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Ta có \(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(2\left(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2}{9}\)
\(2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(2\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{2}{9}\div2\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{18}\left(1\right)\)
Có \(\left(1\right)\Leftrightarrow\left(x+1\right).1=1.18\)
\(\Rightarrow x+1=18\)
\(\Rightarrow x=18-1\)
\(\Rightarrow x=17\)
Ta có \(1\frac{1}{5}x+\frac{2}{3}x=-\frac{56}{125}\)
<=> \(\frac{6}{5}x+\frac{2}{3}x=-\frac{56}{125}\)
<=> \(\frac{28}{15}x=-\frac{56}{125}\)
<=> \(x=-\frac{2}{15}\)
b) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)
<=> \(\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=0\)
<=> \(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)
<=> \(\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
<=> x + 100 = 0
<=> x = -100
\(2\left|x+1\right|=10\)
Th1 :
\(\left|x+1\right|\ge0\Rightarrow x\ge-1\)
\(2\left(x+1\right)=10\)
\(\Rightarrow2x+2=10\)
\(\Rightarrow2x=8\)
\(\Rightarrow x=4\)
Th2 : \(\left|x+1\right|< 0\Rightarrow x< -1\)
\(2\left(-1-x\right)=10\)
\(\Rightarrow-2-2x=10\)
\(\Rightarrow-2x=8\)
\(\Rightarrow x=-4\)
Ta có: 2x + 2x+1 + 2x+2 = 56
=> 2x + 2x . 21 + 2x . 22 = 56
=> 2x . 1 + 2x . 2 + 2x . 4 = 56
=> 2x .(1 + 2 + 4) = 56
=> 2x . 7 = 56
=> 2x = 56 : 7
=> 2x = 8
Mà 8 = 23 = (-2)3
=> x = 2 hoặc x = -2
Vậy x = 2 hoặc x = -2.
\(2^x+2^{x+1}+2^{x+2}=56\\ 2^x\left(1+2^1+2^2\right)=56\\ 2^x.7=56\\ 2^x=56:7\\ 2^x=8=2^3\\ Vậy:x=3\)