\(\frac{90}{x}\)- \(\frac{90}{x+5}\)= \(\frac{1}{4}\)
Giup mình giải với
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TN
24 tháng 11 2016
\(\frac{x+1}{94}+\frac{x+2}{93}+\frac{x+3}{92}=\frac{x+4}{91}+\frac{x+5}{90}+\frac{x+6}{89}\)
\(\Leftrightarrow\frac{x+1}{94}+1+\frac{x+2}{93}+1+\frac{x+3}{92}+1=\frac{x+4}{91}+1+\frac{x+5}{90}+1+\frac{x+6}{89}+1\)
\(\Leftrightarrow\frac{x+95}{94}+\frac{x+95}{93}+\frac{x+95}{92}=\frac{x+95}{91}+\frac{x+95}{90}+\frac{x+95}{89}\)
\(\Leftrightarrow\frac{x+95}{94}+\frac{x+95}{93}+\frac{x+95}{92}-\frac{x+95}{91}-\frac{x+95}{90}-\frac{x+95}{89}=0\)
\(\Leftrightarrow\left(x+95\right)\left(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\right)=0\)
\(\Leftrightarrow x+95=0\).Do \(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\ne0\)
\(\Leftrightarrow x=-95\)
TT
10 tháng 1 2017
(x+1)/94 + ( x+2)/93 + ( x+3)/92.......
= ................ + ( x+6)/89
<=> (x+1)/94 + 1 + ( x+2)/93 +1 .........
=.............. cộng 1 nhá
<=> (x+95)/94 + ( x+96) / 93 + ( x+95)/92
= ( x+95)/91 + ( x+95)/90 + ( x+95)/89
<=> ( x+95) ( 1/94 +1/93 +1/92 )
= ( x+95) ( 1/91 +1/90 +1/89)
<=> ( x+95) ( 1/94 +1/93 +1/92 - 1/91 - 1/90 - 1/89 )
<=> x+95 =0
<=>x = -95
Vậy :x = -95
PV
20 tháng 4 2016
a) (x + 5) / 95 + (x +10)/90 + (x + 15)/85 + (x + 20)/80 = -4
<=> (x + 5)/95 + (x + 5)/90 + 5/90 + (x + 5)/85 + 10/85+ (x + 5)/80 + 15/80 = -4
<=> (x + 5)(1/95+1/90+1/85+1/80) =-4 -5/90-10/85-15/85
<=> (x + 5)(1/95+1/90+1/85+1/80)= -1-(1 + 5/90 )-(1 + 10/85) - (1 + 15/80)
<=>(x + 5)(1/95+1/90+1/85+1/80) = -1 - 95/90 - 95/85 - 95/80
<=>(x + 5)(1/95+1/90+1/85+1/80) = -95 (1/95+1/90+1/85+1/80)
<=> x + 5 = -95 => x = -100
S
6 tháng 2 2019
\(\text{Giải}\)
\(\frac{x+2}{93}+\frac{x+3}{92}=\frac{x+4}{91}+\frac{x+5}{90}\)
\(\Leftrightarrow\frac{x+95}{93}+\frac{x+95}{92}-\frac{x+95}{91}+\frac{x+95}{90}=0\)
\(\Leftrightarrow\left(x+95\right)\left(\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}\right)=0\)
Dễ thấy thừa số thứ 2 khác 0
nên: x+95=0=>x=-95
Vậy: x=-95
6 tháng 2 2019
cộng 2 vế với 2 tức là cộng mỗi phân số với 1.Sau đó được mâu sô chung là 95 rồi khử mẫu và làm như bình thường ,.BẠN NHÉ !
