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x=\(\dfrac{-4.\left(-10\right)}{8}=5\).
y=\(\dfrac{-10.\left(-7\right)}{5}=14.\)
z=\(\dfrac{-7.\left(-24\right)}{14}=12.\)
\(\frac{27}{4}=\frac{-x}{3}=>x=-\frac{81}{4}\notinℤ\)
\(^{y^2=\frac{4}{9}=\left(\frac{2}{3}\right)^2=>y=\pm\frac{2}{3}\notinℤ}\)
\(\frac{27}{4}=\frac{\left(z+3\right)}{-4}=\left(z+3\right)=-27=\left(-3\right)^3=>z+3=-3=>z=-6\)
\(+)|t|-2=-54=>|t|=-52\)(vô lí)
\(+)|t|-2=54=>|t|=56=>t=\pm56\)
\(\dfrac{-4}{8}=\dfrac{x}{-10}=\dfrac{-7}{y}=\dfrac{z}{24}\)
\(\Leftrightarrow\dfrac{-1}{2}=\dfrac{-7}{y}\)
\(\Rightarrow y=14\)
\(\Leftrightarrow\)\(\dfrac{1}{-2}=\dfrac{x}{-10}=\dfrac{z}{24}\)
\(\Rightarrow\) x=5
z=(-12)
Vậy: x=5; y=14; z=(-12)
-x/3=24/4=6
=>x=-18
3/y2=6
=>y2=1/2
hay \(y=\pm\dfrac{\sqrt{2}}{2}\)
\(\dfrac{\left(z+3\right)^3}{-4}=6\)
=>(z+3)3=-24
\(\Leftrightarrow z+3=-\sqrt[3]{24}\)
hay \(z=-\sqrt[3]{24}-3\)
||t|-2|/8=6
=>||t|-2|=48
=>|t|-2=48
=>t=50 hoặc t=-50
a ) \(7x=4y\) hay \(\dfrac{x}{4}=\dfrac{y}{7}\) và \(y-z=24\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :
\(\dfrac{x}{4}=\dfrac{y}{7}=\dfrac{y-z}{7-4}=\dfrac{24}{3}=8\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=32\\y=56\end{matrix}\right.\)
Vậy ............
b ) \(\dfrac{x}{5}=\dfrac{y}{6},\dfrac{y}{8}=\dfrac{z}{7}\)
hay : \(\dfrac{x}{40}=\dfrac{y}{48}=\dfrac{z}{42}\) và \(x+y-z=69\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :
\(\dfrac{x}{40}=\dfrac{y}{48}=\dfrac{z}{42}=\dfrac{x+y-z}{40+48-42}=\dfrac{69}{46}=\dfrac{3}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=60\\y=72\\z=63\end{matrix}\right.\)
Vậy .......
c.
\(\dfrac{x}{y}=\dfrac{2}{5}=\dfrac{x}{2}=\dfrac{y}{5}\)và x - y = 40
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x-y}{2-5}=\dfrac{40}{-3}\)
\(\dfrac{x}{2}=\dfrac{40}{-3}\Rightarrow x=\dfrac{40.2}{-3}=-\dfrac{80}{3}\)
\(\dfrac{y}{5}=\dfrac{40}{-3}\Rightarrow y=\dfrac{40.5}{-3}=-\dfrac{200}{3}\)
Vậy x = \(-\dfrac{80}{3}\), y = \(-\dfrac{200}{3}\)
Tương tự tiếp nghen 
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. \(\Rightarrow\left\{\begin{matrix}\dfrac{-10}{15}=\dfrac{x}{-9}\\\dfrac{-10}{15}=\dfrac{-8}{y}\\\dfrac{-10}{15}=\dfrac{z}{-21}\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=6\\y=12\\z=14\end{matrix}\right.\)
b. \(\Rightarrow\left\{\begin{matrix}\dfrac{-7}{6}=\dfrac{x}{18}\\\dfrac{-7}{6}=\dfrac{-98}{y}\\\dfrac{-7}{6}=\dfrac{-14}{z}\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=-21\\y=84\\z=-12\end{matrix}\right.\)
a) Ta có: \(\dfrac{-10}{15}=\dfrac{x}{-9}\)
\(\Rightarrow15x=-10.\left(-9\right)\)
