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a, Ta có : \(\frac{392-x}{32}+\frac{390-x}{34}+\frac{388-x}{36}+\frac{386-x}{38}+\frac{384-x}{40}=-5\)
=> \(\frac{392-x}{32}+1+\frac{390-x}{34}+1+\frac{388-x}{36}+1+\frac{386-x}{38}+1+\frac{384-x}{40}+1=-5+5=0\)
=> \(\frac{424-x}{32}+\frac{424-x}{34}+\frac{424-x}{36}+\frac{424-x}{38}+\frac{424-x}{40}=0\)
=> \(\left(424-x\right)\left(\frac{1}{32}+\frac{1}{34}+\frac{1}{36}+\frac{1}{38}+\frac{1}{40}\right)=0\)
=> \(424-x=0\)
=> \(x=424\)
Vậy phương trình có nghiệm là x = 424 .
b, Ta có : \(\frac{x+1}{2014}+\frac{x+3}{2012}=\frac{x+5}{2010}+\frac{x+6}{2009}\)
=> \(\frac{x+1}{2014}+1+\frac{x+3}{2012}+1=\frac{x+5}{2010}+1+\frac{x+6}{2009}+1\)
=> \(\frac{x+2015}{2014}+\frac{x+2015}{2012}=\frac{x+2015}{2010}+\frac{x+2015}{2009}\)
=> \(\frac{x+2015}{2014}+\frac{x+2015}{2012}-\frac{x+2015}{2010}-\frac{x+2015}{2009}=0\)
=> \(\left(x+2015\right)\left(\frac{1}{2014}+\frac{1}{2012}-\frac{1}{2010}-\frac{1}{2009}\right)=0\)
=> \(x+2015=0\)
=> \(x=-2015\)
Vậy phương trình có nghiệm là x = -2015 .
a) \(\frac{392-x}{32}+\frac{390-x}{34}+\frac{388-x}{36}+\frac{386-x}{38}+\frac{384-x}{40}=-5\)
<=> \(\frac{392-x}{32}+1+\frac{390-x}{34}+1+\frac{388-x}{36}+1+\frac{386-x}{38}+1+\frac{384-x}{40}=0\)
<=> \(\frac{424-x}{32}+\frac{424-x}{34}+\frac{424-x}{36}+\frac{424-x}{40}=0\)
<=> \(\left(424-x\right)\left(\frac{1}{32}+\frac{1}{34}+\frac{1}{36}+\frac{1}{40}\right)=0\)
<=> 424 - x = 0
<=> x = 424
Vậy S = {424}
b) \(\frac{x+1}{2014}+\frac{x+3}{2012}=\frac{x+5}{2010}+\frac{x+6}{2009}\)
<=> \(\left(\frac{x+1}{2014}+1\right)+\left(\frac{x+3}{2012}+1\right)=\left(\frac{x+5}{2010}+1\right)+\left(\frac{x+6}{2009}+1\right)\)
<=> \(\frac{x+2015}{2014}+\frac{x+2015}{2012}=\frac{x+2015}{2010}+\frac{x+2015}{2009}\)
<=> \(\left(x+2015\right)\left(\frac{1}{2014}+\frac{1}{2012}-\frac{1}{2010}-\frac{1}{2009}\right)=0\)
<=> x + 2015 = 0
<=> x= -2015
Vậy S = {-2015}
ta có : \(\dfrac{392-x}{32}+\dfrac{390-x}{34}+\dfrac{388-x}{36}+\dfrac{386-x}{38}\)+\(\dfrac{384-x}{40}=-5\)
\(\Leftrightarrow\)\(\dfrac{392-x}{32}+1+\dfrac{390-x}{34}+1+\dfrac{388-x}{36}+1\)+\(\dfrac{384-x}{40}+1=0\)
\(\Leftrightarrow\)\(\dfrac{424-x}{32}+\dfrac{424-x}{34}+\dfrac{424-x}{36}+\dfrac{424-x}{38}+\dfrac{424-x}{40}=0\)\(\Leftrightarrow\left(424-x\right)\left(\dfrac{1}{32}+\dfrac{1}{34}+\dfrac{1}{36}+\dfrac{1}{38}+\dfrac{1}{40}\right)=0\)
\(\Leftrightarrow x=424\)(vì \(\dfrac{1}{32}+\dfrac{1}{34}+\dfrac{1}{36}+\dfrac{1}{38}+\dfrac{1}{40}\ne0\))
Vậy tập nghiệm của phương trình là s=\(\left\{424\right\}\)
Câu x ) là bằng - 5 nhé mấy bạn. Làm giúp mình tất cả nhé ! Mình cảm ơn nhiều lắm !
