\(3^{x+2}-5.3^x=36\Rightarrow3^2.3^x-5.3^x=36\Rightarrow4.3^x=36\Rightarrow3^x=9=3^2\Rightarrow x=2\)

\(\left(x^2+2\right)^2-\left(x+2\right)\left(x-2\right)\left(x^2+4\right)=24\)

\(\Leftrightarrow x^4+4x^2+4-\left(x^2-4\right)\left(x^2+4\right)=24\)

\(\Leftrightarrow x^4+4x^2+4-\left(x^4-16\right)=24\)

\(\Leftrightarrow x^4+4x^2+4-x^4+16=24\)

\(\Leftrightarrow4x^2+20=24\)

\(\Leftrightarrow4x^2=4\)

\(\Leftrightarrow x^2=1\)

\(\Leftrightarrow x=\pm1\)

Vậy \(x=\pm1\)

\(\left(x^2+2\right)^2-\left(x+2\right)\left(x-2\right)\left(x^2+4\right)=24\)

\(\Leftrightarrow x^4+4x^2+4-\left(x^2-4\right)\left(x^2+4\right)=24\)

\(\Leftrightarrow x^4+4x^2+4-\left(x^4-16\right)=24\)

\(\Leftrightarrow x^4+4x^2+4x-x^4+16=24\)

\(\Leftrightarrow4x^2+4x+16=24\)

\(\Leftrightarrow\left(2x\right)^2+2.2x+1+15=24\)

\(\Leftrightarrow\left(2x+1\right)^2+15=24\)

\(\Leftrightarrow\left(2x+1\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=3\\2x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

a)( x+ 5)+(x-9)=x+2

x + 5 + x - 9 = x + 2

x + x - x = 2 -5 +9

x = 6

b)( 27 -x) +( 15+x )=x-24

-x + x - x = -24 -27 -15

-x = -66

x=66

hok tốt!

\(a.\left(x+5\right)+\left(x-9\right)=\)\(x+2\)

\(\Rightarrow x+x-x=2+9-5\)

\(\Rightarrow x=6\)

\(b.\left(27-x\right)+\left(15+x\right)=x-24\)

\(\Rightarrow x-x-x=-24-27-15\)

\(\Rightarrow-x=-66\)

\(\Rightarrow x=66\)

Phần 2 là x - ( x + 2 ) > 0 nhé 

a: TH1: x<3

=>3-x-2(5-x)=8

=>3-x-10+2x=8

=>x-7=8

=>x=15(loại)

TH2: 3<=x<5

=>x-3-2(5-x)=8

=>x-3-10+2x=8

=>3x=21

=>x=7(loại)

TH3: x>=5

=>x-3-2x+10=8

=>-x=1

=>x=-1(loại)

b: =>|2x-3|+3|x-4|=8x

TH1: x<3/2

=>3-2x+12-3x=8x

=>8x=-5x+15

=>13x=15

=>x=15/13(nhận)

TH2: 3/2<=x<4

=>2x-3+12-3x=8x

=>8x=-x+9

=>x=1(loại)

TH3: x>=4

=>2x-3+3x-12=8x

=>8x=5x-15

=>3x=-15

=>x=-5(loại)

ủa sao tui ko thấy gì hết vậy nè

=[x(x-2)/2(x²+4)-2x²/(4+x²)(2-x)][x(x-2)(x+1)/x³]

={[x(x-2)(2-x)-4x² ]/2(2-x)(4+x²)} .[x(x-2)(x+1)/x³ ]

=[-x(x²+4)/2(2-x)(4+x²)].[x(x-2)(x+1)/x³ ]

=-x.x(x-2)(x+1)/2(2-x)x³

=(x+1)/2x

\(\dfrac{16}{27}xX=\dfrac{10}{27}-\dfrac{6}{27}\)

\(\dfrac{16}{27}xX=\dfrac{4}{27}\)

\(X=\dfrac{4}{27}:\dfrac{16}{27}\)

X=\(\dfrac{1}{4}\)

16/27x X+2/9=10/27

16/27x X=10/27-2/9

16/27x X=4/27

X=4/27:16/27

X=1/4

Ta có: x + y + 1 = 0

      <=>  x + y = -1

Thay x + y = -1 vào biểu thức N ta được:

N = x²(-1) - y²(-1) + x² - y² + 2(-1) + 3

N = -x² + y² + x² - y² - 2 + 3

N = (-x² + x²) + (y² - y²) + (-2 + 3)

N = -2+3=1

Vậy tại x+y+1=0 thì giá trị của biểu thức N là: 1.

cảm ơn bn

a) \(2\left(2x-7\right)^2=18\)

\(\Leftrightarrow\left(2x-7\right)^2=9\)

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

\(\Leftrightarrow2x-7=3\)

\(\Leftrightarrow2x=10\)

\(\Leftrightarrow x=5\)

 2.(2X-7)^2= 18
 (2X-7)^2=18:2
 (2x-7)^2= 9
 (2x-7)^2 = 3^2 = (-3)^2
 => (2x-7)^2 = 3^2 và (2x-7)^2 =  (-3)^2

Câu 1:

A={1;3;5;7;9}

Câu 2: 

A. 24 x 82 + 24 x 18 - 100 

= 24 x (82 + 18) - 100

= 24 x 100 - 100

= 2400 - 100 = 2300

B.12 + 3 [ 39 - (5 - 2 )² ]

= 12 + 3. [39 - 3² ]

= 12 + 3 . [39-9]

= 12 + 3.30

= 12 + 90 = 102

cảm ơn