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\(ĐKXĐ:\) \(\forall x\in Z\)
\(\frac{x^2}{x^2+2x+2}+\frac{x^2}{x^2-2x+2}-\frac{4\left(x^2-5\right)}{x^4+4}=\frac{322}{65}\)
\(\Leftrightarrow\)\(\frac{x^2\left(x^2-2x+2\right)}{\left(x^2+2x+2\right)\left(x^2-2x+2\right)}+\frac{x^2\left(x^2+2x+2\right)}{\left(x^2-2x+2\right)\left(x^2+2x+2\right)}-\frac{4\left(x^2-5\right)}{\left(x^2-2x+2\right)\left(x^2+2x+2\right)}=\frac{322}{65}\)
\(\Leftrightarrow\)\(\frac{x^4-2x^3+2x^2+x^4+2x^3+2x^2-4x^2+20}{\left(x^2-2x+2\right)\left(x^2+2x+2\right)}=\frac{322}{65}\)
\(\Leftrightarrow\)\(\frac{2x^4+10}{x^4+4}=\frac{322}{65}\)
\(\Rightarrow\)\(65\left(2x^4+10\right)=322\left(x^4+4\right)\)
\(\Leftrightarrow\)\(130x^4+650=322x^4+1288\)
\(\Leftrightarrow\)\(192x^4=-638\) (vô lý)
Vậy pt vô nghiệm
P/S:mk lm bừa thôi, đúng thì you tham khảo, sai thì báo mk biết nha
a/ \(\frac{x^2+2x+1}{x^2+2x+2}+\frac{x^2+2x+2}{x^2+2x+3}=\frac{7}{6}\)
<=> \(\frac{\left(x+1\right)^2}{\left(x+1\right)^2+1}+\frac{\left(x+1\right)^2+1}{\left(x+1\right)^2+2}=\frac{7}{6}\left(1\right)\)
đặt \(\left(x+1\right)^2=a\left(a>0\right)\)
=> \(\left(1\right)\)<=> \(\frac{a}{a+1}+\frac{a+1}{a+2}=\frac{7}{6}\)
<=> \(\frac{a\left(a+2\right)+\left(a+1\right)^2}{\left(a+1\right)\left(a+2\right)}=\frac{7}{6}\)
<=> \(\frac{2a^2+4a+1}{a^2+3a+2}=\frac{7}{6}\)
<=> \(6\left(2a^2+4a+1\right)=7\left(a^2+3a+2\right)\)
<=> \(5a^2+3a-8=0\)
<=> \(5a^2-5a+8a-8=0\)
<=> \(\left(5a+8\right)\left(a-1\right)=0\)
<=> \(a=\frac{-8}{5}\left(h\right)a=1\)
mà \(a>0\)
=> \(a=1\)
=> \(\left(x+1\right)^2=1\)
=> \(x+1=1\left(h\right)x+1=-1\)
=> \(x=0\left(h\right)x=-2\)
vậy ......
chúc bn học tốt
Xét x = 0 và x = -2 , thay vào ta được \(VT=VP\)
Xét x > 0 :
\(VT=\frac{x^2+2x+1}{x^2+2x+2}+\frac{x^2+2x+2}{x^2+2x+3}=1-\frac{1}{x^2+2x+2}+1-\frac{1}{x^2+2x+3}\)
\(=2-\left(\frac{1}{x^2+2x+2}+\frac{1}{x^2+2x+3}\right)>2-\left(\frac{1}{2}+\frac{1}{3}\right)>\frac{7}{6}=VP\) ( loại )
Xét x < -2 :
\(VT=2-\left(\frac{1}{x\left(x+2\right)+2}+\frac{1}{x\left(x+2\right)+3}\right)>2-\left(\frac{1}{2}+\frac{1}{3}\right)=\frac{7}{6}=VP\) ( loại )
Xét -2 < x < 0 :
\(VT=2-\left(\frac{1}{x^2+2x+2}+\frac{1}{x^2+2x+3}\right)>2-\left(\frac{1}{-2}+1\right)=\frac{3}{2}>\frac{7}{6}=VP\) ( loại )
Vậy ...
a) <=> \(6x^2-5x+3-2x+3x\left(3-2x\right)=0\)
<=> \(6x^2-5x+3-2x+9x-6x^2=0\)
<=> \(2x+3=0\)
<=> \(x=\frac{-3}{2}\)
b) <=> \(10\left(x-4\right)-2\left(3+2x\right)=20x+4\left(1-x\right)\)
<=> \(10x-40-6-4x=20x+4-4x\)
<=> \(6x-46-16x-4=0\)
<=> \(-10x-50=0\)
<=> \(-10\left(x+5\right)=0\)
<=> \(x+5=0\)
<=> \(x=-5\)
c) <=> \(8x+3\left(3x-5\right)=18\left(2x-1\right)-14\)
<=> \(8x+9x-15=36x-18-14\)
<=> \(8x+9x-36x=+15-18-14\)
<=> \(-19x=-14\)
<=> \(x=\frac{14}{19}\)
d) <=>\(2\left(6x+5\right)-10x-3=8x+2\left(2x+1\right)\)
<=> \(12x+10-10x-3=8x+4x+2\)
<=> \(2x-7=12x+2\)
<=> \(2x-12x=7+2\)
<=> \(-10x=9\)
<=> \(x=\frac{-9}{10}\)
e) <=> \(x^2-16-6x+4=\left(x-4\right)^2\)
<=> \(x^2-6x-12-\left(x-4^2\right)=0\)
<=> \(x^2-6x-12-\left(x^2-8x+16\right)=0\)
<=> \(x^2-6x-12-x^2+8x-16=0\)
<=> \(2x-28=0\)
<=> \(2\left(x-14\right)=0\)
<=> x-14=0
<=> x=14
a, \(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right).\left(2x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}=0\)
\(\Leftrightarrow\frac{x^2-4x+4}{3}+\frac{9-4x^2}{8}+\frac{x^2-8x+16}{6}=0\)
\(\Leftrightarrow\frac{8\left(x^2-4x+4\right)+3\left(9-4x^2\right)+4\left(x^2-8x+16\right)}{24}=0\)
\(\Leftrightarrow\frac{8x^2-32x+32+27-12x^2+4x^2-32x+64}{24}=0\)
\(\Leftrightarrow\frac{123-64x}{24}=0\Leftrightarrow123-64x=0\Leftrightarrow x=\frac{123}{64}\)
a) \(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}=\frac{x-4}{5}+\frac{x-5}{6}\)
\(\left(\frac{x-1}{2}+1\right)+\left(\frac{x-2}{3}+3\right)+\left(\frac{x-3}{4}+1\right)=\left(\frac{x-4}{5}+1\right)+\left(\frac{x-5}{6}+1\right)\)
\(\frac{x-1}{2}+\frac{x-1}{3}+\frac{x-1}{4}=\frac{x-1}{5}+\frac{x-1}{6}\)
\(\left(x-1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)\)=0
\(x-1=0\)
\(x=1\)
