\(\frac{x^2-3x-x+3}{x-3}=\frac{x\left(x-3\right)-\left(x-3\right)}{x-3}=\frac{\left(x-3\right)\left(x-1\right)}{x-3}=x-1\)( ĐK: \(x\ne3\))

\(\frac{2x^3-5x^2-4x+3}{2x-1}=\frac{\left(2x^3-x^2\right)-\left(4x^2-2x\right)-\left(6x-3\right)}{2x-1}=\frac{x^2\left(2x-1\right)-2x\left(2x-1\right)-3\left(2x-1\right)}{2x-1}=\frac{\left(2x-1\right)\left(x^2-2x-3\right)}{2x-1}=x^2-2x-3\)( ĐK: \(x\ne\frac{1}{2}\))

Tham khảo nhé~

\(\left(x-2\right)^3-x^2\left(x-6\right)=4\)

\(x^3-6x^2+12x-8-x^3+6x^2=4\)

\(12x-8=4\)

\(12x=4+8\)

\(12x=12\)

\(\Rightarrow x=1\)

Vậy \(x=1\)

\(\left(x+1\right)^3-x\left(x-2\right)^2+x-1=0\)

\(x^3+3x^2+3x+1-x^3+4x^2-4x+x-1=0\)

\(7x^2=0\)

\(\Rightarrow x=0\)

Vậy \(x=0\)

Tham khảo nhé~

\(x^3-4x=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^3=0\\4x=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\Rightarrow x=0\\\Rightarrow x=0\end{cases}}\Leftrightarrow x=0\)

\(x^2-x+\frac{1}{4}=0\)

\(x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2=0\)

\(\left(x-\frac{1}{2}\right)^2=0\)

\(\Rightarrow x-\frac{1}{2}=0\)

\(\Rightarrow x=\frac{1}{2}\)

Vậy \(x=\frac{1}{2}\)

Tham khảo nhé~

a) \(x^2-9x+14\)

\(=x^2-2x-7x+14\)

\(=x\left(x-2\right)-7\left(x-2\right)\)

\(=\left(x-2\right)\left(x-7\right)\)

b) \(x^2+17x-18\)

\(=x^2+18x-x-18\)

\(=x\left(x+18\right)-\left(x+18\right)\)

\(=\left(x+18\right)\left(x-1\right)\)

c) \(2x^2-7x+3\)

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

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

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

d) \(x^2-25x+144\)

\(=x^2-9x-16x+144\)

\(=x\left(x-9\right)-15\left(x-9\right)\)

\(=\left(x-9\right)\left(x-15\right)\)

lấy tỉ lệ  nhân 100 ra tỉ suất rồi lấy tỉ suất cộng với tỉ lệ gia tăng

a. 2x²-xy

= x(2x-y)

b. x²-xy-x+y

= (x²-xy)-(x+y)

=x(x-y)-(x-y)

=(x-y)(x-1)

Deep vậy bạn =))

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