(1/4+1/28+1/70) x X=1
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TD
2
11 tháng 6 2021
\(52:x=\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}\)
\(52:x=\frac{1}{1\times4}+\frac{1}{4\times7}+\frac{1}{7\times10}+\frac{1}{10\times13}\)
\(52:x=\frac{1}{3}\times\left(\frac{3}{1\times4}+\frac{3}{4\times7}+\frac{3}{7\times10}+\frac{3}{10\times13}\right)\)
\(52:x=\frac{1}{3}\times\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}\right)\)
\(52:x=\frac{1}{3}\times\left(1-\frac{1}{13}\right)\)
\(52:x=\frac{1}{3}\times\frac{12}{13}\)
\(52:x=\frac{4}{13}\)
\(x=52:\frac{4}{13}\)
\(x=169\)
11 tháng 6 2021
52 :x =1/4 + 1/28 +1/70 + 1/130
52 : x = 4/13
x = 52 : 4/13
x = 169
HC
30 tháng 6 2019
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XO
30 tháng 6 2019
a) Ta có B = \(\left(\frac{2}{15}+\frac{3}{40}+\frac{4}{96}+\frac{5}{204}+\frac{6}{391}\right).x.\left(x-1\right)=\frac{20}{69}\)
=> \(\left(\frac{2}{3.5}+\frac{3}{5.8}+\frac{4}{8.12}+\frac{5}{12.17}+\frac{6}{17.23}\right).x.\left(x-1\right)=\frac{20}{69}\)
=> \(\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{23}\right).x.\left(x-1\right)=\frac{20}{69}\)
=> \(\left(\frac{1}{3}-\frac{1}{23}\right).x.\left(x-1\right)=\frac{20}{69}\)
=> \(\frac{20}{69}.x.\left(x-1\right)=\frac{20}{69}\)
=> \(x.\left(x-1\right)=\frac{20}{69}:\frac{20}{69}\)
=> \(x.\left(x-1\right)=1\)
=> \(x\in\varnothing\)
a) \(\left(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+....+\frac{1}{8554}\right).x=\frac{31}{94}\)
=> \(\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{91.94}\right).x=\frac{31}{94}\)
=> \(\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{91.94}\right)=\frac{31}{94}\)
=> \(\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{91}-\frac{1}{94}\right).x=\frac{31}{94}\)
=> \(\frac{1}{3}.\left(1-\frac{1}{94}\right).x=\frac{31}{94}\)
=> \(\frac{1}{3}.\frac{93}{94}.x=\frac{31}{94}\)
=> \(\frac{31}{94}.x=\frac{31}{94}\)
=> \(x=\frac{31}{94}:\frac{31}{94}\)
=> \(x=1\)
MQ
2
T
17 tháng 8 2016
pt đã cho có dạng \(\frac{1}{\left(x+1\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+10\right)}=\frac{4}{13}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+7}+\frac{1}{x+7}-\frac{1}{x+10}=\frac{4}{13}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+10}=\frac{4}{13}\Leftrightarrow....\)
17 tháng 8 2016
bạn tuấn mình thấy vậy nè
Gỉa sử cho x=1 ta thấy \(\frac{1}{1\times4}\ne\frac{1}{1}-\frac{1}{4}\)
Bạn bấm máy tính thử xem dấu bằng chỉ áp dụng với 2 số tự nhiên liên tiếp thôi còn cái này cách 3 lận
giải thích giúp mình với
20 tháng 1 2018
ta có x2+5x+4
=x2+x+4x+4
=(x2+x)+(4x+4)
=x(x+1)+4(x+1)
=(x+1)(x+4)
tương tự ta đc
x2+11x+28=(x+4)(x+7)
x2+17x+70=(x+7)(x+10)
x2+23x+130=(x+10)(x+13)
=>\(\dfrac{1}{\left(x+1\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+7\right)}+\dfrac{1}{\left(x+7\right)\left(x+10\right)}+\dfrac{1}{\left(x+10\right)\left(x+13\right)}=\dfrac{4}{13}\)\(\dfrac{3}{\left(x+1\right)\left(x+4\right)}+\dfrac{3}{\left(x+4\right)\left(x+7\right)}+\dfrac{3}{\left(x+7\right)\left(x+10\right)}+\dfrac{3}{\left(x+10\right)\left(x+11\right)}=\dfrac{4}{13}\)=>\(\dfrac{1}{x+1}-\dfrac{1}{x+4}+\dfrac{1}{x+4}+....+\dfrac{1}{x+13}=\dfrac{4}{13}\)
