1, Cho 3x=4y. Tính H = 2xy+3x^2/3xy+4y^2
2, Cho a+ b = 3; ab=2
Tính K =\((a)^{2} + 3ab+(b)^{2}+3a+3b + 1\)
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3x=4y
nên x/4=y/3
Đặt x/4=y/3=k
=>x=4k; y=3k
\(H=\dfrac{2xy+3x^2}{3xy+4y^2}=\dfrac{2\cdot4k\cdot3k+3\cdot16k^2}{3\cdot4k\cdot3k+4\cdot9k^2}\)
\(=\dfrac{24k^2+48k^2}{36k^2+36k^2}=1\)
\(a,3x\left(3x+6\right)=9x^2+18x\)
\(b,-\dfrac{1}{2}xy\left(4x^2+6x\right)\)
\(=-2x^3y-3x^2y\)
\(c,-2x^2y^3\left(\dfrac{1}{2}xy+4y^2\right)\)
\(=-x^3y^4-8x^2y^5\)
\(d,-6x^2\left(\dfrac{1}{3}xy^2-\dfrac{1}{2}y\right)\)
\(=-2x^3y^2+3x^2y\)
#\(Urushi\)
a: M=2(-2x-3xy^2+1)-3xy^2+1
=-4x-6xy^2+2-3xy^2+1
=-4x-9xy^2+3
b: Thay x=-2 và y=3 vào M, ta được:
M=2*(-2)-3*(-2)*3^2+1
=-4+1+6*9
=54-3
=51
a. \(2x^2y^3.\frac{1}{4}xy.\left(-3xy\right)=-\frac{3}{2}x^4y^5\text{ đa thức có bậc 4+5 = 9}\)
b. \(\left(-3xy^3\right)^3\left(-\frac{2}{3}x^4y\right)=-27x^3y^9\left(-\frac{2}{3}x^4y\right)=18x^7y^{10}\text{ có bậc 7+10 = 17}\)
c.. \(\frac{2}{3}xy^2-2xy+4x^2y+12+2xy^2-3xy-20-4x^2y=\frac{8}{3}xy^2-5xy-8\) có bậc 3
a) \(M-\left(x^2y-1\right)=-2x^3+x^2y+1\)
\(\Rightarrow M-x^2y+1=-2x^3+x^2y+1\)
\(\Rightarrow M=-2x^3+x^2y+1+x^2y-1\)
\(\Rightarrow M=-2x^3+2x^2y\)
b) \(3x^2+3xy-x^3-M=3x^2+2xy-4y^2\)
\(\Rightarrow-M=3x^2+2xy-4y^2-3x^2-3xy+x^3\)
\(\Rightarrow-M=x^3-4y^2-xy\)
\(\Rightarrow M=-x^3+4y^2+xy\)
a: \(A=-18x^3y^4z\)
Bậc là 8
b: \(M=3x^2+3xy-x^3-3x^2-2xy+4y^2=-x^3+xy+4y^2\)
a) \(3x^2-3xy-5x+5y\)
\(=\left(3x^2-3xy\right)-\left(5x-5y\right)\)
\(=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-5\right)\)
b) \(2x^3y-2xy^3-4xy^2-2xy\)
\(=2xy\left(x^2-y^2-2y-1\right)\)
\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)
\(=2xy\left[x^2-\left(y+1\right)^2\right]\)
\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)
c) \(x^2+1+2x-y^2\)
\(=\left(x^2+2x+1\right)-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x+1+y\right)\left(x+1-y\right)\)
d) \(x^2+4x-2xy-4y+y^2\)
\(=\left(x^2-2xy+y^2\right)+\left(4x-4y\right)\)
\(=\left(x-y\right)^2+4\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y+4\right)\)
e) \(x^3-2x^2+x\)
\(=x\left(x^2-2x+1\right)\)
\(=x\left(x-1\right)^2\)
f) \(2x^2+4x+2-2y^2\)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left[\left(x^2+2x+1\right)+y^2\right]\)
\(=2\left[\left(x+1\right)^2-y^2\right]\)
\(=2\left(x-y+1\right)\left(x+y+1\right)\)
a: =3x(x-y)-5(x-y)
=(x-y)(3x-5)
b: \(=2xy\left(x^2-y^2-2y-1\right)\)
\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)
\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)
d:
Sửa đề: x^2+4x-2xy-4y+y^2
=x^2-2xy+y^2+4x-4y
=(x-y)^2+4(x-y)
=(x-y)(x-y+4)
e: =x(x^2-2x+1)
=x(x-1)^2
f: =2(x^2+2x+1-y^2)
=2[(x+1)^2-y^2]
=2(x+1+y)(x+1-y)
2x(x-2)+2y(x-2)= (x-2)(2x+2y)=2(x-2)(x+y)
b,2(xy+xyz-2x-2z)
c, 3(x^2-xy-x-y)
a) Ta có : 2x2 - 4x + 2xy - 4y
= 2x(x - 2) + 2y(x - 2)
= (x - 2)(2x + 2y)
= 2(x - 2)(x + y)
Bài 1:
3x=4y
nên x/4=y/3
Đặt x/4=y/3=k
=>x=4k; y=3k
\(H=\dfrac{2xy+3x^2}{3xy+4y^2}=\dfrac{2\cdot4k\cdot3k+3\cdot16k^2}{3\cdot4k\cdot3k+9k^2}\)
\(=\dfrac{24k^2+48k^2}{36k^2+9k^2}=\dfrac{72}{45}=\dfrac{8}{5}\)