- (3)^2-(-2^3)^2-(-5^2)^2
- 2^3+3×(-1/2)^0-(1/2)^3×4+[(-2)^2÷1/2]÷8
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2:
a: =4+3/8+5+2/3
=9+3/8+2/3
=216/24+9/24+16/24
=216/24+25/24
=241/24
b; =2+3/8+1+1/4+3+6/7
=6+3/8+1/4+6/7
=6+5/8+6/7
=419/56
c: \(=2+\dfrac{3}{8}-1-\dfrac{1}{4}+5+\dfrac{1}{3}\)
=6+3/8-1/4+1/3
=6+1/8+1/3
=6+11/24
=155/24
d: \(=3+\dfrac{5}{6}+6\cdot\dfrac{13}{6}\)
=3+13+5/6
=16+5/6
=101/6
e: =3+1/2+4+5/7-5-5/14
=3+4-5+1/2+5/7-5/14
=2+7/14+10/14-5/14
=2+12/14
=2+6/7=20/7
f: =9/2+1/2:11/2
=9/2+1/11
=99/22+2/22=101/22
![](https://rs.olm.vn/images/avt/0.png?1311)
2:
a: =4+3/8+5+2/3
=9+3/8+2/3
=216/24+9/24+16/24
=216/24+25/24
=241/24
b; =2+3/8+1+1/4+3+6/7
=6+3/8+1/4+6/7
=6+5/8+6/7
=419/56
c: \(=2+\dfrac{3}{8}-1-\dfrac{1}{4}+5+\dfrac{1}{3}\)
=6+3/8-1/4+1/3
=6+1/8+1/3
=6+11/24
=155/24
d: \(=3+\dfrac{5}{6}+6\cdot\dfrac{13}{6}\)
=3+13+5/6
=16+5/6
=101/6
e: =3+1/2+4+5/7-5-5/14
=3+4-5+1/2+5/7-5/14
=2+7/14+10/14-5/14
=2+12/14
=2+6/7=20/7
f: =9/2+1/2:11/2
=9/2+1/11
=99/22+2/22=101/22
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(1\right)\dfrac{-7}{12}.\dfrac{11}{8}-\dfrac{37}{8}.\dfrac{7}{12}+\dfrac{1}{2}=-\dfrac{7}{12}.\left(\dfrac{11}{8}+\dfrac{37}{8}\right)+\dfrac{1}{2}=-\dfrac{7}{12}.6+\dfrac{1}{2}=-3.\)
\(\left(2\right)\left(\dfrac{2}{3}-\dfrac{1}{4}-\dfrac{5}{6}\right).\left(-2\right)^2+\dfrac{3}{2}:\dfrac{-15}{4}=\dfrac{-5}{12}.4-\dfrac{2}{5}=\dfrac{-5}{3}-\dfrac{2}{5}=\dfrac{-31}{15}.\)
\(\left(3\right)\dfrac{-2}{5}+\dfrac{3}{10}-\dfrac{3}{5}+\dfrac{7}{10}-\dfrac{3}{2}=1-1-\dfrac{3}{2}=-\dfrac{3}{2}.\)
1. \(\dfrac{-7}{12}.\dfrac{11}{8}-\dfrac{37}{8}.\dfrac{7}{12}+\dfrac{1}{2}=\dfrac{-7}{12}\left(\dfrac{11}{8}+\dfrac{37}{8}\right)+\dfrac{1}{2}=\dfrac{-7}{12}.\dfrac{6}{1}+\dfrac{1}{2}=\dfrac{-7}{2}+\dfrac{1}{2}=\dfrac{-6}{2}=-3\)2.
\(\left(\dfrac{2}{3}-\dfrac{1}{4}-\dfrac{5}{6}\right).\left(-2\right)^2+\dfrac{3}{2}:\dfrac{-15}{4}=\dfrac{-5}{12}.4+\dfrac{-2}{5}=\dfrac{-5}{3}+\dfrac{-2}{5}=\dfrac{-31}{15}\)
3.
