Tìm Y
a) (y+1/5)*3/7=5/14
tìm x,y
A) \(\dfrac{x}{y}=\dfrac{7}{4}\) và x+y=33
b) 3.(x-1)+5=-19
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) \(3\left(x-1\right)+5=-19\\ \Rightarrow3\left(x-1\right)=-24\\ \Rightarrow x-1=-8\\ \Rightarrow x=-7\)
tìm y
a)y 56 b) 15 5
__ = __ __ = __
7 49 27 y
\(a,\Rightarrow7y=56\Leftrightarrow y=8\)
b, \(\Rightarrow3y=27\Leftrightarrow y=9\)
a) y.49 = 56.7
y.49=392
y=392:49
y=8
b)15.y=27.5
15.y= 135
y=135:15
y=9
(tick nhé)
a) x.y = -9 và x< y
b)x.y = 7 và x<y
c)x.y = -11 và x>y
a)(x+1)(y-2)=-3
b)(x-3)(y+1)=7
c)(x+5)(y+7)=-5
a: \(\left(x,y\right)\in\left\{\left(-9;1\right);\left(-1;9\right);\left(-3;3\right)\right\}\)
b: \(\left(x,y\right)\in\left\{\left(1;7\right);\left(-7;-1\right)\right\}\)
c: \(\left(x,y\right)\in\left\{\left(11;-1\right);\left(-11;1\right)\right\}\)
a: (x,y)∈{(−9;1);(−1;9);(−3;3)}(x,y)∈{(−9;1);(−1;9);(−3;3)}
b: (x,y)∈{(1;7);(−7;−1)}(x,y)∈{(1;7);(−7;−1)}
c: (x,y)∈{(11;−1);(−11;1)}
a) x.y = -9 và x< y
b)x.y = 7 và x<y
c)x.y = -11 và x>y
a)(x+1)(y-2)=-3
b)(x-3)(y+1)=7
c)(x+5)(y+7)=-5
a: \(\left(x,y\right)\in\left\{\left(-9;1\right);\left(-1;9\right);\left(-3;3\right)\right\}\)
b: \(\left(x,y\right)\in\left\{\left(1;7\right);\left(-7;-1\right)\right\}\)
c: \(\left(x,y\right)\in\left\{\left(11;-1\right);\left(-1;11\right)\right\}\)
Tìm x, y
a) 2 5/x = 21/x (x ∈ N*)
b) x/7 = 5/y (x; y ∈ Z, x > y)
a, bạn viết rõ đề ra nhé
b, \(\Rightarrow xy=35\Rightarrow x;y\inƯ\left(35\right)=\left\{\pm1;\pm5;\pm7;\pm35\right\}\)
x | 1 | -1 | 5 | -5 | 7 | -7 | 35 | -35 |
y | 35 | -35 | 7 | -7 | 5 | -5 | 1 | -1 |
Bài 1: Tìm x và y
a) x/4 = y/-5 và -3x + 2y = 55
b) x/y = -7/4 và 4x - 5y = 72
c) x/ -3 = y/8 và x2 - y2 = -44/5
d) 3x3 + y3 = 64/9
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: \(\left\{{}\begin{matrix}x=\dfrac{21}{2}\\y=\dfrac{-12}{2}=-6\end{matrix}\right.\)
c) \(\dfrac{x}{-3}=\dfrac{y}{8}\)
⇒\(\dfrac{x^2}{-9}=\dfrac{y^2}{64}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x^2}{-9}=\dfrac{y^2}{64}=-\dfrac{44}{\dfrac{5}{-9+64}}=-\dfrac{44}{\dfrac{5}{55}}=-484\)
Tìm x,y,z biết
3 /x/+5/3=3 /y/-1/5=3-z/7 và 2/x/+7/y/+3/z/=-14
Tìm y
a) \(y^{200}=y\)
b)\(y^{2008}=y^{2010}\)
c)\(\left(2y-1\right)^{50}=2y-1\)
d)\(\left(\dfrac{y}{3}-5\right)^{2000}=\left(\dfrac{y}{3}-5\right)^{2008}\)
\(a,\Leftrightarrow y^{200}-y=y\left(y^{199}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=0\\y^{199}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}y=0\\y=1\end{matrix}\right.\)
Vậy ..
\(b,\Leftrightarrow y^{2010}-y^{2008}=y^{2008}\left(y^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y^{2008}=0\\y^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}y=0\\y=1\\y=-1\end{matrix}\right.\)
Vậy ...
\(c,\Leftrightarrow\left(2y-1\right)^{50}-\left(2y-1\right)=\left(2y-1\right)\left(\left(2y-1\right)^{49}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2y-1=0\\\left(2y-1\right)^{49}=1\end{matrix}\right.\)
\(\Leftrightarrow y=\dfrac{1}{2}\)
Vậy ..
\(d,\Leftrightarrow\left(\dfrac{y}{3}-5\right)^{2008}\left(\left(\dfrac{y}{3}-5\right)^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(\dfrac{y}{3}-5\right)^{2008}=0\\\left(\dfrac{y}{3}-5\right)^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{y}{3}-5=0\\\dfrac{y}{3}-5=1\\\dfrac{y}{3}-5=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}y=15\\y=18\\y=12\end{matrix}\right.\)
Vậy ..
Tìm y
a) y x 6/7 = 5/14
b) y : 2/3 = 4/9
c) 3/10 + y = 1/2
d) 5/6 - y = 1/3
a) y = 5/14 : 6/7
y = 5/12
vậy y = 5/12
b) y = 4/9 x 2/3
y = 8/27
vậy t = 8/27
c) y = 1/2 - 3/10
y = 1/5
vậy y = 1/5
d) y = 5/6 - 1/3
y = 1/2
vậy y = 1/2