[x-1]5= -32
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\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
a)
Dãy trên có số số hạng là:
( 20 - 1 ) : 1 + 1 = 20 ( số hạng )
Tổng của dãy trên là:
( 20 + 1 ) x 20 : 2 = 210
Đáp số: 210
b)
Dãy trên có số số hạng là:
( 21 - 1 ) : 2 + 1 = 11 ( số hạng )
Tổng của dãy trên là:
( 21 + 1 ) x 11 : 2 = 121
Đáp số: 121
c) ( 2x - 1 ) x 2 = 13
2x - 1 = \(\dfrac{13}{2}\)
2x = \(\dfrac{15}{2}\)
\(x=\dfrac{15}{4}\)
32 x ( x - 10 ) = 32
( x - 10 ) = 1
x = 11
\(A=1+2+3+...+20\)
Số hạng:
\(\left(20-1\right):1+1=20\) (số hạng)
Tổng: \(\left(20+1\right)\cdot20:2=210\)
\(B=1+3+5+...+21\)
Số hạng:
\(\left(21-1\right):2+1=11\) (số hạng)
Tổng: \(\left(21+1\right)\cdot11:2=121\)
\(\left(2x-1\right)\cdot2=13\)
\(\Rightarrow2x-1=\dfrac{13}{2}\)
\(\Rightarrow2x=\dfrac{15}{2}\)
\(\Rightarrow x=\dfrac{15}{4}\)
\(32\cdot\left(x-10\right)=32\)
\(\Rightarrow x-10=1\)
\(\Rightarrow x=11\)
a, (4 - x )5 +(x - 2)5 =32
(=) 1024 - x5 + x5 - 32 = 32
(=) -x5 + x5 = 32 + 32 - 1024
(=) 0x = -960
=) phương trình vô nghiệm
3/2+5/4+9/8/+17/16+33/32-6+x-1/x+1=31/32-2/2015
=(1+1/2)+(1+1/4)+(1+1/8)+(1+1/16)+(1+1/32-6+x-1/x+1=31/32-2/2015
=(1/2+1/4+1/8+1/16+1/32)+(1+1+1+1+1)-6+x-1/x+1=31/32-2/2015
=31/32+5-6+x-1/x+1=31/32-2/2015
=5-6+x-1/x+1=31/32-2/2015-31/32
=-1+x-1/x+1=-2/2015
=x-1/x+1=-2/2015- -1
=x-1/x+1=2013/2015
=>x=2014
g)
Áp dụng tính chất của dãy tỉ số bằng nhau :
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{x+y}{3+5}=\dfrac{32}{8}=4\)
\(+)\)\(\dfrac{x}{3}=4\Rightarrow x=4\times3=12\)
\(+)\)\(\dfrac{y}{5}=4\Rightarrow y=4\times5=20\)
Vậy \(x=12;y=20\)
h)
\(\dfrac{1}{2}-\left|x-\dfrac{1}{3}\right|=\left|-\dfrac{1}{5}\right|\)
\(\dfrac{1}{2}-\left|x-\dfrac{1}{3}\right|=\dfrac{1}{5}\)
\(\left|x-\dfrac{1}{3}\right|=\dfrac{1}{2}-\dfrac{1}{5}\)
\(\left|x-\dfrac{1}{3}\right|=\dfrac{5}{10}-\dfrac{2}{10}\)
\(\left|x-\dfrac{1}{3}\right|=\dfrac{3}{10}\)
\(\Rightarrow x-\dfrac{1}{3}=\dfrac{3}{10}\) hoặc \(x-\dfrac{1}{3}=\dfrac{-3}{10}\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{3}{10}\\x-\dfrac{1}{3}=\dfrac{-3}{10}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{10}+\dfrac{1}{3}\\x=\dfrac{-3}{10}+\dfrac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{30}+\dfrac{10}{30}\\x=\dfrac{-9}{30}+\dfrac{10}{30}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{19}{30}\\x=\dfrac{1}{30}\end{matrix}\right.\)
4 + x = 7 - 2x
4 + x + 2x = 7
4 + 3x = 7
3x = 7 - 4
3x = 3
x = 3 : 3
x =1
5 + x = 8,3 + 4,7 - x
5 + x + x = 8,3 + 4,7
5 + 2x = 13
2x = 13 - 5
2x = 8
x = 8 : 2
x= 4
d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)
\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)
hay \(x=\dfrac{25}{372}\)
Vậy: \(x=\dfrac{25}{372}\)
e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)
f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)
\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)
\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)
\(\Leftrightarrow3x=\dfrac{1}{9}\)
hay \(x=\dfrac{1}{27}\)
g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)
\(\Leftrightarrow\dfrac{4}{3}x=2\)
hay \(x=\dfrac{3}{2}\)
Vậy: \(x=\dfrac{3}{2}\)
h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)
Vậy ...
i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)
Vậy ...
\(\dfrac{x}{4}=-\dfrac{5}{3}\)\(\Rightarrow x.3=-5.4\Rightarrow3x=-20\Rightarrow x=\dfrac{-20}{3}\)
\(\dfrac{1}{2}+x=1,5.\dfrac{4}{3}\)\(\Rightarrow\dfrac{1}{2}+x=2\Rightarrow x=2-\dfrac{1}{2}\Rightarrow x=\dfrac{3}{2}\)
\(\dfrac{2}{3}-\dfrac{1}{3}.x=\dfrac{1}{5}\Rightarrow\)\(\dfrac{1}{3}.x=\dfrac{2}{3}-\dfrac{1}{5}\Rightarrow\dfrac{1}{3}.x=\dfrac{7}{15}\Rightarrow x=\dfrac{7}{15}:\dfrac{1}{3}\)\(\Rightarrow x=\dfrac{7}{5}\)
\(\left(x-5,2\right).\dfrac{1}{3}=15\Rightarrow x-5,2=15:\dfrac{1}{3}\Rightarrow x-5,2=45\)\(\Rightarrow x=45+5,2\Rightarrow x=50,2\)
P/s Nhớ tick cho mình nha. Thanks bạn
(x-1)5=-32=(-2)5
=>x-1=-2
x=-2+1
x=-1
Vậy x=-1