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a) Ta có \(|5\left(2x+3\right)\ge0\)
\(|2\left(2x+3\right)|\ge0\)
\(|2x+3|\ge0\)
\(\Rightarrow|5\left(2x+3\right)|+|\left(2x+3\right)|+|2x+3|\ge0\)
\(\Rightarrow5\left(2x+3\right)+2\left(2x+3\right)+2x+3=16\)
\(\Rightarrow10x+15+4x+6+2x+3=16\)
\(\Rightarrow\left(10x+4x+2x\right)+\left(15+6+3\right)=16\)
\(\Rightarrow16x+24=16\)
\(\Rightarrow24=16x-16\)
\(\Rightarrow24=x\)
Vậy x=24
\(3^x+3^{x+2}=270\)
\(\Rightarrow3^x+3^x.3^2=270\)
\(\Rightarrow3^x\left(1+3^2\right)=270\)
\(\Rightarrow3^x.10=270\)
\(\Rightarrow3^x=27\)
\(\Rightarrow3^x=3^3\)
\(\Rightarrow x=3\)
Vậy x = 3
c) \(\left|x-1\right|=3\)
\(\Rightarrow\left[\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{4;-2\right\}\)
a ) \(3^x+3^{x+2}=270\)
\(\Leftrightarrow3^x.1+3^x.3^2=270\)
\(\Leftrightarrow3^x.\left(1+3^2\right)=270\)
\(\Leftrightarrow3^x=27\)
\(\Leftrightarrow x=3\)
Vậy x = 3.
b ) \(\left|2x+1\right|+\left|x-3\right|=5\)
\(\Leftrightarrow\left|2x+1\right|+\left|3-x\right|=5\)
\(\Leftrightarrow\left|2x+1+3-x\right|=5\)
\(\Leftrightarrow\left|x+4\right|=5\)
\(\Leftrightarrow x=1\)
Vậy x = 1. ( không chắc lắm )
c ) \(\left|x-1\right|=3\)
\(\Leftrightarrow\left[\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
Vậy x = 4 và x = -2.
a ) \(3-4.\left|5-6x\right|=7\)
\(\Leftrightarrow4.\left|5-6x\right|=-4\)
\(\Leftrightarrow\left|5-6x\right|=-1\)
\(\Leftrightarrow\) Không thõa mãn ( vì \(x\ge0\) )
b) Do \(\left|x+2\right|\ge0;\left|x+\frac{3}{5}\right|\ge0;\left|x+\frac{1}{2}\right|\ge0\)
=> \(4x\ge0\)
=> \(x\ge0\)
Lúc này ta có: \(\left(x+2\right)+\left(x+\frac{3}{5}\right)+\left(x+\frac{1}{2}\right)=4x\)
=> \(\left(x+x+x\right)+\left(2+\frac{3}{5}+\frac{1}{2}\right)=4x\)
=> \(3x+\frac{31}{10}=4x\)
=> \(4x-3x=\frac{31}{10}\)
=> \(x=\frac{31}{10}\)
Vậy \(x=\frac{31}{10}\)
c) Do \(\left|x+\frac{1}{101}\right|\ge0;\left|x+\frac{2}{101}\right|\ge0;\left|x+\frac{3}{101}\right|\ge0;...;\left|x+\frac{100}{101}\right|\ge0\)
=> \(101x\ge0\)
=> \(x\ge0\)
Lúc này ta có: \(\left(x+\frac{1}{101}\right)+\left(x+\frac{2}{101}\right)+\left(x+\frac{3}{101}\right)+...+\left(x+\frac{100}{101}\right)=101x\)
=> \(\left(x+x+x+...+x\right)+\left(\frac{1}{101}+\frac{2}{101}+\frac{3}{101}+...+\frac{100}{101}\right)=101x\)
100 số x
=> \(100x+\frac{\left(1+100\right).100:2}{101}=101x\)
=> \(\frac{101.50}{101}=101x-100x\)
=> \(x=50\)
Vậy x = 50
\(a,\left|5x+4\right|+7=26\\ \left|5x+4\right|=26+7\\ \left|5x+4\right|=33\\ \Rightarrow\left\{{}\begin{matrix}5x+4=33\\5x+4=-33\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}5x=29\\5x=-29\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{29}{5}\\x=-\dfrac{29}{5}\end{matrix}\right.\)
Các câu sau làm tương tự!
a) x= ± 3
b) (x+3)(x-3)=x2-32=x2-9=6x
=>x2=6x+9
=> x2=3(2x+3)
vì 3 khác với 2x+3 mà đề lại cho x^2 => x vô nghiệm