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8)\(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)
=\(\frac{4}{9}:\left(-\frac{1}{7}\right)+\frac{59}{9}:\left(-\frac{1}{7}\right)\)
=\(\left(\frac{4}{9}+\frac{59}{9}\right).\left(-7\right)\)
=7.(-7)
=-49
a)
Ta có:
\(\dfrac{-8}{14}=\dfrac{-4}{7}\): \(\dfrac{2}{27}=\dfrac{2}{27}\) : \(\dfrac{12}{-21}=\dfrac{4}{-7}=\dfrac{-4}{7}\) : \(\dfrac{-36}{63}=\dfrac{-4}{7}\) : \(\dfrac{-12}{-54}=\dfrac{-2}{-9}=\dfrac{2}{9}\) : \(\dfrac{-16}{27}=\dfrac{-16}{27}\)
Vậy trong các phân số trên, các phân số: \(\dfrac{-8}{14};\dfrac{12}{-21};\dfrac{-36}{63}\) biểu diễn cùng 1 số hữu tỉ.
b) Ta có : \(-0,75=\dfrac{-3}{4}\)
\(\Rightarrow3\) phân số cùng biểu diễn số hữu tỉ trên là: \(\dfrac{-6}{8};\dfrac{-9}{12};\dfrac{-12}{16}\)
Giải:
a) \(\dfrac{1}{3}x+\dfrac{1}{5}-\dfrac{1}{2}x=1\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{6}x=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{1}{6}x=\dfrac{-21}{20}\)
\(\Leftrightarrow x=\dfrac{-63}{10}\)
Vậy ...
b) \(\dfrac{3}{2}\left(x+\dfrac{1}{2}\right)-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{3}{2}x+\dfrac{3}{4}-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{11}{8}x=\dfrac{-1}{2}\)
\(\Leftrightarrow x=\dfrac{-4}{11}\)
Vậy ...
Các câu sau làm tương tự câu b)
bài 1) ta có : \(\dfrac{2x-y}{x+y}=\dfrac{2}{3}\Leftrightarrow2\left(x+y\right)=3\left(2x-y\right)\)
\(\Leftrightarrow2x+2y=6x-3y\Leftrightarrow4x=5y\Leftrightarrow\dfrac{x}{y}=\dfrac{5}{4}\)
vậy \(\dfrac{x}{y}=\dfrac{5}{4}\)
bài 1
\(\dfrac{2x-y}{x+y}=\dfrac{2}{3}\Leftrightarrow\dfrac{2.\dfrac{x}{y}-1}{\dfrac{x}{y}+1}=\dfrac{2.\dfrac{x}{y}+2-3}{\dfrac{x}{y}+1}=2-\dfrac{3}{\dfrac{x}{y}+1}=\dfrac{2}{3}\)
\(2-\dfrac{2}{3}=\dfrac{4}{3}=\dfrac{3}{\dfrac{x}{y}+1}\)
\(\left(\dfrac{x}{y}+1\right)=\dfrac{9}{4}\Rightarrow\dfrac{x}{y}=\dfrac{9}{4}-\dfrac{4}{4}=\dfrac{5}{4}\)
\(\dfrac{1}{4}>-\dfrac{3}{4},-1.5< \dfrac{1}{2},\dfrac{2}{-7}< -\dfrac{3}{11}\)
a: \(=\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\cdot\dfrac{18}{5}-\dfrac{6}{5}:\dfrac{-9}{5}+4\)
\(=\dfrac{18}{5}-\dfrac{6}{5}\cdot\dfrac{-5}{9}+4\)
\(=\dfrac{18}{5}+\dfrac{2}{3}+4\)
\(=\dfrac{124}{15}\)
b: \(=\dfrac{9}{25}\cdot\left(\dfrac{3}{5}-\dfrac{1}{5}+\dfrac{1}{2}\right)-\dfrac{3}{8}:\dfrac{9}{8}\)
\(=\dfrac{9}{25}\cdot\dfrac{4}{10}-\dfrac{1}{3}\)
\(=-\dfrac{71}{375}\)
c: \(=\dfrac{7}{10}:\dfrac{4}{5}+\dfrac{2}{9}:\dfrac{5}{9}+\dfrac{1}{8}\)
\(=\dfrac{7}{10}\cdot\dfrac{5}{4}+\dfrac{2}{5}+\dfrac{1}{8}\)
=1+2/5
=7/5
d: \(=\dfrac{3}{7}\left(19+\dfrac{1}{3}-33-\dfrac{1}{3}\right)-\dfrac{2}{7}=\dfrac{3}{7}\cdot\left(-14\right)-\dfrac{2}{7}=-6-\dfrac{2}{7}=\dfrac{-44}{7}\)
e: \(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{-2^{11}\cdot3^{11}-2^{12}\cdot3^{12}}\)
\(=\dfrac{2^{12}\cdot3^{10}\cdot6}{-2^{11}\cdot3^{11}\left(1+2\cdot3\right)}=-\dfrac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\cdot7}=\dfrac{-4}{7}\)
2,
a, Gọi 3 góc của tam giác đó là: A;B;C \(\left(A;B;C>0\right)\)
Theo đề bài ta có:
\(\widehat{\dfrac{A}{1}}=\widehat{\dfrac{B}{2}}=\widehat{\dfrac{C}{3}}\) và \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\) (đ/l tổng 3 góc của tam giác)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{\widehat{A}}{1}=\widehat{\dfrac{B}{2}}=\widehat{\dfrac{C}{3}}=\dfrac{\widehat{A}+\widehat{B}+\widehat{C}}{1+2+3}=\dfrac{180^0}{6}=30^0\)
+) \(\widehat{\dfrac{A}{1}}=30^0\Rightarrow\widehat{A}=30^0.1=30^0\)
+) \(\widehat{\dfrac{B}{2}}=30^0\Rightarrow\widehat{B}=30^0.2=60^0\)
+) \(\widehat{\dfrac{C}{3}}=30^0\Rightarrow\widehat{C}=30^0.3=90^0\)
Vì \(\Delta ABC\) có \(\widehat{C}=90^0\Rightarrow\Delta ABC\) là tam giác vuông.
b, C A B
1) Tính hợp lý :
\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{7}-\dfrac{1}{6}+\dfrac{1}{5}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{2}\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+\left(-\dfrac{1}{5}+\dfrac{1}{5}\right)+\left(\dfrac{1}{6}-\dfrac{1}{6}\right)+\left(-\dfrac{1}{7}+\dfrac{1}{7}\right)+\dfrac{1}{8}\)
\(=\dfrac{1}{8}\)
Các phân số cùng biểu diễn số hữu tỉ là: \(\dfrac{-2}{4};\dfrac{-1}{2};\dfrac{4}{-8}\)