\(\dfrac{-2}{5}+\dfrac{3}{10}-\dfrac{3}{5}+\dfrac{7}{10}-\dfrac{3}{2}=\dfrac{-4}{10}+\dfrac{3}{10}-\dfrac{6}{10}+\dfrac{7}{10}-\dfrac{15}{10}=\dfrac{-15}{10}=\dfrac{-3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{-7}{12}.\dfrac{11}{8}-\dfrac{37}{8}.\dfrac{7}{12}+\dfrac{1}{2}=\dfrac{-7}{12}.\left(\dfrac{11}{8}+\dfrac{37}{8}\right)+\dfrac{1}{2}=-\dfrac{7}{12}.6+\dfrac{1}{2}=-\dfrac{7}{2}+\dfrac{1}{2}=-3.\)
\(\left(\dfrac{2}{3}-\dfrac{1}{4}-\dfrac{5}{6}\right).\left(-2\right)^2+\dfrac{3}{2}:\dfrac{-15}{4}=\dfrac{8-3-10}{12}.4+\dfrac{-2}{5}=\dfrac{-5}{12}.4-\dfrac{2}{5}=\dfrac{-5}{3}-\dfrac{2}{5}=-\dfrac{31}{15}.\)
\(\dfrac{-2}{5}+\dfrac{3}{10}-\dfrac{3}{5}+\dfrac{7}{10}-\dfrac{3}{2}=\left(\dfrac{3}{10}+\dfrac{7}{10}\right)+\left(\dfrac{-2}{5}-\dfrac{3}{5}\right)-\dfrac{3}{2}=1-1-\dfrac{3}{2}=\dfrac{-3}{2}.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1: \(=\dfrac{7}{12}\left(-\dfrac{11}{8}+\dfrac{37}{8}\right)+\dfrac{1}{2}=\dfrac{7}{12}\cdot\dfrac{26}{8}+\dfrac{1}{2}=\dfrac{115}{48}\)
2: \(=\dfrac{8-3-10}{12}\cdot4+\dfrac{3}{2}\cdot\dfrac{-4}{15}\)
\(=\dfrac{-5}{3}+\dfrac{-12}{30}=\dfrac{-5}{3}+\dfrac{-2}{5}=\dfrac{-25-6}{15}=-\dfrac{31}{15}\)
3: \(=\dfrac{-2}{5}-\dfrac{3}{5}+\dfrac{3}{10}+\dfrac{7}{10}-\dfrac{3}{2}=-\dfrac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\frac{22}{5}+\frac{1}{2}\cdot x^2=4\cdot\frac{8}{5}\)
=> \(\frac{22}{5}+\frac{1}{2}\cdot x^2=\frac{32}{5}\)
=> \(\frac{1}{2}\cdot x^2=\frac{32}{5}-\frac{22}{5}\)
=> \(\frac{1}{2}\cdot x^2=2\)
=> \(x^2=2:\frac{1}{2}=4\)
=> x = 2 hoặc x = -2
\(b,\frac{7}{2}-\left|x-\frac{1}{3}\right|=\frac{5}{2}\)
=> \(\left|x-\frac{1}{3}\right|=\frac{7}{2}-\frac{5}{2}\)
=> \(\left|x-\frac{1}{3}\right|=1\)
=> \(x-\frac{1}{3}=1\)hoặc \(x-\frac{1}{3}=-1\)
=> x = 1 + 1/3 hoặc x = -1 + 1/3
=> x = 4/3 hoặc x = -2/3
c, \(\left[x-\frac{1}{2}\right]\left[-3-\frac{x}{2}\right]=0\)
=> x - 1/2 = 0 hoặc -3 - x/2 = 0
=> x = 0 + 1/2 hoặc x/2 = -3
=> x = 1/2 hoặc x = -6
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm
a) x² - 3 = 22
=> x² = 25
=> x = + 5
Vậy x = + 5
b) 2x³ + 5 = -11
2x³ = -16
x³ = -8
x = -2
Vậy x = -2
c) ( x + 2 )² = 81
=> x + 2 = 9
=> x = 7
Vậy x = 7
d) ( 2x + 1 )² = 25
=> 2x + 1 = 5
=> 2x = 4
=> x = 2
Vậy x = 2
e) 5x + 2 = 625
5x = 623 ( vô lí )
g) ( 2x - 3 )² = 36.
=> 2x - 3 = 6
=> 2x = 9
=> x = 4,5
Vậy x = 4,5
h) ( 2x - 1 )³ = -8
=> 2x - 1 = -2
=> 2x = -1
=> x = -1/2
Vậy x = -1/2
i) ( x - 1 )x + 2 = ( x - 1 )x + 6
=> [ (x - 1 )x - ( x - 1 )x ] = 6 - 2
=> 0 = 4 ( vô lí )
Vậy x thuộc rỗng.
k) x² + x = 0
=> x( x + 1 ) = 0
=> x = 0 hoặc x + 1 = 0
=> x = 0 hoặc x = -1
Vậy x = 0 hoặc x = -1
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=\left\{1;2;3;4;5\right\}\)
\(\Rightarrow A=\left\{x\inℕ|1\le x\le5\right\}\)
b) \(B=\left\{0;1;2;3;4\right\}\)
\(\Rightarrow B=\left\{x\inℕ|0\le x\le4\right\}\)
c) \(C=\left\{1;2;3;4\right\}\)
\(\Rightarrow C=\left\{x\inℕ|1\le x\le4\right\}\)
d) \(D=\left\{0;2;4;6;8\right\}\)
\(\Rightarrow D=\left\{x\inℕ|x=2k;0\le k\le4;k\inℕ\right\}\)
e) \(E=\left\{1;3;5;7;9;...49\right\}\)
\(\Rightarrow E=\left\{x\inℕ|x=2k+1;0\le k\le24;k\inℕ\right\}\)
f) \(F=\left\{11;22;33;44;...99\right\}\)
\(\Rightarrow F=\left\{x\inℕ|x=11k;1\le k\le9;k\inℕ\right\}\)