c) Ta có: \(\dfrac{3}{5}+\dfrac{-5}{20}+\dfrac{30}{75}+\dfrac{-7}{4}\)

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

\(=1-2=-1\)

Giải:

a)-1/12+4/3=-1/12+16/12=15/12=5/4

b)(-4/14-3/15)-(1/5-20/35-(-1)).7

=-17/35-22/35.7

=-17/35-22/5

=-171/35

c)3/5+-5/20+30/75+-7/4

=3/5+-1/4+2/5+-7/4

=(3/5+2/5)+(-1/4+-7/4)

=1+-2

=-1

d)5/6.-12/14+7/13

=-5/7+7/13

=-16/91

e)2/-9-5/-36-1/4

=-1/12-1/4

=-1/3

f)2/23+-5/12+7/18+21/23+-7/12

=(2/23+21/23)+(-5/12+-7/12)+7/18

=1+-1+7/18

=7/18

\(\dfrac{4}{5}\)+\(\dfrac{-5}{4}\)=\(\dfrac{0}{4}\)

\(\dfrac{-1}{3}\)+\(\dfrac{2}{5}-\dfrac{5}{6}\)=\(\dfrac{-10}{30}+\dfrac{12}{30}-\dfrac{25}{30}\)=\(\dfrac{-23}{30}\)

\(\dfrac{2}{3}-\dfrac{5}{7}.\dfrac{14}{25}\)=\(\dfrac{2}{3}-\dfrac{2}{5}\)=\(\dfrac{10}{15}-\dfrac{6}{15}\)=\(\dfrac{4}{15}\)

a: \(=\dfrac{16-25}{20}=\dfrac{-9}{20}\)

b: \(=\dfrac{-10}{30}+\dfrac{12}{30}-\dfrac{25}{30}=-\dfrac{23}{30}\)

c: \(=\dfrac{2}{3}-\dfrac{1}{5}\cdot2=\dfrac{2}{3}-\dfrac{2}{5}=\dfrac{10-6}{15}=\dfrac{4}{15}\)

d: \(=\dfrac{12-25}{30}\cdot\dfrac{15}{8}-1=\dfrac{-13}{16}-1=\dfrac{-29}{16}\)

e: \(=\dfrac{-3}{13}-\dfrac{10}{13}-\dfrac{7}{17}+\dfrac{24}{17}-\dfrac{5}{19}-\dfrac{14}{19}=-1\)

\(a,121-\left(115+x\right)=3x-\left(25-9-5x\right)-8\\ 121-115-x=3x-25+9+5x-8\\ 6-x=8x-24\\ 8x+x=-24-6\\ 9x=-30\\ x=-\dfrac{30}{9}=-\dfrac{10}{3}\\ ----\\ b,2^{x+2}.3^{x+1}.5^x=10800\\ \left(2.3.5\right)^x.2^2.3=10800\\ 30^x.12=10800\\ 30^x=\dfrac{10800}{12}=900=30^2\\ Vậy:x=2\)

Bài 1:

a. $=(-25)(-4)(-35)=100(-35)=-3500$

b. $=16-10=6$

c. $=180-(-16)-(-36)=180+16+36=232$

d. $=250-200:[1(-3)^2+(-8)]$

$=250-200:(9-8)=250-200=50$

2.

$60+2(12-x)=-48$

$2(12-x)=60-(-48)=60+48=108$

$12-x=108:2=54$

$x=12-54=-42$
 

Bài 3:

$A=(7+7^2)+(7^3+7^4)+....+(7^{17}+7^{18})$

$=7(1+7)+7^3(7+1)+...+7^{17}(1+7)$

$=(1+7)(7+7^3+....+7^{17})=8(7+7^3+....+7^{17})\vdots 8$

b.

$A=(7+7^2+7^3)+(7^4+7^5+7^6)+...+(7^{16}+7^{17}+7^{18})$

$=7(1+7+7^2)+7^4(1+7+7^2)+....+7^{16}(1+7+7^2)$

$=(1+7+7^2)(7+7^4+....+7^{16})$

$=57(7+7^4+...+7^{16})\vdots 57$

c.

$A=(7+7^2+7^3+7^4)+(7^5+7^6+7^7+7^8)+(7^9+7^{10}+7^{11}+7^{12})+(7^{13}+7^{14}+7^{15}+7^{16})+(7^{17}+7^{18})$
$=7(1+7+7^2+7^3)+7^5(1+7+7^2+7^3)+7^9(1+7+7^2+7^3(+7^{13}(1+7+7^2+7^3)+7^{17}(1+7)$

$=(1+7+7^2+7^3)(7+7^5+7^9+7^{13})+8.7^{17}$

$=400(7+7^5+7^9+7^{13})+8.7^{17}$

Ta thấy $400(7+7^5+7^9+7^{13})\vdots 50$ nhưng $8.7^{17}\not\vdots 50$ nên $A\not\vdots 50$

Bài 1:

a. $=(-25)(-4)(-35)=100(-35)=-3500$

b. $=16-10=6$

c. $180-(-16)-(-36)=180+16+36=196+36=232$

d. $=250-200:[2000.(-3).2-6]$

$=250-200:[2000.(-6)+(-6)]$

$=250-200:[(-6)(2000+1)]=250-200[(-6).2001]$

$=250+200.6.2001=250+2401200=2401450$

Bài 2:

$60+2(12-x)=-48$

$2(12-x)=-48-60=-108$
$12-x=-108:2=-54$

$x=12-(-54)=66$

Bài 3:

Số số hạng của A: $(718-71):1+1=648$

$A=(718+71)\times 648:2=789\times 324$

a. Có: $A=789\times 324=789\times 81\times 4$

$\Rightarrow A\not\vdots 8$ (bạn xem lại đề)

b. $A=789\times 81\times 4\not\vdots 57$ (bạn xem lại đề)

c. $A=789\times 81\times 4\not\vdots 50$ 

Vậy có vẻ đề câu này sai rồi. Bạn xem lại.

3. 3³.19-3³.12=3³.(19-12)=3³.7=189

4. 3.5²-16:2²=3.5²-2⁴:2²=3.25-4=75-4=71

\(a,=\left(\frac{15}{12}-\frac{3}{12}\right)+\left(\frac{5}{13}-\frac{18}{13}\right)\)

\(=1+-1\)

\(=0\)