A (-6.37*0.4)*2.5
B (-0.125*-5.3)*8
C -2.5*(-4)*(-7.9)
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a) \(-\left(315.4+275\right)+4.315-\left(10-275\right)\)
\(=-315.4-275+4.315-10+275\)
\(=315.\left(-4+4\right)+\left(-275+275\right)-10\)
\(=315.0+0-10=0-10=-10\)
b) \(\left(-2,5.0,375.0,4\right)-\left[0,125.3,25.\left(-8\right)\right]\)
\(=\left[\left(-2,5.0,4\right).0,375\right]-\left[\left(-8.0,125\right).3,25\right]\)
\(=\left[-1.0,375\right]-\left[-1.3,25\right]\)
\(=\left[-1.\left(0,375-3,25\right)\right]=-1.\left(-2,875\right)=2,875\)
Cậu có thể tham khảo bài làm trên đây ạ, chúc cậu học tốt ^^
Mk sửa đề nha :
\(C=\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+\frac{12}{7}\)
\(=\frac{-5}{7}.\left(\frac{2}{11}+\frac{9}{11}\right)+\frac{12}{7}\)
\(=\frac{-5}{7}.1+\frac{12}{7}\)
\(=\frac{-5}{7}+\frac{12}{7}\)
\(=1\)
Study well ! >_<
a: =>x+8x+4x=335*4
=>13x=335*4
=>x=1340/13
b: =>10x=2,5
=>x=0,25
=(0,25 nhân 0,4).(0,05.20).(0,125.8).8,76 =1.1.1.8,76 =8,76 cho mình 1 đúng nha
a, 2 x 4 x 8 x 0,5 x 0,125 x 0,25 x ( 0,4321 + 0,5679 )
= ( 2 x 0,5 ) x ( 4 x 0,25 ) x ( 0,125 x 8 ) x ( 0,4321 + 0,5679 )
= 1 x 1 x 1 x ( 0,4321 + 0,5679 )
= 1 x ( 0,4321 + 0,5679 )
= 1 x 1 = 1
b, 3,7 x 3,8 + 8,3 x 3,7 - 0,4 x 3,7 x 5
= 3,7 x ( 3,8 + 8,3 ) - ( 0,4 x 5 ) x 3,7
= 3,7 x 12,1 - 2 x 3,7
= 3,7 x ( 12,1 - 2 )
= 3,7 x 10,1 = 37,37
Ta có: \(A=\frac{3^2}{2\cdot5}+\frac{3^2}{5\cdot8}+\frac{3^2}{8\cdot11}\)
\(=3\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}\right)\)
\(=3\left(\frac12-\frac15+\frac15-\frac18+\frac18-\frac{1}{11}\right)\)
\(=3\left(\frac12-\frac{1}{11}\right)=3\cdot\frac{9}{22}=\frac{27}{22}\) >1
Ta có: \(B=\frac{4}{5\cdot7}+\frac{4}{7\cdot9}+\cdots+\frac{4}{59\cdot61}\)
\(=2\left(\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\cdots+\frac{2}{59\cdot61}\right)\)
\(=2\left(\frac15-\frac17+\frac17-\frac19+\cdots+\frac{1}{59}-\frac{1}{61}\right)\)
\(=2\left(\frac15-\frac{1}{61}\right)=2\cdot\frac{61-5}{305}=2\cdot\frac{56}{305}=\frac{112}{305}<1\)
Ta có: A>1
B<1
Do đó: A>B
Ta có: \(A=\frac{3^2}{2\cdot5}+\frac{3^2}{5\cdot8}+\frac{3^2}{8\cdot11}\)
\(=3\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}\right)\)
\(=3\left(\frac12-\frac15+\frac15-\frac18+\frac18-\frac{1}{11}\right)=3\left(\frac12-\frac{1}{11}\right)=3\cdot\frac{9}{22}=\frac{27}{22}>1\)
TA có: \(B=\frac{4}{5\cdot7}+\frac{4}{7\cdot9}+\cdots+\frac{4}{59\cdot61}\)
\(=2\left(\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\cdots+\frac{2}{59\cdot61}\right)\)
\(=2\left(\frac15-\frac17+\frac17-\frac19+\cdots+\frac{1}{59}-\frac{1}{61}\right)\)
\(=2\left(\frac15-\frac{1}{61}\right)=2\cdot\frac{56}{305}=\frac{112}{305}<1\)
Ta có: B<1
1<A
Do đó: B<A
a; (-2,5.0,38.0,4) - {0,125.3,15.(-8)}
= (-2.5.0,4).0,38 - {-8.0,125.3,15}
= - 1.0,38 - {-1.3,15}
= -0,38 + 3,15
= 2,77
b; {(-20,83).0,2 + (-9,17).0,2 : {2,47.0,5 - (-3,53.0,5}
= {0,2.(-20,83 - 9,17)}: {0,5.(2,47 + 3,53)}
= {- 0,2.30}:{0,5.6}
= - 6:6 : 0,5
= -1 : 0,5
= - 2
A (-6.37*0.4)*2.5=-6.37*(0.4*2.5)=-6.37*1=-6.37
B (-0.125*-5.3)*8=(-0.125*8)*(-5.3)=-1*(-5.3)=5.3
C -2.5*(-4)*(-7.9)=10*(-7.9)=-79