Tìm x biết :
\(3\cdot x-x\cdot y+3\cdot y\)
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\(3\cdot x-x\cdot y+3\cdot y=8\)
\(x\cdot\left(3\cdot y\right)+3\cdot y\)= 8
=>\(x\cdot\left(3-y\right)+3\cdot y-9=8-9\)
=>\(x\cdot\left(3-y\right)=-1\)
Ta có bảng sau
a, 3.x2.y + M - x.y=10x2y - 2xy
(3 x2y-xy) +M= 10x2y -2xy
M=10x2y-2xy+( 3x2y -xy)
M=(10x2y+3x2y)-(2xy+xy)
M=13 x2y-3xy
b,(6xy-5y2)-N=x2-2xy+4 y2
N= 6xy -5y2-( x2-2xy+4y2)
N= 6xy -5y2-x2 +2xy -4y2
N= (6xy +2xy)- (5y2+4y2)-x2
N= 8xy -9y2-x2
hok tốt
boy with luv
kt
\(25\%.y+50\%.y-\frac{3}{4}.y+4.y=10\)
\(y.\left(\frac{1}{4}+\frac{1}{2}-\frac{3}{4}+4\right)=10\)
\(y.4=10\)
\(y=\frac{5}{2}\)
\(x.\frac{1}{4}-\frac{3}{4}=6:\frac{3}{4}\)
\(x.\frac{1}{4}-\frac{3}{4}=6.\frac{4}{3}\)
\(x.\frac{1}{4}=8+\frac{3}{4}\)
\(x.\frac{1}{4}=\frac{35}{4}\)
\(x=\frac{35}{4}:\frac{1}{4}\)
\(x=35\)
25% x y + 50% x y - 3/4 x y + 4 x y = 10
1/4 x y + 1/2 x y - 3/4 x y + 4 x y = 10
y x ( 1/4 + 1/2 - 3/4 + 4 ) = 10
y x 4 = 10
y = 10 : 4
y = 2.5
\(A=\left(\dfrac{-3}{7}.x^3.y^2\right).\left(\dfrac{-7}{9}.y.z^2\right).\left(6.x.y\right)\)
\(A=\left(\dfrac{-3}{7}x^3y^2\right).\left(\dfrac{-7}{9}yz^2\right).6xy\)
\(A=\left(\dfrac{-3}{7}.\dfrac{-7}{9}.6\right).\left(x^3.x\right)\left(y^2.y.y\right).z^2\)
\(A=2x^4y^4z^2\)
\(B=-4.x.y^3\left(-x^2.y\right)^3.\left(-2.x.y.z^3\right)^2\)
\(B=\left[\left(-4\right).\left(-2\right)\right].\left(x.x^6.x^2\right)\left(y^3.y^3.y^2\right)\left(z^6\right)\)
\(B=8x^7y^{y^8}z^6\)
\(\left(x+1\right)\left(y-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\y-2=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0-1\\y=0+2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-1\\y=2\end{cases}}\)
Vậy x = - 1 ; y = 2