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\(10x=14y=15z\)
\(BCNN\left(10;14;15\right)=2.3.5.7=210\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{210}{10}=21\\y=\dfrac{210}{14}=15\\z=\dfrac{210}{15}=14\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)=\left(21;15;14\right)\)
a, 7x + 10x = 5x
17x = 5x
17x - 5x = 0
12x = 0
x =0
2;
a, 4x + 7x = 22
11x = 22
x = 2
b, 12x - 8x = 25
4x = 25
x = \(\dfrac{25}{4}\)
c, \(\dfrac{1}{2}\)x - \(\dfrac{1}{3}\)x = \(\dfrac{4}{5}\)
(\(\dfrac{1}{2}-\dfrac{1}{3}\))x = \(\dfrac{4}{5}\)
\(\dfrac{1}{6}\)x = \(\dfrac{4}{5}\)
x = \(\dfrac{4}{5}\) : \(\dfrac{1}{6}\)
x = \(\dfrac{24}{5}\)
a) 2018 - 7x + 5 = 4x + 3 - 10x
10x - 7x - 4x = 3 - 2018 - 5
x(10 - 7 - 4) = -2020
x . (-1) = -2020
x = (-2020) : (-1)
x = 2020
a, M = 7x - 7y + 4 ax - 4ay - 5
= 7 ( x - y ) + 4a ( x - y ) - 5
= 0 + 0 - 5
= -5
b, N = x ( x2 + y2 ) - y ( x2 + y2 ) + 3
= ( x - y ) ( x2 + y2 ) + 3
=0 + 3
=3
a, M= 7(x-y)+4a(x-y)-5 = 7.0+4a.0-5=-5
b, N=(x2 +y2 ).(x-y) +3=(x2 +y2 ).0+3=3
a) thay x,y=3 ta đc
A=7\(\times\)3 + 7\(\times\)3
=7\(\times\)(3+3)
=7\(\times\)6
= 42
b)thay x,y=3 ta có :
B=10\(\times\)3+14\(\times\)3+4\(\times\)3
=3\(\times\)(10+14+4)
=3\(\times\)28
=84