\(\frac{x+17}{83}+\frac{x+27}{73}+\frac{x+37}{63}+3=0\)0
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N
30 tháng 12 2018
\(\Leftrightarrow\left(\frac{x+14}{86}+1\right)+\left(\frac{x+15}{85}+1\right)+\left(\frac{x+16}{84}+1\right)+\left(\frac{x+17}{83}+1\right)+\left(\frac{166}{4}-4\right)=0\)
\(\Leftrightarrow\frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)
\(\Leftrightarrow\left(x+100\right).\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\left(x+100\right)=0\Rightarrow x=-100\left(\text{vì }\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)\ne0\)
30 tháng 1 2017
a) (x-1)x(x+1)(x+2) = 24
<=> [(x-1)(x+2)][x(x+1) = 24
<=> (x^2+x-2)(x^2+x) = 24 (1)
Đặt t=x^2+x-1 = (x+1/2)^2 - 5/4 (*)
(1) trở thành (t-1)(t+1) = 24
<=> t^2 - 1 - 24 = 0
<=> t^2 - 25 = 0
<=> t^2 = 25
<=> t=5 hoặc t=-5
Mà t >= -5/4 ( từ *) => t = (x+1/2)^2-5/4 = 5
<=> (x+1/2)^2 = 25/4
Đến đây dễ r`
30 tháng 1 2017
c) x^4 + 3x^3 + 4x^2 + 3x + 1 = 0
<=> x^4 + x^3 + 2x^3 + 2x^2 + 2x^2 + 2x + x + 1 = 0
<=> (x+1)(x^3 + 2x^2 + 2x + 1) = 0
<=> (x +1)(x^3 + x^2 + x^2 + x + x + 1) = 0
<=> (x+1)^2.(x^2+x+1) = 0
Mà x^2+x+1 = (x+1/2)^2 + 3/4 > 0
Nên x+1=0 <=> x=-1
Vậy ...
6 tháng 10 2016
b) \(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=0\)
\(\Rightarrow\)
| x+1=0 | x+2=0 | x+4=0 | x+5=0 |
17 tháng 10 2016
số đó là
0 nhé bn
mình chỉ
chăc thế thôi
hihi
T
9 tháng 2 2019
c)Ta có: \(x^4+3x^3+4x^2+3x+1=0\)
\(\Leftrightarrow x\left(x^3+2x^2+2x+1\right)+1\left(x^3+2x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3+2x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\left(x^2+x+1\right)=0\)
Ta có: \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\) nên vô nghiệm
Suy ra x + 1 =0 hay x = -1
KT
22 tháng 2 2018
a) \(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+17}{83}+\frac{x+116}{4}=0\)
\(\Leftrightarrow\)\(\frac{x+14}{86}+1+\frac{x+15}{85}+1+\frac{x+16}{84}+1+\frac{x+17}{83}+1+\frac{x+116}{4}-4=0\)
\(\Leftrightarrow\)\(\frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)
\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\)\(x+100=0\)
\(\Leftrightarrow\)\(x=-100\)
Vậy...
