sao lại đọc mà cũng không biết đọc à bn ?-?

Ta có: \(3x=4y\Rightarrow\dfrac{x}{4}=\dfrac{y}{3}\Rightarrow\dfrac{x}{20}=\dfrac{y}{15}\)

            \(2y=5z\Rightarrow\dfrac{y}{5}=\dfrac{z}{2}\Rightarrow\dfrac{y}{15}=\dfrac{z}{6}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{6}=\dfrac{x+z}{20+6}=\dfrac{52}{26}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=20.2=40\\y=15.2=30\\z=6.2=12\end{matrix}\right.\)

b)\(3x\left(x+3y\right)-6xy\left(x+3y\right)\)

\(=\left(3x-6xy\right)\left(x+3y\right)\)

c)\(x\left(x+y\right)-5x-5y\)

\(=x\left(x+y\right)-5\left(x+y\right)\)

\(=\left(x-5\right)\left(x+y\right)\)

Bài 1: 

b. \(3x\left(x+3y\right)-6xy\left(x+3y\right)\)

= (3x - 6xy)(x + 3y)

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

c. \(x\left(x+y\right)-5x-5y\)

= x(x + y) - 5(x + y)

= (x - 5)(x + y)

d. \(3\left(x-y\right)-5x\left(y-x\right)\)

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

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

Bài 3:

a. x + 6x² = 0

<=> x(1 + 6x) = 0

<=> \(\left[{}\begin{matrix}x=0\\1+6x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{6}\end{matrix}\right.\)

b. 2(x + 3) - x(x + 3) = 0

<=> (2 - x)(x + 3) = 0

<=> \(\left[{}\begin{matrix}2-x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

c. 5x(x - 2) - (2 - x) = 0

<=> 5x(x - 2) + (x - 2) = 0

<=> (5x + 1)(x - 2) = 0

<=> \(\left[{}\begin{matrix}5x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{5}\\x=2\end{matrix}\right.\)

d. (x + 1) = (x + 1)²

<=> (x + 1) - (x + 1)² = 0

<=> (1 - x - 1)(x + 1) = 0

<=> -x(x + 1) = 0

<=> \(\left[{}\begin{matrix}-x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)