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Bài 1:
\(a,\dfrac{x}{3}=\dfrac{y}{7}\) và \(x+y=20\)
\(=\dfrac{x+y}{3+7}=\dfrac{20}{10}=2\)
\(\Rightarrow x=2.3=6\)
\(y=2.7=14\)
Vậy \(x=6\) và \(y=14\)
\(b,\dfrac{x}{5}=\dfrac{y}{2}\) và \(x-y=6\)
\(=\dfrac{x-y}{5-2}=\dfrac{6}{3}=2\)
\(\Rightarrow x=2.5=10\)
\(y=2.2=4\)
Vậy \(x=10\) và \(y=4\)
\(c,\dfrac{x}{7}=\dfrac{18}{14}\)
Từ tỉ lệ thức trên ta có:
\(14x=7.18\)
\(x=\dfrac{7.18}{14}\)
\(x=9\)
Vậy \(x=9\)
\(d,6:x=1\dfrac{3}{4}:5\)
\(6:x=\dfrac{7}{20}\)
\(x=6:\dfrac{7}{20}\)
\(x=\dfrac{120}{7}\)
Vậy \(x=\dfrac{120}{7}\)
\(e,\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\) và \(x-y+z=8\)
\(=\dfrac{x-y+z}{2-4+6}=\dfrac{8}{4}=2\)
\(\Rightarrow x=2.2=4\)
\(y=2.4=8\)
\(z=2.6=12\)
Vậy \(x=4;y=8;z=12\)
a, \(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x+y}{3+7}=\dfrac{1}{2}\)
Từ đó suy ra x=1,5; y=3,5
b,\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{x-y}{5-2}=\dfrac{1}{2}\)
Từ đó suy ra x=2,5; y=1
c,\(\dfrac{x}{7}=\dfrac{18}{14}\Leftrightarrow\dfrac{x}{7}=\dfrac{9}{7}\Rightarrow x=9\)
d,\(\dfrac{6}{x}=\dfrac{\dfrac{7}{4}}{5}\Leftrightarrow\dfrac{6}{x}=\dfrac{24}{7}\left(\dfrac{\dfrac{7}{4}}{5}\right)\Leftrightarrow\dfrac{6}{x}=\dfrac{6}{\dfrac{120}{7}}\Rightarrow x=\dfrac{120}{7}\)
e,\(\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{8}=\dfrac{x-y+z}{2-4+8}=\dfrac{4}{3}\)
Từ đó suy ra x=\(\dfrac{8}{3}\); y=\(\dfrac{16}{3}\); z=\(\dfrac{32}{3}\)
1) \(M=\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{2^{30}+2^{20}}{2^{12}+2^{22}}=\dfrac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}=\dfrac{2^{20}}{2^{12}}=256\)
2) \(\dfrac{2x-y}{x+y}=\dfrac{2}{3}\Leftrightarrow2x+2y=6x-3y\Leftrightarrow4x=5y\Leftrightarrow\dfrac{x}{y}=\dfrac{5}{4}\)
\(a,A=\dfrac{\dfrac{3}{4}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}-\dfrac{5}{6}+\dfrac{5}{8}}\\ A=\dfrac{\dfrac{405}{572}}{\dfrac{645}{1001}}+\dfrac{\dfrac{5}{12}}{\dfrac{25}{24}}\\ A=\dfrac{189}{172}+\dfrac{2}{5}\\ A=\dfrac{1289}{860}\)
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
\(\dfrac{2x-y}{x+y}=\dfrac{2}{3}\Leftrightarrow\dfrac{2.\dfrac{x}{y}-1}{\dfrac{x}{y}+1}=\dfrac{2.\dfrac{x}{y}+2-3}{\dfrac{x}{y}+1}=2-\dfrac{3}{\dfrac{x}{y}+1}=\dfrac{2}{3}\)
\(2-\dfrac{2}{3}=\dfrac{4}{3}=\dfrac{3}{\dfrac{x}{y}+1}\)
\(\left(\dfrac{x}{y}+1\right)=\dfrac{9}{4}\Rightarrow\dfrac{x}{y}=\dfrac{9}{4}-\dfrac{4}{4}=\dfrac{5}{4}\)