=>\(\dfrac{1}{x+1}-\dfrac{1}{x+13}=\dfrac{4}{13}\)
=>\(\dfrac{13\left(x+13\right)}{13\left(x+1\right)\left(x+13\right)}-\dfrac{13\left(x+1\right)}{13\left(x+1\right)\left(x+13\right)}=\dfrac{4\left(x+1\right)\left(x+13\right)}{13\left(x+1\right)\left(x+13\right)}\)
=> 13(x+13)-13(x+1)=4(x+1)(x+13)
=> 13[(x+13)-(x+1)]=(4x+4)(x+13)
=>13(x+13-x-1)=4x2+52x+4x+52
=13.12=4x2+56x+52
=>4x2+56x+52=156
=>4x2+56x-104=0
NT
2
1 tháng 8 2017
a) (2x + 1)(3x - 2) = (5x - 8)(2x + 1)
<=> 6x2 - x - 2 = 10x2 - 11x - 8
<=> 6x2 - 10x2 - x + 11x -2 + 8 = 0
<=> -4x2 + 10x + 6 = 0
<=> -2 (2x2 - 5x - 3) = 0
<=> 2x2 - 5x - 3 = 0
<=> 2x2 - 6x + x - 3 = 0
<=> x (2x + 1) - 3 (2x + 1) = 0
<=> (x - 3) (2x + 1) = 0
* x - 3 = 0 => x = 3
* 2x + 1 = 0 => x = -1/2
S = {-1/2; 3}
b) 4x2 – 1 = (2x +1)(3x -5)
<=> 4x2 – 1 - (2x +1)(3x -5) = 0
<=> (2x - 1) (2x + 1) - (2x + 1)(3x - 5) = 0
<=> (2x + 1) (2x - 1 - 3x + 5) = 0
<=> (2x + 1) (-x + 4) = 0
* 2x + 1 = 0 <=> x = -1/2
* -x + 4 = 0 <=> x = 4
S = {-1/2; 4}
c) (x + 1)2 = 4(x2 – 2x + 1)
<=> (x + 1)2 - 4(x2 – 2x + 1) = 0
<=> (x + 1)2 - 4(x2 – 1)2 = 0
* (x + 1)2 = 0 <=> x = -1
* 4(x2 - 1)2 = 0 <=> x = 1 và x = -1
S = {-1; 1}
d) 2x3 + 5x2 – 3x = 0
<=> x (2x2 + 5x - 3) = 0
<=> x (2x2 + 6x - x - 3) = 0
<=> x [x(2x - 1) + 3 (2x - 1)] = 0
<=> x (2x - 1) (x + 3) = 0
* x = 0
* 2x - 1 = 0 <=> x = 1/2
* x + 3 = 0 <=> x = -3
S = { -3; 0; 1/2}
1 tháng 8 2017
\(\frac{1}{x^2+5x+4}+\frac{1}{x^2+11x+28}+\frac{1}{x^2+17x+70}=\frac{3}{4x-2}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+10\right)}=\frac{3}{4x-2}\)
\(\Leftrightarrow3x^2+21x+36=0\)
\(\Leftrightarrow x=-3\)
H9
HT.Phong (9A5)
CTVHS
24 tháng 9 2023
a) \(\dfrac{6}{13}:\left(\dfrac{1}{2}-x\right)=\dfrac{15}{39}\)
\(\dfrac{1}{2}-x=\dfrac{6}{13}:\dfrac{15}{39}\)
\(\dfrac{1}{2}-x=\dfrac{6}{5}\)
\(x=\dfrac{1}{2}-\dfrac{6}{5}\)
\(x=-\dfrac{7}{10}\)
b) \(3\times\left(\dfrac{x}{4}+\dfrac{x}{28}+\dfrac{x}{70}+\dfrac{x}{130}\right)=\dfrac{60}{13}\)
\(3\times x\times\left(\dfrac{1}{4}+\dfrac{1}{28}+\dfrac{1}{70}+\dfrac{1}{130}\right)=\dfrac{60}{13}\)
\(x\times\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+\dfrac{3}{7\times13}\right)=\dfrac{60}{13}\)
\(x\times\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}\right)=\dfrac{60}{13}\)
\(x\times\left(1-\dfrac{1}{13}\right)=\dfrac{60}{13}\)
\(x\times\dfrac{12}{13}=\dfrac{60}{13}\)
\(x=\dfrac{60}{13}:\dfrac{12}{13}\)
\(x=5\)
NT
1
AL
19 tháng 2 2019
a)
\(\dfrac{1}{x+1}+\dfrac{2}{x^3-x^2-x+1}+\dfrac{3}{x^2-1}=0\) (\(x\ne\pm1\))
\(\Rightarrow\dfrac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)^2}+\dfrac{2}{\left(x+1\right)\left(x-1\right)^2}+\dfrac{3\left(x-1\right)}{\left(x+1\right)\left(x-1\right)^2}=0\)
\(\Rightarrow\dfrac{x^2-2x+1+2+3x-3}{\left(x+1\right)\left(x-1\right)^2}=0\)
\(\Rightarrow\dfrac{x^2+x-2}{\left(x+1\right)\left(x-1\right)^2}=0\)
\(\Rightarrow x^2-x+2=0\)
\(\Rightarrow\left(x-1\right)\left(x+2\right)=0\)
=> Th1 :
x- 1 =0
=> x = 1 ( hư cấu vì không thỏa mãn ĐK )
Th2 :
x+2 = 0
=> x = -2 ( hợp lí )
Vậy nghiệm của phương trình là x = -2
(1/4+1/28+1/70) x X=1
= 3/10 x X = 1
= X = 1 : 3/10
= X = 10/3
(1/4 + 1/28 + 1/70) x X = 1
(35/140 + 5/140 + 2/140) x X = 1
42/140 x X = 1
3/10 x X =1
X = 1 : 3/10
X = 1. 10/3
X = 10/3