a)\(3^9:3^7+5\times2^3\)

\(9+40=49\)

b)\(2^3\times3^2-5^{16}:5^{14}\)

\(72-25=47\)

a)   \(3^9:3^7+5\text{×}2^3=3^2+5\text{×}8=9+40=49\)

b)   \(2^3\text{×}3^2-5^{16}:5^{14}=8\text{×}9-5^2=72-25=47\)

\(-\dfrac{7}{2}-\left(\dfrac{5}{3}\right)^2:\dfrac{10}{27}\\ =-\dfrac{7}{2}-\dfrac{25}{9}:\dfrac{10}{27}\\ =-\dfrac{7}{2}-\dfrac{25}{9}.\dfrac{27}{10}\\ =-\dfrac{7}{2}-\dfrac{675}{90}\\ =-\dfrac{7}{2}-\dfrac{75}{9}\\ =-\dfrac{63}{18}-\dfrac{150}{18}\\ =-\dfrac{213}{18}\)

`a) -3/5 xx 7/9 + (-3/5) xx 2/9 + 3/5`

`=-3/5 xx (7/9+2/9) + 3/5`

`=-3/5 xx 1 + 3/5`

`=-3/5+3/5`

`=0`

Bổ sung : \(\dfrac{1}{2}x-\dfrac{2}{3}=\left|-2\dfrac{1}{3}\right|\\ \dfrac{1}{2}x-\dfrac{2}{3}=\left|-\dfrac{7}{3}\right|\\ \dfrac{1}{2}x-\dfrac{2}{3}=\dfrac{7}{3}\\ \dfrac{1}{2}x=\dfrac{7}{3}+\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{9}{3}\\ \dfrac{1}{2}x=3\\ x=3:\dfrac{1}{2}\\ x=3.2\\ x=6\)

A= 4/7.

Biết có cái

a: \(\left(2x-y+7\right)^{2022}>=0\forall x,y\)

\(\left|x-1\right|^{2023}>=0\forall x\)

=>\(\left(2x-y+7\right)^{2022}+\left|x-1\right|^{2023}>=0\forall x,y\)

mà \(\left(2x-y+7\right)^{2022}+\left|x-1\right|^{2023}< =0\forall x,y\)

nên \(\left(2x-y+7\right)^{2022}+\left|x-1\right|^{2023}=0\)

=>\(\left\{{}\begin{matrix}2x-y+7=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2x+7=9\end{matrix}\right.\)

\(P=x^{2023}+\left(y-10\right)^{2023}\)

\(=1^{2023}+\left(9-10\right)^{2023}\)

=1-1

=0

c: \(\left|x-3\right|>=0\forall x\)

=>\(\left|x-3\right|+2>=2\forall x\)

=>\(\left(\left|x-3\right|+2\right)^2>=4\forall x\)

mà \(\left|y+3\right|>=0\forall y\)

nên \(\left(\left|x-3\right|+2\right)^2+\left|y+3\right|>=4\forall x,y\)

=>\(P=\left(\left|x-3\right|+2\right)^2+\left|y-3\right|+2019>=4+2019=2023\forall x,y\)

Dấu '=' xảy ra khi x-3=0 và y-3=0

=>x=3 và y=3

a: Thay x=16 vào A, ta được:

\(A=\dfrac{2\cdot4}{4+3}=\dfrac{8}{7}\)

a) 6x^2+27x+4x+18-[ 6x^2+x+12x+2]=7

6x^2+31x+18-6x^2-13x-2=7

18x+16=7

18x=-9

x=-1/2

(6x-3)(3x-1)-[18x^2-2x-27x+3]=0

18x^2-6x-9x+3-18x^2+2x+27x-3=0

14x=0

x=0

A=\(2^2-9^3+4^{-2}.16-2.5^2\)
\(=4-729+1-50=-774\)
B=\(\left(2^3.2\right).\dfrac{1}{2}+3^{-2}.3^2-7.1+5\)
\(B=2^4.\dfrac{1}{2}+1-7+5=8+1-7+5=7\)
 

 C = 2^-3 + (5²)³.5^-3 + 4^-3.16 - 2.3² - 105.(\(\dfrac{24}{51}\))⁰

C =  \(\dfrac{1}{8}\) + 5⁶.5^-3 + 4^-3.4² - 2.9 - 105.1

C =  \(\dfrac{1}{8}\) + 5³ + \(\dfrac{1}{4}\) - 18 - 105

C =  (\(\dfrac{1}{8}\) + \(\dfrac{1}{4}\))  - (105 - 125 + 18)

C = \(\dfrac{3}{8}\) - (-20 + 18)

C = \(\dfrac{3}{8}\)  + 2

C = \(\dfrac{19}{8}\)

D = 2^-3 . \(\dfrac{1}{2^{-2}}\).\(\dfrac{2}{3}\) + 4^-2.8 - 7. (\(\dfrac{17}{23}\))⁰ + 19

D = 2^-1.\(\dfrac{2}{3}\) + \(\dfrac{1}{16}\).8 - 7.1 + 19

D = \(\dfrac{1}{3}\) + \(\dfrac{1}{2}\) - 7 + 19

D = \(\dfrac{5}{6}\)  + (19 - 7)

D = \(\dfrac{5}{6}\) + 12

D = \(\dfrac{77}{6}\)

Bài 1:

a) \(\frac{16}{15}.\frac{\left(-5\right)}{14}.\frac{54}{24}.\frac{56}{21}\)

\(=\frac{4.2.2}{5.3}.\frac{\left(-5\right)}{2.7}.\frac{3.3}{4}.\frac{8}{3}\)

\(=\frac{4.2.2.\left(-5\right).3.3.8}{5.3.2.7.4.3}\)

\(=\frac{-16}{7}\)

b) \(\frac{7}{3}.\frac{\left(-5\right)}{2}.\frac{15}{21}.\frac{4}{\left(-5\right)}\)

\(=\frac{7}{3}.\frac{\left(-5\right)}{2}.\frac{5}{7}.\frac{2.2}{\left(-5\right)}\)

\(=\frac{7.\left(-5\right).5.2.2}{3.2.7.\left(-5\right)}\)

\(=\frac{10}{3}\)

Bài 2:

a) \(\frac{21}{24}.\frac{11}{9}.\frac{5}{7}=\frac{7}{8}.\frac{11}{9}.\frac{5}{7}=\frac{11.5}{8.9}=\frac{55}{72}\)

b) \(\frac{5}{23}.\frac{17}{26}+\frac{5}{23}.\frac{9}{26}\)

\(=\frac{5}{23}.\left(\frac{17}{26}+\frac{9}{26}\right)=\frac{5}{23}.1=\frac{5}{23}\)

c) \(\left(\frac{3}{29}-\frac{1}{5}\right).\frac{29}{3}=\frac{3}{29}.\frac{29}{3}-\frac{1}{5}.\frac{29}{3}\)

\(=1-1\frac{14}{15}=\frac{14}{15}\)

Bài 3:

a) x/5 = 2/5

=> x =2

b) -4/x = 20/14 = 10/7

=> -4/x = 10/7

=> x.10 = (-4).7

x.10 = - 28

x= -28 :10

x= -2,8

c) 4/7 = 12/x = 12/ 21

=> 12/x = 12/21

=> x = 21

d) 3/7 = x / 21 = 9/21

=> x/21 = 9/21

=> x= 9