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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
\(\dfrac{2x-y}{x+y}=\dfrac{2}{3}\Leftrightarrow\dfrac{2.\dfrac{x}{y}-1}{\dfrac{x}{y}+1}=\dfrac{2.\dfrac{x}{y}+2-3}{\dfrac{x}{y}+1}=2-\dfrac{3}{\dfrac{x}{y}+1}=\dfrac{2}{3}\)
\(2-\dfrac{2}{3}=\dfrac{4}{3}=\dfrac{3}{\dfrac{x}{y}+1}\)
\(\left(\dfrac{x}{y}+1\right)=\dfrac{9}{4}\Rightarrow\dfrac{x}{y}=\dfrac{9}{4}-\dfrac{4}{4}=\dfrac{5}{4}\)
a,\(\dfrac{3}{5}+\dfrac{2}{7}=\dfrac{31}{35}\)
b,\(\dfrac{-4}{3}:\dfrac{2}{15}=\dfrac{-60}{6}=-10\)
c,\(\dfrac{3}{7}.\dfrac{2}{9}+\dfrac{7}{9}.\dfrac{3}{7}=\dfrac{3}{7}.\left(\dfrac{2}{9}+\dfrac{7}{9}\right)=\dfrac{3}{7}\)
d,\(\left(\dfrac{-2}{3}\right)^3.9^2+\left(\dfrac{-3}{4}\right)^2.32\)
\(=\dfrac{\left(-2\right)^3}{3^3}.3^4+\dfrac{\left(-3\right)^2}{4^2}.2^5\)
\(=\left(-8\right).3+\dfrac{3^2}{4^2}.2^5\)
\(=\left(-24\right)+2.9\)
\(=\left(-24\right)+18\)
\(=-6\)
a. = \(\dfrac{-1}{24}-\left\{\dfrac{1}{4}-\dfrac{-3}{8}\right\}\)
= \(\dfrac{-1}{24}-\left\{\dfrac{1}{4}+\dfrac{3}{8}\right\}\)
= \(\dfrac{-1}{24}-\dfrac{5}{8}\)
= \(\dfrac{-2}{3}\)
b. = \(12\dfrac{7}{88}-3\dfrac{5}{11}\)
= \(8\dfrac{5}{8}\)
c. = \(\dfrac{-28}{9}+\dfrac{-413}{9}\)
= \(-49\)
d. = \(\dfrac{8}{35}:\dfrac{2}{11}+\dfrac{-8}{35}:\dfrac{2}{11}\)
= \(\dfrac{2}{11}:\left(\dfrac{8}{35}+\dfrac{-8}{35}\right)\)
= 0
a) \(\dfrac{-2}{3}+\dfrac{3}{4}-\dfrac{-1}{6}+\dfrac{-2}{5}=\dfrac{1}{12}-\dfrac{-1}{6}+\dfrac{-2}{5}=\dfrac{1}{4}+\dfrac{-2}{5}=\dfrac{-3}{20}\)
b) \(\dfrac{-2}{3}+\dfrac{-1}{5}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{-7}{10}=\left(\dfrac{-2}{3}-\dfrac{5}{6}\right)+\left(\dfrac{-1}{5}-\dfrac{-7}{10}\right)+\dfrac{3}{4}\)
\(=\dfrac{-3}{2}+\dfrac{1}{2}-\dfrac{3}{4}\)
= \(=-1-\dfrac{3}{4}\)
\(=\dfrac{-1}{4}\)
c)\(\dfrac{1}{2}-\dfrac{-2}{5}+\dfrac{1}{3}+\dfrac{5}{7}-\dfrac{-1}{6}+\dfrac{-4}{35}+\dfrac{1}{41}\)
= \(\left(\dfrac{1}{2}-\dfrac{-1}{6}+\dfrac{1}{3}\right)+\left(\dfrac{-4}{35}+\dfrac{5}{7}-\dfrac{-2}{5}\right)+\dfrac{1}{41}\)
