a) \(\frac{3x}{-4}=\frac{-9}{6}\)

=> \(\frac{-3x}{4}=\frac{-9}{6}\)

=> \(\frac{-3x}{4}=\frac{-3}{2}\)

=> \(\frac{-3x}{4}=\frac{-6}{4}\)

=> \(-3x=-6\)

=> \(x=\left(-6\right):\left(-3\right)=2\)

b) \(\frac{2}{x}=\frac{y}{10}=\frac{4}{-5}\)

=> \(\frac{2}{x}=\frac{y}{10}=\frac{-4}{5}\)

+) \(\frac{2}{x}=\frac{-4}{5}\)

=> \(\left(-4\right)\cdot x=2\cdot5\)

=> \(\left(-4\right)\cdot x=10\)

=> \(x=\frac{10}{-4}=\frac{-10}{4}=\frac{-5}{2}\)

+) \(\frac{y}{10}=\frac{-4}{5}\)

=> \(5\cdot y=-40\)

=> \(y=\left(-40\right):5=-8\)

Vậy \(x=-\frac{5}{2},y=-8\)

c) \(\frac{7}{-x}=\frac{5}{y}=\frac{2}{34}\)

=> \(\frac{-7}{x}=\frac{5}{y}=\frac{1}{17}\)

+) \(\frac{-7}{x}=\frac{1}{17}\Rightarrow x=\left(-7\right)\cdot17=-119\)

+) \(\frac{5}{y}=\frac{1}{17}\Rightarrow y=5\cdot17=85\)

Vậy \(x=-119,y=85\)

d) \(\frac{x^2}{4}=16\)

=> \(x^2=16\cdot4=64\)

=> \(x\in\left\{8;-8\right\}\)

e) \(\frac{3}{x^3}=\frac{4}{-36}\)

=> \(\frac{3}{x^3}=\frac{1}{-9}=\frac{-1}{9}\)

=> \(x^3=3:\left(-\frac{1}{9}\right)=-27\)

=> \(x=-3\)

Vậy x = -3

a: \(\Leftrightarrow x\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;9;-9;12;-12;18;-18;36;-36\right\}\)

mà -3<x<30

nên \(x\in\left\{-2;-1;1;2;3;4;6;9;12;18\right\}\)

b: \(\Leftrightarrow x\in\left\{0;4;-4;8;-8;12;-12;...\right\}\)

mà -16<=x<20

nên \(x\in\left\{-16;-12;-8;-4;0;4;8;12;16\right\}\)

c: \(\Leftrightarrow x-1+4⋮x-1\)

\(\Leftrightarrow x-1\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(x\in\left\{2;0;3;-1;5;-3\right\}\)

d: \(\Leftrightarrow2x+4-5⋮x+2\)

\(\Leftrightarrow x+2\in\left\{1;-1;5;-5\right\}\)

hay \(x\in\left\{-1;-3;3;-7\right\}\)

a, 8 x 6/32 x 1/5 = 3/10

    8 x 6/32 x 1/5 = 8 x 3/16 x 1/5 = 3/10

b, 15/16 x 36/35 x 28 = 27

a=3/10

b=27

chúc em học tốt OwO

a) Cách 1:  8 x \(\frac{6}{32}\)x\(\frac{1}{5}\)=\(\frac{48}{32}\)x\(\frac{1}{5}\)=\(\frac{48}{160}\)=\(\frac{3}{10}\)

    Cách 2 : 8 x \(\frac{6}{32}\)x\(\frac{1}{5}\)= 8 x 6 x 1 / 32 x 5 = 8 x 2 x 3 / 8 x 2 x 2 x 5 = \(\frac{3}{10}\)( Cùng chia cả tử và mẫu cho 8 và 2 )

dễ vãi lol ra bn ngu vãi đái

a, \(\left(x-y+4\right)^2-\left(2x+3y-1\right)^2=\left(x-y+4+2x+3y-1\right)\left(x-y+4-2x-3y+1\right)\)

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

b, \(x^6+y^6=\left(x^2\right)^3+\left(y^2\right)^3=\left(x^2+y^2\right)\left(x^4-x^2y^2+y^4\right)\)

c, \(x^{16}-1=\left(x^2\right)^8-1=\left[\left(x^2\right)^4\right]^2-1=\left(x^8-1\right)\left(x^8+1\right)\)

Phân tích à :)

a) ( x - y + 4 )² - ( 2x + 3y - 1 )²

= [ ( x - y + 4 ) - ( 2x + 3y - 1 ) ][ ( x - y + 4 ) + ( 2x + 3y - 1 ) ]

= ( x - y + 4 - 2x - 3y + 1 )( x - y + 4 + 2x + 3y - 1 )

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

b) x⁶ + y⁶ = ( x² )³ + ( y² )³ = ( x² + y² )( x⁴ - x²y² + y⁴ )

c) x¹⁶ - 1 = ( x⁸ )² - 1² = ( x⁸ - 1 )( x⁸ + 1 )

= ( x⁴ - 1 )( x⁴ + 1 )( x⁸ + 1 )

= ( x² - 1 )( x² + 1 )( x⁴ + 1 )( x⁸ + 1 )

= ( x - 1 )( x + 1 )( x² + 1 )( x⁴ + 1 )( x⁸ + 1 )

d) x³⁶ - 64 = ( x¹⁸ )² - 8²

= ( x¹⁸ - 8 )( x¹⁸ + 8 )

