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\(2x-3+3|x-1|=4x+1.\)
\(\Leftrightarrow3|x-1|=2x+4\)
*Với x < 1 ta có phương trình:
\(3\left(-x+1\right)=2x+4\)
\(\Leftrightarrow-3x+3=2x+4\)
\(\Leftrightarrow5x+1=0\)
\(\Leftrightarrow x=-\frac{1}{5}\)(TM)
*Với \(x\ge1\)ta có phương trình:
\(2x-3+3\left(x-1\right)=4x+1\)
\(\Leftrightarrow2x-3+3x-3=4x+1\)
\(\Leftrightarrow x-7=0\)
\(\Leftrightarrow x=7\)(TM)
Vậy ............
a)\(1\frac{5}{6}=\frac{-x}{5}\)
\(\Leftrightarrow\frac{11}{6}=\frac{-x}{5}\)\(\Rightarrow-6x=55\)
\(\Rightarrow x=-\frac{55}{6}\)
b)\(\frac{4\cdot25}{8}=\frac{-3\cdot5}{x}\)
\(\Leftrightarrow\frac{100}{8}=\frac{-15}{x}\)\(\Rightarrow100x=-15\cdot8\)
\(\Rightarrow100x=-120\)
\(\Rightarrow x=-\frac{6}{5}\)
\(\dfrac{-4}{x}=\dfrac{x}{-49}\\ \Rightarrow x^2=\left(-4\right)\left(-49\right)\\ \Rightarrow x^2=196\\ \Rightarrow x=\pm14\)
\(\dfrac{3.6}{x-3}=\dfrac{5}{3}\\ \Rightarrow5\left(x-3\right)=3.3.6\\ \Rightarrow5\left(x-3\right)=54\\ \Rightarrow x-3=\dfrac{54}{5}\\ \Rightarrow x=\dfrac{54}{5}+3\\ \Rightarrow x=\dfrac{69}{15}\)
\(\left(2x+1\right):2=12:3\\ \left(2x+1\right):2=4\\2x+1=2\\ 2x=1\\ x=\dfrac{1}{2} \)
\(\left(2x-14\right):3=12:9\\ \left(2x-14\right):3=\dfrac{4}{3}\\ 2x-14=4\\ 2x=16\\ x=8\)
\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{X\left(X+2\right)}\)
\(\frac{1}{2}.\left(\frac{1}{1.3}+...+\frac{1}{X\left(X+2\right)}\right)\)= \(\frac{16}{34}\)
\(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+...+\frac{1}{X}-\frac{1}{X+2}\right)\)
=15
TA CÓ : 1/1.3 + 1/3.5 + 1/5.7 +... + 1/X(X+2) = 8/17
=> 2/1.3 + 2/3.5 + 2/5.7 +... + 2/X(X+2) = 8/17 . 2 = 16/17
<=> 1 - 1/X+2 = 16/17
X+2/X+2 - 1/X+2 = 16/17
X+2 -1/X+2 = 16/17
=> X+2 -1 =16 VÀ X+2 = 17
=> X = 15
Ta có: \(5^{x-1}+3\cdot5^{x+1}=1900\)
\(\Leftrightarrow5^x:5+3\cdot5^x\cdot5=1900\)
\(\Leftrightarrow5^x\cdot\frac{1}{5}+15\cdot5^x=1900\)
\(\Leftrightarrow5^x\cdot\left(\frac{1}{5}+15\right)=1900\)
\(\Leftrightarrow5^x\cdot15,2=1900\)
\(\Leftrightarrow5^x=\frac{1900}{15,2}=125\)
\(\Leftrightarrow5^x=5^3\)
hay x=3
Vậy: x=3