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\(a,\dfrac{x}{5}=\dfrac{-18}{10}\\ \Rightarrow x=-\dfrac{18}{10}.5\\ \Rightarrow x=-9\\ b,\dfrac{6}{x-1}=\dfrac{-3}{7}\\ \Rightarrow6.7=-3\left(x-1\right)\\ \Rightarrow42=-3x+3\\ \Rightarrow42+3x-3=0\\ \Rightarrow3x+39=0\\ \Rightarrow3x=-39\\ \Rightarrow x=-13\\ c,\dfrac{y-3}{12}=\dfrac{3}{y-3}\\ \Rightarrow\left(y-3\right)^2=36\\ \Rightarrow\left[{}\begin{matrix}y-2=6\\y-2=-6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}y=8\\y=-4\end{matrix}\right.\)
\(d,\dfrac{x}{25}=\dfrac{-5}{x^2}\\ \Rightarrow x^3=-125\\ \Rightarrow x^3=\left(-5\right)^3\\ \Rightarrow x=-5\)
a) \(\dfrac{5}{x}=\dfrac{-10}{12}.\Rightarrow x=-6.\)
b) \(\dfrac{4}{-6}=\dfrac{x+3}{9}.\Rightarrow x+3=-6.\Leftrightarrow x=-9.\)
c) \(\dfrac{x-1}{25}=\dfrac{4}{x-1}.\left(đk:x\ne1\right).\Leftrightarrow\dfrac{x-1}{25}-\dfrac{4}{x-1}=0.\)
\(\Leftrightarrow\dfrac{x^2-2x+1-100}{25\left(x-1\right)}=0.\Leftrightarrow x^2-2x-99=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=11.\\x=-9.\end{matrix}\right.\) \(\left(TM\right).\)
a) \(\dfrac{-5}{6}.\dfrac{120}{25}< x< \dfrac{-7}{15}.\dfrac{9}{14}\)
\(\Rightarrow-4< x< \dfrac{-3}{10}\)
\(\Rightarrow\dfrac{-40}{10}< x< \dfrac{-3}{10}\)
\(\Rightarrow x\in\left\{\dfrac{-39}{10};\dfrac{-38}{10};\dfrac{-37}{10};...;\dfrac{-5}{10};\dfrac{-4}{10}\right\}\)
b) \(\left(\dfrac{-5}{3}\right)^2< x< \dfrac{-24}{35}.\dfrac{-5}{6}\)
\(\Rightarrow\dfrac{25}{9}< x< \dfrac{4}{7}\)
\(\Rightarrow\dfrac{175}{63}< x< \dfrac{36}{63}\)
\(\Rightarrow x=\varnothing\)
c) \(\dfrac{1}{18}< \dfrac{x}{12}< \dfrac{y}{9}< \dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{2}{36}< \dfrac{3x}{36}< \dfrac{4y}{36}< \dfrac{9}{36}\)
\(\Rightarrow x\in\left\{1;2\right\}\)
+) Với \(x=1\)
\(\Rightarrow y\in\left\{1;2\right\}\)
+) Với \(x=2\)
\(\Rightarrow y=2\)
Vậy \(x=1\) thì \(y\in\left\{1;2\right\}\); \(x=2\) thì \(y=8\).
Bài 2:
\(a,\dfrac{2}{x}=\dfrac{x}{8}\\ \Rightarrow x.x=8.2\\ \Rightarrow x^2=16\\ \Rightarrow x=\pm4\)
\(b,\dfrac{2x-9}{240}=\dfrac{39}{80}\\ \Rightarrow80\left(2x-9\right)=240.39\\ \Rightarrow160x-720=9360\\ \Rightarrow160x=10080\\ \Rightarrow x=63\)
\(c,\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Rightarrow3\left(x-1\right)=8.9\\ \Rightarrow3\left(x-1\right)=72\\ \Rightarrow x-1=24\\ \Rightarrow x=25\)
Lời giải:
a. $\frac{x}{7}=\frac{6}{21}$
$x=\frac{6}{21}.7$
$x=2$
b.
$\frac{-5}{y}=\frac{20}{28}$
$y=-5:\frac{20}{28}$
$y=-7$
c.
$\frac{-4}{8}=\frac{-7}{y}$
$y=-7:\frac{-4}{8}$
$y=14$
a, \(\dfrac{x}{7}=\dfrac{6}{21}\Leftrightarrow\dfrac{3x}{21}=\dfrac{6}{21}\Rightarrow x=2\)
b, \(\dfrac{-5}{y}=\dfrac{20}{28}\Leftrightarrow\dfrac{20}{-4y}=\dfrac{20}{28}\Leftrightarrow y=-7\)
c, \(\dfrac{-4}{8}=-\dfrac{7}{y}\Rightarrow-4y=-56\Leftrightarrow y=14\)
Giải:
a) \(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\)
\(\Rightarrow5x\in\left\{0;\pm5;10\right\}\)
\(\Rightarrow x\in\left\{0;\pm1;2\right\}\)
b) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\)
\(\Rightarrow-12.\left(x-6\right)=4.18\)
\(\Rightarrow-12x+72=72\)
\(\Rightarrow-12x=72-72\)
\(\Rightarrow-12x=0\)
\(\Rightarrow x=0:-12\)
\(\Rightarrow x=0\)
\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow y=\dfrac{-12.24}{18}=-16\)
c) \(\dfrac{x+46}{20}=x.\dfrac{2}{5}\)
\(\dfrac{x+46}{20}=\dfrac{2x}{5}\)
\(\Rightarrow5.\left(x+46\right)=2x.20\)
\(\Rightarrow5x+230=40x\)
\(\Rightarrow5x-40x=-230\)
\(\Rightarrow-35x=-230\)
\(\Rightarrow x=-230:-35\)
\(\Rightarrow x=\dfrac{46}{7}\)
Chúc bạn học tốt!
