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4 câu đầu hìn như sai đề :v
`m)(3/2-2/(-5)):x-1/2=3/2`
`<=>(3/2+2/5):x=3/2+1/2=2`
`<=>19/10:x=2`
`<=>x=19/10:2=19/20`
`n)(3/2-5/11-3/13)(2x-2)=(-3/4+5/22+3/26)`
`<=>(3/2-5/11-3/13)(2x-2)+3/4-5/22-3/26=0`
`<=>(3/2-5/11-3/13)(2x-2)+1/2(3/2-5/11-3/13)=0`
`<=>(3/2-5/11-3/13)(2x-2+1/2)=0`
Mà `3/2-5/11-3/13>0`
`<=>2x-2+1/2=0`
`<=>2x-3/2=0`
`<=>2x=3/2<=>x=3/4`
![](https://rs.olm.vn/images/avt/0.png?1311)
`h)x/2-1/x=1/12(x ne 0)`
`<=>6x^2-12=x`
`<=>6x^2-x-12=0`
`<=>6x^2-9x+8x-12=0`
`<=>3x(2x-3)+4(2x-3)=0`
`<=>(2x-3)(3x+4)=0`
`<=>` \(\left[ \begin{array}{l}x=\dfrac32\\x=-\dfrac43\end{array} \right.\)
`i)x^2-7/6x+1/3=0`
`<=>6x^2-7x+2=0`
`<=>6x^2-3x-4x+2=0`
`<=>3x(2x-1)-2(2x-1)=0`
`<=>(2x-1)(3x-2)=0`
`<=>` \(\left[ \begin{array}{l}x=\dfrac12\\x=\dfrac23\end{array} \right.\)
Câu cuối không có dấu "=" nên không tìm được x :v
- Hai câu h, i bấm nốt đáp án để đẹp nha ;-; câu k thiếu đề :v
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a: \(=\dfrac{3}{4}-\dfrac{5}{6}+\dfrac{3}{2}=\dfrac{9-10+18}{12}=\dfrac{17}{12}\)
b: \(=\left(\dfrac{1}{9}+\dfrac{6}{9}\right)^2-\dfrac{1}{3}=\dfrac{49}{81}-\dfrac{27}{81}=\dfrac{22}{81}\)
c; \(=\dfrac{5}{11}\left(-\dfrac{3}{7}-\dfrac{5}{7}\right)+\dfrac{-8}{7}\cdot\dfrac{6}{11}=\dfrac{-8}{7}\left(\dfrac{5}{11}+\dfrac{6}{11}\right)=-\dfrac{8}{7}\)
d: \(=\dfrac{2^{26}}{2^{15}\cdot2^{12}}=\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(H=\left[0,\left(32\right).1,\left(5\right)-0,\left(25\right)\right].\dfrac{11}{83}\)
\(\Leftrightarrow H=\left(\dfrac{32}{99}.\dfrac{14}{9}-\dfrac{25}{99}\right).\dfrac{11}{83}\)
\(\Leftrightarrow H=\left(\dfrac{448}{891}-\dfrac{25}{99}\right).\dfrac{11}{83}\)
\(\Leftrightarrow H=\left(\dfrac{448}{891}-\dfrac{225}{891}\right).\dfrac{11}{83}\)
\(\Leftrightarrow H=\dfrac{448-225}{891}.\dfrac{11}{83}\)
\(\Leftrightarrow H=\dfrac{223}{891}.\dfrac{11}{83}\)
\(\Leftrightarrow H=\dfrac{2453}{73953}\)
\(\Leftrightarrow H=\dfrac{223}{6723}\)
2) \(A=\dfrac{0,5+0,\left(3\right)-0,1\left(6\right)}{2,5+1,\left(6\right)-0,8\left(3\right)}\)
\(\Leftrightarrow A=\dfrac{\dfrac{3}{6}+\dfrac{2}{6}-\dfrac{1}{6}}{\dfrac{15}{6}+\dfrac{10}{6}-\dfrac{5}{6}}\)
\(\Leftrightarrow A=\dfrac{\dfrac{3+2-1}{6}}{\dfrac{15+10-5}{6}}\)
\(\Leftrightarrow A=\dfrac{\dfrac{4}{6}}{\dfrac{20}{6}}\)
\(\Leftrightarrow A=\dfrac{4}{6}.\dfrac{6}{20}\)
\(\Leftrightarrow A=\dfrac{24}{120}\)
\(\Leftrightarrow A=\dfrac{1}{5}\)
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a: \(=\dfrac{3}{4}+\dfrac{1}{4}\cdot\dfrac{-3}{2}+5=\dfrac{23}{4}-\dfrac{3}{8}=\dfrac{43}{8}\)
b: \(=12\cdot\left(\dfrac{4}{6}-\dfrac{5}{6}\right)^2=12\cdot\left(\dfrac{1}{6}\right)^2=12\cdot\dfrac{1}{36}=\dfrac{1}{3}\)
c: =3-6+1/4:2
=-3+1/8
=-23/8
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{2^{10}.3^8-2.3^9.2^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(=\dfrac{2^{10}.\left(3^8-3^9\right)}{2^{10}.3^8.\left(1+5\right)}=\dfrac{3^8-3^9}{3^8.6}=\dfrac{3^8.\left(1-3\right)}{3^8.6}=\dfrac{-2}{6}=-\dfrac{1}{3}\)
