\(\left(3-x\right)^3=-\dfrac{27}{64}\)

\(\left(3-x\right)^3=\left(\dfrac{-3}{4}\right)^3\)

\(=>3-x=\dfrac{-3}{4}\)

\(x=3-\dfrac{-3}{4}=\dfrac{12}{4}+\dfrac{3}{4}\)

\(x=\dfrac{15}{4}\)

________

\(\left(x-5\right)^3=\dfrac{1}{-27}\)

\(\left(x-5\right)^3=\left(\dfrac{-1}{3}\right)^3\)

\(=>x-5=\dfrac{-1}{3}\)

\(x=\dfrac{-1}{3}+5=\dfrac{-1}{3}+\dfrac{15}{3}\)

\(x=\dfrac{14}{3}\)

_____________

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8}\)

\(\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{3}{2}\right)^3\)

\(=>x-\dfrac{1}{2}=\dfrac{3}{2}\)

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

\(x=2\)

________

\(\left(2x-1\right)^2=\dfrac{1}{4}\)            

\(\left(2x-1\right)^2=\left(\dfrac{1}{2}\right)^2\)           hoặc              \(\left(2x-1\right)^2=\left(\dfrac{-1}{2}\right)^2\)

\(=>2x-1=\dfrac{1}{2}\)                                       \(2x-1=\dfrac{-1}{2}\)

\(2x=\dfrac{1}{2}+1=\dfrac{1}{2}+\dfrac{2}{2}\)                               \(2x=\dfrac{-1}{2}+1=\dfrac{-1}{2}+\dfrac{2}{2}\)

\(2x=\dfrac{3}{2}\)                                                     \(2x=\dfrac{1}{2}\)

\(x=\dfrac{3}{2}:2=\dfrac{3}{2}.\dfrac{1}{2}\)                                     \(x=\dfrac{1}{2}:2=\dfrac{1}{2}.\dfrac{1}{2}\)

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

____________

\(\left(2-3x\right)^2=\dfrac{9}{4}\)

\(\left(2-3x\right)^2=\left(\dfrac{3}{2}\right)^2\)                hoặc                  \(\left(2-3x\right)^2=\left(\dfrac{-3}{2}\right)^2\)

\(=>2-3x=\dfrac{3}{2}\)                                               \(2-3x=\dfrac{-3}{2}\)

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

\(3x=\dfrac{1}{2}\)                                                            \(3x=\dfrac{7}{2}\)

\(x=\dfrac{1}{2}.\dfrac{1}{3}\)                                                          \(x=\dfrac{7}{2}.\dfrac{1}{3}\)

\(x=\dfrac{1}{6}\)                                                               \(x=\dfrac{7}{6}\)

______________

\(\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\) -> Kiểm tra đề câu này

(3-x)^{3_{=(-\(\dfrac{3}{4}\))}3}

^{3-x=-\(\dfrac{3}{4}\)}

  ^{x=3-(-\(\dfrac{3}{4}\))}

  ^{x=\(\dfrac{15}{4}\)}

a) \(A=\dfrac{2}{3}+\dfrac{3}{4}.\left(\dfrac{-4}{9}\right)=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)

b) \(B=2\dfrac{3}{11}.1\dfrac{1}{12}.\left(-2,2\right)=\dfrac{25}{11}.\dfrac{13}{12}.\dfrac{-11}{5}=-\dfrac{65}{12}\)

c) \(C=\left(\dfrac{3}{4}-0,2\right)\left(0,4-\dfrac{4}{5}\right)=\left(\dfrac{3}{4}-\dfrac{1}{5}\right)\left(\dfrac{2}{5}-\dfrac{4}{5}\right)=\dfrac{11}{20}\left(\dfrac{-2}{5}\right)=\dfrac{-11}{50}\)

A = 2/3 + -1/3

    = 1/3

B = 25/11 . 13/12 . (-2,2)

    = 325/132 . (-2,2)

    = -65/12

C = 11/20 . -2/5

    = -11/50

Chúc bạn học tốt!! ^^

    = -

a, A = \(\dfrac{2}{3}+\dfrac{3}{4}.\left(\dfrac{-4}{9}\right)\)

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

       =\(\dfrac{1}{3}\)

b, B = \(2\dfrac{3}{11}.1\dfrac{1}{12}.\left(-2,2\right)\)

       = \(\dfrac{23}{11}.\dfrac{13}{12}.\dfrac{-11}{5}\)

       = \(\dfrac{23.13.\left(-11\right)}{11.12.5}\)

       = \(\dfrac{-299}{60}\)

c, C =\(\left(\dfrac{3}{4}-0,2\right)\)\(.\left(0,4\right)-\dfrac{4}{5}\)

        = \(\left(\dfrac{3}{4}-\dfrac{1}{2}\right)\)\(.\dfrac{2}{5}-\dfrac{4}{5}\)

        = \(\left(\dfrac{3}{4}-\dfrac{2}{4}\right)\)\(.\dfrac{-2}{5}\)

        = \(\dfrac{1}{4}.\dfrac{-2}{5}\)

