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a: \(=\dfrac{5}{9}-\dfrac{12+5}{15}:\dfrac{6}{5}\)
\(=\dfrac{5}{9}-\dfrac{17}{15}\cdot\dfrac{5}{6}=\dfrac{5}{9}-\dfrac{17}{18}=-\dfrac{7}{18}\)
b: \(=\left[\dfrac{6}{5}\left(3+\dfrac{3}{8}-6-\dfrac{5}{8}\right)\right]\cdot\dfrac{5}{6}=-3-\dfrac{1}{4}=-\dfrac{13}{4}\)
a)-1-2-3-4-5-6-....-80
=(-1)+(-2)+(-3)+(-4)+(-5)+(-6)+...+(-80)
Khoảng cách giữa các số:(-1)-(-2)=1
Tổng các số hạng:(-1)-(-80)+1=80 số
Tổng:[(-1)+(-80)].80:2= -3240
=>-1-2-3-4-5-6+......-80=-3240
b,1-2+3-4+5-6+......+2021-2022
=(1-2)+(3-4)+(5-6)+...+(2021-2022)
=(-1)+(-1)+(-1)+...+(-1)
Tổng số cặp là:
(2022-1+1):2=1011 cặp
-1.1011=-1011
=>1-2+3-4+5-6+......+2021-2022= -1011
c, Đề bài sai
d,-4-8-12-16-.......-2020
=-4+(-8)+(-12)+(-16)+...+(-2020)
Khoảng cách giữa các số:-4-(-8)=4
Tổng các số hạng:[-4-(-2020]:4+1=505 số
Tổng:[-4+(-2020)].505:2=-511060
=>-4-8-12-16-.......-2020=-511060
`@` `\text {Ans}`
`\downarrow`
`a.`
\(0,3-\dfrac{4}{9}\div\dfrac{4}{3}\cdot\dfrac{6}{5}+1\)
`=`\(0,3-\dfrac{1}{3}\cdot\dfrac{6}{5}+1\)
`=`\(0,3-0,4+1\)
`= -0,1 + 1`
`= 0,9`
`b.`
\(1+2\div\left(\dfrac{2}{3}-\dfrac{1}{6}\right)\cdot\left(-2,25\right)\)
`=`\(1+2\div\dfrac{1}{2}\cdot\left(-2,25\right)\)
`=`\(1+4\cdot\left(-2,25\right)\)
`= 1+ (-9) = -8`
`c.`
\(\left[\left(\dfrac{1}{4}-0,5\right)\cdot2+\dfrac{8}{3}\right]\div2\)
`=`\(\left(-\dfrac{1}{4}\cdot2+\dfrac{8}{3}\right)\div2\)
`=`\(\left(-\dfrac{1}{2}+\dfrac{8}{3}\right)\div2\)
`=`\(\dfrac{13}{6}\div2\)
`=`\(\dfrac{13}{12}\)
`d.`
\(\left[\left(\dfrac{3}{8}-\dfrac{5}{12}\right)\cdot6+\dfrac{1}{3}\right]\cdot4\)
`=`\(\left(-\dfrac{1}{24}\cdot6+\dfrac{1}{3}\right)\cdot4\)
`=`\(\left(-\dfrac{1}{4}+\dfrac{1}{3}\right)\cdot4\)
`=`\(\dfrac{1}{12}\cdot4=\dfrac{1}{3}\)
`e.`
\(\left(\dfrac{4}{5}-1\right)\div\dfrac{3}{5}-\dfrac{2}{3}\cdot0,5\)
`=`\(-\dfrac{1}{5}\div\dfrac{3}{5}-\dfrac{1}{3}\)
`=`\(-\dfrac{1}{3}-\dfrac{1}{3}=-\dfrac{2}{3}\)
`f.`
\(0,8\div\left\{0,2-7\left[\dfrac{1}{6}+\left(\dfrac{5}{21}-\dfrac{5}{14}\right)\right]\right\}\)
`=`\(0,8\div\left[0,2-7\left(\dfrac{1}{6}-\dfrac{5}{42}\right)\right]\)
`=`\(0,8\div\left(0,2-7\cdot\dfrac{1}{21}\right)\)
`=`\(0,8\div\left(0,2-\dfrac{1}{3}\right)\)
`= 0,8 \div (-2/15)`
`=-6`
`@` `yHGiangg.`
a) Ta có: \(A=\dfrac{2}{3}-\left(-\dfrac{5}{7}+\dfrac{2}{3}\right)\)
\(=\dfrac{2}{3}+\dfrac{5}{7}-\dfrac{2}{3}\)
\(=\dfrac{5}{7}\)
a)
\(M=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{4+4\sqrt{5}+5}-\sqrt{4-4\sqrt{5}+5}\)
\(=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(=\left|2+\sqrt{5}\right|-\left|2-\sqrt{5}\right|\)
\(=2+\sqrt{5}-\left(\sqrt{5}-2\right)\) (vì \(2+2\sqrt{5}>0;2-\sqrt{5}< 0\) )
\(=2+\sqrt{5}-\sqrt{5}+2\\ =4\)
b)
\(N=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\left|\sqrt{7}-1\right|-\left|\sqrt{7}+1\right|\)
