a) \(25.8^3-23.8^3=8^3\left(25-23\right)\)

                                 \(=8^3.2\)

                                  \(=2^9.2=2^{10}\)  

b) \(5^4-2.5^3=5^3.5-2.5^3\)

                        \(=5^3\left(5-2\right)\)

                        \(=5^3.3=375\)

c)\(2.4^3-4^3.7-6.4^3=4^3\left(2-7-6\right)\)

                                       \(=4^3.-11=-704\)

d)\(3^2.10^3-\left[13^2-\left(5^2.4+2^2.15\right)\right].10^3\)

\(=3^2.10^3-\left[13^2-2^2\left(5^2+15\right)\right].10^3\)

\(=3^2.10^3-\left[13^2-2^2.40\right].10^3\)

\(=10^3\left[3^2-9\right]\)

\(=0\)

Bài 1 :

a) \(25.8^3-23.8^3\)

\(=8^3.\left(25-23\right)\)

\(=512.2=1024\)

b) \(5^4-2.5^3\)

\(=5^3.5-2.5^3\)

\(=5^3\left(5-2\right)\)

\(=125.3\)

\(=375\)

c) \(2.4^3-4^3.7-6.4^3\)

\(=4^3.\left(2-7-6\right)\)

\(=64.\left(-11\right)=-704\)

d) \(3^2.10^3-\left[13^2-\left(5^2.4+2^2.15\right)\right]\)

\(=9.1000-\left[169-\left(25.4+4.15\right)\right]\)

\(=9000-\left[169-4\left(25+15\right)\right]\)

\(=9000-\left[169-4.40\right]\)

\(=9000-\left[169-160\right]\)

\(=9000-9=8991\)

a.\(25.8^3-23.8^3=8^3.\left(25-23\right)=512.2=1024\)

b. \(5^4-2.5^3=5^3.\left(5-2\right)=125.3=375\)

c. \(2.4^3-4^3.7-6.4^3=4^3.\left(2-7-6\right)\)\(=64.\left(-11\right)=-704\)

d. \(3^2.10^3-\left[13^2-\left(5^2.4+2^2.15\right)\right]=\)\(3^2.10^3-\left[13^2-\left(100+60\right)\right]\) \(=3^2.10^3-\left(13^2-160\right)=9000-9=8991\)

\(a, \)\(\left(3^2\right)^2-\left(2^3\right)^2-\left(-5^2\right)^2=9^2-8^2-10^2\)

= \(81-64-100\)

\(=-83\)

\(b,\)\(2^3+3.\left(-\dfrac{1}{2}\right)^0-\left(\dfrac{1}{2}\right)^2.4+\left(\left(-2\right)^2:\dfrac{1}{2}\right):8=8+3.1-\dfrac{1}{4}.4+\left(4:\dfrac{1}{2}\right):8\) \(=8+3-1+8:8\)

\(=8+3-1+1\)

\(=11\)

a,

\(\dfrac{4^2\cdot4^3}{2^{10}}=\dfrac{4^5}{2^{10}}=\dfrac{\left(2^2\right)^5}{2^{10}}=\dfrac{2^{10}}{2^{10}}=1\)

b,

\(\dfrac{\left(0,6\right)^5}{\left(0,2\right)^6}=\dfrac{\left(0,2\cdot3\right)^5}{\left(0,2\right)^5\cdot0,2}=\dfrac{\left(0,2\right)^5\cdot3^5}{\left(0,2\right)^5\cdot0,2}=\dfrac{243}{0,2}=\dfrac{243}{\dfrac{1}{5}}=243\cdot5=1215\)

c,

\(\dfrac{2^7\cdot9^3}{6^5\cdot8^2}=\dfrac{2^7\cdot\left(3^2\right)^3}{\left(2\cdot3\right)^5\cdot\left(2^3\right)^2}=\dfrac{2^6\cdot2\cdot3^6}{2^5\cdot3^5\cdot2^6}=\dfrac{3}{2^4}=\dfrac{3}{16}\)

d,

\(\dfrac{6^3+3\cdot6^2+3^3}{-13}=\dfrac{\left(2\cdot3\right)^3+3\cdot\left(2\cdot3\right)^2+3^3}{-13}=\dfrac{2^3\cdot3^3+3\cdot2^2\cdot3^2+3^3}{-13}=\dfrac{2^3\cdot3^3+2^2\cdot3^3+3^3}{-13}\dfrac{3^3\left(2^3+2^2+1\right)}{-13}=\dfrac{3^3\cdot13}{-13}=-3^3=-27\)

a.

