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\(\left(\frac{1}{2}\right)^{3x-1}=\left(\frac{1}{2}\right)^5\)
\(\Rightarrow3x-1=5\)
3x \(=6\)
x \(=2\)
Vậy x=2
\(\frac{1}{5}-\frac{5}{2}\left|3x-\frac{1}{5}\right|=\frac{2}{3}\left|3x-\frac{1}{5}\right|-\frac{2}{3}\)
Đặt \(a=\left|3x-\frac{1}{5}\right|\) ta đc:
\(\frac{1}{5}-\frac{5}{2}a=\frac{2}{3}a-\frac{2}{3}\)
\(\Leftrightarrow-\frac{5}{2}a=\frac{2}{3}a-\frac{13}{15}\)
\(\Leftrightarrow-\frac{5a}{2}=\frac{10a}{15}-\frac{13}{15}\)
\(\Leftrightarrow-\frac{5a}{2}=\frac{10a-13}{15}\Rightarrow-5a\cdot15=2\left(10a-13\right)\)
\(\Leftrightarrow-75a=20a-26\)
\(\Leftrightarrow-95a=-26\Leftrightarrow a=\frac{26}{95}\)
\(\Leftrightarrow\left|3x-\frac{1}{5}\right|=\frac{26}{95}\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}3x-\frac{1}{5}=\frac{26}{95}\\3x-\frac{1}{5}=-\frac{26}{95}\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}3x=\frac{9}{19}\\3x=-\frac{7}{95}\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{3}{19}\\x=-\frac{7}{285}\end{array}\right.\)
f(x)=9x3-1/3x+3x2-3x+1/3x2-1/9x3-3x2-9x+27+3x
= 9x3-1/9x3+3x2+1/3x2-3x2-1/3-3x-9x+3x+27
= 80/9x3+1/3x2-28/3x+27
\(\Rightarrow32\left(2x+3\right)=24\left(3x-1\right)\)
\(\Rightarrow64x+96=72x-24\)
\(\Rightarrow96+24=72x-64x\)
\(\Rightarrow120=8x\)
\(\Rightarrow x=\frac{120}{8}\)
\(\Rightarrow x=15\)
Vậy x=15
\(\frac{2x+3}{24}=\frac{3x-1}{32}\)
\(\Leftrightarrow32\left(2x+3\right)=24\left(3x-1\right)\)
\(\Leftrightarrow64x+96=72x-24\)
\(\Leftrightarrow64x-72x=-96-24\)
\(\Leftrightarrow-8x=-120\)
\(\Leftrightarrow x=15\)
Vậy x = 15
Đặt bt trên là A nha
Đổi |x-1|=|1-x|
Suy ra A=|1-x|+x-2|+|x-3|
Áp dụng BĐTGTTĐ ta có
A=|1-x|+x-2|+|x-3|\(\ge\)|1-x+x-3|=2
Dấu = xảy ra khi \(\hept{\begin{cases}x-2=0\\1< x< 3\end{cases}}\)đồng thời xảy ra
Vậy x =2
b,
\(\left|3x+\frac{1}{2}\right|\ge0\)
\(\left|3x+\frac{1}{6}\right|\ge0\)
..........
\(\left|3x+380\right|\ge0\)
Suy ra đề bài \(\ge\)0
suy ra 58x \(\ge\)0
Suy ra \(3x+\frac{1}{2}+3x+\frac{1}{6}+......+3x+380=58x\)
Tự tính nhé hok tốt
\(\frac{1}{2}^{3x-1}=\frac{1}{32}\)
\(\Leftrightarrow\frac{1}{2}^{3x-1}=\frac{1}{2}^5\)
\(\Leftrightarrow3x-1=5\)
\(\Leftrightarrow3x=6\)
\(\Leftrightarrow x=2\)
Tìm x
\(\left(\frac{1}{2}\right)^{3x-1}=\frac{1}{32}\)
\(\left(\frac{1}{2}\right)^{3x-1}=\left(\frac{1}{2}\right)^5\)
\(\Rightarrow3x-1=5\)
\(\Rightarrow3x=6\)
\(\Rightarrow x=2\)
Vậy \(x=2\).