5,4999055

5,4999055

\(\frac{428517}{571428}+\frac{30.63.65.8}{117.200.49}+\frac{-525}{100}+\frac{36992}{4046}\)

\(=\)\(\frac{428517}{571428}+\frac{6}{7}+\frac{-21}{4}+\frac{64}{7}\)

Quy đồng mẫu số đc kq cuổi cùng là : 5,4999055

Lời giải:

Gọi giá trị trên là $A$

$A=\frac{4}{15}+\frac{6}{7}+4+3+\frac{1}{7}$
$=\frac{4}{15}+(4+3)+(\frac{6}{7}+\frac{1}{7})$

$=\frac{4}{15}+7+1=8\frac{4}{15}$

 \(A=\frac{8056}{2012.16-1982}\)= \(\frac{2014.4}{2012.15+2012-1982}\)=\(\frac{2014.4}{2012.15+30}\)=\(\frac{2014.4}{2012.15+2.15}\)=\(\frac{2014.4}{15.\left(2012+2\right)}=\frac{2014.4}{15.2014}=\frac{4}{15}\)

B = \(\frac{1.2.6+2.4.12+4.8.24+7.14.42}{1.6.9+2.12.18+4.24.36+7.42.63}\)

= \(\frac{1.2.3.2+2.2.2.12+4.4.2.24+7.7.2.42}{1.2.3.9+2.12.2.9+4.24.4.9+7.42.7.9}\)

= \(\frac{2\left(1.2.3+2.2.12+4.4.24+7.7.42\right)}{9\left(1.2.3+2.2.12+4.4.24+7.7.42\right)}\)

= \(\frac{2}{9}\)

Ta có: \(\frac{4}{15}=\frac{4.3}{15.3}=\frac{12}{45};\frac{2}{9}=\frac{2.5}{9.5}=\frac{10}{45}\)

Vì \(\frac{12}{45}>\frac{10}{45}\Rightarrow\frac{4}{15}>\frac{2}{9}\Rightarrow A>B\)

Vậy A > B

\(A=\dfrac{8056}{2012.16-1982}\)

\(A=\dfrac{8056}{32192-1982}\)

\(A=\dfrac{8056}{30210}=\dfrac{12}{45}\)

\(B=\dfrac{1.2.6+2.4.12+4.8.24+7.14.42}{1.6.9+2.12.18+4.24.36+7.42.63}\)

\(B=\dfrac{12+96+768+4116}{54+432+3456+18522}\)

\(B=\dfrac{4992}{22464}=\dfrac{10}{45}\)
Vậy: \(\dfrac{12}{45}>\dfrac{10}{45}\Rightarrow A>B\)

\(A=\frac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)

\(A=\frac{1.2.\left(1+2^2+3^2+4^2+5^2\right)}{3.4.\left(1+2^2+3^2+4^2+5^2\right)}\)

\(A=\frac{1.2}{3.4}\)

\(A=\frac{1}{6}\)

Ta thấy : \(B=\frac{111111}{666665}>\frac{111111}{666666}=\frac{1}{6}\)

Vậy B > A

Theo đề bài, ta có:

\(A=\frac{1\times2+2\times4+3\times6+4\times8+5\times10}{3\times4+6\times8+9\times12+12\times16+15\times20}\)

\(A=\frac{1\times2\times\left(1+2^2+3^2+4^2+5^2\right)}{3\times4\times\left(1+2^2+3^2+4^2+5^2\right)}\)

\(A=\frac{1\times2}{3\times4}\)

\(A=\frac{1}{6}\)

Ta thấy rằng: \(B=\frac{111111}{666665}>\frac{111111}{666666}=\frac{1}{6}\)

Vậy \(B>A\)

a, =2.0,1-0,5                                  b,=\(\left(15\dfrac{1}{4}-25\dfrac{1}{4}\right):\left(-\dfrac{5}{7}\right)\)

    =0,2-0,5                                        =  -10:(\(\dfrac{-5}{7}\))

    =-0,3                                             = 14

           c, \(=\dfrac{5^4.\left(4.5\right)^4}{\left(5^2\right)^4.4^5}=\dfrac{5^4.4^4.5^4}{5^8.4^5}=\dfrac{1}{4^{ }}\)

\(\dfrac{1\cdot2\cdot3+2\cdot4\cdot6+4\cdot8\cdot12}{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20}\\ =\dfrac{1\cdot2\cdot3+2\cdot1\cdot2\cdot2\cdot2\cdot3+4\cdot1\cdot4\cdot2\cdot4\cdot3}{1\cdot3\cdot5+2\cdot1\cdot2\cdot3\cdot2\cdot5+4\cdot1\cdot4\cdot3\cdot4\cdot5}\\ =\dfrac{1\cdot2\cdot3\cdot\left(1+2^3+4^3\right)}{1\cdot3\cdot5\cdot\left(1+2^3+4^3\right)}\\ =\dfrac{1\cdot2\cdot3}{1\cdot3\cdot5}\\ =\dfrac{6}{15}\)

Bạn ghi đề sai rồi, mình sửa lại đề ở phần (*) rồi nhé!

Ta có: \(\dfrac{1.2.3+2.4.6+4.8.12}{1.3.5+2.6.10+4.12.20}\) (*)

= \(\dfrac{1.2.3\left(1+2^3+4^3\right)}{1.3.5\left(1+2^3+4^3\right)}\) = \(\dfrac{1.2.3}{1.3.5}\) = \(\dfrac{2}{5}\)