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1) \(\left(+15\right)+\left(+17\right)=15+17=32\)
2) \(\left(-3\right)+\left(-7\right)=-3-7=-\left(3+7\right)=-10\)
3) \(\left(-25\right)+\left(+4\right)=-25+4=-\left(25-4\right)=-21\)
4) \(\left(-6\right)+\left(-54\right)=-6-54=-\left(6+54\right)=-60\)
5) \(\left(-15\right)+20=20-15=5\)
6) \(\left(-5\right)+8+7+5\)
\(=\left(-5+5\right)+\left(8+7\right)\)
\(=15\)
7) \(\left(-8\right)+\left(-11\right)+\left(-2\right)\)
\(=\left[\left(-8\right)+\left(-2\right)\right]+\left(-11\right)\)
\(=\left(-10\right)+\left(-11\right)\)
\(=-21\)
8) \(15+\left(-5\right)+\left(-14\right)+\left(-16\right)\)
\(=\left[15+\left(-5\right)\right]+\left[\left(-14\right)+\left(-16\right)\right]\)
\(=10+\left(-30\right)\)
\(=-20\)
9) \(\left(-20\right)+\left(-14\right)+3+\left(-86\right)\)
\(=\left[\left(-20\right)+3\right]+\left[\left(-14\right)+\left(-86\right)\right]\)
\(=\left(-17\right)+\left(-100\right)\)
\(=-117\)
10) \(\left(-136\right)+123+\left(-264\right)+\left(-83\right)+240\)
\(=\left[\left(-136\right)+\left(-264\right)\right]+\left[123+\left(-83\right)\right]+240\)
\(=\left(-400\right)+40+240\)
\(=\left(-360\right)+240\)
\(=-120\)
11) \(\left(-596\right)+2001+1999+\left(-404+189\right)\)
\(=\left(-596\right)+2001+1999-404+189\)
\(=\left[\left(-596\right)-404\right]+\left(2001+189\right)+1999\)
\(=\left(-1000\right)+2190+1999\)
\(=1190+1999\)
\(=3189\)
12) \(314+\left(-153\right)+64+121+\left(-247\right)+218\)
\(=\left(314+64+121\right)+\left[\left(-153\right)+\left(-247\right)\right]+218\)
\(=\left(378+121\right)+\left(-400\right)+218\)
\(=499-400+218\)
\(=99+218\)
\(=317\)
\(\text{#}Toru\)
a; \(\dfrac{3}{11}\) + \(\dfrac{5}{-9}\) + \(\dfrac{4}{11}\) - \(\dfrac{4}{9}\) + \(\dfrac{3}{17}\) + \(\dfrac{15}{11}\)
= (\(\dfrac{3}{11}\) + \(\dfrac{4}{11}\) + \(\dfrac{15}{11}\)) - (\(\dfrac{5}{9}\) + \(\dfrac{4}{9}\)) + \(\dfrac{3}{17}\)
= 2 - 1 + \(\dfrac{3}{17}\)
= 1 + \(\dfrac{3}{17}\)
= \(\dfrac{20}{17}\)
c; N = \(\dfrac{\dfrac{5}{7}-\dfrac{5}{9}-\dfrac{5}{11}}{\dfrac{15}{7}+\dfrac{15}{9}+\dfrac{15}{11}}\)
Phải là - \(\dfrac{5}{7}\) chỗ tử số mới đúng em nhé!
\(\dfrac{-5}{9}+\dfrac{8}{15}+\dfrac{-2}{11}+\dfrac{4}{-9}+\dfrac{7}{15}\)
=\(\left(\dfrac{-5}{9}+\dfrac{-4}{9}\right)+\left(\dfrac{8}{15}+\dfrac{7}{15}\right)-\dfrac{2}{11}\)
=\(-1+1-\dfrac{2}{11}=\dfrac{-2}{11}\)
\(=\left(\dfrac{-5}{9}+\dfrac{-4}{9}\right)+\left(\dfrac{8}{15}+\dfrac{7}{15}\right)+\dfrac{-2}{11}\\ =-1+1+\dfrac{-2}{11}\\ =0+\dfrac{-2}{11}=\dfrac{-2}{11}\)
2. Tính (+5) + ( +4) = 9
3.Tính (- 9) + ( - 1) = -10
4.Tính : 28 + (-15) = 13
5.Tính (-15) + 12 = - 3
6.Tính (-25) + 25 = 0
7.Tính : 13 - 17 = - 4
8.Tính : (-78) - 9 = - 87
9.Tính : ( -24) - ( - 16) = - 8
10.Tính : 17 - (-3) = 20
11.Tính nhanh : (-21) - ( 48 + 52 - 21) = -100
2, (+5)+(+4)=+9 hoạc 9 cũng được
3, (-9)+(-1)=-10
4, 28+(-15)=13
5, (-15)+12=-3
6, (-25)+25=0
7, 13-17=4
8, (-78)-9=-87
9, (-24)-(-16)=-40
10, 17-(-3)=20
11, (-21+21)-(48-52)=-4
câu a :
\(\dfrac{-8}{24}+\dfrac{-4}{12}=\dfrac{-1}{3}+\dfrac{-1}{3}=\dfrac{-2}{3}\)
câu b :
\(\dfrac{-20}{35}+\dfrac{16}{24}=\dfrac{-4}{7}+\dfrac{2}{3}=\dfrac{2}{21}\)
câu c :
\(\dfrac{-3}{9}+\dfrac{-6}{15}=\dfrac{-1}{3}+\dfrac{-2}{5}=\dfrac{-11}{15}\)
câu d :
\(\dfrac{3}{13}-\dfrac{4}{10}=\dfrac{3}{13}-\dfrac{2}{5}=\dfrac{-11}{65}\)
câu e :
\(\dfrac{5}{17}-\dfrac{9}{15}=\dfrac{5}{17}-\dfrac{3}{5}=\dfrac{-26}{85}\)
câu g :
\(\dfrac{9}{18}-\dfrac{6}{15}+\dfrac{3}{-9}=\dfrac{9}{18}-\dfrac{6}{15}+\dfrac{-3}{9}\\ =\dfrac{1}{2}-\dfrac{2}{5}+\dfrac{-1}{3}=\dfrac{-7}{30}\)
câu h :
\(\dfrac{5}{4}-\dfrac{1}{2}+\dfrac{-7}{8}=\dfrac{10}{8}-\dfrac{4}{8}+\dfrac{-7}{8}=\dfrac{-1}{8}\)
a: =-1+17/20=-3/20
b: =(28/60-33/60)*(-25/3)
=(-1/12)*(-25/3)=1/12*25/3=25/36
c: \(=\dfrac{1}{3}\cdot\dfrac{1}{3}=\dfrac{1}{9}\)
\(\dfrac{1}{15}-\dfrac{2}{15}+\dfrac{3}{15}-\dfrac{4}{15}+\dfrac{5}{15}-\dfrac{6}{15}+\dfrac{7}{15}-\dfrac{8}{15}+\dfrac{9}{15}\)
\(=\dfrac{1-2+3-4+5-6+7-8+9}{15}\)
\(=\dfrac{\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+\left(7-8\right)+9}{15}\) => có 4 cặp
\(=\dfrac{\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)+9}{15}\)
\(=\dfrac{\left(-1\right).4+9}{15}\)
\(=\dfrac{\left(-4\right)+9}{15}\)
\(=\dfrac{5}{15}=\dfrac{1}{3}\)