\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
=> 2(2x+1) = 6.7
4x+2=42
4x=40
x=10
Vậy x=10
a)\(\dfrac{6}{2x+1}=\dfrac{2}{7}\\ =>6.7=2.\left(2x+1\right)\\ =>2x+1=\dfrac{6.7}{2}=\dfrac{42}{2}=21\\ =>2x=21-1=20\\ =>x=\dfrac{20}{2}=10\)
b) \(\dfrac{24}{7x-3}=-\dfrac{4}{25}\\ =>24.25=-4.\left(7x-3\right)\\ =>7x-3=\dfrac{24.25}{-4}=-150\\ =>7x=-150+3=-147\\ =>x=\dfrac{-147}{7}=-21\)
c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=-\dfrac{12}{18}\\ =>x-6=\dfrac{4.18}{-12}=-6\\ =>x=-6+6=0\\ y=\dfrac{-12.24}{18}=-16\)
d) \(-\dfrac{1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\\ < =>-\dfrac{8}{40}\le-\dfrac{5x}{40}\le\dfrac{10}{40}\\ =>-8\le-5x\le10\\ Mà:-8< -5.1< -5.0< -5.\left(-1\right)< -5.\left(-2\right)=10\\ =>x\in\left\{-2;-1;0;1\right\}\)
e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\\ < =>\dfrac{x+46}{20}=\dfrac{5x+2}{5}\\ =>5\left(x+46\right)=20\left(5x+2\right)\\ < =>5x+230=100x+40\\ < =>230-40=100x-5x\\ < =>190=95x\\ =>x=\dfrac{190}{95}=2\)
f) \(y\dfrac{5}{y}=\dfrac{56}{y}\\ < =>\dfrac{y^2+5}{y}=\dfrac{56}{y}\\ =>y\left(y^2+5\right)=56y\\ =>y^2+5=\dfrac{56y}{y}=56\\ =>y^2=56-5=51\\ =>y=\sqrt{51}\)
1.
\(\left(\dfrac{-1}{8}+\dfrac{-5}{6}\right)\cdot\dfrac{6}{23}\\ =-\dfrac{23}{24}\cdot\dfrac{6}{23}\\ =-\dfrac{6}{24}=-\dfrac{1}{4}\)
2. Xem lại đề nha!
4.
\(x+0,75=-1\dfrac{1}{4}\\ x+\dfrac{3}{4}=-\dfrac{3}{4}\\ x=-\dfrac{3}{4}-\dfrac{3}{4}\\ x=-\dfrac{3}{4}+\left(-\dfrac{3}{4}\right)=-\dfrac{6}{4}=-\dfrac{3}{2}\)
5.
\(\dfrac{x}{28}=-\dfrac{4}{7}\\ \Leftrightarrow7x=-4.28\\ \Rightarrow7x=-112\\ \Rightarrow x=-112:7=-16\)
6.
\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\\ \Leftrightarrow\left(3x-y\right).4=3\left(x+y\right)\\ \Rightarrow12x-4y=3x+3y\\ \Rightarrow12x-3x=4y+3y\\ \Rightarrow9x=7y\\ \Leftrightarrow\dfrac{x}{7}=\dfrac{y}{9}\Leftrightarrow\dfrac{x}{y}=\dfrac{7}{9}\)
Vậy giá trị của tỉ số \(\dfrac{x}{y}=\dfrac{7}{9}\).
Có
z= 2016 - (x+ y)
x= 2016- (y+ z)
y= 2016- (x+ z)
Thế vào ta được
[2016 - (x+ y)]/ (x+ y) + [2016 - (z+ y)]/ (z+ y) + [2016 - (x+ z)]/ (x+ z)
=2016/ (x+ y) - 1 + ( 2016)/ (z+ y)- 1 + ( 2016)/ (x+ z)-1
= 2016/ 8 - 3= 249
\(\dfrac{x}{8}-\dfrac{1}{4}=\dfrac{1}{y}\)
\(\Leftrightarrow\dfrac{x-2}{8}=\dfrac{1}{y}\)
\(\Leftrightarrow x-2=\dfrac{8}{y}\)
Do \(x-2\in Z\Rightarrow\dfrac{8}{y}\in Z\)
\(\Rightarrow y=Ư\left(8\right)\)
\(\Rightarrow y=\left\{-8;-4;-2;-1;1;2;4;8\right\}\)
\(\Rightarrow x=\left\{1;0;-2;-6;10;6;4;3\right\}\)