Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 2:
a: =>11/13-5/42+x=15/18+11/13
=>x-5/42=15/18
=>x=5/6+5/42=35/42+5/42=40/42=20/21
b: 2x-3=x+1/2
=>2x-x=3+1/2
=>x=7/2
Theo bài ra ta có:
\(\dfrac{a}{b}=\dfrac{3}{4};\dfrac{b}{c}=\dfrac{8}{9}\)
\(\Rightarrow\dfrac{a}{b}.\dfrac{b}{c}=\dfrac{3}{4}.\dfrac{8}{9}\Rightarrow\dfrac{a}{c}=\dfrac{2}{3}\)
Vậy tỉ số giữa a và c là \(\dfrac{2}{3}\).
Chúc bạn học tốt!!!
Bài làm
Theo đề bài ta có :
\(\dfrac{a}{b}=\dfrac{3}{4};\dfrac{b}{c}=\dfrac{8}{9}\)
=> \(\dfrac{a}{b}.\dfrac{b}{c}=\dfrac{3}{4}.\dfrac{8}{9}\)
=> \(\dfrac{a}{c}=\dfrac{2}{3}\)
Vậy tỉ số giữa a và c là \(\dfrac{2}{3}\)
Giải:
Theo đề ra, ta có:
\(\dfrac{a}{b}=\dfrac{2}{7}\) và \(\dfrac{b}{c}=\dfrac{21}{26}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=\dfrac{b}{7}\\\dfrac{b}{21}=\dfrac{c}{26}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{6}=\dfrac{b}{21}\\\dfrac{b}{21}=\dfrac{c}{26}\end{matrix}\right.\Leftrightarrow\dfrac{a}{6}=\dfrac{b}{21}=\dfrac{c}{26}\)
\(\Leftrightarrow\dfrac{a}{6}=\dfrac{c}{26}\)
\(\Leftrightarrow\dfrac{a}{c}=\dfrac{6}{26}=\dfrac{3}{13}\)
Vậy tỉ số của hai số a vad b là \(\dfrac{3}{13}\).
Chúc bạn học tốt!!!
Tỉ số của a và b là \(\dfrac{2}{7}\)
\(\Rightarrow\) Tỉ số của a và b là \(\dfrac{6}{21}\)
Tỉ số của a và b là \(\dfrac{6}{21}\), tỉ số của b và c là \(\dfrac{21}{26}\). Vậy tỉ số của a và c là \(\dfrac{6}{26}\) hay \(\dfrac{3}{13}\)
Bài 1:
a)
\(\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Leftrightarrow\dfrac{x-1}{9}=\dfrac{24}{9}\\ \Leftrightarrow x-1=24\\ x=24+1\\ x=25\)
b)
\(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{8}\\ \dfrac{3x}{7}+1=\dfrac{-1}{8}\cdot\left(-4\right)\\ \dfrac{3x}{7}+1=\dfrac{1}{2}\\ \dfrac{3x}{7}=\dfrac{1}{2}-1\\ \dfrac{3x}{7}=\dfrac{-1}{2}\\ 3x=\dfrac{-1}{2}\cdot7\\ 3x=\dfrac{-7}{2}\\ x=\dfrac{-7}{2}:3\\ x=\dfrac{-7}{6}\)
c)
\(x+\dfrac{7}{12}=\dfrac{17}{18}-\dfrac{1}{9}\\ x+\dfrac{7}{12}=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{12}\\ x=\dfrac{1}{4}\)
d)
\(0,5x-\dfrac{2}{3}x=\dfrac{7}{12}\\ \dfrac{1}{2}x-\dfrac{2}{3}x=\dfrac{7}{12}\\ x\cdot\left(\dfrac{1}{2}-\dfrac{2}{3}\right)=\dfrac{7}{12}\\ \dfrac{-1}{6}x=\dfrac{7}{12}\\ x=\dfrac{7}{12}:\dfrac{-1}{6}\\ x=\dfrac{-7}{2}\)
e)
\(\dfrac{29}{30}-\left(\dfrac{13}{23}+x\right)=\dfrac{7}{46}\\ \dfrac{29}{30}-\dfrac{13}{23}-x=\dfrac{7}{46}\\ \dfrac{277}{690}-x=\dfrac{7}{46}\\ x=\dfrac{277}{690}-\dfrac{7}{46}\\ x=\dfrac{86}{345}\)
f)
\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\\ \left(x-\dfrac{1}{12}\right):\dfrac{23}{12}=\dfrac{7}{46}\\ x-\dfrac{1}{12}=\dfrac{7}{46}\cdot\dfrac{23}{12}\\ x-\dfrac{1}{12}=\dfrac{7}{24}\\ x=\dfrac{7}{24}+\dfrac{1}{12}\\ x=\dfrac{3}{8}\)
g)
\(\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{13}{15}-\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{1}{6}\\ \dfrac{13}{21}+x=\dfrac{1}{6}:\dfrac{7}{12}\\ \dfrac{13}{21}+x=\dfrac{2}{7}\\ x=\dfrac{2}{7}-\dfrac{13}{21}\\ x=\dfrac{-1}{3}\)
