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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Bài lớp \(6\) chưa sử dụng dấu \(\Leftrightarrow\) chị nhé ! Vẫn phải sử dụng dấu \(\Rightarrow\) Khi nào bài lớp \(8\) trở lên thì cj hãy dùng \(\Leftrightarrow\) ạ

23/12

1/3

a.x=23/12

b.x=1/3

\(\dfrac{2x}{14}-\dfrac{x}{14}=-\dfrac{21}{14}\)

\(2x-x=-21\)

\(x=-21\)

x=-21

1) \(\Rightarrow16x^2+24x+9+9x^2-24x+16+4-25x^2=x\)

\(\Rightarrow x=29\)

2)

a) \(=x^2-9-x^2+6x-9=6x-18\)

b) \(=\left(3x-1+2x+1\right)^2=\left(5x\right)^2=25x^2\)

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

1: \(-3x^3\left(2a\cdot x^2-bx+c\right)=-6x^5+9x^4-3x^3\)

=>\(-3x^3\left(2a\cdot x^2-bx+c\right)=-3x^3\left(2x^2-3x+1\right)\)

=>\(2a\cdot x^2-bx+c=2x^2-3x+1\)

=>2a=2; -b=-3; c=1

=>a=1; b=3; c=1

2: \(\left(ax+b\right)\left(x^2-cx+2\right)=x^3+x^2-2\)

=>\(a\cdot x^3-ac\cdot x^2+2a\cdot x+bx^2-bc\cdot x+2b=x^3+x^2-2\)

=>a=1; -ac+b =1; 2a-bc=0; 2b=-2

=>a=1; b=-1; -c+(-1)=1; 2-(-1)*c=0

=>a=1; b=-1; -c=2; 2+c=0

=>a=1; b=-1; c=-2