dfrac{-25}{12}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=-\dfrac{1}{6}\). \(min_B=\dfrac{-25}{12}\)

c) \(C=\dfrac{4}{x^2}-\dfrac{3}{x}-1=\left(\dfrac{4}{x^2}-\dfrac{3}{x}+\dfrac{9}{16}\right)-\dfrac{25}{16}=\left(\dfrac{2}{x}+\dfrac{2}{3}\right)^2-\dfrac{25}{16}\ge\dfrac{-25}{16}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=-3\). \(min_C=\dfrac{-25}{16}\)

d) \(D=x^2+y^2-x+3y+7=\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+3y+\dfrac{9}{4}\right)+\dfrac{9}{2}=\left(x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{3}{2}\right)^2+\dfrac{9}{2}\ge\dfrac{9}{2}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-3}{2}\end{matrix}\right.\). \(min_D=\dfrac{9}{2}\)

18 - 10 : ( +2 ) - 7 

= 18 - 5 - 7

= 13 - 7

= 16

( 6 . 8 - 10 : 5 ) + 3 . ( -7 )

= ( 48 - 2 ) + -21

= 46 + ( - 21 )

= 25

( 6 . 8  - 10 : 5 ) + 3 . ( - 7 )

= ( 48 - 2 ) + ( - 21 )

= 46 + ( - 21 )

= 25

Giải nhanh:Nhìn đề ta thấy đáp án là 2017 hoặc 2018

con lạy bố, bố ghi vào bài thi như thế chắc đc 10 điểm nhỉ????

\(\left(x-7\right)\left(x+2019\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-7=0\\x+2019=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-2019\end{cases}}\)

\(9-25=\left(7-x\right)-\left(25+7\right)\)

\(\Leftrightarrow-16=7-x-25-7\)

\(\Leftrightarrow-x=-16+25\)

\(\Leftrightarrow-x=9\)

\(\Leftrightarrow x=-9\)

\(2\left(4x-2x\right)-7x=15\)

\(\Leftrightarrow4x-7x=15\)

\(\Leftrightarrow x=-5\)

a ) 9 - 25 = ( 7 - x ) - ( 25 + 7 )

    9 - 25  =  7 - x - 25 - 7 

  9 - 25 - 7 + 25 + 7 = -x 

9        = - x

 => x = -9

Vậy x = -9

b) 2 . ( 4x - 2x ) - 7x = 15

    8x  - 4x - 7x          = 15 

-3x = 15

  x   =  15 : ( - 3 ) 

  x = -5 

Vậy x = -5

c ) ( x - 7 ). ( x + 2019 ) = 0

 => x - 7 = 0 hoặc x + 2019 = 0

   => x    = 7 hoặc x = - 2019 

 vậy x \(\in\){ 7 ; -2019 }

a. 9 - 25 = (7 - x) - (25 + 7)

=> -16 = 7 - x - 25 - 7

=> -16 = -x - 25

=> -x = 9

=> x = -9

b. 2 . (4x - 2x) - 7x = 15

=> 2 . 2x - 7x = 15

=> 4x - 7x = 15

=> -3x = 15

=> x = -5

c. (x - 7) . (x + 2019) = 0

=> x - 7 = 0 hoặc x + 2019 = 0

=> x = 7 hoặc x = -2019

e) Ta có: \(2\left|x-\dfrac{1}{2}\right|\ge0\forall x\)

\(\Leftrightarrow2\left|x-\dfrac{1}{2}\right|+2021\ge2021\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

 x,y = ( 6,5);(10,30

b,

b.a=30=1.30=2.15=3.10=5.6

=>(b,a)={(1,30),(2,15),(3,10),(5,6)}

c,

(x+1)(y+2)=10=1.10=2.5

TH1:x+1=1;y+2=10=>x=0,y=8

tuong tu=>(x,y)={(0,8),(1,3),(4,0)}