1)Tìm y
a) y*3+y*5+y-289=7829
b)4*8+19-y=28
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a,Ta có:
\(\dfrac{x}{y}=\dfrac{7}{4}=\dfrac{x}{7}=\dfrac{y}{4}\)
ÁP dụng tcdtsbn , ta có:
\(\dfrac{x}{7}=\dfrac{y}{4}=\dfrac{x+y}{7+4}=\dfrac{33}{11}=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=21\\y=12\end{matrix}\right.\)
b,
\(\Rightarrow3.\left(x-1\right)=-24\)
\(\Rightarrow x-1=-8\)
\(\Rightarrow x=-7\)
A)\(\dfrac{x}{y}=\dfrac{7}{4}\Rightarrow\dfrac{x}{7}=\dfrac{y}{4}\)
Áp dụng t/c dtsbn ta có:
\(\dfrac{x}{7}=\dfrac{y}{4}=\dfrac{x+y}{7+4}=\dfrac{33}{11}=3\)
\(\dfrac{x}{7}=3\Rightarrow x=21\\ \dfrac{y}{4}=3\Rightarrow y=12\)
B) \(3\left(x-1\right)+5=-19\\ \Rightarrow3\left(x-1\right)=-24\\ \Rightarrow x-1=-8\\ \Rightarrow x=-7\)
a: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{4}=\dfrac{y}{-5}=\dfrac{-3x+2y}{-12-10}=\dfrac{55}{-22}=\dfrac{-5}{2}\)
Do đó: \(\left\{{}\begin{matrix}x=\dfrac{-20}{2}=-10\\y=\dfrac{25}{2}\end{matrix}\right.\)
b: Ta có: \(\dfrac{x}{y}=\dfrac{-7}{4}\)
nên \(\dfrac{x}{-7}=\dfrac{y}{4}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{-7}=\dfrac{y}{4}=\dfrac{4x-5y}{-28-20}=\dfrac{72}{-48}=\dfrac{-3}{2}\)
Do đó: \(\left\{{}\begin{matrix}x=\dfrac{21}{2}\\y=\dfrac{-12}{2}=-6\end{matrix}\right.\)
\(\dfrac{19}{20}-y=\dfrac{8}{5}-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{19}{20}-y=\dfrac{32-15}{20}\)
\(\Leftrightarrow\dfrac{19}{20}-y=\dfrac{17}{20}\Leftrightarrow y=\dfrac{19}{20}-\dfrac{17}{20}=\dfrac{2}{20}=\dfrac{1}{10}\)
\(y:\dfrac{2}{3}=\dfrac{4}{9}.3\)
\(\Leftrightarrow y:\dfrac{2}{3}=\dfrac{12}{9}\Leftrightarrow y=\dfrac{12}{9}.\dfrac{2}{3}=\dfrac{8}{9}\)
\(a.\dfrac{19}{20}-y=\dfrac{17}{20}\)
\(y=\dfrac{2}{17}\)
\(b.y:\dfrac{2}{3}=\dfrac{4}{3}\)
\(y=\dfrac{8}{9}\)
a/\(\dfrac{19}{20}-y=\dfrac{8}{5}-\dfrac{3}{4}\)
\(\dfrac{19}{20}-y=\dfrac{17}{20}\)
\(y=\dfrac{19}{20}-\dfrac{17}{20}\)
\(y=\dfrac{2}{20}=\dfrac{1}{10}\)
b/\(y:\dfrac{2}{3}=\dfrac{4}{9}\times3\)
\(y:\dfrac{2}{3}=\dfrac{4}{3}\)
\(y=\dfrac{4}{3}\times\dfrac{2}{3}\)
\(y=\dfrac{8}{9}\)
#データネ
a) 19/20 - y = 8/5 - 3/4
19/20 - y = 17/20
y = 17/20 + 19/20
y = 36/20
b) y : 2/3 = 4/9 x 3
y : 2/3 = 12/9
y = 12/9 x 2/3
y = 24/27
tích cho mik nhaaaaaaaaaaa
Giải:
a) \(y^2=3-\left|2x-3\right|\)
Vì \(-\left|2x-3\right|\le0\forall x\) nên \(3-\left|2x-3\right|\le3\forall x\) nên \(y^2\le3\rightarrow y^2\in\left\{0;1\right\}\) (vì \(y\in Z\) )
TH1:
\(y^2=0\)
\(\Rightarrow y=0\)
\(\Rightarrow\left|2x-3\right|=3\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
TH2:
\(y^2=1\)
\(\Rightarrow y=\pm1\)
\(\dfrac{1}{5}+y=7\)
\(y=7-\dfrac{1}{5}\)
\(y=\dfrac{35}{5}-\dfrac{1}{5}\)
\(y=\dfrac{34}{5}\)
_________________________
\(y:\dfrac{3}{4}=\dfrac{5}{7}\)
\(y=\dfrac{5}{7}\times\dfrac{3}{4}\)
\(y=\dfrac{15}{28}\)
Chúc bạn học tốt
a)y x (3+5+1)=7829+289=8118
y x 9=8118
y=902
Vậy y=902
b)32+19-y=28
51-y=28
y=51-28
y=23
Vậy y=23
Học giỏi nhé!!! ^.^
a ) y x ( 3+5+1) = 7829 + 289
y x 9 = 8118
y = 8118 : 9 902
b ) 32 + 19 - y = 28
51 - y =28
y = 28 + 51
y = 79