2x + 6 . 2x+1 = 26 . 218
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a) \(\frac{-x}{2}+\frac{2x}{3}+x+\frac{1}{4}+2x+\frac{1}{6}=\frac{3}{8}.\)
\(\frac{-x}{2}+\frac{2x}{3}+3x+\frac{5}{12}=\frac{3}{8}\)
\(x.\left(-\frac{1}{2}+\frac{2}{3}+3\right)+\frac{5}{12}=\frac{3}{8}\)
\(x\cdot\frac{19}{6}=-\frac{1}{24}\)
x = -1/76
b) \(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{12}{26}\)
\(\frac{3}{2x+1}+\frac{2.5}{2.\left(2x+1\right)}-\frac{2.3}{3.\left(2x+1\right)}=\frac{6}{13}\)
\(\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)
\(\frac{3+5-2}{2x+1}=\frac{6}{13}\)
\(\frac{6}{2x+1}=\frac{6}{13}\)
=> 2x + 1 = 13
2x = 12
x = 6
\(a,-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+1}{6}=\dfrac{8}{3}\)
\(\Rightarrow-\dfrac{6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{8}{3}\)
\(\Rightarrow\dfrac{-6x+8x+3x+3+4x+2}{12}=\dfrac{8}{3}\)
\(\Rightarrow\dfrac{9x+5}{12}=\dfrac{8}{3}\)
\(\Rightarrow27x+15=96\)
\(\Rightarrow27x=81\)
\(\Rightarrow x=3\left(tm\right)\)
\(b,\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\)
\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)
\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)
\(\Rightarrow\dfrac{3+5-2}{2x+1}=\dfrac{6}{13}\)
\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)
\(\Rightarrow2x+1=13\)
\(\Rightarrow2x=12\)
\(\Rightarrow x=6\left(tm\right)\)
#Toru
a) \(-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+2}{6}=\dfrac{8}{3}\)
\(\Rightarrow\dfrac{-6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{4\cdot8}{12}\)
\(\Rightarrow-6x+8x+3x+3+4x+2=32\)
\(\Rightarrow9x+5=32\)
\(\Rightarrow9x=32-5\)
\(\Rightarrow9x=27\)
\(\Rightarrow x=\dfrac{27}{9}\)
\(\Rightarrow x=3\)
b) \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\) (ĐK: \(x\ne-\dfrac{1}{2}\))
\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)
\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)
\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)
\(\Rightarrow2x+1=13\)
\(\Rightarrow2x=12\)
\(\Rightarrow x=\dfrac{12}{2}\)
\(\Rightarrow x=6\left(tm\right)\)
1. \(2^x-26=6\)
\(\Rightarrow2^x=6+26\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
2. \(64\cdot4^x=16^8\)
\(\Rightarrow4^3\cdot4^x=4^{16}\)
\(\Rightarrow4^x=4^{16}:4^3\)
\(\Rightarrow4^x=4^{13}\)
\(\Rightarrow x=13\)
3. \(\left(2x-1\right)^4=16\)
\(\Rightarrow\left(2x-1\right)^4=2^4\)
\(\Rightarrow2x-1=2\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=\dfrac{3}{2}\)
4. \(\left(2x+1\right)^3=125\)
\(\Rightarrow\left(2x+1\right)^3=5^3\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
\(\Leftrightarrow x^2-6x+8=6\sqrt{2x+1}-18\left(Đk:x\ge-\dfrac{1}{2}\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)=\dfrac{12\left(x-4\right)}{\sqrt{2x+1}+3}\left(\sqrt{2x+1}+3>0\right)\)
+) \(x=4\left(TM\right)\)
+) \(x\ne4\Rightarrow x-2=\dfrac{12}{\sqrt{2x+1}+3}\)
\(\Leftrightarrow x-4=\dfrac{12-2\left(\sqrt{2x+1}+3\right)}{\sqrt{2x+1}+3}\)
\(\Leftrightarrow x-4+\dfrac{2\left(x-4\right)}{\left(\sqrt{2x+1}+3\right)^2}=0\)
\(\Leftrightarrow1+\dfrac{2}{\left(\sqrt{2x+1}+3\right)^2}=0\left(x\ne4\right)\)
Vì \(\dfrac{2}{\left(\sqrt{2x+1}+3\right)^2}>0\forall x\) => VT>0
=> phương trình vô nghiệm
Vậy \(S=\left\{4\right\}\)
2^x+6.2^x+1=26.2^18
2^x+6.2^x.2=26.2^18
2^x.(6.2+1)=13.2.2^18
2^x.13=13.2^19
=>2^x=2^19
=>x=19
Vậy x=19