Tính một cách hợp lí:
\(\dfrac{-5}{6}.\)\(\dfrac{\text{7}}{\text{1}\text{1}}\)+\(\dfrac{-5}{\text{1}\text{1}}\).\(\text{ }\dfrac{\text{4}}{\text{6}}\)+\(\dfrac{\text{5}}{\text{6}}\)
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\(\dfrac{38}{11}+\left(\dfrac{16}{13}+\dfrac{6}{11}\right)\\ =\dfrac{38}{11}+\dfrac{16}{13}+\dfrac{6}{11}\\ =\left(\dfrac{38}{11}+\dfrac{6}{11}\right)+\dfrac{16}{13}\\ =\dfrac{44}{11}+\dfrac{16}{13}\\ =4+\dfrac{16}{13}\\ =\dfrac{52}{13}+\dfrac{16}{13}\\ =\dfrac{68}{13}\\ \dfrac{3}{4}:\dfrac{3}{5}-\dfrac{1}{5}\\ =\dfrac{3}{4}\times\dfrac{5}{3}-\dfrac{1}{5}\\ =\dfrac{5}{4}-\dfrac{1}{5}\\ =\dfrac{25-4}{20}\\ =\dfrac{21}{20}\)
Phép tính 1:
\(\dfrac{38}{11}+\left(\dfrac{16}{13}+\dfrac{6}{11}\right)\)
\(=\left(\dfrac{38}{11}+\dfrac{6}{11}\right)+\dfrac{16}{13}\)
\(=4+\dfrac{16}{13}=\dfrac{4\cdot13+16}{13}\)(Dấu "." trong phép tính là dấu nhân)
\(=\dfrac{68}{13}\)
Phép tính 2:
\(\dfrac{3}{4}:\dfrac{3}{5}-\dfrac{1}{5}\)
\(=\dfrac{3}{4}\cdot\dfrac{5}{3}-\dfrac{1}{5}\)
\(=\dfrac{\left(3\cdot5\right):3\cdot5}{\left(4\cdot3\right):3\cdot5}-\dfrac{1\cdot4}{5\cdot4}\)
\(=\dfrac{25}{20}-\dfrac{4}{20}\)
\(=\dfrac{21}{20}\)
b)
\(=\frac{3}{7}\left(\frac{9}{11}+\frac{2}{11}\right)\)
\(=\frac{3}{7}.1=\frac{3}{7}\)
bài 1.tính bằng cách hợp lí nhất :
a) 1/7 + 2/7 + 3/7 + ..... 3 5/7 + 3 6/7
b) 3/5x 3 5/9 + 6/5 x 11/9
a)=18.17-18.7
=18.(17-7)=18.10=180
b)=54-(6.17+6.9)=54-102-54=(54-54)-102=0-102=-102
c)=(33.17-33.5)-(17.33-17.5)=33.17-33.5-17.33+17.5=(33.17-17.33)-5.(33-17)=0-80=-80
1, \(xy^3-x^3y=xy\left(y^2-x^2\right)=xy\left(y-x\right)\left(x+y\right)\)
2, \(5x\left(3y+4x-6\right)\)
3, \(3x\left(2-y\right)\)
4, \(x\left(x^2+2x+1\right)=x\left(x+1\right)^2\)
5, \(x\left(4x^2-12x+9\right)=x\left(2x-3\right)^2\)
6, \(2xy\left(x+2y-5x^2y\right)\)
7, \(x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)
11, \(\left(x+y\right)\left(x-1\right)\)
\(1,xy^3-x^3y=xy\left(y^2-x^2\right)=xy\left(y-x\right)\left(y+x\right)\\ 2,15xy+20x^2-30x=5x\left(3y+4x-6\right)\\ 3,6x-3xy=3x\left(2-y\right)\\ 4,x^3+2x^2+x=x\left(x^2+2x+1\right)=x\left(x+1\right)^2\\ 5,4x^3-12x^2+9x=x\left(4x^2-12x+9\right)=x\left(2x-3\right)^2\\ 6,2x^2y+4xy^2-10x^3y^2=2xy\left(x+2y-5x^2y\right)\\ 11,x\left(x-1\right)-y\left(1-x\right)=x\left(x-1\right)+y\left(x-1\right)=\left(x-1\right)\left(x+y\right)\)
A=(6 : 35−116 x 67) : (415 x 1011+5211)A=(6 : 35−116 x 67) : (415 x 1011+5211)
A=(6 : 35−76 x 67) : (215 x 1011+5711)A=(6 : 35−76 x 67) : (215 x 1011+5711)
A=(10−1) : (4211+5711)A=(10−1) : (4211+5711)
A=9 : 9A=9 : 9
A=1
- \(\dfrac{5}{6}\).\(\dfrac{7}{11}\) + \(\dfrac{-5}{11}\).\(\dfrac{4}{6}\) + \(\dfrac{5}{6}\)
= - \(\dfrac{5}{6}\).\(\dfrac{7}{11}\) \(-\dfrac{5}{6}\).\(\dfrac{4}{11}\) + \(\dfrac{5}{6}\) x 1
= - \(\dfrac{5}{6}\).(\(\dfrac{7}{11}\) + \(\dfrac{4}{11}\) - 1)
= - \(\dfrac{5}{6}\).(1 - 1)
= - \(\dfrac{5}{6}\).0
= 0