Bài 4:
a) Ta có: \(4x-4-10+6x=-12\)
\(\Leftrightarrow10x=2\)
hay \(x=\dfrac{1}{5}\)
b) Ta có: \(2x\left(x-1\right)-3\left(x^2-4x\right)+x\left(x+2\right)=-3\)
\(\Leftrightarrow2x^2-2x-3x^2+12x+x^2+2x=-3\)
\(\Leftrightarrow12x=-3\)
hay \(x=-\dfrac{1}{4}\)
Bài 2 :
a, \(x\left(x+y\right)-y\left(x+y\right)=\left(x-y\right)\left(x+y\right)=x^2-y^2\)
Thay x = -1/2 ; y = -2 vào biểu thức trên ta được :
\(\dfrac{1}{4}-4=\dfrac{1}{4}-\dfrac{16}{4}=-\dfrac{15}{4}\)
b, \(xy\left(x+y\right)-x^2\left(x+y\right)-y^2\left(x-y\right)=\left(xy-x^2\right)\left(x+y\right)-y^2\left(x-y\right)\)
\(=x\left(y-x\right)\left(x+y\right)+y^2\left(y-x\right)=\left(y-x\right)\left(x^2+xy+y^2\right)=y^3-x^3\)
Thay x = -2 ; y = -3 vào biểu thức trên ta được :
\(=-27-\left(-8\right)=-19\)
Bài 3 a
2(3x - 1) - 3(2x+3)
➞ 6x -2- 6x -9= 0
➞ -2 -9 =0 ( vô lý )
<=> phương trình vô ngiệm
Bài 5:
Ta có: \(x^{n-2}\cdot\left(x^2-1\right)-x\left(x^{n-1}-x^{n-3}\right)\)
\(=x^n-x^{n-2}-x^n+x^{n-2}\)
=0
Bài 2:
a) Ta có: \(x\left(x+y\right)-y\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y\right)\)
\(=x^2-y^2\)
\(=\left(-\dfrac{1}{2}\right)^2-\left(-2\right)^2\)
\(=\dfrac{1}{4}-4=\dfrac{-15}{4}\)
b) Ta có: \(xy\left(x+y\right)-x^2\left(x+y\right)-y^2\left(x-y\right)\)
\(=\left(x+y\right)\left(xy-x^2\right)-y^2\left(x-y\right)\)
\(=x\left(x+y\right)\left(y-x\right)-y^2\left(x-y\right)\)
\(=-x\left(x+y\right)\left(x-y\right)-y^2\left(x-y\right)\)
\(=\left(x-y\right)\left(-x^2-xy-y^2\right)\)
\(=\left[\left(-2\right)-\left(-3\right)\right]\cdot\left[-\left(-2\right)^2-\left(-2\right)\cdot\left(-3\right)-\left(-3\right)^2\right]\)
\(=-4-6-9=-19\)
Bài 3:
a) Ta có: \(2\left(3x-1\right)-3\left(2x+3\right)\)
\(=6x-2-6x-9\)
=-11
b) Ta có: \(5x\left(x+1\right)-2\left(3x+1\right)-\left(7-x\right)\)
\(=5x^2+5x-6x-2-7+x\)
\(=5x^2-9\)
