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a: x<5 thì 5-x>0
A=5x+5-x+5=4x+10
b: Khi x>=0 thì \(B=5x+10+3x=8x+10\)
Khi x<0 thì B=5x+10-3x=2x+10
d: Khi x>=3 thì \(D=x-3-3x+15=-2x+12\)
Khi x<3 thì D=3-x-3x+15=-4x+18
\(\dfrac{x}{x-5}+\dfrac{4x}{x+5}+\dfrac{x\left(x-15\right)}{x^2-25}\)
= \(\dfrac{x\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}+\dfrac{4x\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}+\dfrac{x\left(x-15\right)}{\left(x+5\right)\left(x-5\right)}\)
= \(\dfrac{x^2+5x+4x^2-20x+x^2-15x}{\left(x-5\right)\left(x+5\right)}\)
= \(\dfrac{6x^2-30x}{\left(x-5\right)\left(x+5\right)}\)
= \(\dfrac{6x\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}\)
= \(\dfrac{6x}{x+5}\)
a, \(\left(x-5\right)\left(x+2\right)+\left(x+1\right)\left(2-x\right)=15\)
\(\Leftrightarrow x^2+2x-5x-10+2x-x^2+2-x=15\Leftrightarrow-2x-23=0\)
\(\Leftrightarrow x=-\frac{23}{2}\)
b, \(\left(2x-3\right)\left(x+5\right)-\left(x-2\right)\left(2x+1\right)=3\)
\(\Leftrightarrow2x^2+10x-3x-15-\left(2x^2+x-4x-2\right)=3\)
\(\Leftrightarrow10x-16=0\Leftrightarrow x=\frac{8}{5}\)
a) \(\left(x-5\right)\left(x+2\right)+\left(x+1\right)\left(2-x\right)=15\)
\(\Leftrightarrow x^2+2x-5x-10+2x-x^2+2-x=15\)
\(\Leftrightarrow-2x-8=15\)
\(\Leftrightarrow-2x=23\)
\(\Leftrightarrow x=-\frac{23}{2}\)
Vậy...
b) \(\left(2x-3\right)\left(x+5\right)-\left(x-2\right)\left(2x+1\right)=3\)
\(\Leftrightarrow2x^2+10x-3x-15-2x^2-x+4x+2=3\)
\(\Leftrightarrow10x-13=3\)
\(\Leftrightarrow10x=16\)
\(\Leftrightarrow x=\frac{8}{5}\)
Vậy...
5x(x-3)^2-5(x-1)^3+15(x+2)(x-2)=5
5x(x-3)^2-5(x-1)^3+15(x^2-2^2)=5
5x(x^2-6x+9)-5(x^3-3x^2+3x-1)+15x^2-60=5
5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-60=5
30x-55=5
30x=60
x=2