= ( x⁶ - 2 )( x¹² + 2x + 4 )( x⁶ + 2 )( x¹² - 2x + 4 )

e) 9x² + 90x + 225 - ( x - 7 )²

= ( 3x + 15 )² - ( x - 7 )²

= ( 3x + 15 - x + 7 )( 3x + 15 + x - 7 )

= ( 2x + 22 )( 2x - 8 )

= 2( x + 11 ).2( x - 4 )

= 4( x + 11 )( x - 4 )

f) 49( y - 4 )² - 9y² - 36y - 36

= 7²( y - 4 )² - ( 9y² + 36y + 36 )

= ( 7y - 28 )² - ( 3y + 6 )² 

= ( 7y - 28 - 3y - 6 )( 7y - 28 + 3y + 6 )

= ( 4y - 34 )( 10y - 22 )

= 2( 2y - 17 ).2( 5y - 11 )

= 4( 2y - 17 )( 5y - 11 )

\(a)3-\left(17-x\right)=-24+12\)

\(\Leftrightarrow3-17+x=-12\)
\(\Leftrightarrow-14+x=-12\)

\(\Leftrightarrow x=-12+14=2\)

a)  3 - (17 - x) = -24 + 12

=> 3 - 17 + x = -12

=> -14 + x = -12

=> x = -12 + 14

=> x = 2

b) -5 + (16 - x) = 2².3 - 5²

=> -5 + 16 - x = 12 - 25

=> 11 - x = -13

=> x = 11 + 13

=> x = 24

c) -26 - (x - 7) = 3⁴ - 9²

=> -26 - x + 7 = 81 - 81

=> -19 - x = 0

=> x = -19 - 0

=> x = -19

d) -36 - (4 - x) = 37 + (-5)

=> -36 -  4 + x = 32

=> -40 + x = 32

=> x = 32 + 40

=> x = 72

\(b)-5+\left(16-x\right)=2^2.3-5^2\)

\(\Leftrightarrow-5+16-x=4.3-25\)

\(\Leftrightarrow11-x=-13\Leftrightarrow x=24\)

\(\left(\dfrac{1}{2}-\dfrac{x}{3}\right)^2=\dfrac{36}{49}\\ \Rightarrow\left(\dfrac{1}{2}-\dfrac{x}{3}\right)^2=\left(\dfrac{6}{7}\right)^2\\ \Rightarrow\dfrac{1}{2}-\dfrac{x}{3}=\pm\dfrac{6}{7}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}-\dfrac{x}{3}=\dfrac{6}{7}\\\dfrac{1}{2}-\dfrac{x}{3}=-\dfrac{6}{7}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\dfrac{x}{3}=-\dfrac{5}{14}\\\dfrac{x}{3}=\dfrac{19}{14}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{14}\times3\\x=\dfrac{19}{14}\times3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{15}{14}\\x=\dfrac{57}{14}\end{matrix}\right.\)

\(\left(3-\dfrac{2}{3}x\right)^3=-\dfrac{1}{64}\\ \Rightarrow\left(3-\dfrac{2}{3}x\right)^3=\left(-\dfrac{1}{4}\right)^3\\ \Rightarrow3-\dfrac{2}{3}x=-\dfrac{1}{4}\\ \Rightarrow\dfrac{2}{3}x=3-\left(-\dfrac{1}{4}\right)\\ \Rightarrow\dfrac{2}{3}x=\dfrac{13}{4}\\ \Rightarrow x=\dfrac{13}{4}:\dfrac{2}{3}\\ \Rightarrow x=\dfrac{13}{4}\times\dfrac{3}{2}\\ \Rightarrow x=\dfrac{39}{8}\)

`2/9 + 6/27 + 8/36 + 12/54 + 16/72 + 18/81`

`= 2/9 + 2/9 + 2/9 + 2/9 + 2/9 + 2/9`

`= 2/9 xx 6`

`= 12/9 = 4/3`

               Để olm giúp em em nhé!

a,   \(\dfrac{x+2}{7x+42}\) = \(\dfrac{x+2}{7.\left(x+6\right)}\) = \(\dfrac{\left(x+2\right)\left(x-6\right)}{7\left(x-6\right)\left(x+6\right)}\) (đk \(x\ne\) \(\mp\) 6)

 \(\dfrac{-13x}{x^2-36}\) = \(\dfrac{-13x}{\left(x-6\right)\left(x+6\right)}\) = \(\dfrac{-7.13.x}{7.\left(x-6\right).\left(x+6\right)}\) = \(\dfrac{-91x}{7.\left(x-6\right)\left(x+6\right)}\)

b, \(\dfrac{7}{4x+16}\) = \(\dfrac{7\left(x-4\right)}{4.\left(x+4\right).\left(x-4\right)}\) (đk \(x\ne\) \(\pm\) 4)

     \(\dfrac{15}{x^2-16}\) = \(\dfrac{15.4}{\left(x-4\right)\left(x+4\right).4}\) = \(\dfrac{60}{4.\left(x-4\right).\left(x+4\right)}\)

c, \(\dfrac{12}{x^2-4}\) = \(\dfrac{12}{\left(x-2\right).\left(x+2\right)}\)  Đk \(x\) \(\ne\)  \(\pm\) 2

    \(\dfrac{2}{x-3}\) đk \(x\) \(\ne\) 3

\(\dfrac{12}{x^2-4}\) = \(\dfrac{12.\left(x-3\right)}{\left(x^2-4\right).\left(x-3\right)}\) = \(\dfrac{12x-36}{\left(x^2-4\right).\left(x-3\right)}\)

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