\(\Rightarrow15x=90\)
\(\Rightarrow x=6\)
Khi đó: \(\dfrac{6}{-9}=\dfrac{-8}{y}=\dfrac{z}{-21}\)
\(\Rightarrow y=\dfrac{-8\left(-9\right)}{6}=12\)
và \(z=\dfrac{-8\left(-21\right)}{12}\) \(=14\)
Vậy \(\left[{}\begin{matrix}x=6\\y=12\\z=14\end{matrix}\right.\)
b) Lại có: \(\dfrac{-7}{6}=\dfrac{x}{18}\)
\(\Rightarrow6x=-7.18\)
\(\Rightarrow6x=-126\)
\(\Rightarrow x=-21\)
Khi đó \(\dfrac{-21}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}\)
\(\Rightarrow y=\dfrac{-98.18}{-21}=84\)
và \(z=\dfrac{-14.84}{-98}=12\)
Vậy \(\left[{}\begin{matrix}x=-21\\y=84\\z=12\end{matrix}\right.\)
b, \(\dfrac{x-3}{4}=\dfrac{15}{20}\)
<=> \(\dfrac{x-3}{4}=\dfrac{3}{4}\)
=> x-3=3
<=> x=6
Vậy x=6
\(a,\dfrac{x}{15}=\dfrac{4}{y}=\dfrac{-2}{5}\)
* \(\dfrac{x}{15}=\dfrac{-2}{5}\)
\(\Rightarrow\dfrac{x}{15}=\dfrac{-6}{15}\)
\(\Rightarrow x=-6\)
*\(\dfrac{4}{y}=\dfrac{-2}{5}\)
\(\Rightarrow\dfrac{4}{y}=\dfrac{4}{-10}\)
\(\Rightarrow y=-10\)
Vậy x = - 6 ; y = - 10
\(b,\dfrac{x-3}{4}=\dfrac{15}{20}\)
=> ( x - 3 ) . 20 = 4. 15
=> 20x - 60 = 60
=> 20x = 60 + 60
=> 20x = 120
=> x = 120 : 20
=> x = 6
Vậy x = 6
\(c,\dfrac{-5}{9}+\dfrac{-8}{15}+\dfrac{22}{-9}+\dfrac{-7}{15}< x\le\dfrac{-1}{3}+\dfrac{-1}{4}+\dfrac{-5}{12}\)
\(\Rightarrow\dfrac{-5}{9}+\dfrac{-8}{15}+\dfrac{-22}{9}+\dfrac{-7}{15}< x\le\dfrac{-4}{12}+\dfrac{-3}{12}+\dfrac{-5}{12}\)
\(\Rightarrow\left(\dfrac{-5}{9}+\dfrac{-22}{9}\right)+\left(\dfrac{-8}{15}+\dfrac{-7}{15}\right)< x\le-1\)
\(\Rightarrow-3+\left(-1\right)< x\le-1\)
\(\Rightarrow-4< x\le-1\)
\(\Rightarrow x=-3;-2;-1\)
Bài 2:
\(a,\dfrac{2}{x}=\dfrac{x}{8}\\ \Rightarrow x.x=8.2\\ \Rightarrow x^2=16\\ \Rightarrow x=\pm4\)
\(b,\dfrac{2x-9}{240}=\dfrac{39}{80}\\ \Rightarrow80\left(2x-9\right)=240.39\\ \Rightarrow160x-720=9360\\ \Rightarrow160x=10080\\ \Rightarrow x=63\)
\(c,\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Rightarrow3\left(x-1\right)=8.9\\ \Rightarrow3\left(x-1\right)=72\\ \Rightarrow x-1=24\\ \Rightarrow x=25\)
Ta có: \(-\dfrac{12}{16}=-\dfrac{3}{4}\\ Vì:-\dfrac{12}{16}=\dfrac{15}{x}=\dfrac{y}{-24}=\dfrac{-z}{8}\left(gt\right)\\ < =>-\dfrac{3}{4}=\dfrac{15}{x}=\dfrac{y}{-24}=\dfrac{-z}{8}\\ =>x=\dfrac{15.4}{-3}=-20\\ y=\dfrac{-3.\left(-24\right)}{4}=18\\ z=-\dfrac{\left(-3\right).8}{4}=6\)
Ta có:\(\dfrac{-12}{16}\) =\(\dfrac{-3}{4}\)
mà \(\dfrac{-12}{16}=\dfrac{15}{x}=\dfrac{y}{-24}=\dfrac{-Z}{8}\)
=>\(\dfrac{-3}{4}=\dfrac{15}{x}=\dfrac{y}{-24}=\dfrac{-Z}{8}\)
=> -3.x=15.4=60
4.y=-3.(-24)=72
Z.(-4)=-3.8=-24
=>x=-20
y=18
Z=6
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