\(\left(8x^3-60x^2+150x-125\right)-\left(27x^3-108x^2+144x-64\right)+\left(x^3+3x^2+3x+1\right)=0\)
\(-18x^3+51x^2+9x-60=0\)
\(\left(2x-5\right)\left(x+1\right)\left(3x-4\right)=0\)
\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-1\\x=\frac{4}{3}\end{array}\right.\)
a, \(\frac{x+16}{49}+\frac{x+18}{47}=\frac{x+20}{45}-1\)
\(\Leftrightarrow1+\frac{x+16}{49}+1+\frac{x+18}{47}=\frac{x+20}{45}-1+2\)
\(\Leftrightarrow\frac{x+16+49}{49}+\frac{x+18+47}{47}=\frac{x+20+45}{45}\)
\(\Leftrightarrow\frac{x+65}{49}+\frac{x+65}{47}-\frac{x+65}{45}=0\)
\(\Leftrightarrow\left(x+65\right)\left(\frac{1}{49}+\frac{1}{47}-\frac{1}{45}\right)=0\)
Ta có: \(\frac{1}{49}+\frac{1}{47}-\frac{1}{45}\)>0
\(\Rightarrow x+65=0\)
\(\Leftrightarrow x=-65\)
Vậy x = -65
b, \(\frac{x-69}{30}+\frac{x-67}{32}+\frac{x-65}{34}=\frac{x-63}{36}+\frac{x-61}{38}+\frac{x-59}{40}\)
\(\Leftrightarrow\frac{x-69}{30}-1+\frac{x-67}{32}-1+\frac{x-65}{34}-1+\frac{x-63}{36}-1+\frac{x-61}{38}-1+\frac{x-59}{40}-1\)
\(\Leftrightarrow\frac{x-99}{30}+\frac{x-99}{32}+\frac{x-99}{34}-\frac{x-99}{36}-\frac{x-99}{38}-\frac{x-99}{40}=0\)
\(\Leftrightarrow\left(x-99\right)\left(\frac{1}{30}+\frac{1}{32}+\frac{1}{34}-\frac{1}{36}-\frac{1}{38}-\frac{1}{40}\right)=0\)
Vì \(\frac{1}{30}+\frac{1}{32}+\frac{1}{34}-\frac{1}{36}-\frac{1}{38}-\frac{1}{40}\)>0
\(\Rightarrow x-99=0\)
\(\Leftrightarrow x=99\)
Vậy x =99
a, \(\frac{5}{x+7}+\frac{8}{2x+14}=\frac{3}{2}\) Đkxđ : \(x\ne-7\)
⇔ \(\frac{5}{x+7}+\frac{8}{2\left(x+7\right)}=\frac{3}{2}\)
⇔ \(\frac{10}{2\left(x+7\right)}+\frac{8}{2\left(x+7\right)}=\frac{3\left(x+7\right)}{2\left(x+7\right)}\)
⇒ \(10+8=3\left(x+7\right)\)
⇔ \(10+8=3x+21\)
⇔ \(-3x=21-10-8\)
⇔ \(-3x=3\)
⇔ \(x=-1\) ( tm )
Ptr có tập nhiệm : S \(=\left\{-1\right\}\)
b, \(\frac{x+3}{x-3}-\frac{1}{x}=\frac{3}{x\left(x-3\right)}\) Đkxđ : \(x\ne3;x\ne0\)
⇔ \(\frac{x\left(x+3\right)}{x\left(x-3\right)}-\frac{1\left(x-3\right)}{x\left(x-3\right)}=\frac{3}{x\left(x-3\right)}\)
⇒ \(x\left(x-3\right)-1\left(x-3\right)=3\)
⇔ \(x^2-3x-x+3=3\)
⇔ \(x^2-4x=0\)
⇔ \(x\left(x-4\right)=0\)
⇔ \(\left\{{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=0\left(ktm\right)\\x=4\left(tm\right)\end{matrix}\right.\)
Ptr có tập nhiệm : S \(=\left\{4\right\}\)
a) \(\frac{36\left(x-2\right)}{32-16x}=\frac{36\left(x-2\right)}{16\left(2-x\right)}=-\frac{36\left(2-x\right)}{16\left(2-x\right)}=-\frac{36}{16}=-\frac{9}{4}\)
b) \(\frac{3x^2-12x+12}{x^4-8x}=\frac{3\left(x^2-4x+4\right)}{x\left(x^3-8\right)}=\frac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}=\frac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}=\frac{3x-6}{x^3+2x^2+4x}\)
c) \(\frac{7x^2+14x+7}{3x^2+3x}=\frac{7\left(x^2+2x+1\right)}{3x\left(x+1\right)}=\frac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\frac{7\left(x+1\right)}{3x}=\frac{7x+7}{3x}\)
d) \(\frac{x^4-5x^2+4}{x^4-10x^2+9}=\frac{x^4-x^2-4x^2+4}{x^4-x^2-9x^2+9}=\frac{x^2\left(x^2-1\right)-4\left(x^2-1\right)}{x^2\left(x^2-1\right)-9\left(x^2-1\right)}=\frac{\left(x^2-4\right)\left(x^2-1\right)}{\left(x^2-9\right)\left(x^2-1\right)}=\frac{\left(x-2\right)\left(x+2\right)}{\left(x-3\right)\left(x+3\right)}\)
