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\(x+\frac{1}{5}+x+\frac{1}{6}+x+\frac{1}{7}=x+\frac{1}{8}+x+\frac{1}{9}\)
\(3x+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}=2x+\frac{1}{8}+\frac{1}{9}\)
\(3x+\frac{107}{210}=2x+\frac{17}{72}\)
\(3x-2x=\frac{17}{72}-\frac{107}{210}\)
\(x=\frac{-689}{2520}\)
Vậy \(x=\frac{-689}{2520}\).
a) \(6.8^{x-1}+8^{x+1}=6.8^{19}+8^{21}\)
\(\Rightarrow x-1+x+1=19+21\)
\(=2x=40\)
\(\Rightarrow x=20\)
b) \(4.3^{x-1}+2.3^{x+2}=4.3^6+2.3^9\)
\(\Rightarrow x-1+x+2=6+9\)
\(\Rightarrow2x+1=15\)
\(\Rightarrow2x=14\)
\(\Rightarrow x=7\)
Từ đề bài=>(x+1)/10+(x+2)/9+(x+3)/8-(-3)=0
<=>......................................(như trên)+3=0
<=>(x+1)/10+1+(x+2)/9+1+(x+3)/8+1=0( vì 1+1+1=3)
<=>(x+1+10)/10+(x+2+9)/9+(x+3+8)/8=0
<=>(x+11)/10+(x+11)/9+(x+11)/8=0
<=>(x+11).(1/10+1/9+1/8)=0
Mà 1/10<1/9<1/8
=>1/10+1/9+1/8 khác 0
<=>x+11=0<=>x=-11
Vậy x=-11
Ta có: \(x+\frac{1}{9}=\frac{8}{x}-1\)
\(\Leftrightarrow x-\frac{8}{x}=-1-\frac{1}{9}\)
\(\Leftrightarrow\frac{x^2-8}{x}=-\frac{10}{9}\)
\(\Leftrightarrow9.\left(x^2-8\right)=-10x\)
\(\Leftrightarrow9x^2+10x-72=0\)
\(\Leftrightarrow\left(9x^2+10x+\frac{25}{9}\right)-\frac{673}{9}=0\)
\(\Leftrightarrow\left(3x+\frac{5}{3}\right)^2=\frac{673}{9}\)
\(\Leftrightarrow3x+\frac{5}{3}=\pm\frac{\sqrt{673}}{3}\)
+ \(3x+\frac{5}{3}=\frac{\sqrt{673}}{3}\)\(\Leftrightarrow\)\(3x=\frac{\sqrt{673}-5}{3}\)
\(\Leftrightarrow\)\(x=\frac{\sqrt{673}-5}{9}\)
+ \(3x+\frac{5}{3}=-\frac{\sqrt{673}}{3}\)\(\Leftrightarrow\)\(3x=\frac{-\sqrt{673}-5}{3}\)
\(\Leftrightarrow\)\(x=\frac{-\sqrt{673}-5}{9}\)
Vậy \(x\in\left\{\frac{\sqrt{673}-5}{9};\frac{-\sqrt{673}-5}{9}\right\}\)
Ta có : \(\frac{x+1}{9}=\frac{8}{x-1}\left(x\ne1\right)\)
\(\Rightarrow\left(x+1\right).\left(x-1\right)=8.9\)
\(\Rightarrow x^2-1=72\Rightarrow x^2=73\Rightarrow x=\pm\sqrt{73}\)