= \(1+1+\dfrac{1}{41}\)
= \(\dfrac{83}{41}\)
d)\(\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
= \(\dfrac{1}{100}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{98}+...+\dfrac{1}{3}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{1}\)
= \(\dfrac{1}{100}-\dfrac{1}{1}\)
= \(\dfrac{-99}{100}\)
d đảo 1/1.2.1/2.3 ... 1/99.1000
=1/1 -1/2 +1/2-1/3 ... -1/99 - 1/1000
=1/1 -1/1000
=999/1000
a: \(\Leftrightarrow\left(3x-2\right):\dfrac{7}{5}=\dfrac{17}{7}:\dfrac{13}{5}=\dfrac{85}{91}\)
\(\Leftrightarrow3x-2=\dfrac{85}{91}\cdot\dfrac{7}{5}=\dfrac{17}{13}\)
=>3x=43/13
hay x=43/39
b: \(\Leftrightarrow9x+207=121-8x\)
=>19x=-86
hay x=-86/19
c: \(\Leftrightarrow x^2-9=16\)
=>x2=25
=>x=5 hoặc x=-5
d: \(\Leftrightarrow\left|x\right|=\dfrac{1.64\cdot3.11}{8.51}\simeq0,6\)
=>x=0,6 hoặc x=-0,6
a) \(x+\dfrac{1}{3}=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}-\dfrac{1}{3}\Leftrightarrow x=\dfrac{5}{12}\) vậy \(x=\dfrac{5}{12}\)
b) \(x-\dfrac{2}{5}=\dfrac{5}{7}\Leftrightarrow x=\dfrac{5}{7}+\dfrac{2}{5}\Leftrightarrow x=\dfrac{39}{35}\) vậy \(x=\dfrac{39}{35}\)
c) \(-x-\dfrac{2}{3}=\dfrac{-6}{7}\Leftrightarrow x=\dfrac{-2}{3}+\dfrac{6}{7}\Leftrightarrow x=\dfrac{4}{21}\) vậy \(x=\dfrac{4}{21}\)
d) \(\dfrac{4}{7}-x=\dfrac{1}{3}\Leftrightarrow x=\dfrac{4}{7}-\dfrac{1}{3}\Leftrightarrow x=\dfrac{5}{21}\) vậy \(x=\dfrac{5}{21}\)
a) x + \(\dfrac{1}{3}\) = \(\dfrac{3}{4}\)
x = \(\dfrac{3}{4}\) - \(\dfrac{1}{3}\)
x = \(\dfrac{5}{12}\)
Vậy x = \(\dfrac{5}{12}\)
b) x - \(\dfrac{2}{5}\) = \(\dfrac{5}{7}\)
x = \(\dfrac{5}{7}\) + \(\dfrac{2}{5}\)
x = \(\dfrac{39}{35}\)
Vậy x = \(\dfrac{39}{35}\)
c) -x - \(\dfrac{2}{3}\) = \(-\dfrac{6}{7}\)
- x = \(-\dfrac{6}{7}\) + \(\dfrac{2}{3}\)
- x = \(-\dfrac{4}{21}\)
⇒ x = \(\dfrac{4}{21}\)
Vậy x = \(\dfrac{4}{21}\)
d) \(\dfrac{4}{7}\) - x = \(\dfrac{1}{3}\)
x = \(\dfrac{4}{7}\) - \(\dfrac{1}{3}\)
x = \(\dfrac{5}{21}\)
Vậy x = \(\dfrac{5}{21}\)
a, 1/3-3/4+3/5+1/4-2/9-1/36+1/15
