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DA
0
PH
4
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 !
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
NH
1
KN
4 tháng 3 2020
\(\frac{x+2}{2002}+\frac{x+5}{1999}+\frac{x+201}{1803}=-3\)
\(\Rightarrow\frac{x+2}{2002}+1+\frac{x+5}{1999}+1+\frac{x+201}{1803}+1=0\)
\(\Rightarrow\frac{x+2004}{2002}+\frac{x+2004}{1999}+\frac{x+2004}{1803}=0\)
\(\Rightarrow\left(x+2004\right)\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)=0\)
Dễ thấy \(\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)>0\)nên x + 2004 = 0
Vậy x = -2004
26 tháng 1 2018
\(\dfrac{x+1}{94}+\dfrac{x+2}{93}+\dfrac{x+3}{92}=\dfrac{x+4}{91}+\dfrac{x+5}{90}+\dfrac{x+6}{89}\)
<=>\(\dfrac{x+1}{94}+1+\dfrac{x+2}{93}+1+\dfrac{x+3}{92}+1=\dfrac{x+4}{91}+1+\dfrac{x+5}{90}+1+\dfrac{x+6}{89}+1\)
<=>\(\dfrac{x+95}{94}+\dfrac{x+95}{93}+\dfrac{x+95}{92}=\dfrac{x+95}{91}+\dfrac{x+95}{90}+\dfrac{x+95}{89}\)
<=>\(\dfrac{x+95}{94}+\dfrac{x+95}{93}+\dfrac{x+95}{92}-\dfrac{x+95}{91}-\dfrac{x+95}{90}-\dfrac{x+95}{89}=0\)
<=> (x+95)(\(\dfrac{1}{94}+\dfrac{1}{93}+\dfrac{1}{92}-\dfrac{1}{91}-\dfrac{1}{90}-\dfrac{1}{89}\))
vì \(\dfrac{1}{94}+\dfrac{1}{93}+\dfrac{1}{92}-\dfrac{1}{91}-\dfrac{1}{90}-\dfrac{1}{89}\) khác 0
=>x+95=0=>x=-95
chức bạn học tốt ^ ^
NH
4
4 tháng 3 2020
\(\frac{x+2}{2002}+\frac{x+5}{1999}+\frac{x+201}{1803}=-3\)
\(\Leftrightarrow\frac{x+2}{2002}+1+\frac{x+5}{1999}+1+\frac{x+201}{1803}+1=-3+1+1+1\)
\(\Leftrightarrow\frac{x+2004}{2002}+\frac{x+2004}{1999}+\frac{x+2004}{1803}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)=0\)
\(\Leftrightarrow x+2004=0\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\ne0\right)\)
<=> x=-2004
4 tháng 3 2020
a,\(\frac{x+2}{2002}+\frac{x+5}{1999}+\frac{x+201}{1803}=-3\)
\(< =>\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+5}{1999}+1\right)+\left(\frac{x+201}{1803}+1\right)=0\)
\(< =>\frac{x+2004}{2002}+\frac{x+2004}{1999}+\frac{x+2004}{1803}=0\)
\(< =>\left(x+2004\right).\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)=0\)
Do \(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\ne0\)
\(=>x+2004=0\)
\(=>x=-2004\)
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
16 tháng 8 2020
a) Ta có: \(85^2-15^2\)
\(=\left(85-15\right)\left(85+15\right)\)
\(=70\cdot100=7000\)
b) Ta có: \(93^3+21\cdot93^2+3\cdot49\cdot93+343\)
\(=93^3+3\cdot93^2\cdot7+3\cdot93+7^2+7^3\)
\(=\left(93+7\right)^3\)
\(=100^3=1000000\)
c) Ta có: \(73^2-13^2-10^2+20\cdot13\)
\(=73^2-\left(13^2+10^2-20\cdot13\right)\)
\(=73^2-\left(13^2-2\cdot13\cdot10+10^2\right)\)
\(=73^2-\left(13-10\right)^2\)
\(=73^2-3^2=\left(73-3\right)\left(73+3\right)\)
\(=70\cdot76=5320\)
TG
16 tháng 8 2020
a) \(85^2-15^2=\left(85-15\right)\left(85+15\right)=70.100=7000\)
b) \(93^3+21.93^2+3.49.93+343\)
\(=93^3+3.7.93^2+3.7^2.93+7^3\)
\(=\left(93+7\right)^3\)
\(=100^3=1000000\)
c) \(73^2-13^2-10^2+20.13\)
\(=73^2-\left(13^2+10^2-20.13\right)\)
\(=73^2-\left(13-10\right)^2\)
\(=73^2-3^2\)
\(=\left(73+3\right)\left(73-3\right)\)
\(=76.70=5320\)
d) Viết = Latex hộ mình
\(\frac{x+91}{81}+\frac{x+92}{82}+\frac{x+93}{83}=3\)
\(\frac{x+91}{81}-1+\frac{x+92}{82}-1+\frac{x+93}{83}-1=0\)
\(\frac{x+10}{81}+\frac{x+10}{82}+\frac{x+10}{83}=0\)
\(\left(x+10\right)\left(\frac{1}{81}+\frac{1}{82}+\frac{1}{83}\right)=0\)
=> x + 10 = 0 (vì \(\left(\frac{1}{81}+\frac{1}{82}+\frac{1}{83}\right)\ne0\))=> x = -10