bài 3:
a, đặt \(\dfrac{x}{12}=\dfrac{y}{9}=\dfrac{z}{5}=k\)
=>x=12k,y=9k,z=5k
ta có: ayz=20=> 12k.9k.5k=20
=> (12.9.5)k^3=20
=>540.k^3=20
=>k^3=20/540=1/27
=>k=1/3
=>x=12.1/3=4
y=9.1/3=3
z=5.1/3=5/3
vậy x=4,y=3,z=5/3
b,ta có: \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}\)
A/D tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x^2}{25}=\dfrac{y^2}{49}=\dfrac{z^2}{9}=\dfrac{x^2+y^2-z^2}{25+49-9}=\dfrac{585}{65}=9\)
=>x=5.9=45
y=7.9=63
z=3*9=27
vậy x=45,y=63,z=27
a: =>x-3=9
=>x=12
b: =>10-x=-26
=>x=36
c: =>x:4-1=2
=>x:4=3
=>x=12
d: =>x^2=4
=>x=2 hoặc x=-2
e: =>(x-2)^2=100
=>x-2=10 hoặc x-2=-10
=>x=12 hoặc x=-8
a, \(\dfrac{42}{54}=\dfrac{7}{x}\)
Ta có: \(x.42=7.54\)
\(=>x.42=378\)
\(=>x=378:42\)
\(=>x=9\)
Vậy x = 9
b, \(\dfrac{-2}{3}=\dfrac{y}{15}\)
Ta có: \(y.3=\left(-2\right).15\)
\(=>y.3=-30\)
\(=>y=\left(-30\right):3\)
\(=>y=-10\)
Vậy y = -10
c, \(\dfrac{6}{10}=\dfrac{3}{x}=\dfrac{y}{-20}\)
* Ta có: \(x.6=3.10\)
\(=>x.6=30\)
\(=>x=30:6\)
\(=>x=5\)
Vì x = 5 \(\Rightarrow\dfrac{3}{5}=\dfrac{y}{-20}\)
Ta có: \(y.5=3.\left(-20\right)\)
\(=>y.5=-60\)
\(=>y=\left(-60\right):5\)
\(=>y=-12\)
Vậy x = 5 ; y = -12
d, \(\dfrac{-x}{-6}=\dfrac{-5}{6}\Rightarrow\dfrac{x}{6}=\dfrac{-5}{6}\Rightarrow x=-5\) ( Cùng mẫu số )
Vậy x = -5
\(#NqHahh\)
\(a.\) \(\dfrac{42}{54}=\dfrac{7}{x}\)
\(\Rightarrow x\cdot42=7\cdot54\)
\(\Rightarrow x\cdot42=378\)
\(\Rightarrow x=378:42\)
\(\Rightarrow x=9\)
Vậy \(\dfrac{42}{54}=\dfrac{7}{9}.\)
\(b.\) \(\dfrac{-2}{3}=\dfrac{y}{15}\)
\(\Rightarrow y\cdot3=\left(-2\right)\cdot15\)
\(\Rightarrow y\cdot3=\left(-30\right)\)
\(\Rightarrow y=\left(-30\right):3\)
\(\Rightarrow y=\left(-10\right)\)
Vậy \(\dfrac{-2}{3}=\dfrac{-10}{15}\)
\(c.\) \(\dfrac{6}{10}=\dfrac{3}{x}=\dfrac{y}{-20}\)
\(\Rightarrow x\cdot6=3\cdot10\)
\(\Rightarrow x\cdot6=30\)
\(\Rightarrow x=30:6\)
\(\Rightarrow x=5\)
Vậy: \(\dfrac{6}{10}=\dfrac{3}{5}=\dfrac{y}{-20}\)
Mặt khác: \(\dfrac{3}{5}=\dfrac{y}{-20}\)
\(\Rightarrow y\cdot5=3\cdot\left(-20\right)\)
\(\Rightarrow y\cdot5=\left(-60\right)\)
\(\Rightarrow y=\left(-60\right):5\)
\(\Rightarrow y=\left(-12\right)\)
Vậy \(\dfrac{6}{10}=\dfrac{3}{5}=\dfrac{-12}{-20}\)
\(d.\) \(\dfrac{-x}{-6}=\dfrac{-5}{6}\)
\(\Rightarrow\dfrac{x}{6}=\dfrac{-5}{6}\)
Do cùng mẫu số nên ta xét tử, ta thấy:
\(x=\left(-5\right)\)
Vậy \(\dfrac{-5}{6}=\dfrac{-5}{6}\)