\(\dfrac{1}{2}.\left(4x-5\right)-2^2-18=0\\ \Rightarrow\dfrac{1}{2}.\left(4x-5\right)-22=0\\ \Rightarrow\dfrac{1}{2}.\left(4x-5\right)=22\\ \Rightarrow4x-5=22:\dfrac{1}{2}\\ \Rightarrow4x-5=44\\ \Rightarrow4x=49\\ \Rightarrow x=\dfrac{49}{4}\)

    \(Vậy...\)

Cảm ơn anh !!

x đầu tên là nhân hay x cần tìm vậy ạ

1.

1) \(M=\left[6.\left(-\dfrac{1}{3}\right)^2-3\left(-\dfrac{1}{3}\right)+1\right]:\left(-\dfrac{1}{3}-1\right)\)

\(=\left[6.\dfrac{1}{9}+2\right]:\left(-\dfrac{1}{3}-\dfrac{3}{3}\right)\\ =\left(\dfrac{2}{3}+2\right):\left(-\dfrac{4}{3}\right)\\ =\dfrac{8}{3}.\left(-\dfrac{3}{4}\right)\\ =-2\)

    \(Vậy...\)

2) \(\left(2x-3\right)^2=25\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=5\\2x-3=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)

    \(Vậy...\)

\(x:y:z=3:5:\left(-2\right)\\ \Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{-2}\)

          Và \(x+y+z=36\)

Áp dụng tính chất dãy tỷ số bằng nhau ta có:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{-2}=\dfrac{x+y+z}{3+5+\left(-2\right)}=\dfrac{36}{6}=6\\ \Rightarrow\left\{{}\begin{matrix}x=6.3=18\\y=6.5=30\\z=6.\left(-2\right)=-12\end{matrix}\right.\)

    \(Vậy...\)

a, \(9x^2-6xy+y^2\)

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

b, \(9x^2-6x+1\)

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

c, \(x^2-xy+x-y\)

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

d, \(2x^2-2xy-3x+3y\)

\(=2x\left(x-y\right)-3\left(x-y\right)\\ =\left(2x-3\right)\left(x-y\right)\)

e, \(x^2+3x-xy-3y\)

\(=x\left(x+3\right)-y\left(x+3\right)\\ =\left(x-y\right)\left(x+3\right)\)

f, \(a^2-2ab+b^2-4\)

\(=\left(a-b\right)^2-2^2\\ =\left(a-b-2\right)\left(a-b+2\right)\)

Gửi từng câu 1 đi 

Dài.....