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\(=\dfrac{x^2+2x-x^2+4x-4+6-5x}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x-2}\)
5) Ta có: \(\dfrac{x^3-x^2-2x-20}{x^2-4}-\dfrac{5}{x+2}+\dfrac{3}{x-2}\)
\(=\dfrac{x^3-x^2-2x-20}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^3-x^2-2x-20-5x+10+3x+6}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^3-x^2-4x-4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2\left(x-1\right)-4\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{\left(x-1\right)\left(x^2-4\right)}{\left(x^2-4\right)}\)
\(=x-1\)
6) Ta có: \(\dfrac{x-1}{x^3}-\dfrac{x+1}{x^3-x^2}+\dfrac{3}{x^3-2x^2+x}\)
\(=\dfrac{x-1}{x^3}-\dfrac{x+1}{x^2\left(x-1\right)}+\dfrac{3}{x\left(x-1\right)^2}\)
\(=\dfrac{\left(x-1\right)^3}{x^3\cdot\left(x-1\right)^2}-\dfrac{x\left(x+1\right)\left(x-1\right)}{x^3\cdot\left(x-1\right)^2}+\dfrac{3x^2}{x^3\cdot\left(x-1\right)^2}\)
\(=\dfrac{x^3-3x^2+3x-1-x\left(x^2-1\right)+3x^2}{x^3\cdot\left(x-1\right)^2}\)
\(=\dfrac{x^3+3x-1-x^3+x}{x^3\cdot\left(x-1\right)^2}\)
\(=\dfrac{4x-1}{x^3\cdot\left(x-1\right)^2}\)
\(a,2x^2+6x=2x\left(x+3\right)\\ b,x^2+2xy+y^2-9z^2\\ =\left(x^2+2xy+y^2\right)-\left(3z\right)^2\\ =\left(x+y\right)^2-\left(3z\right)^2\\ =\left(x+y-3z\right)\left(x+y+3z\right)\\ b,x^3-2x^2+x\\ =x\left(x^2+2x+1\right)\\ =x\left(x+1\right)^2\\ d,x^2-2x-15=x^2-5x+3x-15\\ =x\left(x-5\right)+3\left(x-5\right)\\ =\left(x+3\right)\left(x-5\right)\)
\(2x\left(x-3\right)+\left(x-1\right)\left(x+1\right)\)
\(=2x^2-6x+x^2-1\)
\(=3x^2-6x+1\)
a) \(2x\left(x+5\right)-2x^2=2x^2+10x-2x^2=10x\)
b) \(\left(x+3\right)^2+\left(x-1\right)\left(3+2x\right)=x^2+6x+9+3x+2x^2-3-2x\)
\(=3x^2+7x+6\)
a: \(2x\left(x+5\right)-2x^2=2x^2+10x-2x^2=10x\)
b: \(\left(x+3\right)^2+\left(2x+3\right)\left(x-1\right)\)
\(=x^2+6x+9+2x^2-2x+3x-3\)
\(=3x^2+7x+6\)
2x ( x - 5 ) x . ( 3 - 2x ) = 26
2x\(^2\)- 10x . 3x - 2x\(^2\)= 26
2x\(^2\). ( 10x - 3x ) = 26
2x\(^2\). 7x = 26
14x\(^3\) = 26
x\(^3\) = 26 : 14
x\(^3\) = \(\frac{13}{7}\)
→ X = 1.229.... \(\approx\)1,3
\(2x\left(x-5\right)\cdot x\left(3-2x\right)=26\)
\(\Leftrightarrow\left(2x^2-10x\right)\left(3x-6x\right)=26\)
\(\Leftrightarrow6x^3-30x^2-12x^3+60x^2=26\)
\(\Leftrightarrow-12x^3+30x^2=26\)
\(\Leftrightarrow2\left(-6x^3+15x^2\right)=26\)
\(\Leftrightarrow-6x^3+15x^2=13\)
\(\Leftrightarrow-6x^3+15x^2-13=0\)
...............mình chỉ làm được đến đây thôi!
x(x – y) + y(x + y)
= x.x – x.y + y.x + y.y
= x2 – xy + xy + y2
= x2 + y2.
Tại x = –6 ; y = 8, giá trị biểu thức bằng : (–6)2 + 82 = 36 + 64 = 100.
x.(x2 – y) – x2.(x + y) + y.(x2 – x)
= x.x2 – x.y – (x2.x + x2.y) + y.x2 – y.x
= x3 – xy – x3 – x2y + x2y – xy
= (x3 – x3) + (x2y – x2y) – xy – xy
= –2xy
Tại và y = –100, giá trị biểu thức bằng:
a) x(x – y) + y (x + y) = x2 – xy +yx + y2= x2+ y2
với x = -6, y = 8 biểu thức có giá trị là (-6)2 + 82 = 36 + 64 = 100
c: \(=\dfrac{x^2+x-x^2+x+2}{\left(x-1\right)\left(x+1\right)}=\dfrac{2}{x-1}\)
a) \(\dfrac{x^2+xy}{x^2-y^2}=\dfrac{x\left(x+y\right)}{\left(x+y\right)\left(x-y\right)}=\dfrac{x}{x-y}\)
b)\(\dfrac{x}{x-1}-\dfrac{x}{x+1}+\dfrac{2}{x^2-1}=\dfrac{x\left(x+1\right)-x\left(x-1\right)+2}{\left(x-1\right)\left(x+1\right)}=\dfrac{2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{2}{x-1}\)