25 tháng 3 2020
a) Ta có: \(\frac{x-91}{37}+\frac{x-86}{42}+\frac{x-78}{50}+\frac{x-49}{79}=4\)
\(\Leftrightarrow\frac{x-91}{37}-1+\frac{x-86}{42}-1+\frac{x-78}{50}-1+\frac{x-49}{79}-1=0\)
\(\Leftrightarrow\frac{x-91-37}{37}+\frac{x-86-42}{42}+\frac{x-78-50}{50}+\frac{x-49-79}{79}=0\)
\(\Leftrightarrow\frac{x-128}{37}+\frac{x-128}{42}+\frac{x-128}{50}+\frac{x-128}{79}=0\)
\(\Leftrightarrow\left(x-128\right)\left(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}\right)=0\)
Vì \(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}>0\)
nên x-128=0
hay x=128
Vậy: x=128
b) Ta có: \(\frac{x-29}{1970}+\frac{x-27}{1972}+\frac{x-25}{1974}+\frac{x-23}{1976}+\frac{x-1970}{29}+\frac{x-1972}{27}+\frac{x-1974}{25}+\frac{x-1976}{23}-8=0\)
\(\Leftrightarrow\frac{x-29}{1970}-1+\frac{x-27}{1972}-1+\frac{x-25}{1974}-1+\frac{x-23}{1976}-1+\frac{x-1970}{29}-1+\frac{x-1972}{27}-1+\frac{x-1974}{25}-1+\frac{x-1976}{23}-1=0\)
\(\Leftrightarrow\frac{x-29-1970}{1970}+\frac{x-27-1972}{1972}+\frac{x-25-1974}{1974}+\frac{x-23-1976}{1976}+\frac{x-1970-29}{29}+\frac{x-1972-27}{27}+\frac{x-1974-25}{25}+\frac{x-1976-23}{23}=0\)
\(\Leftrightarrow\left(x-1999\right)\left(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{29}+\frac{1}{27}+\frac{1}{25}+\frac{1}{23}\right)=0\)
Vì \(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{29}+\frac{1}{27}+\frac{1}{25}+\frac{1}{23}>0\)
nên x-1999=0
hay x=1999
Vậy: x=1999
25 tháng 3 2020
a) Ta có \(\frac{x-91}{37}+\frac{x-86}{42}+\frac{x-78}{50}+\frac{x-49}{79}\)=4
<=>\(\frac{x-91}{37}+\frac{x-86}{42}+\frac{x-78}{50}+\frac{x-49}{79}-4=0\)
<=>\(\frac{x-91}{37}-1+\frac{x-86}{42}-1+\frac{x-78}{50}-1+\frac{x-49}{79}-1=0\)
<=>\(\frac{x-128}{37}+\frac{x-128}{42}+\frac{x-128}{50}+\frac{x-128}{79}=0\)
<=>(x-128)\(\left(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}\right)=0\)
Vì \(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}>0\)=>x-128=0<=>x=128
b)Tương tự
<=>x-128=0
<=>x=128
Chú ý \(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}\)>0
b)tương tự
S
31 tháng 12 2018
\(\frac{x-241}{17}+\frac{x-220}{19}+\frac{x-195}{21}+\frac{x-166}{23}=0\)
\(\Leftrightarrow\frac{x-258}{17}+\frac{x-258}{19}+\frac{x-258}{21}+\frac{x-258}{23}=-10\)
\(\Leftrightarrow\left(x-258\right)\left(\frac{1}{17}+\frac{1}{19}+\frac{1}{21}+\frac{1}{23}\right)=-10\)
\(.....................\)
đến đây thì dễ rồi :)
1 tháng 4 2020
Bài 15:
Ta có: \(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)
\(\Leftrightarrow\frac{x+2}{2008}+1+\frac{x+3}{2007}+1+\frac{x+4}{2006}+1+\frac{x+2028}{6}-3=0\)
\(\Leftrightarrow\frac{x+2+2008}{2008}+\frac{x+3+2007}{2007}+\frac{x+4+2006}{2006}+\frac{x+2028-18}{6}=0\)
\(\Leftrightarrow\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\right)=0\)
Vì \(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}>0\)
nên x+2010=0
hay x=-2010
Vậy: x=-2010
Bài 17:
Ta có: \(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)
\(\Leftrightarrow\frac{x+1}{65}+1+\frac{x+3}{63}+1=\frac{x+5}{61}+1+\frac{x+7}{59}+1\)
\(\Leftrightarrow\frac{x+1+65}{65}+\frac{x+3+63}{63}=\frac{x+5+61}{61}+\frac{x+7+59}{59}\)
\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}=\frac{x+66}{61}+\frac{x+66}{59}\)
\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}-\frac{x+66}{61}-\frac{x+66}{59}=0\)
\(\Leftrightarrow\left(x+66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)=0\)
Vì \(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\ne0\)
nên x+66=0
hay x=-66
Vậy: x=-66
x=- 100