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`(3x^3 -2x^2 +3x-2):(x^2+1)`
`=(3x^3 +3x)-(2x^2-2):(x^2+1)`
`=(3x^3 +3x)-(2x^2+2) :(x^2+1)`
`=3x(x^2+1)-2(x^2+1):(x^2+1)`
`=(x^2+1)(3x-2):(x^2+1)`
`=3x-2`
`@ Ariko`
![](https://rs.olm.vn/images/avt/0.png?1311)
`@` `\text {Ans}`
`\downarrow`
`30 \div { 12*5^2 - 2*[567 - (2*7^2 + 8^2*5)]}`
`= 30 \div {300 - 2*[567 -(98 + 320)]} `
`= 30 \div [300 - 2*(567 - 418)]`
`= 30 \div (300 - 2*149)`
`= 30 \div (300 - 298)`
`= 30 \div 2`
`= 15`
\(30:\left\{12\cdot5^2-2\cdot\left[567-\left(2\cdot7^2+8^2\cdot5\right)\right]\right\}\)
\(=30:\left\{12\cdot5^2-2\cdot\left[567-\left(98+320\right)\right]\right\}\)
\(=30:\left[300-2\cdot\left(567-418\right)\right]\)
\(=30:\left(300-2\cdot158\right)\)
\(=30:\left(300-316\right)\)
\(=30:-16\)
\(=-\dfrac{15}{8}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3^2\cdot5^2-\left\{4^3\cdot25-3\cdot\left[3\cdot\left(17+3\right)^3-7\cdot10^2\right]\right\}\)
\(=\left(3\cdot5\right)^2-\left[64\cdot25-3\cdot\left(3\cdot20^3-7\cdot100\right)\right]\)
\(=15^2-\left[8^2\cdot5^2-3\cdot\left(3\cdot8000-700\right)\right]\)
\(=225-\left[\left(5\cdot8\right)^2-3\cdot\left(24000-700\right)\right]\)
\(=225-\left(40^2-3\cdot23300\right)\)
\(=225-\left(1600-69900\right)\)
\(=225-\left(-68300\right)\)
\(=68525\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left[\left(\dfrac{5}{2}\right)^2-2\left(\dfrac{-1}{3}\right)^2+2\right]-\left(\dfrac{-1}{3}\right)^3=\left[\dfrac{25}{4}-\dfrac{2}{9}+2\right]+\dfrac{1}{27}=\dfrac{289}{36}+\dfrac{1}{27}=\dfrac{871}{108}\)
`@` `\text {Ans}`
`\downarrow`
`c)`
`[ (5/2)^2 - 2*(-1/3)^2 + 2] - (-1/3)^3`
`= (25/4 - 2*1/9 + 2) - (-1/27)`
`= (25/4 - 2/9 + 2) + 1/27`
`= 217/36 + 2 + 1/27`
`= 289/36 + 1/27`
`= 871/108`
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\dfrac{-1}{3}\right)^2-\left[\left(-\dfrac{5}{2}\right)-\left(1,5-\dfrac{8}{3}\right)\right]=\dfrac{1}{9}-\left[\left(\dfrac{-5}{2}\right)-\left(\dfrac{-7}{6}\right)\right]=\dfrac{1}{9}-\left(\dfrac{-4}{3}\right)=\dfrac{13}{9}\)
(-1/3)² - [(-5/2) - (1,5 - 8/3)]
= 1/9 - (-5/2 + 7/6)
= 1/9 + 4/3
= 13/9
![](https://rs.olm.vn/images/avt/0.png?1311)
\(34.5^2-4^3.3^2-37.2+50\\ =34.25-64.9-37.2\\ =850-576-74\\ =274-74\\ =200\\ 100-\left\{20.\left[3^2.10-3.\left(35-8\right)\right]\right\}\\ =100-\left[20.\left(9.10-3.27\right)\right]\\ =100-\left[20.\left(90-81\right)\right]\\ =100-\left(20.9\right)\\ =100-180\\ =-80\)
b) 34.5² - 4³.3² - 37.2 + 50
= 34.25 - 64.9 - 74 + 50
= 850 - 576 - 74 + 50
= (850 + 50) - (576 + 74)
= 900 - 650
= 250
d) 100 - {20.[3².10 - 3.(35 - 8)]}
= 100 - [20.(9.10 - 3.27)]
= 100 - [20.(90 - 81)]
= 100 - 20.9
= 100 - 180
= -80
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{3}-\left[\dfrac{2}{3}-\left(0,5-\dfrac{3}{4}\right)^2\right]\)
\(=\dfrac{1}{3}-\left[\dfrac{2}{3}-\left(\dfrac{1}{2}-\dfrac{3}{4}\right)^2\right]\)
\(=\dfrac{1}{3}-\left[\dfrac{2}{3}-\left(\dfrac{2}{4}-\dfrac{3}{4}\right)^2\right]\)
\(=\dfrac{1}{3}-\left[\dfrac{2}{3}-\left(-\dfrac{1}{4}\right)^2\right]\)
\(=\dfrac{1}{3}-\left(\dfrac{2}{3}-\dfrac{1}{16}\right)\)
\(=\dfrac{1}{3}-\dfrac{29}{48}\)
\(=-\dfrac{13}{48}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(-\dfrac{1}{3}\right)^2-\left[\left(-\dfrac{5}{2}\right)-\left(1,5-\dfrac{8}{3}\right)\right]\)
\(=\dfrac{1}{9}-\left[\left(-\dfrac{5}{2}\right)-\left(-\dfrac{7}{6}\right)\right]\)
\(=\dfrac{1}{9}-\left(-\dfrac{4}{3}\right)\)
\(=\dfrac{1}{9}+\dfrac{4}{3}\)
\(=\dfrac{1}{9}+\dfrac{12}{9}\)
\(=\dfrac{13}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(30:\left\{12.5^2-2.\left[567-\left(2.7^2+8^2.5\right)\right]\right\}\\ =30:\left\{12.25-2.\left[567-\left(2.49+64.5\right)\right]\right\}\\ =30:\left\{12.25-2.\left[567-\left(98+320\right)\right]\right\}\\ =30:\left[12.25-2.\left(567-418\right)\right]\\ =30:\left(12.25-2.149\right)\\ =30:\left(300-298\right)\\ =30:2\\ =15\)
\(\frac{8^{15}.3^{16}}{4^{22}.9^8}=\frac{\left(2^3\right)^{15}.3^{16}}{\left(2^2\right)^{22}.\left(3^2\right)^8}=\frac{2^{45}.3^{16}}{2^{44}.3^{16}}=2\)
\(\frac{8^{15}.3^{16}}{4^{22}.9^8}=\frac{\left(2^3\right)^{15}.3^{16}}{\left(2^2\right)^{22}.\left(3^2\right)^8}=\frac{2^{45}.3^{16}}{2^{44}.3^{16}}=2\)