\(a,A=\left(x+y\right)^2-9z^2=\left(x+y-3z\right)\left(x+y+3z\right)\\ A=\left(5+7-36\right)\left(5+7+36\right)=-24\cdot48=-1152\\ b,B=\left(2x-y\right)\left(2x+y\right)+\left(2x+y\right)=\left(2x+y\right)\left(2x-y-1\right)\\ B=\left(2+2\right)\left(2-2-1\right)=4\cdot\left(-1\right)=-4\)

\(\dfrac{3}{x-5}=\dfrac{-4}{x+2}\left(x\ne5;-2\right).\\ \Leftrightarrow\dfrac{3}{x-5}+\dfrac{4}{x+2}=0.\\ \Leftrightarrow\dfrac{3x+6+4x-20}{\left(x-5\right)\left(x+2\right)}=0.\\ \Rightarrow7x=14.\\ \Leftrightarrow x=2\left(TM\right).\)

\(\dfrac{2}{3}+\dfrac{1}{3}:x=\dfrac{1}{2}\)
\(\dfrac{1}{3}:x=\dfrac{1}{2}-\dfrac{2}{3}\)
\(\dfrac{1}{3}:x=-\dfrac{1}{6}\)
\(x=\dfrac{1}{3}:\left(-\dfrac{1}{6}\right)\)
\(x=-2\)
Vậy ...
#AvoidMe

`2/3 +1/3 : x=1/2`

`=> 1/3 : x= 1/2 -2/3`

`=> 1/3 : x= 3/6 -4/6`

`=> 1/3 : x=-1/6`

`=> x=1/3 :(-1/6)`

`=> x=1/3 xx (-6)`

`=> x= -2`

Vậy `x=-2`

(x+2).5=210

23/12

1/3

a.x=23/12

b.x=1/3

x = 5/3+1/4

x = 23/12

x = 1/2 x 2/3

x= 1/3

giúp mình đi mọi người

5/6-x=2/3

x=5/6-2/3

x=1/6

4/3:x=2

x=4/3:2

x=2/3

a) x = 1/6

b) x = 2/3

\(a,50\%x-0,2+x=\dfrac{4}{5}\)

\(\Leftrightarrow\dfrac{1}{2}x-0,2+x=\dfrac{4}{5}\)

\(\Leftrightarrow\dfrac{1}{2}x+x=\dfrac{4}{5}+0,2\)

\(\Leftrightarrow\dfrac{3}{2}x=\dfrac{4}{5}+\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{3}{2}x=1\)

\(\Leftrightarrow x=\dfrac{2}{3}\)

\(b,\left(x-\dfrac{3}{4}\right):\dfrac{1}{2}+\dfrac{3}{2}=\dfrac{25}{2}\)

\(\Leftrightarrow\left(x-\dfrac{3}{4}\right).2=\dfrac{25}{2}-\dfrac{3}{2}\)

\(\Leftrightarrow\left(x-\dfrac{3}{4}\right).2=\dfrac{22}{2}\)

\(\Leftrightarrow x-\dfrac{3}{4}=11:2\)

\(\Leftrightarrow x=\dfrac{11}{2}+\dfrac{3}{4}\)

\(\Leftrightarrow x=\dfrac{25}{4}\)

a: \(M=\left(\dfrac{-3}{7}x^3y\right)\cdot\dfrac{7xy^3}{12}-x^2y^2\cdot\left(-\dfrac{3}{4}x^2y^2\right)\)

\(=\dfrac{-1}{4}x^4y^4+\dfrac{3}{4}x^4y^4\)

\(=\dfrac{1}{2}x^4y^4\)

b: Hệ số là 1/2

Biến là \(x^4;y^4\)

bậc là 4+4=8

c: Thay x=-1 và y=-2 vào M, ta được:

\(M=\dfrac{1}{2}\left(-1\right)^4\cdot\left(-2\right)^4=\dfrac{1}{2}\cdot16=8\)

Câu a,b  hình như nhầm đề mình tự sửa nha ;-;

a, Ta có : \(\left(x^2-x-6\right)^2+\left(x-3\right)^2\)

\(=\left(x^2-3x+2x-6\right)^2+\left(x-3\right)^2\)

\(=\left(x-3\right)^2\left(x+2\right)^2+\left(x-3\right)^2\)

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

b, Ta có : \(\left(x^2-x-20\right)^2+\left(x+4\right)^2\)

\(=\left(x^2+4x-5x-20\right)^2+\left(x+4\right)^2\)

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

\(=\left(x+4\right)^2\left(\left(x-5\right)^2+1\right)\)