\(\left(x+4\right)^2-81=0\Leftrightarrow\left(x+4\right)^2-9^2=0\)

\(\Leftrightarrow\left(x+4+9\right)\times\left(x+4-9\right)=0\)

\(\Leftrightarrow\left(x+13\right)\times\left(x-5\right)=0\)

\(\left[{}\begin{matrix}x+13=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=5\end{matrix}\right.\)

\(25:\left(x+1\right)=5\)

\(x+1=5\)

\(x=4\)

Thay x=28 và A=x+|10|

ta có 

A=28+|10|

A=28+10

A=38

a)  (-37 - 17) . (-9) + 35 . (-9 - 11)

=-54*(-9)+35*(-20)

=486+(-700)

=-214

b) (-25) . (75 - 45) - 75 . ( 45 - 25)

=(-25)*30-75*20

=-750-1500

=-2250

bài 2 phần a sai đề r` tui lm phần b thôi

Bài 2:

B) x(y + 2)= 5

 =>x và y+2 thuộc Ư(5)={1;-1;5;-5}

Xét x=1 =>y+2=5 <=>y=3

Xét x=-1 =>y+2=-5 <=>y=-7

Xét x=5 =>y+2=1 =>y=-1

Xét x=-5 =>y+2=-1 =>y=-3

Vậy các cặp (x;y) thỏa mãn là (1;3),(-1;-7),(5;-1),(-5;-3)

Bài 1:

\(N=2x^2+4y^2-2x-4y+15=2\left(x^2-x+\dfrac{1}{4}\right)+\left(4y^2-4y+1\right)+\dfrac{27}{2}=2\left(x-\dfrac{1}{2}\right)^2+\left(2y-1\right)^2+\dfrac{27}{2}\ge\dfrac{27}{2}\)

\(minN=\dfrac{27}{2}\Leftrightarrow x=y=\dfrac{1}{2}\)

Bài 2:

\(\Leftrightarrow4x^2+12x+9-25x^2+50x-25=0\)

\(\Leftrightarrow21x^2-62x+16=0\)

\(\Leftrightarrow\left(3x-8\right)\left(7x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=\dfrac{2}{7}\end{matrix}\right.\)

a  x = \(\dfrac{-1}{12}\)

b  x = \(\dfrac{-4}{3}\)

c  x = \(\dfrac{-1}{6}\)

d  x = \(\dfrac{-1}{4}\)

\(\left(4x+1\right)^2=\dfrac{4}{9}\)

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

\(\text{Vậy }4x+1=\dfrac{2}{3}\)

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

        \(x\)         \(=\left(\dfrac{-1}{3}\right).\dfrac{1}{4}=\dfrac{-1}{12}\)

\(\text{hoặc }4x+1=\dfrac{-2}{3}\)

        \(4x\)        \(=\left(\dfrac{-2}{3}\right)+\left(-1\right)=\dfrac{-5}{3}\)

         \(x\)         \(=\left(\dfrac{-5}{3}\right).\dfrac{1}{4}=\dfrac{-5}{12}\)

\(\Rightarrow x\in\left\{\dfrac{-1}{12};\dfrac{-5}{12}\right\}\)

\(\left(3x-1\right)^2=25\)

\(\left(3x-1\right)^2=\perp\left(5\right)^2\)

\(\text{Vậy }3x-1=5\)

       \(3x\)        \(=5+1=6\)

        \(x\)         \(=6:3=2\)

\(\text{hoặc }3x-1=-5\)

        \(3x\)       \(=\left(-5\right)+1=-4\)

         \(x\)        \(=\left(-4\right):3=\dfrac{-4}{3}\)

\(\Rightarrow x\in\left\{2;\dfrac{-4}{3}\right\}\)

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

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

\(\text{Vậy }x-\dfrac{1}{3}=\dfrac{1}{2}\)

       \(x\)         \(=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)

\(\text{hoặc }x-\dfrac{1}{3}=\dfrac{-1}{2}\)

        \(x\)         \(=\left(\dfrac{-1}{2}\right)+\dfrac{1}{3}=\dfrac{-1}{6}\)

\(\Rightarrow x\in\left\{\dfrac{5}{6};\dfrac{-1}{6}\right\}\)

\(\left(4x-3\right)^2=16\)

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

\(\text{Vậy }4x-3=4\) 

        \(4x\)       \(=4+3=7\)

          \(x\)       \(=7:4=\dfrac{7}{4}\)

\(\text{hoặc }4x-3=-4\)

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

          \(x\)        \(=\left(-1\right):4=\dfrac{-1}{4}\)

\(\Rightarrow x\in\left\{\dfrac{7}{4};\dfrac{-1}{4}\right\}\)

\(\text{a)}\left|x-6\right|=6-x=-x+6=-\left(x-6\right)\Leftrightarrow x-6<0\Leftrightarrow x<6\)

Vậy x<6

\(\text{b)}x-25<15\Leftrightarrow x<15+25=40\)

Vậy x<40

a, 11 - ( 15 + 11 ) = x - ( 25 - 9 )

11-15-11=x-25+9

-15=x-16

x=1

b, 2 - x = 17 - ( - 5 )

2-x=22

x=-20

c, x - 12 = ( - 9 ) - 15

x-12=-24

x=-12

d, 9 - 25 = ( 7 - x ) - ( 25 + 7 )

(7-x)-32=-16

7-x=16

x=-9

#H