1 tháng 3 2017
a) \(\frac{x+1}{94}+\frac{x+2}{93}+\frac{x+3}{92}=\frac{x+4}{91}+\frac{x+5}{90}+\frac{x+6}{89}\)
\(\Leftrightarrow\left(\frac{x+1}{94}+1\right)+\left(\frac{x+2}{93}+1\right)+\left(\frac{x+3}{92}+1\right)=\left(\frac{x+4}{91}+1\right)+\left(\frac{x+5}{90}+1\right)+\left(\frac{x+6}{89}+1\right)\)
\(\Leftrightarrow\frac{x+95}{94}+\frac{x+95}{93}+\frac{x+95}{92}-\frac{x+95}{91}-\frac{x+95}{90}-\frac{x+95}{89}=0\)
\(\Leftrightarrow\) \(\left(x+95\right)\left(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\right)=0\)
Vì \(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\ne0\)
\(\Rightarrow x+95=0\)
\(\Leftrightarrow x=-95\)
Vậy phương trình có một nghiệm x = -95
b) \(\frac{x-1}{59}+\frac{x-2}{58}+\frac{x-3}{57}=\frac{x-4}{56}+\frac{x-5}{55}+\frac{x-6}{54}\)
\(\Leftrightarrow\left(\frac{x-1}{59}-1\right)+\left(\frac{x-2}{58}-1\right)+\left(\frac{x-3}{57}-1\right)=\left(\frac{x-4}{56}-1\right)+\left(\frac{x-5}{55}-1\right)+\left(\frac{x-6}{54}-1\right)\)
\(\Leftrightarrow\frac{x-60}{59}+\frac{x-60}{58}+\frac{x-60}{57}-\frac{x-60}{56}-\frac{x-60}{55}-\frac{x-60}{54}=0\)
\(\Leftrightarrow\left(x-60\right)\left(\frac{1}{59}+\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}-\frac{1}{54}\right)=0\)
Vì \(\frac{1}{59}+\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}-\frac{1}{54}\ne0\)
\(\Rightarrow x-60=0\)
\(\Leftrightarrow x=60\)
Vậy phương trình có một nghiệm x = 60
1 tháng 3 2017
a) \(\frac{x+1}{94}+\frac{x+2}{93}+\frac{x+3}{92}=\frac{x+4}{91}+\frac{x+5}{90}+\frac{x+6}{89}\)
\(\Rightarrow\left(\frac{x+1}{94}+1\right)+\left(\frac{x+2}{93}+1\right)+\left(\frac{x+3}{92}+1\right)=\left(\frac{x+4}{91}+1\right)+\left(\frac{x+5}{90}+1\right)+\left(\frac{x+6}{89}+1\right)\)
\(\Rightarrow\frac{x+95}{94}+\frac{x+95}{93}+\frac{x+95}{92}=\frac{x+95}{91}+\frac{x+95}{90}+\frac{x+95}{89}\)
\(\Rightarrow\frac{x+95}{94}+\frac{x+95}{93}+\frac{x+95}{92}-\frac{x+95}{91}-\frac{x+95}{90}-\frac{x+95}{89}=0\)
\(\Rightarrow\left(x+95\right)\left(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\right)=0\)
Mà \(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\ne0\)
\(\Rightarrow x+95=0\)
\(\Rightarrow x=-95\)
Vậy x = -95
b) \(\frac{x-1}{59}+\frac{x-2}{58}+\frac{x-3}{57}=\frac{x-4}{56}+\frac{x-5}{55}+\frac{x-6}{54}\)
\(\Rightarrow\left(\frac{x-1}{59}-1\right)+\left(\frac{x-2}{58}-1\right)+\left(\frac{x-3}{57}-1\right)=\left(\frac{x-4}{56}-1\right)+\left(\frac{x-5}{55}-1\right)+\left(\frac{x-6}{54}-1\right)\)
\(\Rightarrow\frac{x-60}{59}+\frac{x-60}{58}+\frac{x-60}{57}-\frac{x-60}{56}-\frac{x-5}{55}-\frac{x-6}{54}=0\)
\(\Rightarrow\left(x-60\right)\left(\frac{1}{59}+\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}-\frac{1}{54}\right)=0\)
Mà \(\frac{1}{59}+\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}-\frac{1}{54}\ne0\)
\(\Rightarrow x-60=0\)
