Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) 2x(x+3) – 3x2(x+2) + x(3x2 + 4x – 6)
= (2x . x + 2x . 3) – (3x2 . x + 3x2 . 2) + (x . 3x2 + x . 4x – x . 6)
= 2x2 + 6x – (3x3 + 6x2) + (3x3 + 4x2 - 6x)
= 2x2 + 6x – 3x3 – 6x2 + 3x3 + 4x2 - 6x
= (– 3x3 + 3x3 ) + (2x2 - 6x2 + 4x2 ) + (6x – 6x)
= 0 + 0 + 0
= 0
b) 3x(2x2 – x) – 2x2(3x+1) + 5(x2 – 1)
= [3x . 2x2 + 3x . (-x)] – (2x2 . 3x + 2x2 . 1) + [5x2 + 5 . (-1)]
= 6x3 – 3x2 – (6x3 +2x2) + 5x2 – 5
= 6x3 – 3x2 – 6x3 - 2x2 + 5x2 – 5
= (6x3 – 6x3 ) + (-3x2 – 2x2 + 5x2) – 5
= 0 + 0 – 5
= - 5
a)
`4*(2y+3x)-3(x-3y)`
`=8y+12x-3x+9y`
`=8y+9y+12x-3x`
`=17y+9x`
b)
`x^2 +2x-x(7x-3)`
`=x^2 +2x-7x^2 +3x`
`=x^2 -7x^2 +2x+6x`
`= -6x^2 +8x`
a) 4x2(5x2 + 3) – 6x(3x3 – 2x + 1) – 5x3 (2x – 1)
= 4x2 . 5x2 + 4x2 . 3 – [6x . 3x3 + 6x . (-2x) + 6x . 1] – [5x3 . 2x + 5x3 . (-1)]
= 20x4 + 12x2 – (18x4 – 12x2 + 6x) – (10x4 – 5x3)
= 20x4 + 12x2 - 18x4 + 12x2 - 6x - 10x4 + 5x3
= (20x4 – 18x4 - 10x4 ) + 5x3 + (12x2 + 12x2 ) – 6x
= -8x4 + 5x3 + 24x2 – 6x
\(\begin{array}{l}b)\dfrac{3}{2}x\left( {{x^2} - \dfrac{2}{3}x + 2} \right) - \dfrac{5}{3}{x^2}(x + \dfrac{6}{5})\\ = \dfrac{3}{2}x.{x^2} + \dfrac{3}{2}x.( - \dfrac{2}{3}x) + \dfrac{3}{2}x.2 - (\dfrac{5}{3}{x^2}.x + \dfrac{5}{3}{x^2}.\dfrac{6}{5})\\ = \dfrac{3}{2}{x^3} - {x^2} + 3x - (\dfrac{5}{3}{x^3} + 2{x^2})\\ = \dfrac{3}{2}{x^3} - {x^2} + 3x - \dfrac{5}{3}{x^3} - 2{x^2}\\ = (\dfrac{3}{2}{x^3} - \dfrac{5}{3}{x^3}) + ( - {x^2} - 2{x^2}) + 3x\\ = \dfrac{{ - 1}}{6}{x^3} - 3{x^2} + 3x\end{array}\)
Bài 1 :
a) \(M=\dfrac{1}{2}x^2y.\left(-4\right)y\)
\(\Rightarrow M=-2x^2y^2\)
Khi \(x=\sqrt[]{2};y=\sqrt[]{3}\)
\(\Rightarrow M=-2.\left(\sqrt[]{2}\right)^2.\left(\sqrt[]{3}\right)^2\)
\(\Rightarrow M=-2.2.3=-12\)
b) \(N=xy.\sqrt[]{5x^2}\)
\(\Rightarrow N=xy.\left|x\right|\sqrt[]{5}\)
\(\Rightarrow\left[{}\begin{matrix}N=xy.x\sqrt[]{5}\left(x\ge0\right)\\N=xy.\left(-x\right)\sqrt[]{5}\left(x< 0\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}N=x^2y\sqrt[]{5}\left(x\ge0\right)\\N=-x^2y\sqrt[]{5}\left(x< 0\right)\end{matrix}\right.\)
Khi \(x=-2< 0;y=\sqrt[]{5}\)
\(\Rightarrow N=-x^2y\sqrt[]{5}=-\left(-2\right)^2.\sqrt[]{5}.\sqrt[]{5}=-4.5=-20\)
2:
Tổng của 4 đơn thức là;
\(A=11x^2y^3+\dfrac{10}{7}x^2y^3-\dfrac{3}{7}x^2y^3-12x^2y^3=0\)
=>Khi x=-6 và y=15 thì A=0
a.\(x=0;y=-1\)
\(\Rightarrow2.0-\dfrac{-1\left(0^2-2\right)}{0.-1-1}=0-\dfrac{2}{-1}=2\)
b.\(x=2\)
\(\Rightarrow4.2^2-3\left|2\right|-2=16-6-2=8\)
\(x=-3\)
\(\Rightarrow4.\left(-3\right)^2-3\left|-3\right|-2=36-9-2=25\)
c.\(x=-\dfrac{1}{5};y=-\dfrac{3}{7}\)
\(\Rightarrow5.\left(-\dfrac{1}{5}\right)^2-7.\left(-\dfrac{3}{7}\right)+6=5.\dfrac{1}{25}+3+6=\dfrac{1}{5}+3+6=\dfrac{46}{5}\)
thay x=2 và biểu thức A ta đc
\(A=4.2^2-3.\left|2\right|-2=4.4-6-2=16-6-2=8\)
thay x=-3 biểu thức A ta đc
\(A=4.\left(-3\right)^2-3.\left|-3\right|-2=4.9-9-2=36-9-2=25\)
thay x=-1/5 ; y=-3/7 biểu thức B ta đc
\(B=5.\left(-\dfrac{1}{5}\right)^2-7.\left(-\dfrac{3}{7}\right)+6\)
\(B=5\cdot\dfrac{1}{25}+3+6\)
\(B=\dfrac{1}{5}+3+6=\dfrac{46}{5}\)
a) \(A=4x\left(x+y\right)-5y\left(x-y\right)-4x^2=4x^2+4xy-5xy+5y^2-4x^2=5y^2-xy\)
Với x = -5; y = 2 thì: \(A=5\cdot2^2-\left(-5\right)\cdot2=20+10=30\)
b) \(B=-3x\left(x^2+y^2\right)+2y\left(x^2-y\right)=-3x^3-3xy^2+2yx^2-2y^2=-3x^3+2x^2y-3xy^2-2y^2\)
Với x = 1; y = 2 thì: \(B=-3\cdot1^3+2\cdot1^2\cdot2-3\cdot1\cdot2^2-2\cdot2^2=-3+4-12-8=-19\)
a) Cách 1:
\(6(y - x) - 2(x - y)\)
\( = 6y - 6x - 2x + 2y\)
\( = 8y - 8x\)
Cách 2:
\(6(y - x) - 2(x - y)\\= 6(y-x)+2(y-x)\\=(6+2).(y-x)\\=8.(y-x)\\=8y-8x\)
b) \(3{x^2} + x - 4x - 5{x^2}\)
\( = (3{x^2} - 5{x^2}) + (x - 4x)\)
\( = - 2{x^2} - 3x\)