e) \(\cdot\frac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}=\frac{x^3\left(x+1\right)+\left(x+1\right)}{x^4-x^3+x^2+x^2-x+1}=\frac{\left(x^3+1\right)\left(x+1\right)}{x^2\left(x^2-x+1\right)+\left(x^2-x+1\right)}=\frac{\left(x+1\right)^2\left(x^2-x+1\right)}{\left(x^2+1\right)\left(x^2-x+1\right)}=\frac{\left(x+1\right)^2}{x^2+1}=\frac{x^2+2x+1}{x^2+1}\)
ĐKXĐ: x khác 2 và -2
Ta có : \(\frac{x-2}{x+2}\)- \(\frac{x+2}{x-2}\)= \(\frac{-24}{5}\)
<=> \(\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\)- \(\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}\)= \(\frac{-24}{5}\)
<=> \(\frac{\left(x-2\right)^2-\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}\)= \(\frac{-24}{5}\)
<=> \(\frac{\left(x-2+x+2\right)\left(x-2-x-2\right)}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{-24}{5}\)
<=> \(\frac{2x.\left(-4\right)}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{-24}{5}\)
<=> -40x= -24(x^2-4)
<=> -40x= -24x^2+96
<=> 24x^2-40x-96=0
<=> 24x^2-72x+32x-96=0
<=> 24x(x-3)+32(x-3)=0
<=> (x-3)(24x+32)=0
=> \(\orbr{\begin{cases}x-3=0\\24x+32=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=3\\x=\frac{-4}{3}\end{cases}}\)
Vậy S=\(\hept{\begin{cases}\\\end{cases}}3;\frac{-4}{3}\)
\(a.\frac{x+5}{2021}+\frac{x+6}{2020}+\frac{x+7}{2019}=-3\\ \Leftrightarrow\frac{x+5}{2021}+1+\frac{x+6}{2020}+1+\frac{x+7}{2019}+1=0\\ \Leftrightarrow\frac{x+2026}{2021}+\frac{x+2026}{2020}+\frac{x+2026}{2019}=0\\ \Leftrightarrow\left(x+2026\right)\left(\frac{1}{2021}+\frac{1}{2020}+\frac{1}{2019}\right)=0\\\Leftrightarrow x+2026=0\left(Vi\frac{1}{2021}+\frac{1}{2020}+\frac{1}{2019}\ne0\right)\\ \Leftrightarrow x=-2026\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{-2026\right\}\)
\(b.\frac{2-x}{100}-1=\frac{1-x}{101}-\frac{x}{102}\\ \Leftrightarrow\frac{2-x}{100}+1=\frac{1-x}{101}+1+1-\frac{x}{102}\\\Leftrightarrow \frac{102-x}{100}-\frac{102-x}{101}-\frac{102-x}{102}=0\\ \Leftrightarrow\left(102-x\right)\left(\frac{1}{100}-\frac{1}{101}-\frac{1}{102}\right)=0\\ \Leftrightarrow102-x=0\left(Vi\frac{1}{100}-\frac{1}{101}-\frac{1}{102}\ne0\right)\\ \Leftrightarrow x=102\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{102\right\}\)
c/ PT tương đương
\(\frac{x+1}{93}-1+\frac{x-2}{45}-2+\frac{x+4}{32}-3=0\)
\(\Leftrightarrow\frac{x-92}{93}+\frac{x-92}{45}+\frac{x-92}{32}=0\)
\(\Leftrightarrow\left(x-92\right)\left(\frac{1}{93}+\frac{1}{45}+\frac{1}{32}\right)=0\Rightarrow x=92\)
Ta có :
\(\frac{392-x}{32}+\frac{390-x}{34}+\frac{388-x}{36}+\frac{386-x}{38}+\frac{384-x}{40}=-5\)
\(\Leftrightarrow\left(\frac{392-x}{32}+1\right)+\left(\frac{390-x}{34}+1\right)+\left(\frac{388-x}{36}+1\right)+\left(\frac{386-x}{38}+1\right)+\left(\frac{384-x}{40}\right)=0\)
\(\Leftrightarrow\frac{424-x}{32}+\frac{424-x}{34}+\frac{424-x}{36}+\frac{424-x}{38}+\frac{424-x}{40}=0\)
\(\Leftrightarrow\left(424-x\right)\left(\frac{1}{32}+\frac{1}{34}+\frac{1}{36}+\frac{1}{38}+\frac{1}{40}\right)=0\)
Mà : \(\frac{1}{32}+\frac{1}{34}+\frac{1}{36}+\frac{1}{38}+\frac{1}{40}\ne0\)
\(\Leftrightarrow424-x=0\)
\(\Leftrightarrow x=424\)
Vậy x = 424