Bài 1:

a, \(\left(x-2\right)^2=9\)

\(\Rightarrow x-2\in\left\{-3;3\right\}\Rightarrow x\in\left\{-1;5\right\}\)

b, \(\left(3x-1\right)^3=-8\)

\(\Rightarrow3x-1=-2\Rightarrow3x=-1\)

\(\Rightarrow x=-\dfrac{1}{3}\)

c, \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)

\(\Rightarrow x+\dfrac{1}{2}\in\left\{-\dfrac{1}{4};\dfrac{1}{4}\right\}\)

\(\Rightarrow x\in\left\{-\dfrac{3}{4};-\dfrac{1}{4}\right\}\)

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

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

Vì \(\dfrac{2}{3}\ne\pm1;\dfrac{2}{3}\ne0\) nên \(x=2\)

e, \(\left(\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{16}\)

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

Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(x-1=4\Rightarrow x=5\)

a) \(5\dfrac{3}{8}-1\dfrac{9}{10}=\dfrac{43}{8}-\dfrac{19}{10}=\dfrac{215}{40}-\dfrac{76}{40}=\dfrac{139}{40}\)

b) \(\left(-3\dfrac{1}{4}\right)+\left(-2\dfrac{1}{3}\right)=-\dfrac{13}{4}+\left(-\dfrac{7}{3}\right)=-\dfrac{39}{12}+\left(-\dfrac{28}{12}\right)=\dfrac{-67}{12}\)

c) \(\left(-5\dfrac{1}{8}\right)+3\dfrac{2}{4}=\left(-\dfrac{41}{8}\right)+\dfrac{14}{4}=\left(-\dfrac{41}{8}\right)+\dfrac{28}{8}=-\dfrac{13}{8}\)

d)\(\left(-3\right)-\left(-2\dfrac{2}{5}\right)=\left(-3\right)-\left(-\dfrac{12}{5}\right)=\left(-\dfrac{15}{5}\right)+\left(-\dfrac{12}{5}\right)=-\dfrac{27}{5}\)

1: \(=-\dfrac{7}{80}\cdot\dfrac{4}{7}-\dfrac{2}{9}:\dfrac{16}{3}+\dfrac{5}{24}\cdot\left(\dfrac{-50+38}{15}\right)^2\)

\(=\dfrac{-1}{20}-\dfrac{2}{9}\cdot\dfrac{3}{16}+\dfrac{5}{24}\cdot\dfrac{16}{25}\)

\(=\dfrac{-1}{20}-\dfrac{1}{24}+\dfrac{2}{15}\)

\(=\dfrac{-6-5+16}{120}=\dfrac{5}{120}=\dfrac{1}{24}\)

2: \(=1500-\left\{10^3-11\cdot\left[49-5\cdot8\right]\right\}\)

\(=1500-\left\{1000-11\cdot9\right\}\)

\(=1500-1000+99=599\)

a: \(=\dfrac{5\cdot\left(8-6\right)}{10}=\dfrac{5\cdot2}{10}=1\)

b: \(\dfrac{\left(-4\right)^2}{5}=\dfrac{16}{5}\)

\(B=\dfrac{3}{7}-\dfrac{1}{5}-\dfrac{3}{7}=-\dfrac{1}{5}\)

c: \(C=\left(6-2.8\right)\cdot\dfrac{25}{8}-\dfrac{8}{5}\cdot4\)

\(=\dfrac{16}{5}\cdot\dfrac{25}{8}-\dfrac{32}{5}\)

\(=5\cdot2-\dfrac{32}{5}=10-\dfrac{32}{5}=\dfrac{18}{5}\)

d: \(D=\left(\dfrac{-5}{24}+\dfrac{18}{24}+\dfrac{14}{24}\right):\dfrac{-17}{8}\)

\(=\dfrac{27}{24}\cdot\dfrac{-8}{17}=\dfrac{-9}{8}\cdot\dfrac{8}{17}=\dfrac{-9}{17}\)

 Theo  thứ tự nhé

a) <

b) <

c) >

1. Ta có : Dùng phân số \(\frac{84}{71}\)làm phân số trung gian ta được : \(\frac{84}{81}>\frac{73}{71}\)

2. Ta có : \(\left[\frac{3}{8}\right]^5=\left[\frac{3}{2^3}\right]^5=\frac{243}{2^{15}};\left[\frac{5}{243}\right]^3=\left[\frac{5}{3^5}\right]^3=\frac{125}{3^{15}}\)

Chọn phân số \(\frac{243}{3^{15}}\)làm phân số trung gian để so sánh ta được : \(\frac{243}{2^{15}}>\frac{125}{3^{15}}\)

Vậy : \(\left[\frac{3}{8}\right]^5>\left[\frac{5}{243}\right]^3\)

Còn một cái nữa là ta có thể so sánh với số trung gian như 1 và 0 .

Vì a/b=c/d khi a.c=b.d

Giờ thì mình kết thúc bài làm.