Tìm giá trị cực đại M của hàm 2 biến \[f(x,y) = 4(x - y) - {x^2} - {y^2}\]

A. \[y = C\left( x \right)sinx\]

B. \[y = \frac{{C\left( x \right)}}{{sinx}}\]

C. \[y = C\left( x \right) + sinx\]

D. \[y = C\left( x \right) - sinx\]

A. x=1, y=1

B. x=0, y=0

C. x=1, y=0

D. x=0, y=1

A. M= 9, m= 2

B. M= 8, m=

C. M= 10, m= 2

D. M= 12, m= -2

A. \[\frac{{\partial z}}{{\partial x}} = \frac{1}{{\sqrt {{x^2} + {y^2}} }}\]

B. \[\frac{{\partial z}}{{\partial x}} = \frac{{ - 1}}{{\sqrt {{x^2} + {y^2}} }}\]

C. \[\frac{{\partial z}}{{\partial x}} = \frac{{2x}}{{\sqrt {{x^2} + {y^2}} }}\]

D. \[\frac{{\partial z}}{{\partial x}} = \frac{x}{{\sqrt {{x^2} + {y^2}} }}\]

A. \[y = {C_1}{e^{2x}} + {C_2}x{e^{2x}};C1,C2 \in R\]

B. \[y = {e^{2x}} + ({C_1}\cos (2x) + {C_2}\sin (2x));C1,C2 \in R\]

C. \[y = {C_1}{e^{2x}} + {C_2}{e^{2x}};C1,C2 \in R\]

D. \[y = {C_1}{e^{ - 2x}} + {C_2}x{e^{ - 2x}};C1,C2 \in R\]

A. \[y = {C_1}{e^{ - x}} + {C_2}{e^{3x}}\]

B. \[y = {C_1}{e^x} + {C_2}{e^{3x}}\]

C. \[y = {C_1}{e^x} + {C_2}{e^{ - 3x}}\]

D. \[y = {C_1}{e^{2x}} + {C_2}{e^{3x}}\]

A. \[\sqrt {1 + {x^2}} + ln\left| {lny} \right| = C\]

B. \[arctanx + ln\left| {lny} \right| = C\]

C. \[\sqrt {1 + {x^2}} + ylny = Carcsinx + ln\left| {lny} \right| = C\]