=(1/3+3/5+1/15)-(3/4-1/4+2/9+1/36)
=1 - 3/4
=1/4
b, 3-1/4+2/3-5-1/3+6/5-6+7/4-3/2
=(3-5-6)-(1/4-7/4)+(2/3-1/3)+(6/5-3/2)
=-8 +3/2 +1/3 -3/10
=-97/15
Bài1:
Ta có:
a)\(\sqrt{\dfrac{3^2}{5^2}}=\sqrt{\dfrac{9}{25}}=\dfrac{3}{5}\)
b)\(\dfrac{\sqrt{3^2}+\sqrt{42^2}}{\sqrt{5^2}+\sqrt{70^2}}=\dfrac{\sqrt{9}+\sqrt{1764}}{\sqrt{25}+\sqrt{4900}}=\dfrac{3+42}{5+70}=\dfrac{45}{75}=\dfrac{3}{5}\)
c)\(\dfrac{\sqrt{3^2}-\sqrt{8^2}}{\sqrt{5^2}-\sqrt{8^2}}=\dfrac{\sqrt{9}-\sqrt{64}}{\sqrt{25}-\sqrt{64}}=\dfrac{3-8}{5-8}=\dfrac{-5}{-3}=\dfrac{5}{3}\)
Từ đó, suy ra: \(\dfrac{3}{5}=\sqrt{\dfrac{3^2}{5^2}}=\dfrac{\sqrt{3^2}+\sqrt{42^2}}{\sqrt{5^2}+\sqrt{70^2}}\)
Bài 2:
Không có đề bài à bạn?
Bài 3:
a)\(\sqrt{x}-1=4\)
\(\Rightarrow\sqrt{x}=5\)
\(\Rightarrow x=\sqrt{25}\)
\(\Rightarrow x=5\)
b)Vd:\(\sqrt{x^4}=\sqrt{x.x.x.x}=x^2\Rightarrow\sqrt{x^4}=x^2\)
Từ Vd suy ra:\(\sqrt{\left(x-1\right)^4}=16\)
\(\Rightarrow\left(x-1\right)^2=16\)
\(\Rightarrow\left(x-1\right)^2=4^2\)
\(\Rightarrow x-1=4\)
\(\Rightarrow x=5\)
Bài 1:
a) \(x^2-3=1\)
\(\Rightarrow x^2=1+3=4\)
\(\Rightarrow x=\pm2\)
b)\(2x^3+12=-4\)
\(\Rightarrow2x^3=-4-12=-16\)
\(\Rightarrow x^3=-8\)
\(\Rightarrow x=-2\)
c)\(\left(2x-3\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=\dfrac{7}{2}\\-\dfrac{1}{2}\end{matrix}\right.\)
a) \(x^2-3=1\Rightarrow x^2=4\Rightarrow x=\pm2\)
b) \(2x^3+12=-4\Rightarrow2x^3=-16\)
\(\Rightarrow x^3=-\dfrac{16}{2}=-8=-2^3\)
\(\Rightarrow x=-2\)
c) \(\left(2x-3\right)^2=16\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
d,h,i,k cững tương tự....
Thực hiện phép tính
a) \(1-\dfrac{1}{2}:\dfrac{-3}{2}+\dfrac{2}{3}=1-\dfrac{1}{2}.\dfrac{2}{-3}+\dfrac{2}{3}=\)
\(=1+\dfrac{1}{3}+\dfrac{2}{3}=\dfrac{3+1+2}{3}=\dfrac{6}{3}=2\)
b) \(\dfrac{-3}{4}-\dfrac{-7}{2}+\dfrac{2}{-3}=\dfrac{-3}{4}+\dfrac{7}{2}+\dfrac{-2}{3}=\)
\(=\dfrac{-9+42-8}{12}=\dfrac{25}{12}\)
c) \(0,25-3\dfrac{1}{2}:\dfrac{1}{2}+\dfrac{-3}{4}.\dfrac{2}{-3}=0,25-\dfrac{7}{2}.2+\dfrac{1}{2}=\)
\(=0,25-7+0,5=-\dfrac{25}{4}\)