a) \(9^3\cdot3^2\)

\(=\left(3^2\right)^3\cdot3^2\)

\(=3^6\cdot3^2\)

\(=3^8\)

b) \(x^7\cdot x:x^4\)

\(=x^8:x^4\)

\(=x^4\)

c) \(7\cdot3^9+3^{10}+51\cdot3^8\)

\(=3^8\cdot\left(7\cdot3+3^2+51\right)\)

\(=3^8\cdot81\)

\(=3^8\cdot3^4\)

\(=3^{12}\)

d) \(5^{15}\cdot125^3\cdot625\)

\(=5^{15}\cdot5^9\cdot5^4\)

\(=5^{28}\)

a) \(7^5:343\)

\(=7^5:7^3\)

\(=7^{5-3}\)

\(=7^2\)

b) \(a^{12}:a^{18}\)

\(=\dfrac{a^{12}}{a^{18}}\)

\(=\dfrac{a^0}{a^6}\)

\(=\dfrac{1}{a^6}\)

c) \(x^7\cdot x^4\cdot x\)

\(=x^{7+4+1}\)

\(=x^{12}\)

a) Sữa đề: \(x^2+2x-3=0\)

\(\Rightarrow x^2-x+3x-3=0\)

\(\Rightarrow x\left(x-1\right)+3\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

b) \(x^2-3x=0\)

\(\Rightarrow x\left(x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

c) \(2x-8x^3=0\)

\(\Rightarrow2x\left(1-4x^2\right)=0\)

\(\Rightarrow2x\left(1-2x\right)\left(1+2x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x=0\\1-2x=0\\1+2x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

d) \(\dfrac{2}{3}-6x^2=0\)

\(\Rightarrow\dfrac{2}{3}\left(1-9x^2\right)=0\)

\(\Rightarrow\dfrac{2}{3}\left(1-3x\right)\left(1+3x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}1-3x=0\\1+3x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

Nếu cắt đi mỗi sợi 1200cm thì sợi thứ nhất vẫn dài hơn sợi thứ hai 54m

Ta có sợi thứ nhất gấp 4 lần sợi thứ hai nên hiệu số phần bằng nhau:

\(4-1=3\) (phần)

Độ dài của sợi thứ nhất sau khi đã cắt 1200cm:

\(54:3\cdot4=72\left(m\right)\)

Độ dài của sợi thứ hai sau khi đã cắt 1200cm:

\(72-54=18\left(m\right)\)

Đổi: 1200cm = 12m

Độ dài ban đầu của sợi thứ nhất:

\(72+12=84\left(m\right)\)

Độ dài ban đầu của sợi thứ hai:

\(18+12=30\left(m\right)\)

Đáp số: ..

Ta có \(\widehat{xOy}\) và \(\widehat{yOz}\) là hai góc kề nhau nên:

\(\widehat{xOy}+\widehat{yOz}=\widehat{xOz}\)

\(\Rightarrow\widehat{xOz}=130^o+60^o=190^o\)

a) \(-\dfrac{1}{24}-\left[\dfrac{1}{4}-\left(\dfrac{1}{2}-\dfrac{7}{8}\right)\right]\)

\(=-\dfrac{1}{24}-\left(\dfrac{1}{4}+\dfrac{3}{8}\right)\)

\(=-\dfrac{1}{24}-\dfrac{5}{8}\)

\(=-\dfrac{1}{24}-\dfrac{15}{24}\)

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

b) \(\left(\dfrac{5}{7}-\dfrac{7}{5}\right)-\left[-\dfrac{1}{2}-\left(-\dfrac{2}{7}-\dfrac{1}{10}\right)\right]\)

\(=-\dfrac{24}{35}-\left(-\dfrac{1}{2}+\dfrac{27}{70}\right)\)

\(=-\dfrac{24}{35}+\dfrac{4}{35}\)

\(=-\dfrac{20}{35}\)

\(=-\dfrac{4}{7}\)

c) \(\left(-\dfrac{1}{2}\right)-\left(-\dfrac{3}{5}\right)+\left(-\dfrac{1}{9}\right)+\dfrac{1}{71}-\left(-\dfrac{2}{7}\right)+\dfrac{4}{35}-\dfrac{7}{18}\)

