2/3×x-0,125=15/8
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\(a,\left(-\dfrac{2}{3}\right)^8+\left(-\dfrac{2}{3}\right)^8=2\left(-\dfrac{2}{3}\right)^8=2\cdot\dfrac{\left(-2\right)^8}{3^8}=\dfrac{2\cdot2^8}{3^8}=\dfrac{2^9}{3^8}=\dfrac{512}{6561}\)
bài 1)
a) \(\dfrac{\left(-3\right)^{10}.15^5}{25^3.\left(-9\right)^7}\)
\(=\dfrac{\left(-3\right)^{10}.\left(3.5\right)^5}{\left(5^2\right)^3.\left(-3.3\right)^7}\)
\(=\dfrac{\left(-3\right)^{10}.3^5.5^5}{5^6.\left(-3\right)^7.3^7}\)
\(=\dfrac{\left(-3\right)^3.1.1}{5.1.3^2}\)
\(=\dfrac{-27.1.1}{5.1.9}\)
\(=\dfrac{-27}{45}\)
\(=\dfrac{-9}{15}\)
b)\(2^3+3.\left(\dfrac{1}{9}\right)^0-2^{-2}.4\left[\left(-2\right)^2:\dfrac{1}{2}\right].8\)
\(=8+3.1-\dfrac{1}{2^2}.4+\left[\left(4:\dfrac{1}{2}\right)\right].8\)
\(=8+3.1-\dfrac{1}{4}.4+\left[4.\dfrac{2}{1}\right].8\)
\(=8+3.1-\dfrac{1}{4}.4+8.8\)
\(=8+3-1+64\)
\(=11-1+64\)
\(=10+64\)
\(=74\)
Ta có: \(3x^2-x+2\)
\(=3\left(x^2-\dfrac{1}{3}x+\dfrac{2}{3}\right)\)
\(=3\left(x^2-2\cdot x\cdot\dfrac{1}{6}+\dfrac{1}{36}+\dfrac{23}{36}\right)\)
\(=3\left(x-\dfrac{1}{6}\right)^2+\dfrac{23}{12}\ge\dfrac{23}{12}>0\forall x\)(đpcm)
\(\dfrac{7}{15}và\dfrac{5}{13}=\dfrac{7.13}{15.13}=\dfrac{91}{195};\dfrac{5}{13}=\dfrac{5.15}{13.15}=\dfrac{75}{195}\)
\(\dfrac{11}{12}và\dfrac{7}{48}=\dfrac{11.4}{12.4}=\dfrac{44}{48};\dfrac{7}{48}\)
\(\dfrac{24}{35}và\dfrac{21}{7}=\dfrac{24}{35};\dfrac{21}{7}=\dfrac{21.5}{7.5}=\dfrac{105}{35}\)
\(\dfrac{15}{16}và\dfrac{7}{8}=\dfrac{15}{16};\dfrac{7}{8}=\dfrac{7.2}{8.2}=\dfrac{14}{16}\)
\(\frac{2}{3}\cdot x-0,125=\frac{15}{8}\)
\(\frac{2}{3}\cdot x=\frac{15}{8}+0,125=\frac{15}{8}+\frac{1}{8}=2\)
\(x=2:\frac{2}{3}=2\cdot\frac{3}{2}=3\)
Vậy x=3
2/3.x - 0,125 = 15/8
2/3.x - 1/8 = 15/8
2/3.x = 15/8 +1/8
2/3.x = 2
x = 2:2/3
x = 3