h)
\(2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}+\dfrac{3}{2}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}:2\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{8}\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\end{matrix}\right.\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\ \dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{29}{24}\\ x=\dfrac{29}{24}:\dfrac{1}{2}\\ x=\dfrac{29}{12}\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\\ \dfrac{1}{2}x=\dfrac{-7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{-13}{24}\\ x=\dfrac{-13}{24}:\dfrac{1}{2}\\ x=\dfrac{-13}{12}\)
i)
\(3\cdot\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=0-\dfrac{1}{9}\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}:3\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{27}\\ \left(3x-\dfrac{1}{2}\right)^3=\left(\dfrac{-1}{3}\right)^3\\ \Leftrightarrow3x-\dfrac{1}{2}=\dfrac{-1}{3}\\ 3x=\dfrac{-1}{3}+\dfrac{1}{2}\\ 3x=\dfrac{1}{6}\\ x=\dfrac{1}{6}:3\\ x=\dfrac{1}{18}\)
Bài 1:
a, Ta có:
\(\dfrac{-8}{15}=-\dfrac{5}{18}+-\dfrac{1}{6}\)
b, Ta có:
\(-\dfrac{8}{15}=\dfrac{11}{15}-\dfrac{19}{15}\)
Bài 2:
a, \(\dfrac{11}{13}-\left(\dfrac{5}{12}-x\right)=-\left(\dfrac{15}{18}-\dfrac{11}{13}\right)\)
\(\Rightarrow\dfrac{11}{13}-\dfrac{5}{12}+x=-\dfrac{15}{18}+\dfrac{11}{13}\)
\(\Rightarrow x=-\dfrac{15}{18}+\dfrac{11}{13}+\dfrac{5}{12}-\dfrac{11}{13}\)
\(\Rightarrow x=-\dfrac{15}{8}+\dfrac{5}{12}=-\dfrac{35}{24}\)
b, \(2x-3=x+\dfrac{1}{2}\)
\(\Rightarrow2x-x=\dfrac{1}{2}+3\Rightarrow x=\dfrac{7}{2}\)
Chúc bạn học tốt!!!
a) \(\dfrac{1,28}{3,15}=\dfrac{128}{315}\)
b) \(\dfrac{2}{5}:3\dfrac{1}{4}=\dfrac{2}{5}:\dfrac{13}{4}=\dfrac{2}{5}\cdot\dfrac{4}{13}=\dfrac{8}{65}\)
c) \(\left(-1\dfrac{3}{7}\right):1,24=\dfrac{-10}{7}:\dfrac{31}{25}=\dfrac{-10}{7}\cdot\dfrac{25}{31}=\dfrac{-250}{217}\)
d) \(2\dfrac{1}{5}:3\dfrac{1}{7}=\dfrac{11}{5}:\dfrac{22}{7}=\dfrac{11}{5}\cdot\dfrac{7}{22}=\dfrac{7}{10}\)
Khi đem mỗi phân số đã cho cộng với phân số \(\dfrac{A}{B}\) thì được hai phân số mới có hiệu không thay đổi và bằng:
\(\dfrac{7}{11}-\dfrac{1}{5}=\dfrac{24}{55}\)
Phân số lớn mới là:
\(\dfrac{24}{55}:\left(3-1\right).3=\dfrac{36}{55}\)
Phân số \(\dfrac{A}{B}=\dfrac{36}{55}-\dfrac{7}{11}=\dfrac{1}{55}\)
Đ/S: \(\dfrac{1}{55}\)
B2:
a, \(25\times(-\dfrac{1}{5})^2+8^3:\left(\dfrac{4}{3}\right)^3\)
= \(25\times\dfrac{1}{25}+512:\dfrac{64}{3}\)
= \(1+24\)
= 25
b, \(27:\left(\dfrac{3}{2}\right)^3-4^2\times\left(-\dfrac{1}{2}\right)^2\)
= \(27:\dfrac{27}{8}-16\times\dfrac{1}{4}\)
= \(8-4\)
= 4
\(\dfrac{a}{b+c}=\dfrac{b}{a+c}=\dfrac{c}{a+b}\)
\(\Rightarrow\dfrac{a}{b+c}+1=\dfrac{b}{a+c}+1=\dfrac{c}{a+b}+1\)
\(\Rightarrow\dfrac{a+b+c}{b+c}=\dfrac{a+b+c}{a+c}=\dfrac{a+b+c}{a+b}\)
\(\Rightarrow b+c=a+c=b+a\)
\(\Rightarrow a=b=c\)
\(\dfrac{a}{b+c}=\dfrac{b}{a+c}=\dfrac{c}{a+b}=\dfrac{a}{a+a}=\dfrac{1}{2}\)
Cậu chưa xét a+b+c=0