\(=-\dfrac{1}{2}+\dfrac{3}{5}-\dfrac{1}{9}+\dfrac{1}{71}+\dfrac{2}{7}+\dfrac{4}{35}-\dfrac{7}{18}\)

\(=\dfrac{1}{10}-\dfrac{62}{639}+\dfrac{2}{5}-\dfrac{7}{18}\)

\(=\dfrac{19}{6390}+\dfrac{1}{90}\)

\(=\dfrac{1}{71}\)

d) \(\left(3-\dfrac{1}{4}+\dfrac{2}{3}\right)-\left(5-\dfrac{1}{3}-\dfrac{6}{5}\right)-\left(6-\dfrac{7}{4}+\dfrac{3}{2}\right)\)

\(=\dfrac{63}{20}-\dfrac{52}{15}-\dfrac{23}{4}\)

\(=-\dfrac{19}{60}-\dfrac{23}{4}\)

\(=-\dfrac{91}{15}\)

a) \(5\left(x+7\right)-10=2^3\cdot5\)

\(\Rightarrow5\left(x+7\right)-10=40\)

\(\Rightarrow5\left(x+7\right)=40+10\)

\(\Rightarrow x+7=\dfrac{50}{5}\)

\(\Rightarrow x+7=10\)

\(\Rightarrow x=10-7\)

\(\Rightarrow x=3\)

b) \(9x-2\cdot3^2=3^4\)

\(\Rightarrow9x-18=81\)

\(\Rightarrow9x=81+18\)

\(\Rightarrow9x=99\)

\(\Rightarrow x=\dfrac{99}{9}\)

\(\Rightarrow x=11\)

c) \(5^{25}\cdot5^{x-1}=5^{25}\)

\(\Rightarrow5^{x-1}=5^{25}:5^{25}\)

\(\Rightarrow5^{x-1}=1\)

\(\Rightarrow5^{x-1}=5^0\)

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

\(-\dfrac{1}{24}-\left[\dfrac{1}{4}-\left(\dfrac{1}{2}-\dfrac{7}{8}\right)\right]\)

\(=-\dfrac{1}{24}-\left[\dfrac{1}{4}-\left(\dfrac{4}{8}-\dfrac{7}{8}\right)\right]\)

\(=-\dfrac{1}{24}-\left(\dfrac{1}{4}+\dfrac{3}{8}\right)\)

\(=-\dfrac{1}{24}-\left(\dfrac{2}{8}+\dfrac{3}{8}\right)\)

\(=-\dfrac{1}{24}-\dfrac{5}{8}\)

\(=-\dfrac{1}{24}-\dfrac{15}{24}\)

\(=\dfrac{-16}{24}\)

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

a) Ta có: \(\dfrac{15}{7}>1\) (tử lớn hơn mẫu)

\(\dfrac{9}{14}< 1\) (tử nhỏ hơn mẫu)

Vậy: \(\dfrac{15}{7}>\dfrac{9}{14}\)

b) Ta có: 

\(\dfrac{899}{900}=1-\dfrac{1}{900}\)

\(\dfrac{1235}{1236}=1-\dfrac{1}{1236}\)

Mà: \(\dfrac{1}{900}>\dfrac{1}{1236}\)

Vậy: \(\dfrac{1235}{1236}>\dfrac{899}{900}\)

c) Ta có:

\(\dfrac{77}{75}=1+\dfrac{2}{75}\)

\(\dfrac{37}{35}=1+\dfrac{2}{35}\)

Mà: \(\dfrac{2}{75}< \dfrac{2}{35}\)

Vậy: \(\dfrac{37}{35}>\dfrac{77}{75}\)

(*phải là cạnh chứ nhỉ*)

Đổi: 45000dm=4500m

Cạnh thứ nhất dài:

4500+3,69=4503,69(m)

Cạnh thứ ba dài:

4503,69+4500=9003,69

Đáp số: ...