\(-2^3+2^2+\left(-1\right)^{2013}=-8+4-1=-5\)

b.

\(\left(3^3\right)^2-\left[\left(-2\right)^3\right]^2-\left(-5\right)^2=27^2-\left(-8\right)^2-25=729-64-25=640\)

c.

\(2^3+3\times\left(-\frac{1}{2016}\right)^0-\left(\frac{1}{2}\right)^2\times4-\left[\left(-2\right)^2\div\frac{1}{2}\right]=8+3\times0-\frac{1}{4}\times4-\left(4\times2\right)=8+3-1-8=2\)

\(8,1-\left(x-6\right)=4\left(2-2x\right)\)

\(\Leftrightarrow1-x+6=8-8x\)

\(\Leftrightarrow-x+8x=8-1-6\)

\(\Leftrightarrow7x=1\)

\(\Leftrightarrow x=\dfrac{1}{7}\)

\(9,\left(3x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)

\(10,\left(x+3\right)\left(x^2+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)

`8)1-(x-5)=4(2-2x)`

`<=>1-x+5=8-6x`

`<=>5x=2<=>x=2/5`

`9)(3x-2)(x+5)=0`

`<=>[(x=2/3),(x=-5):}`

`10)(x+3)(x^2+2)=0`

  Mà `x^2+2 > 0 AA x`

 `=>x+3=0`

`<=>x=-3`

`11)(5x-1)(x^2-9)=0`

`<=>(5x-1)(x-3)(x+3)=0`

`<=>[(x=1/5),(x=3),(x=-3):}`

`12)x(x-3)+3(x-3)=0`

`<=>(x-3)(x+3)=0`

`<=>[(x=3),(x=-3):}`

`13)x(x-5)-4x+20=0`

`<=>x(x-5)-4(x-5)=0`

`<=>(x-5)(x-4)=0`

`<=>[(x=5),(x=4):}`

`14)x^2+4x-5=0`

`<=>x^2+5x-x-5=0`

`<=>(x+5)(x-1)=0`

`<=>[(x=-5),(x=1):}`

\(11,=>\left[{}\begin{matrix}5x-1=0\\x^2-9=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=3\\x=-3\end{matrix}\right.\\ 12,=>\left(x+3\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\\ 13,=>x\left(x-5\right)-4\left(x-5\right)=0\\ =>\left(x-4\right)\left(x-5\right)=0\\ =>\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

\(14,=>x^2+5x-x-5=0\\ =>x\left(x+5\right)-\left(x+5\right)=0\\ =>\left(x-1\right)\left(x+5\right)=0\\ =>\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

a)

\(\begin{array}{l}\left( {9x - {2^3}} \right):5 = 2\\9x - {2^3} = 2.5\\9x - 8 = 10\\9x = 18\\x = 2\end{array}\)

Vậy \(x = 2\)

b)

\(\begin{array}{l}\left[ {{3^4} - \left( {{8^2} + 14} \right):13} \right]x = {5^3} + {10^2}\\\left[ {81 - \left( {64 + 14} \right):13} \right]x = 125 + 100\\\left[ {81 - 78:13} \right]x = 125 + 100\\\left[ {81 - 6} \right]x = 225\\75x = 225\\x = 3\end{array}\)

Vậy \(x = 3\)

\(\left(\frac{1}{3}-1\frac{5}{6}\right)^2\)

\(=\left(\frac{1}{3}-\frac{11}{6}\right)^2\)

\(=\left(\frac{2}{6}-\frac{11}{6}\right)^2=\left(-\frac{9}{6}\right)^2\)

\(=\left(-\frac{3}{2}\right)^2=\frac{9}{4}\)