\(\Rightarrow x=60\)
Vậy x = 60
12 tháng 3 2017
\(S=\frac{1}{4}+\frac{2}{4}+\frac{3}{4}+....+\frac{90}{4}\)
\(=\frac{1+2+3+...+90}{4}\)
\(=\frac{\frac{90\left(90+1\right)}{2}}{4}\)
\(=\frac{4095}{4}\)
2 tháng 2 2019
a,\(\left(\frac{x}{x+1}\right)^2+\left(\frac{x}{x-1}\right)^2=90\)\(\Leftrightarrow\left(\frac{x}{x+1}\right)^2+2.\frac{x}{x+1}.\frac{x}{x-1}+\left(\frac{x}{x-1}\right)^2-\frac{2x^2}{x^2-1}=90\)
\(\Leftrightarrow\left(\frac{x}{x+1}+\frac{x}{x-1}\right)^2-\frac{2x^2}{x^2-1}=90\)\(\Leftrightarrow\left(\frac{x^2-x+x^2+x}{x^2-1}\right)^2-\frac{2x^2}{x^2-1}=90\)
\(\Leftrightarrow\left(\frac{2x^2}{x^2-1}\right)^2-\frac{2x^2}{x^2-1}-90=0\)\(\Leftrightarrow\left(\frac{2x^2}{x^2-1}-10\right)\left(\frac{2x^2}{x^2-1}+9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{2x^2}{x^2-1}=10\\\frac{2x^2}{x^2-1}=-9\end{cases}\Leftrightarrow......}\)
b,Đặt \(\frac{x-2}{x+1}=a;\frac{x+2}{x-1}=b\Rightarrow ab=\frac{\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left(x-1\right)}=\frac{x^2-4}{x^2-1}\)
Từ đó ta có phương trình:\(20a^2-5b^2+48ab=0\Leftrightarrow20a^2-2ab-5b^2+50ab=0\)
\(\Leftrightarrow2a\left(10a-b\right)+5b\left(10a-b\right)=0\Leftrightarrow\left(2a+5b\right)\left(10a-b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2a=-5b\\10a=b\end{cases}}\)
TH1:\(2a=-5b\Leftrightarrow\frac{2\left(x-2\right)}{x+1}=\frac{-5\left(x+2\right)}{x-1}\)\(\Rightarrow2\left(x-2\right)\left(x-1\right)=-5\left(x+2\right)\left(x+1\right)\)\(\Leftrightarrow2x^2-6x+4=-5x^2-15x-10\)\(\Leftrightarrow7x^2+9x+14=0\)
\(\Leftrightarrow7\left(x^2+\frac{9}{7}x+2\right)=0\Leftrightarrow7\left(x^2+2.\frac{9}{14}+\frac{81}{196}\right)+\frac{311}{28}=0\)
\(\Leftrightarrow7\left(x+\frac{9}{14}\right)^2+\frac{311}{28}=0\),vô lí
TH2:Tự làm nhé ,tương tự
KT
12 tháng 8 2018
\(\frac{x+5}{95}+\frac{x+10}{90}+\frac{x+15}{85}+\frac{x+20}{80}=-4\)
<=> \(\frac{x+5}{95}+1+\frac{x+10}{90}+1+\frac{x+15}{85}+1+\frac{x+20}{80}+1=0\)
<=> \(\frac{x+100}{95}+\frac{x+100}{90}+\frac{x+100}{85}+\frac{x+100}{80}=0\)
<=> \(\left(x+100\right)\left(\frac{1}{95}+\frac{1}{90}+\frac{1}{85}+\frac{1}{80}\right)=0\)
<=> \(x+100=0\) (do 1/95 + 1/90 + 1/85 + 1/80 khác 0)
<=> \(x=-100\)
Vậy...
ĐKXĐ: \(\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}\)
pt đã cho \(\Leftrightarrow\frac{90\left(x+5\right)-90x}{x\left(x+5\right)}=\frac{1}{4}\)\(\Leftrightarrow\frac{90x+450-90x}{x^2+5x}=\frac{1}{4}\)\(\Leftrightarrow\frac{450}{x^2+5x}=\frac{1}{4}\)\(\Rightarrow x^2+5x=1800\)\(\Leftrightarrow x^2+5x-1800=0\)\(\Leftrightarrow x^2+45x-40x-1800=0\)\(\Leftrightarrow x\left(x+45\right)-40\left(x+45\right)=0\)\(\Leftrightarrow\left(x+45\right)\left(x-40\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=-45\left(nhận\right)\\x=40\left(nhận\right)\end{cases}}\)
Vậy tập nghiệm của pt đã cho là \(S=\left\{-45;40\right\}\)