        = \(\dfrac{-2}{20}=\dfrac{-1}{10}\)

Chúc bạn học tốt!

a) Ta có: \(\left(\dfrac{9}{25}-2\cdot18\right):\left(3\dfrac{4}{5}+0.2\right)\)

\(=\left(\dfrac{9}{25}-36\right):\left(\dfrac{19}{5}+\dfrac{1}{5}\right)\)

\(=\left(\dfrac{9}{25}-\dfrac{900}{25}\right):\dfrac{20}{5}\)

\(=\dfrac{-891}{25}\cdot\dfrac{1}{4}\)

\(=-\dfrac{891}{100}\)

b) Ta có: \(\dfrac{3}{8}\cdot19\dfrac{1}{3}+\dfrac{3}{8}\cdot33\dfrac{1}{3}\)

\(=\dfrac{3}{8}\cdot\dfrac{58}{3}+\dfrac{3}{8}\cdot\dfrac{100}{3}\)

\(=\dfrac{58}{8}+\dfrac{100}{8}\)

\(=\dfrac{158}{8}=\dfrac{79}{4}\)

c) Ta có: \(15\cdot\left(-\dfrac{2}{3}\right)^2-\dfrac{7}{3}\)

\(=15\cdot\dfrac{4}{9}-\dfrac{7}{3}\)

\(=\dfrac{20}{3}-\dfrac{7}{3}\)

\(=\dfrac{13}{3}\)

d) Ta có: \(\dfrac{1}{2}\sqrt{64}-\sqrt{\dfrac{4}{25}}+\left(-1\right)^{2007}\)

\(=\dfrac{1}{2}\cdot8-\dfrac{2}{5}-1\)

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

\(=3-\dfrac{2}{5}\)

\(=\dfrac{15}{5}-\dfrac{2}{5}=\dfrac{13}{5}\)

e) Ta có: \(\left(-\dfrac{5}{2}\right)^2:\left(-15\right)-\left(0.45+\dfrac{3}{4}\right)\cdot\left(-1\dfrac{5}{9}\right)\)

\(=\dfrac{25}{4}\cdot\dfrac{-1}{15}-\left(\dfrac{9}{20}+\dfrac{15}{20}\right)\cdot\dfrac{-14}{9}\)

\(=\dfrac{-25}{60}-\dfrac{24}{20}\cdot\dfrac{-14}{9}\)

\(=\dfrac{-25}{60}+\dfrac{28}{15}\)

\(=\dfrac{-25}{60}+\dfrac{112}{60}\)

\(=\dfrac{87}{60}=\dfrac{29}{20}\)

f) Ta có: \(\left(-\dfrac{1}{3}\right)-\left(-\dfrac{3}{5}\right)^0+\left(1-\dfrac{1}{2}\right)^2:2\)

\(=-\dfrac{1}{3}-1+\left(\dfrac{1}{2}\right)^2\cdot\dfrac{1}{2}\)

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

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

\(=\dfrac{-32}{24}+\dfrac{3}{24}=\dfrac{-29}{24}\)

g) Ta có: \(\left(\dfrac{1}{2}\right)^{15}\cdot\left(\dfrac{1}{4}\right)^{20}\)

\(=\left(\dfrac{1}{2}\right)^{15}\cdot\left(\dfrac{1}{2}\right)^{40}\)

\(=\left(\dfrac{1}{2}\right)^{55}\)

\(=\dfrac{1}{2^{55}}\)

h) Ta có: \(\dfrac{5^4\cdot20}{25^5\cdot4^5}\)

\(=\dfrac{5^4\cdot5\cdot2^2}{5^{10}\cdot2^{10}}\)

\(=\dfrac{5^5}{5^{10}}\cdot\dfrac{2^2}{2^{10}}\)

\(=\dfrac{1}{5^5}\cdot\dfrac{1}{2^8}\)

\(=\dfrac{1}{800000}\)

a) \(=\dfrac{\left(-1\right)^4}{3^4}=\dfrac{1}{81}\)

b) \(=\dfrac{\left(-9\right)^3}{4^3}=\dfrac{-729}{64}\)

c) \(=\left(-\dfrac{2}{10}\right)^2=\left(-\dfrac{1}{5}\right)^2=\dfrac{1}{25}\)

d) \(=1\)

\(a,=\dfrac{1}{81}\\ b,=\dfrac{729}{64}\\ c,=0,04\\ d,=1\)

Bài 1:
$(y+\frac{1}{3})+(y+\frac{1}{9})+(y+\frac{1}{27})+(y+\frac{1}{81})=\frac{56}{81}$

$(y+y+y+y)+(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81})=\frac{56}{81}$
$4\times y+\frac{40}{81}=\frac{56}{81}$

$4\times y=\frac{56}{81}-\frac{40}{81}=\frac{16}{81}$
$y=\frac{16}{81}:4=\frac{4}{81}$

Bài 2:

$18: \frac{x\times 0,4+0,32}{x}+5=14$

$18: \frac{x\times 0,4+0,32}{x}=14-5=9$

$\frac{x\times 0,4+0,32}{x}=18:9=2$

$x\times 0,4+0,32=2\times x$

$2\times x-x\times 0,4=0,32$

$x\times (2-0,4)=0,32$
$x\times 1,6=0,32$
$x=0,32:1,6=0,2$

Bài 3:

