a.

$42(-45)-55(-42)=42(-45)+55.42=42(-45+55)=42.10=420$

b.

$(-2-3)^2-(-2)^8:(-2)^5=(-5)^2-(-2)^3=25-(-8)=33$
c.

$=\frac{1}{5}+\frac{3}{10}+\frac{3}{2}+\frac{4}{7}$

$=\frac{2}{10}+\frac{3}{10}+\frac{3}{2}+\frac{4}{7}$

$=\frac{5}{10}+\frac{3}{2}+\frac{4}{7}$
$=\frac{1}{2}+\frac{3}{2}+\frac{4}{7}=2+\frac{4}{7}=\frac{18}{7}$

d.

$=\frac{3}{22}-\frac{7}{15}-\frac{3}{22}+\frac{7}{15}-\frac{1}{2}$

$=(\frac{3}{22}-\frac{3}{22})+(\frac{-7}{15}+\frac{7}{15})-\frac{1}{2}$

$=0+0-\frac{1}{2}=\frac{-1}{2}$

e.

$=\frac{77}{12}: \frac{11}{4}+\frac{45}{4}.\frac{2}{15}$

$=\frac{7}{3}+\frac{3}{2}=\frac{23}{6}$

f.

$=\frac{-7}{11}(\frac{11}{19}+\frac{12}{19}-\frac{4}{19})$

$=\frac{-7}{11}.\frac{19}{19}=\frac{-7}{11}$
 

Lời giải:

$11\frac{3}{13}-(2\frac{4}{7}+5\frac{3}{13})=11\frac{3}{13}-5\frac{3}{13}-2\frac{4}{7}$

$=6-2\frac{4}{7}=4-\frac{4}{7}=\frac{24}{7}$

b) 146/13 - ( 18/7 + 68/13)

 = 146/13 - 710/91

 = 24/7 :))

a: \(42\cdot\left(-45\right)-55\cdot\left(-42\right)\)

\(=-42\cdot45+55\cdot42\)

\(=42\left(55-45\right)=42\cdot10=420\)

b: \(\left(-2-3\right)^2-\left(-2\right)^8:\left(-2\right)^5\)

\(=\left(-5\right)^2-\left(-2\right)^3\)

\(=25-\left(-8\right)=33\)

c: \(\dfrac{1}{5}-\dfrac{-3}{2}+\dfrac{4}{7}-\dfrac{3}{-10}\)

\(=\dfrac{1}{5}+\dfrac{3}{2}+\dfrac{3}{10}+\dfrac{4}{7}\)

\(=\dfrac{2+15+3}{10}+\dfrac{4}{7}=2+\dfrac{4}{7}=\dfrac{18}{7}\)

d: \(\dfrac{3}{22}-\left(\dfrac{7}{15}-\dfrac{-3}{22}\right)+\dfrac{7}{15}-\dfrac{1}{2}\)

\(=\dfrac{3}{22}-\dfrac{7}{15}+\dfrac{3}{22}+\dfrac{7}{15}-\dfrac{1}{2}\)

\(=\dfrac{3}{11}-\dfrac{1}{2}=\dfrac{-5}{22}\)

e: \(6\dfrac{5}{12}:2\dfrac{3}{4}+11\dfrac{1}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)

\(=\dfrac{77}{12}:\dfrac{11}{4}+\dfrac{45}{4}\cdot\dfrac{2}{15}\)

\(=\dfrac{77}{12}\cdot\dfrac{4}{11}+\dfrac{45}{15}\cdot\dfrac{2}{4}\)

\(=\dfrac{7}{3}+\dfrac{3}{2}=\dfrac{23}{6}\)

f: \(\dfrac{-7}{11}\cdot\dfrac{11}{19}+\dfrac{-7}{11}\cdot\dfrac{12}{19}-\dfrac{4}{19}\cdot\dfrac{-7}{11}\)

\(=\dfrac{-7}{11}\left(\dfrac{11}{19}+\dfrac{12}{19}-\dfrac{4}{19}\right)\)

\(=\dfrac{-7}{11}\cdot1=-\dfrac{7}{11}\)

C = \(\dfrac{2}{3.5}\) + \(\dfrac{2}{5.7}\) + ... + \(\dfrac{2}{99.101}\)

C = \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + ... + \(\dfrac{1}{99}\) - \(\dfrac{1}{101}\)

C = \(\dfrac{1}{3}\) - \(\dfrac{1}{101}\)

C = \(\dfrac{98}{303}\)

Để \(\dfrac{5}{3n+1}\) là số nguyên thì \(5⋮3n+1\)

=>\(3n+1\in\left\{1;-1;5;-5\right\}\)

=>\(3n\in\left\{0;-2;4;-6\right\}\)

=>\(n\in\left\{0;-\dfrac{2}{3};\dfrac{4}{3};-2\right\}\)

mà n nguyên

nên \(n\in\left\{0;-2\right\}\)

Vì tử số là 5, nên mẫu số phải là 1 hoặc 5 (vì 5 chỉ có 2 ước là 1 và 5). 
Vậy ta có hai trường hợp:
1) Nếu $3n+1 = 1$ thì $n = 0$.
2) Nếu $3n+1 = 5$ thì $n = \frac{4}{3}$.
Vì $n$ phải là số nguyên, nên giá trị duy nhất của $n$ là $n = 0$.

\(\dfrac{7}{8}\) : \(x\) + \(\dfrac{3}{4}\) = - \(\dfrac{1}{2}\)

\(\dfrac{7}{8}\) : \(x\)         = - \(\dfrac{1}{2}\) - \(\dfrac{3}{4}\)

\(\dfrac{7}{8}\) : \(x\)       = - \(\dfrac{5}{4}\)

\(x\) = \(\dfrac{7}{8}\) : (- \(\dfrac{5}{4}\))

\(x\) = - \(\dfrac{7}{10}\)

A = \(\dfrac{3n-1}{n+2}\) (n \(\in\) z; n ≠ -2)

A \(\in\) Z ⇔ 3n -  1 ⋮ n + 2

        3n + 6 - 7 ⋮ n + 2

    3.(n + 2) - 7 ⋮ n + 2

                    7 ⋮ n + 2

n + 2 \(\in\) Ư(7) = {-7; -1; 1; 7}

Lập bảng ta có:

Theo bảng trên ta có: 

 n \(\in\) {-9; -3; -1; 5}

Kết luận để A = \(\dfrac{3n-1}{n+2}\) là số nguyên thì n \(\in\) {-9; -3;  -1; 5}

sau 12 tháng, số tiền bố của Tuấn lấy ra là khoảng 1,072,482.164 đồng, bao gồm cả vốn và lãi.

\(\dfrac{5}{2\cdot4}+\dfrac{5}{4\cdot6}+...+\dfrac{5}{48\cdot50}\)

\(=\dfrac{5}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{48\cdot50}\right)\)

\(=\dfrac{5}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)

\(=\dfrac{5}{2}\left(\dfrac{1}{2}-\dfrac{1}{50}\right)=\dfrac{5}{2}\cdot\dfrac{24}{50}=\dfrac{120}{100}=\dfrac{6}{5}\)

961 số 0