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làm lại
ta có : \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)
=>\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)
=>\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+....+\frac{3}{\left(x+2\right)\left(x+5\right)}=\frac{9}{20}\)
=>\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{x+2}-\frac{1}{x+5}=\frac{9}{20}\)
=>\(\frac{1}{2}-\frac{1}{x+5}=\frac{9}{20}\)
=>\(\frac{1}{x+5}=\frac{1}{2}-\frac{9}{20}\)
=>\(\frac{1}{x+5}=\frac{1}{20}\)
=>\(x+5=20\)
=>\(x=20-5\)
=>\(x=15\)
ta có : \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+....+\frac{1}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)
=>\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(x+2\right)\left(x+3\right)}=\frac{3}{20}\)
=>\(3.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+....+\frac{1}{\left(x+2\right)\left(x+3\right)}\right)=3.\frac{3}{20}\)
=>\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+....+\frac{3}{\left(x+2\right)\left(x+3\right)}=\frac{9}{20}\)
=>\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{x+2}-\frac{1}{x+3}=\frac{9}{20}\)
=>\(\frac{1}{2}-\frac{1}{x+3}=\frac{9}{20}\)
=>\(\frac{1}{x+3}=\frac{1}{2}-\frac{9}{20}\)
=>\(\frac{1}{x+3}=\frac{1}{20}\)
=>\(x+3=20\)
=>\(x=20-3\)
=>\(x=17\)
1) \(\left|x-2\right|=x\)
th1 : \(x\ge2\) thì \(\left|x-2\right|=x\Leftrightarrow x-2=x\Leftrightarrow-2=0x\)(vô lí)
th2 : \(x< 2\) thì \(\left|x-2\right|=x\Leftrightarrow2-x=x\Leftrightarrow2=2x\Leftrightarrow x=1\)(tmđk)
vậy \(x=1\)
b)
\(\left(x+\dfrac{1}{2}\right)^{20}\ge0\)
\(\left(x-\dfrac{1}{3}\right)^{40}\ge0\)
Mà:
\(\left(x+\dfrac{1}{2}\right)^{20}+\left(x-\dfrac{1}{3}\right)^{40}< 1\)
\(\Leftrightarrow x\in\left\{\varnothing\right\}\)
(2^20 +1).(2^40 - 2^20 +1) -2^60
= 2^20 . 2^40 - 2^40 + 2^20 + 2^40 -2^20 + 1 -2^60
=2^60 +1 - 2^60
= 1
\(\hept{\begin{cases}x+y=40\\x-y=20\end{cases}}\)
\(\Rightarrow x+y+x-y=60\)
\(\Rightarrow2x=60\)
\(\Rightarrow x=30\)
\(\Rightarrow y=10\)
1.(x+1)+(x+2)+(x+3)+.......+(x+19)+(x+20)=40
⇒20x+(1+2+...+20)=40
⇒20x+210=40
⇒20x=40-210=-170
⇒x=-8.5
1. (x+1)+(x+2)+...+(x+20)=40
x+1+x+2+...+x+20 =40
20x+(1+2+...+20) =40
20x+210 =40
20x =40-210
20x =-170
x =-170:20
x =-8,5
Vậy x=-8,5
a) (-5)2 hay là -(52). 2 cái khác nhau đó nha
b)(2x+1)2+40=(-7)20:718
⇔(2x+1)2+40=(-7)2
⇔(2x+1)2=49-40
⇔(2x+1)2=32
⇔2x+1=3
2x+1=-3
⇔x=1
x=-2
vậy x ϵ {1;-2}
c)(-2)x+1+(-2)x=-1024
⇔(-2)x.(2+1)=-1024
⇔(-2)x.3=-1024
⇔(-2)x=-1024/3 (vô lí)
⇒ ko có giá trị của x
vậy x ϵϕ
x+20:1/2=40
=>x+20x2=40
=>x+40=40
=>x=40-40
=>x=0
x+20:1/2=40
=>x+20x2=40
=>x+40=40
=>x=40-40
=>x=0