a,

\(\left(\dfrac{-2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)

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

\(=\left(-1+1\right):\dfrac{4}{5}=0\)

b,

a) \(\left(-\dfrac{2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)

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

= \(0:\dfrac{4}{5}\)

= 0

b) \(\dfrac{5}{9}\left(\dfrac{1}{11}-\dfrac{5}{22}\right)+\dfrac{5}{9}\left(\dfrac{1}{15}-\dfrac{2}{3}\right)\)

= \(\dfrac{5}{9}:\left(\dfrac{1}{11}-\dfrac{5}{22}+\dfrac{1}{15}-\dfrac{2}{3}\right)\)

= \(\dfrac{5}{9}:-\dfrac{81}{110}\)

= \(-\dfrac{550}{729}\)

Giải:

a) \(\left(\dfrac{-2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(\dfrac{-1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)

\(=\left[\left(\dfrac{-2}{3}+\dfrac{2}{7}\right)+\left(\dfrac{-1}{3}+\dfrac{4}{7}\right)\right]:\dfrac{4}{5}\)

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

\(=\left(-1+1\right):\dfrac{4}{5}\)

\(=0:\dfrac{4}{5}\)

\(=0.\dfrac{4}{5}\)

\(=0\)

b) \(\dfrac{5}{9}:\left(\dfrac{1}{11}-\dfrac{5}{22}\right)+\dfrac{5}{9}:\left(\dfrac{1}{15}-\dfrac{2}{3}\right)\)

\(=\dfrac{5}{9}:\left(\dfrac{2}{22}-\dfrac{5}{22}\right)+\dfrac{5}{9}:\left(\dfrac{1}{15}-\dfrac{10}{15}\right)\)

\(=\dfrac{5}{9}:\dfrac{-3}{22}+\dfrac{5}{9}:\dfrac{-3}{5}\)

\(=\dfrac{5}{9}:\left(\dfrac{-3}{22}-\dfrac{3}{5}\right)\)

\(=\dfrac{5}{9}:\left(\dfrac{-3}{22}-\dfrac{3}{5}\right)\)

\(=\dfrac{5}{9}:\dfrac{-81}{110}\)

\(=-\dfrac{550}{729}\)

Chúc bạn học tốt!!!

 @ mãi ms dc

a: if (a=b) or (b=c) or (a=c) then writeln('Tam giac can');

b: if (a=c) and (b=c) then writeln('Tam giac deu');

c: if (x mod 2<>0) and (x mod 3=0) then writeln('x la so nguyen le va chia het cho 3');

Chỉ có thể là số nguyên dương và số 0 vì GTTĐ của 1 số luôn lớn hơn hoặc bằng 0

giá trị của một số luôn là số nguyên dương. kể cả trường hợp số nguyên a là số âm hay là số dương thì cũng vậy

Giá trị tuyệt đối của số 0 vân là số 0

= a^nx2

=a\(^{n\cdot2}\)

KHO THE

\(A=\frac{\left[\left(25-1\right):1+1\right]\left(25+1\right)}{2}=325.\)

\(B=\frac{\left[\left(51-3\right):2+1\right]\left(51+3\right)}{2}=675\)

\(C=\frac{\left[\left(81-1\right):4+1\right]\left(81+1\right)}{2}=861\)

a,\(5x+10y=5\left(x+2y\right)\)

b,\(=3xy\left(x+3yz\right)\)

c,\(=\left(x+1\right)\left(5-3x-3\right)=\left(x+1\right)\left(2-3x\right)\)

d,\(=\left(x-y\right)\left(2+3x\right)\)

e,\(=y\left(xy+3x^2+y^2\right)\)

f,\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\)

g,\(=x^2-x+\dfrac{1}{4}-\left(\dfrac{5}{2}\right)^2=\left(x-\dfrac{1}{2}\right)^2-\left(\dfrac{5}{2}\right)^2=\left(x-3\right)\left(x+2\right)\)

h,\(=x^2+9x+\dfrac{81}{4}-\left(\dfrac{7}{2}\right)^2=\left(x+\dfrac{9}{2}\right)^2-\left(\dfrac{7}{2}\right)^2=\left(x+1\right)\left(x+8\right)\)

i,\(=x^2-10x+25-4^2=\left(x-5\right)^2-4^2=\left(x-1\right)\left(x-9\right)\)

k,\(=x^2+x+\dfrac{1}{4}-\left(\dfrac{7}{2}\right)^2=\left(x+\dfrac{1}{2}\right)^2-\left(\dfrac{7}{2}\right)^2=\left(x-3\right)\left(x+4\right)\)

l.\(=3\left(x^2+\dfrac{8}{3}x+\dfrac{4}{3}\right)=3\left(x^2+\dfrac{8}{3}x+\dfrac{16}{9}-\dfrac{4}{9}\right)\)

\(=3[\left(x+\dfrac{4}{3}\right)^2-\left(\dfrac{2}{3}\right)^2]=3\left(x+\dfrac{2}{3}\right)\left(x+2\right)\)

A, 5(x+2y)

B, 3xy(x+3yz)

C, (x+1)[5-3(x+1)]

D, (x-y)(2+3x)

E, y(xy+3x^2+y^2)

F ,(x+y)(x-y-1)

G, (x-3)(x+2)

H, (x+1)(x+8)

I, (x-9)(x-1)

K, (x-3)(x+4)

L, (x+2)(3x+2)