tính nhẩm
96:8
giúp mình i
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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)
a: 13/45=13/45
12/3=4=180/45
b: 55/45=11/9=88/72
1/8=9/72
96 : 8 = ( 80 + 16 ) : 8 = 80 : 8 + 16 : 8 = 10 + 2 = 12
96 : 8 = 12