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\(\frac{27}{4}=\frac{-x}{3}=>x=-\frac{81}{4}\notinℤ\)
\(^{y^2=\frac{4}{9}=\left(\frac{2}{3}\right)^2=>y=\pm\frac{2}{3}\notinℤ}\)
\(\frac{27}{4}=\frac{\left(z+3\right)}{-4}=\left(z+3\right)=-27=\left(-3\right)^3=>z+3=-3=>z=-6\)
\(+)|t|-2=-54=>|t|=-52\)(vô lí)
\(+)|t|-2=54=>|t|=56=>t=\pm56\)
x=\(\dfrac{-4.\left(-10\right)}{8}=5\).
y=\(\dfrac{-10.\left(-7\right)}{5}=14.\)
z=\(\dfrac{-7.\left(-24\right)}{14}=12.\)
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
Từ \(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}\) và \(\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{y}{12}=\dfrac{z}{15}\)
\(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8+12-15}=\dfrac{10}{5}=2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=2\Rightarrow x=2\cdot8=16\\\dfrac{y}{12}=2\Rightarrow y=2\cdot12=24\\\dfrac{z}{15}=2\Rightarrow z=2\cdot15=30\end{matrix}\right.\)
Theo bài ta có :
\(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}\)
\(x+y-z=10\)
\(\Leftrightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8-12+15}=\dfrac{10}{5}=2\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=2\Leftrightarrow x=16\\\dfrac{y}{12}=2\Leftrightarrow y=24\\\dfrac{z}{15}=2\Leftrightarrow z=30\end{matrix}\right.\)
Vậy ....
2)\(x+y+z=9^2=81\)
Ta có:\(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}\left(1\right)\)
\(\dfrac{y}{5}=\dfrac{z}{7}\Rightarrow\dfrac{y}{20}=\dfrac{z}{28}\left(2\right)\)
Từ (1) và (2)\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)
\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{x+y+z}{15+20+28}=\dfrac{81}{63}=\dfrac{9}{7}\)
\(\Rightarrow x=\dfrac{135}{7};y=\dfrac{180}{7};z=36\)
-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, \(\dfrac{3}{x}+\dfrac{y}{3}=\dfrac{5}{6}\)
ta có: \(\dfrac{3}{x}+\dfrac{y}{3}=\dfrac{5}{6}=>\dfrac{3}{x}=\dfrac{5}{6}-\dfrac{y}{3}=\dfrac{5-2y}{6}\)
=>\(\dfrac{3}{x}=\dfrac{5-2y}{6}=>x.\left(5-2y\right)=3.6=18\)
=> x và 5-2y thuộc Ư của 18={1,-1,2,-2,3,-3,6,-6}
vì 5-2y là số lẻ=> 5-2y= +-1 hoặc 5-2y=+-3
xét bảng
| 5-2y | 1 | -1 | 3 | -3 |
| y | 2 | 3 | 1 | 4 |
| x | 18 | -18 | 6 | -6 |
vậy giá trị x,y cần tìm là: {x=18.y=2}
{x=-18.y=3}
{x=6, y=1}Ư
{x=-6,y=4}
Bài 1:
a: =>3x-3-4=0
=>3x=7
hay x=7/3
b: =>2x-2+3x+6=0
=>5x+4=0
hay x=-4/5
c: =>\(4x^2+4x-1=0\)
hay \(x\in\left\{\dfrac{-1+\sqrt{2}}{2};\dfrac{-1-\sqrt{2}}{2}\right\}\)
d: \(\Leftrightarrow3x-3+2x-4+6=0\)
=>5x+1=0
hay x=-1/5
Ta có :
\(-\dfrac{24}{-6}=\dfrac{x}{3}\)
\(\Rightarrow x=\dfrac{-24\cdot3}{-6}=12\)
=> TA CÓ :
\(\dfrac{12}{3}=\dfrac{4}{y^2}\)
\(\Rightarrow y^2=\dfrac{4\cdot3}{12}=1\)
\(\Rightarrow y=\pm1\)
=> Ta có :
\(\dfrac{4}{1}=\dfrac{z^3}{-2}\)
\(\Rightarrow z^3=\dfrac{4\cdot\left(-2\right)}{1}=-8\)
\(\Rightarrow z=-2\)
Vậy x= 12 ; y = \(\pm1\) ;z=-2