46 : 44 - 2 x y - 15 với x = 2 : y = 1
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Giải:
a) \(\dfrac{-5}{8}=\dfrac{x}{16}\)
\(\Rightarrow x=\dfrac{16.-5}{8}=-10\)
\(\dfrac{3x}{9}=\dfrac{2}{6}\)
\(\Rightarrow3x=\dfrac{2.9}{6}=3\)
\(\Rightarrow x=1\)
b) \(\dfrac{x+3}{15}=\dfrac{1}{3}\)
\(\Rightarrow x+3=\dfrac{1.15}{3}=5\)
\(\Rightarrow x=2\)
\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
\(\Rightarrow2x+1=\dfrac{6.7}{2}=21\)
\(\Rightarrow x=10\)
c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\)
\(\Rightarrow x-6=\dfrac{18.4}{-12}=-6\)
\(\Rightarrow x=0\)
\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow y=\dfrac{-12.24}{18}=-16\)
\(\dfrac{3-x}{-12}=\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow\dfrac{3-x}{-12}=\dfrac{192}{-72}\)
\(\Rightarrow3-x=\dfrac{192.-12}{-72}=32\)
\(\Rightarrow x=-29\)
\(\Rightarrow\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow y+1=\dfrac{16.-72}{192}=-6\)
d) \(\dfrac{-2}{3}< \dfrac{x}{5}< \dfrac{-1}{6}\)
\(\Rightarrow\dfrac{-20}{30}< \dfrac{6x}{30}< \dfrac{-5}{30}\)
\(\Rightarrow6x\in\left\{-18;-12;-6\right\}\)
\(\Rightarrow x\in\left\{-3;-2;-1\right\}\)
\(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\)
\(\Rightarrow5x\in\left\{-5;0;5;10\right\}\)
\(\Rightarrow x\in\left\{-1;0;1;2\right\}\)
e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=x+\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=\dfrac{5x+2}{5}\)
\(\Rightarrow5.\left(x+46\right)=20.\left(5x+2\right)\)
\(\Rightarrow5x+230=100x+40\)
\(\Rightarrow5x-100x=40-230\)
\(\Rightarrow-95x=-190\)
\(\Rightarrow x=-190:-95\)
\(\Rightarrow x=2\)
\(y\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y+\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow\dfrac{y^2+5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y^2+5=86\)
\(\Rightarrow y^2=86-5\)
\(\Rightarrow y^2=81\)
\(\Rightarrow\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\)
Chúc bạn học tốt!
a: =>x/15=-3/5
=>x=-9
b: =>36/y=4/7
=>y=36:4/7=63
c: =>xy=-12
=>(x,y) thuộc {(-1;12); (12;-1); (1;-12); (-12;1); (2;-6); (-6;2); (6;-2); (-2;6); (3;-4); (-4;3); (-3;4); (4;-3)}
d: =>xy=-18
=>(x,y) thuộc {(1;-18); (-18;1); (-1;18);(18;-1); (2;-9); (-9;2); (-2;9); (9;-2); (3;-6); (-6;3); (-3;6); (6;-3)}
(x+y+z)+(x+z+t)+(y+z+t)+(x+y+t)=46+41+44+37=168
=>3x+3y+3z+3t=168
=>3(x+y+z+t)=168
=>x+y+z+t=56
Thay x+y+z+t=168 vào rồi tính ta được
x=x+y+z+t-(y+z+t)=56-44=12
y=x+y+z+t-(x+z+t)=56-41=15
z=x+y+z+t-(x+y+t)=56-37=19
t=x+y+z+t-(x+y+z)=56-46=10
a/
\(3x=4z\Rightarrow x=\frac{4z}{3};2y-3z=4z\Rightarrow y=\frac{7z}{2}\)
\(\Rightarrow x+y-z=\frac{4z}{3}+\frac{7z}{2}-z=46\)
Giải r tìm z từ đó tìm được x và y
b/ Tương tự câu a
a) \(\dfrac{x}{y}=\dfrac{9}{7}\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}\) (1)
\(\dfrac{y}{z}=\dfrac{7}{3}\Rightarrow\dfrac{z}{3}=\dfrac{y}{7}\) (2)
Từ (1) và (2) suy ra: \(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x-y-z}{9-7-3}=\dfrac{-15}{-1}=15\)
\(\Rightarrow\left\{{}\begin{matrix}x=15.9\\y=15.7\\z=15.3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=135\\y=105\\z=45\end{matrix}\right.\)
Vậy, x = 135, y = 105, z = 45
b, \(\dfrac{x}{-3}=\dfrac{y}{-8}\Leftrightarrow\dfrac{x^2}{9}=\dfrac{y^2}{64}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x^2}{9}=\dfrac{y^2}{64}=\dfrac{x^2-y^2}{9-64}=-\dfrac{44}{\dfrac{5}{-55}}=-\dfrac{44}{5}:\left(-55\right)=-\dfrac{44}{5}.-\dfrac{1}{55}=\dfrac{44}{275}=0,16\)
+) \(\dfrac{x^2}{9}=0,16\Rightarrow x^2=1,44\Rightarrow x=\pm1,2\)
+) \(\dfrac{y^2}{64}=0,16\Rightarrow y^2=10,24\Rightarrow y=\pm3,2\)
Vậy ...
\(=4^2-2xy-15=16-2.2.1-15=-3\)