$\frac{3\times x}{2}=\frac{2}{5}+x+\frac{1}{3}$

$1,5\times x=x+\frac{11}{15}$

$1,5\times x-x=\frac{11}{15}$

$x\times (1,5-1)=\frac{11}{15}$

$x\times 0,5=\frac{11}{15}$

$x=\frac{11}{15}: 0,5=\frac{22}{15}$

a: \(A=3^{\dfrac{2}{5}}\cdot3^{\dfrac{1}{5}}\cdot3^{\dfrac{1}{5}}=3^{\dfrac{2}{5}+\dfrac{1}{5}+\dfrac{1}{5}}=3^{\dfrac{4}{5}}\)

b: \(B=\left(-27\right)^{\dfrac{1}{3}}=\left[\left(-3\right)^3\right]^{\dfrac{1}{3}}=\left(-3\right)^{\dfrac{1}{3}\cdot3}=\left(-3\right)^1=-3\)

c: \(C=\sqrt[3]{-64}\cdot\left(\dfrac{1}{2}\right)^3\)

\(=\sqrt[3]{\left(-4\right)^3}\cdot\dfrac{1}{2^3}=-4\cdot\dfrac{1}{8}=-\dfrac{4}{8}=-\dfrac{1}{2}\)

d: \(D=\left(-27\right)^{\dfrac{1}{3}}\cdot\left(\dfrac{1}{3}\right)^4\)

\(=\left[\left(-3\right)^3\right]^{\dfrac{1}{3}}\cdot\dfrac{1}{3^4}\)

\(=\left(-3\right)^{3\cdot\dfrac{1}{3}}\cdot\dfrac{1}{81}=\dfrac{-3}{81}=\dfrac{-1}{27}\)

e: \(E=\left(\sqrt{3}+1\right)^{106}\cdot\left(\sqrt{3}-1\right)^{106}\)

\(=\left[\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\right]^{106}\)

\(=\left(3-1\right)^{106}=2^{106}\)

f: \(F=360^{\sqrt{5}+1}\cdot20^{3-\sqrt{5}}\cdot18^{3-\sqrt{5}}\)

\(=360^{\sqrt{5}+1}\cdot\left(20\cdot18\right)^{3-\sqrt{5}}\)

\(=360^{\sqrt{5}+1}\cdot360^{3-\sqrt{5}}=360^{\sqrt{5}+1+3-\sqrt{5}}=360^4\)

g: \(G=2023^{3+2\sqrt{2}}\cdot2023^{2\sqrt{2}-3}\)

\(=2023^{3+2\sqrt{2}+2\sqrt{2}-3}\)

\(=2023^{4\sqrt{2}}\)

a) Ta có: \(\dfrac{5}{8}+\dfrac{3}{17}+\dfrac{4}{18}+\dfrac{20}{-17}+\dfrac{-2}{9}+\dfrac{21}{56}\)

\(=\left(\dfrac{3}{17}-\dfrac{20}{17}\right)+\left(\dfrac{2}{9}-\dfrac{2}{9}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)\)

\(=-1+1=0\)

b) Ta có: \(\left(\dfrac{9}{16}+\dfrac{8}{-27}\right)+\left(1+\dfrac{7}{16}+\dfrac{-19}{27}\right)\)

\(=\left(\dfrac{9}{16}+\dfrac{7}{16}\right)+\left(\dfrac{-8}{27}-\dfrac{19}{27}\right)+1\)

=1-1+1=1

e: \(=\left(\dfrac{18}{37}+\dfrac{19}{37}\right)+\left(\dfrac{8}{24}+\dfrac{2}{3}\right)-\dfrac{47}{24}=2-\dfrac{47}{24}=\dfrac{1}{24}\)

f: \(=-8\cdot\dfrac{1}{2}:\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)

\(=-4:\dfrac{13}{12}=\dfrac{-48}{13}\)

g: \(=\dfrac{4}{25}+\dfrac{11}{2}\cdot\dfrac{5}{2}-\dfrac{8}{4}=\dfrac{4}{25}+\dfrac{55}{4}-2=\dfrac{1191}{100}\)

1: \(\left(\dfrac{1}{2}\right)^{27}=\left(\dfrac{1}{8}\right)^9\)

\(\left(\dfrac{1}{3}\right)^{18}=\left(\dfrac{1}{9}\right)^9\)

mà 1/8>1/9

nên \(\left(\dfrac{1}{2}\right)^{27}>\left(\dfrac{1}{3}\right)^{18}\)

2: \(\dfrac{-4}{7}=\dfrac{-32}{56}\)

\(\dfrac{-5}{8}=\dfrac{-35}{56}\)

mà -32>-35

nên \(\dfrac{-4}{7}>\dfrac{-5}{8}\)

hay \(\left(\dfrac{-4}{7}\right)^{205}>\left(-\dfrac{5}{8}\right)^{205}\)