y : 2 + y : 3 + y : 4 = 13
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\(\dfrac{y}{2}+\dfrac{y+y}{3}+\dfrac{y+y+y}{4}=\dfrac{13}{12}\)
\(\dfrac{1}{2}y+\dfrac{2}{3}y+\dfrac{3}{4}y=\dfrac{13}{12}\)
\(y\left(\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}\right)=\dfrac{13}{12}\)
\(\dfrac{23}{12}y=\dfrac{13}{12}\)
\(y=\dfrac{13}{12}\div\dfrac{23}{12}=\dfrac{13}{23}\)
1) \(x+y=10\) mà \(x=y\) nên: \(x=y=\dfrac{10}{2}=5\)
2) \(2x+3y=180\) mà \(x=y\)
Ta có: \(2y+3y=180\Rightarrow5y=180\Rightarrow y=180:5=36\)
Vậy \(x=y=36\)
3) \(x+y=180\) mà \(x=y\) nên: \(x=y=\dfrac{180}{2}=90\)
4) \(3x+5y=13\) mà \(y=2x\) ta có:
\(3x+5\cdot2x=13\Rightarrow13x=13\Rightarrow x=1\)
\(y=2x=2\cdot1=2\)
Các câu còn lại bạn làm tương tự
\(a)(2{y^4} - 13{y^3} + 15{y^2} + 11y - 3):({y^2} - 4y - 3)=2y^2-5y+1\)
b) \((5{x^3} - 3{x^2} + 10):({x^2} + 1)=5x-3+\dfrac{-5x+13}{x^2+1}\)
a) 0,5 x - \(\frac{2}{3}\)x = \(\frac{7}{12}\)
x . ( 0,5 - \(\frac{2}{3}\)) = \(\frac{7}{12}\)
x . \(\frac{-1}{6}\) = \(\frac{7}{12}\)
x = \(\frac{7}{12}\): \(\frac{-1}{6}\)
x = \(\frac{-7}{2}\)
b) x : \(\frac{25}{3}\) = -25
x = -25 . \(\frac{25}{3}\)
x = \(\frac{-625}{3}\)
c) 5,5 x = \(\frac{13}{15}\)
x = \(\frac{13}{15}\): 5,5
x = \(\frac{26}{165}\)
d) (\(\frac{3x}{7}\)+ 1 ) : ( -4) = \(\frac{-1}{28}\)
( \(\frac{3x}{7}\)+ 1 ) = \(\frac{-1}{28}\). ( -4)
\(\frac{3x}{7}\)+ 1 = \(\frac{1}{7}\)
\(\frac{3x}{7}\) = \(\frac{1}{7}\)- 1
\(\frac{3x}{7}\) = \(\frac{-6}{7}\)
=> x = -2
e) y + 30% y = -1,3
y + \(\frac{3}{10}\)y = -1,3
y . ( 1 + \(\frac{3}{10}\)) = -1,3
y . \(\frac{13}{10}\) = -1,3
y = -1,3 : \(\frac{13}{10}\)
y = -1
g) y - 25% y = \(\frac{1}{2}\)
y - \(\frac{1}{4}\)y = \(\frac{1}{2}\)
y . ( 1 - \(\frac{1}{4}\)) = \(\frac{1}{2}\)
y . \(\frac{3}{4}\) = \(\frac{1}{2}\)
y = \(\frac{1}{2}\): \(\frac{3}{4}\)
y = \(\frac{2}{3}\)
ai tốt bụng thì tk cho mk nha, mk đg âm điểm đây huhu
2: Tọa độ giao điểm là:
\(\left\{{}\begin{matrix}2x-1=x+1\\y=x+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
sửa đề x/2 = y/3 = z/4
Theo tc dãy tỉ số bằng nhau
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+2z}{2+3+8}=\dfrac{13}{13}=1\Rightarrow x=2;y=3;z=4\)
a) \(\dfrac{10^{12}+5^{11}.2^9-5^{13}.2^8}{4.5^5.10^6}\)
\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^2.5^5.2^6.5^6}\)
\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^8.5^{11}}\)
\(=\dfrac{\left(2^8.5^{11}\right)\left(2^4.5+2-5^2\right)}{2^8.5^{11}}\)
\(=2^4.5+2-5^2\)
\(=57\)
b) \(\dfrac{\left[5\left(x-y\right)^4-3\left(x-y\right)^3+4\left(x-y\right)^2\right]}{\left(y-x\right)^2}\)
\(=\dfrac{\left(x-y\right)^2\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y-x\right)^2}\)
\(=\dfrac{\left(x^2+y^2-2xy\right)\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y^2+x^2-2xy\right)}\)
\(=5\left(x-y\right)^2-3\left(x-y\right)+4\)
c) \(\dfrac{\left(x+y\right)^5-2\left(x+y\right)^4+3\left(x+y\right)^3}{-5\left(x+y\right)^3}\)
\(=\dfrac{\left(x+y\right)^3\left[5\left(x+y\right)^2-2\left(x+y\right)+3\right]}{-5\left(x+y\right)^3}\)
\(=\dfrac{5\left(x+y\right)^2-2\left(x+y\right)+3}{-5}\)
y : 2 + y : 3 + y : 4 = 13
y : ( 2+3+4 ) = 13
y : 9 = 13
y = 13 x 9
y = 117
k cho mình nhé !!
y : 2 + y : 3 + y : 4 = 13
y : (2 + 3 + 4) = 13
y : 9 = 13
y = 13 x 9
y = 117
Học tốt
Akina