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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
\(\Leftrightarrow\left(\frac{x+14}{86}+1\right)+\left(\frac{x+15}{85}+1\right)+\left(\frac{x+16}{84}+1\right)+\left(\frac{x+17}{83}+1\right)+\left(\frac{166}{4}-4\right)=0\)
\(\Leftrightarrow\frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)
\(\Leftrightarrow\left(x+100\right).\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\left(x+100\right)=0\Rightarrow x=-100\left(\text{vì }\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)\ne0\)
\(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+14}{83}+\frac{x+116}{4}=0\)
\(\frac{x+14}{86}+1+\frac{x+15}{85}+1+\frac{x+16}{84}+1+\frac{x+14}{83}+1+\frac{x+116}{4}-4=0\)
\(\frac{x+14+86}{86}+\frac{x+15+85}{85}+\frac{x+16+84}{84}+\frac{x+14+83}{83}+\frac{x+116-16}{4}=0\)
\(\frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)
\(\left(x+100\right)\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)
Vì \(\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)\ne0\)
\(\Rightarrow x+100=0\)
\(\Rightarrow x=-100\)
Vậy........
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 |
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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...
<=>\(\left(x+2\right)^2-\left(x-2\right)^3=-\left(x-6\right)\left(x^2-x+2\right)\)
=>\(12x\left(x-1\right)-8=4\left(3x^2-3x-2\right)\)
=>\(-\left(x-6\right)\left(x^2-x+2\right)=4\left(3x^2-3x-2\right)\)
=>x=-5;-2 hoặc 2
<=>\(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+17}{83}+\frac{x+116}{4}=\frac{1894289\left(x+100\right)}{6370665}\)
=>\(\frac{1894289\left(x+100\right)}{6370665}=0\)(rút gọn)
=>\(\frac{1894289x}{6370665}+\frac{37885780}{1274133}=0\)
=>\(\frac{1894289\left(x+100\right)}{6370665}=0\)(giải PT này )
=>x=-100
\(\dfrac{x+14}{86}+\dfrac{x+15}{85}+\dfrac{x+16}{84}+\dfrac{x+17}{83}+\dfrac{x+116}{4}=0\)
\(\Leftrightarrow\left(\dfrac{x+14}{86}+1\right)+\left(\dfrac{x+15}{85}+1\right)+\left(\dfrac{x+16}{84}+1\right)+\left(\dfrac{x+17}{83}+1\right)+\left(\dfrac{x+116}{4}-1\right)=0\)
=>x+100=0
hay x=-100
\(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+17}{83}+\frac{x+116}{4}=0\)
\(\Leftrightarrow\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+17}{83}+\frac{x+100}{4}+4=0\)
\(\Leftrightarrow\left(\frac{x+14}{86}+1\right)+\left(\frac{x+15}{85}+1\right)+\left(\frac{x+14}{86}+1\right)+\left(\frac{x+13}{87}+1\right)+\frac{x+100}{4}=0\)
\(\Leftrightarrow\frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow x+100=0\left(vì\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\ne0\right)\)
\(\Leftrightarrow x=-100\)
vậy.............................
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`
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 ...