~ Học tốt ~
Bài 1:
1) \(3^2.\dfrac{1}{243}.81^2.\dfrac{1}{3^3}\)
\(=3^2.\left(\dfrac{1}{3}\right)^5.\left(3^4\right)^2.\dfrac{1}{3^3}\)
\(=3^2.\dfrac{1}{3^5}.3^8.\dfrac{1}{3^3}\)
\(=3^2=9\)
2) \(\left(4.2^5\right):\left(2^3.\dfrac{1}{16}\right)\)
\(=\left(2^2.2^5\right):[2^3.\left(\dfrac{1}{2}\right)^4]\)
\(=2^7:2^3:\dfrac{1}{2^4}\)
\(=2^4.2^4=256\)
3)\(\left(2^{-1}+3^{-1}\right)+\left(2^{-1}.2^0\right):2^3\)
\(=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2}.1:2^3\)
\(=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2^4}\)
\(=\dfrac{43}{48}\)
4)\(\left(-\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(=-3-1+\dfrac{1}{4}.\dfrac{1}{2}\)
\(=-3-1+\dfrac{1}{8}\)
\(=-4+\dfrac{1}{8}\\ \)
\(=-\dfrac{31}{8}\)
5)\([\left(0,1\right)^2]^0+[\left(\dfrac{1}{7}\right)^{-1}]^2.\dfrac{1}{49}.[\left(2^2\right)^3:2^5]\\ =1+7^2.\dfrac{1}{7^2}.2^6:2^5\\ =1+1.2\\ =3\)
Chúc bạn học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(\dfrac{1}{2}{x^2}.\dfrac{6}{5}{x^3} = \dfrac{1}{2}.\dfrac{6}{5}.{x^2}.{x^3} = \dfrac{3}{5}{x^5}\);
b)
\(\begin{array}{l}{y^2}(\dfrac{5}{7}{y^3} - 2{y^2} + 0,25) = {y^2}.\dfrac{5}{7}{y^3} - {y^2}.2{y^2} + {y^2}.0,25)\\ = \dfrac{5}{7}{y^5} - 2{y^4} + 0,25{y^2}\end{array}\);
c)
\(\begin{array}{l}(2{x^2} + x + 4)({x^2} - x - 1) \\= 2{x^2}({x^2} - x - 1) + x({x^2} - x - 1) + 4({x^2} - x - 1)\\ = 2{x^4} - 2{x^3} - 2{x^2} + {x^3} - {x^2} - x + 4{x^2} - 4x - 4 \\= 2{x^4} - {x^3} + {x^2} - 5x - 4\end{array}\);
d)
\(\begin{array}{l}(3x - 4)(2x + 1) - (x - 2)(6x + 3) \\= 3x(2x + 1) - 4(2x + 1) - x(6x + 3) + 2(6x + 3)\\ = 6{x^2} + 3x - 8x - 4 - 6{x^2} - 3x + 12x + 6\\ = 4x + 2\end{array}\).
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a) \(5^6:5^5+\left(\dfrac{4}{9}\right)^0=5^{6-5}+1=5+1=6\)
b) \(\left(\dfrac{3}{7}\right)^{21}:\left(1-\dfrac{40}{49}\right)^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left[\left(\dfrac{3}{7}\right)^2\right]^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^6\)
\(=\left(\dfrac{3}{7}\right)^{21-6}=\left(\dfrac{3}{7}\right)^{15}\)
c) \(\left(\dfrac{2}{3}\right)^3-\left(\dfrac{-52}{3}\right)^0+\dfrac{4}{9}\)
\(=\dfrac{8}{27}-1+\dfrac{4}{9}\)
\(=\dfrac{8-27+12}{27}=-\dfrac{7}{27}\)
\(a)5^6:5^5+\left(\dfrac{4}{9}\right)^0=5^1+1=6\)
\(b,\left(\dfrac{3}{7}\right)^{21}:\left(1-\dfrac{40}{49}\right)^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{49-40}{49}\right)^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^3=\left(\dfrac{3}{7}\right)^{21}:[\left(\dfrac{3}{7}\right)^2]^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^6=\left(\dfrac{3}{7}\right)^{21-6}\)
\(=\left(\dfrac{3}{7}\right)^{15}\)
\(c,3.\left(\dfrac{2}{3}\right)^3-\left(\dfrac{-52}{3}\right)^0+\dfrac{4}{9}\)
\(=3.\dfrac{8}{27}-1+\dfrac{4}{9}\)
\(=\dfrac{8}{9}-1+\dfrac{4}{9}\)
\(=\dfrac{8-9+4}{9}=\dfrac{1}{3}\)
\(a,=\dfrac{4}{33}+\dfrac{1}{3}=\dfrac{5}{11}\\ b,=\dfrac{1}{6}-\dfrac{5}{3}+\dfrac{1}{2}=-1\)