\(=\sqrt{7}-1-\left(\sqrt{7}+1\right)\) (vì \(\sqrt{7}-1>0;\sqrt{7}+1>0\) )
\(=\sqrt{7}-1-\sqrt{7}-1\\ =-2\)
1) 5 + (-4) = 1
2) (-8) + 2 = -6
3) 8 + (-2) = 6
4) 11 + (-3) = 8
5) (-11) + 2 = -9
6) (-7) + 3 = -4
7) (-5) + 5 = 0
8) 11 + (-12) = -1
9) (-18) + 20 = 2
10) (15) + (-12) = 3
11) (-17) + 17 = 0
12) 16 + (-2) = 14
13) (30) + (-14) = 16
14) (-19) + 20 = 1
15) (-18) + 15 = -3
16) (10) + (-6) = 4
17) (-28) + 14 = -14
18) 15 + (-30) = -15
19) (15) + (-4) = 11
20) (-21) + 11 = -10
21) 8 + (-22) = -14
22) (-15) + 4 = -11
23) (-3) + 2 = -1
24) 17 + (-14) = 3
25) 17 + (-14) = 3
\(a,\left(x-2\right)\left(x+3\right)-x\left(x-5\right)=x^2-2x+3x-6-x^2+5x=6x-6\)
\(b,\dfrac{1}{x-2}+\dfrac{-2}{x+2}+\dfrac{2x-8}{x^2-4}=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{2x-8}{\left(x+2\right)\left(x-2\right)}=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-4}{\left(x+2\right)\left(x-2\right)}+\dfrac{2x-8}{\left(x+2\right)\left(x-2\right)}=\dfrac{x+2-2x+4+2x-8}{\left(x+2\right)\left(x-2\right)}=\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{1}{x+2}\)
a: \(145+43+\left(-145\right)+57\)
\(=\left(145-145\right)+\left(43+57\right)\)
=0+100
=100
b: \(\left(-123\right)+101+\left(-777\right)+99\)
\(=\left(-123-777\right)+\left(101+99\right)\)
=-900+200
=-700
c: \(\left(-8955\right)+33+\left(-45\right)+\left(-133\right)\)
\(=\left(-8955-45\right)+\left(33-133\right)\)
\(=\left(-9000\right)+\left(-100\right)=-9100\)
d: \(\left(-2\right)+\left(-4\right)+\left(-6\right)+\left(-8\right)+\left(-10\right)+8+10+12\)
\(=\left(12-2\right)+\left(-4-6\right)+\left(-8+8\right)+\left(-10+10\right)\)
=10-10+0+0
=0
a: Ta có: \(\left(8\cdot5^7+5^6-5^5\right):5^5\)
\(=8\cdot5^2+5-1\)
\(=200+4=204\)
b: Ta có: \(\left(9^{30}-27^{19}\right):3^{57}+\left(125^9-25^{12}\right):5^{24}\)
\(=3^{60}:3^{57}-3^{57}:3^{57}+5^{27}:5^{24}-5^{24}:5^{24}\)
\(=27-1+125-1\)
=150
a. (8,57 - 55 + 56) : 55
= (8,57 : 55) - (55 : 55) + (56 : 55)
= 1,72 - 1 + 5
= 2,89 - 1 + 5
= 6,89
b. (930 - 2719) : 357 + (1259 - 2512) : 524
= (930 : 357) - (2719 : 357) + (1259 : 524) - (2512 : 524)
= 33 - 1 + 125 - 1
= 27 - 1 + 125 - 1
= 150
c. (1012 + 511 . 29 - 513 - 28) : 4 . 55 . 106
= (1012 + 2,5 , 1010 - 513 - 28) : 1,25 . 1010
= (1012 : 1,25 . 1010) + (2,5 . 1010 : 1,25 . 1010) - (513 : 1,25 . 1010) - (28 : 1,25 . 1010)
= 80 + 2 - \(\dfrac{25}{256}\) - \(\dfrac{1}{48828125}\)
= 81,90234373 \(\approx\) 82
a) \(\left( { - 3} \right).\left( { - 2} \right).\left( { - 5} \right).4\)\( = \left[ {\left( { - 3} \right).\left( { - 2} \right)} \right].\left( { - 5} \right).4\)\( = 6.\left( { - 5} \right).4 = - 30.4 = - 120\).
b) \(3.2.\left( { - 8} \right).\left( { - 5} \right)\)\( = 3.2.\left[ {\left( { - 8} \right).\left( { - 5} \right)} \right] = 6.40\)\( = 240\).
