Phương pháp giải:

- Đặt tính: Viết phép tính sao cho các chữ số cùng hàng thẳng cột với nhau.

- Tính:  Trừ các số lần lượt từ phải sang trái.

Lời giải chi tiết:

= 49629,83799- 19741266= - 19691636,16

D

D

D

=1420

\(142\times4+142\times5+142=142\times\left(4+5+1\right)=142\times10=1420\)

=1420

5,5 + \(\dfrac{3}{4}\) - 5 + \(\dfrac{1}{4}\)

(5,5 - 5) + (\(\dfrac{3}{4}\) + \(\dfrac{1}{4}\))

= 0,5  + 1

= 1,5 

\(\dfrac{5}{2}\) x  \(\dfrac{2}{3}\) + \(\dfrac{1}{4}\) : \(\dfrac{3}{2}\)

= \(\dfrac{5}{2}\) x \(\dfrac{2}{3}\) + \(\dfrac{1}{4}\) x \(\dfrac{2}{3}\)

=  (\(\dfrac{5}{2}\) + \(\dfrac{1}{4}\)) x \(\dfrac{2}{3}\)

= (\(\dfrac{10}{4}\) + \(\dfrac{1}{4}\)) x \(\dfrac{2}{3}\)

= \(\dfrac{11}{4}\) x \(\dfrac{2}{3}\)

= \(\dfrac{11}{6}\) 

Tính \(x\): 

  435  - [\(x\) + 16] = 425 : 17

  435 - [\(x\) + 16] = 25

          [\(x\) + 16] = 435 - 25

           \(x\) + 16 = 410

           \(x\)        = 410 - 16

          \(x\)         = 394

\(x\) x \(\dfrac{1}{4}\) - \(\dfrac{3}{4}\) = 6 : \(\dfrac{3}{4}\)

\(x\) \(\times\) \(\dfrac{1}{4}\) - \(\dfrac{3}{4}\) = 6 x \(\dfrac{4}{3}\)

\(x\) x \(\dfrac{1}{4}\) - \(\dfrac{3}{4}\)  = 8

\(x\) x \(\dfrac{1}{4}\)        = 8 + \(\dfrac{3}{4}\)

\(x\) x \(\dfrac{1}{4}\)        = \(\dfrac{32}{4}\) + \(\dfrac{3}{4}\)

\(x\) x \(\dfrac{1}{4}\)       = \(\dfrac{35}{4}\)

\(x\)              =  \(\dfrac{35}{4}\) : \(\dfrac{1}{4}\)

\(x\)              = \(\dfrac{35}{4}\) x 4

\(x\)             = 35

a) \(435-\left(x+16\right)=425:17\)

\(435-\left(x+16\right)=25\)

\(x+16=435-25\)

\(x+16=410\)

\(x=410-16\)

\(x=394\)

b) \(\left(x+\frac{3}{4}\right)\times\frac{7}{4}+\frac{7}{6}=5\)

\(\left(x+\frac{3}{4}\right)\times\frac{7}{4}=5-\frac{7}{6}\)

\(\left(x+\frac{3}{4}\right)\times\frac{7}{4}=\frac{23}{6}\)

\(x+\frac{3}{4}=\frac{23}{6}:\frac{7}{4}\)

\(x+\frac{3}{4}=\frac{46}{21}\)

\(x=\frac{46}{21}-\frac{3}{4}\)

\(x=\frac{121}{84}\)

435 - ( x + 16 ) = 425 : 17

435 - ( x + 16 ) = 25

           x + 16   = 435 - 25

           x + 16   = 410

           x           = 410 - 16

           x           = 394

\(435-\left(x+16\right)=425\div17\)

\(435-\left(x+16\right)=25\)

\(x+16=435-25\)

\(x+16=410\)

\(x=410-16\)

\(x=394\)

\(\left(x+\frac{3}{4}\right)\times\frac{7}{4}+\frac{7}{6}=5\)

\(\left(x+\frac{3}{4}\right)\times\frac{7}{4}=5-\frac{7}{6}\)

\(\left(x\times\frac{3}{4}\right)\times\frac{7}{4}=\frac{23}{6}\)

\(x+\frac{3}{4}=\frac{23}{6}\div\frac{7}{4}\)

\(x\times\frac{3}{4}=\frac{46}{21}\)

\(x=\frac{46}{21}-\frac{3}{4}\)

\(x=\frac{121}{84}\)

\(\)

a) \(435-\left(x+16\right)=425:17\)

\(435-\left(x+16\right)=25\)

\(\Rightarrow x+16=435-25=410\)

\(\Rightarrow x=410-16=394\)

b) \(\left(x+\frac{4}{3}\right)\cdot\frac{7}{4}=5-\frac{7}{6}\)

\(\left(x+\frac{4}{3}\right)\cdot\frac{7}{4}=\frac{23}{6}\)

\(\Rightarrow x+\frac{4}{3}=\frac{23}{6}:\frac{7}{4}=\frac{46}{21}\)

\(\Rightarrow x=\frac{46}{21}-\frac{4}{3}=\frac{6}{7}\)

Ủng hộ mk nka!!!^_^^_^^_^

a) 435 - ( x + 16 ) = 25

x+ 16 = 435 - 25

x + 16 = 410

x = 410 - 16

x = 394

b) (x+4/3 ) * 7/4 = 23/6

x+4/3= 23/6 : 7/4

x+4/3 = 46/21

x = 46/21  - 4/3

x = 6/7

435-(x+16)=425:17

435-(x+16)=25

       x+16 =435-25

       x+16 =410

       x       =394

b) \(\left(x+\frac{4}{3}\right)\times\frac{7}{4}=5-\frac{7}{6}\)

    \(\left(x+\frac{4}{3}\right)\times\frac{7}{4}=\frac{23}{6}\)

               \(\left(x+\frac{4}{3}\right)=\frac{23}{6}:\frac{7}{4}\)

                \(\left(x+\frac{4}{3}\right)=\frac{46}{21}\)

                                \(x=\frac{46}{21}-\frac{4}{3}\)

                                 \(x=\frac{6}{7}\)

T mk nha bạn ^...^ !!!

a) \(S=1^5+3^5+....+75^5+99^5\)

\(\left(2a+1\right)^5-\left(2a+1\right)=2a\left(2a+1\right)\left(2a+2\right)\left[\left(2a+1\right)^2+1\right]\)

\(\left(2a+1\right)^5-\left(2a+1\right)=4a\left(2a+1\right)\left(a+1\right)\left[\left(2a+1\right)^2+1\right]⋮4\)

\(S=\left(1^5-1\right)+\left(3^5-3\right)+....+\left(75^5-75\right)+\left(99^5-99\right)+\left(1+3+5+...+75+99\right)\)

\(\Leftrightarrow\begin{matrix}1^5-1⋮4\\3^5-3⋮4\\5^5-5⋮4\\...........\\75^5-5⋮4\\99^5-99⋮4\end{matrix}\)

\(S_1=1+3+5+7+...+75+99=\frac{\left(1+75\right)\left[\frac{75-1}{2}+1\right]}{2}+99=38.38+96+3\)

\(\Rightarrow S_1:4\) dư 3

\(\Leftrightarrow S\) chia 4 dư 3