Rút gọn: A=x*/x-2//x2+8x-20
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\(A=\frac{x\left|x-2\right|}{x^2+8x-20}=\frac{x\left|x-2\right|}{x^2-2x+10x-20}=\frac{x\left|x-2\right|}{x\left(x-2\right)+10\left(x-2\right)}=\frac{x\left|x-2\right|}{\left(x+10\right)\left(x-2\right)}\)
Xét \(x-2\ge0\Leftrightarrow x\ge2\) ta có :
\(A=\frac{x\left(x-2\right)}{\left(x+10\right)\left(x-2\right)}=\frac{x}{x+10}\)
Xét \(x-2< 0\Leftrightarrow x< 2\) ta có :
\(A=\frac{x\left(2-x\right)}{\left(x+10\right)\left(x-2\right)}=\frac{-x}{x+10}\)
\(\sqrt{\left(x-4\right)^2}+\frac{x-4}{\sqrt{x^2-8x+16}}\)
\(=x-4+\frac{x-4}{\sqrt{\left(x-4\right)^2}}\)
\(=x-4+\frac{x-4}{x-4}\)
\(=x-4+1\)
\(=x-3\)
\(\sqrt{\left(x-4\right)^2}+\frac{x-4}{\sqrt{x^2-8x+16}}\)
\(=x-4+\frac{x-4}{\sqrt{\left(x+4\right)^2}}\)
\(=x-4+\frac{x-4}{x-4}\)
\(=x-4+1\)
= x - 3
B= \(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).....\left(1-\dfrac{1}{20}\right)\)
B= \(\dfrac{1}{2}.\dfrac{2}{3}.....\dfrac{19}{20}\)
B= \(\dfrac{1.2.....19}{2.3.....20}\)
B= \(\dfrac{1}{20}\)
Câu 1:
\(\left(a-b\right)\left(a^2+ab+b^2\right)-\left(a^3+b^3\right)\)
\(=a^3-b^3-a^3-b^3\)
\(=-2b^3\)
Câu 2:
a: \(x^2-6x+9=0\)
\(\Leftrightarrow\left(x-3\right)^2=0\)
=>x-3=0
hay x=3
b: \(x^2-\dfrac{2}{5}x+\dfrac{1}{25}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{1}{5}+\dfrac{1}{25}=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{5}\right)^2=0\)
=>x-1/5=0
hay x=1/5
8x3+36x2+54x+27
tại x =-4
=>8×(-4)3+36×(-4)2+54×(-4)+27
=8×(-64)+36×16+54×(-4)+27
=-512+576-216+27
=-125
(4x-3)(16x2+12x+9)-x2(64x-4)
=4x(16x2+12x+9)- 3(16x2+12x+9)-x2(64x-4)
=(64x3+48x2+36x)-(48x2+36x+27)-(64x3-4x2)
=64x3+48x2+36x-48x2-36x-27-64x3+4x2
=(64x3-64x3)+(48x2-48x2+4x2)+(36x-36x)-27
=4x2-27
tại x=-1/4
=> 4×(-1/4)2-27
=4×1/16-27
=1/4-27
=-107/4
(ko bt cs đúng ko nx )