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a) \(A=5xy-3,5y^2-2xy+1,3xy+3x-2y\)
\(=\left(5xy-2xy+1,3xy\right)-3,5y^2+3x-2y\)
\(=-3,5y^2+4,3xy+3x-2y\)
b) \(B=\dfrac{1}{2}ab^2-\dfrac{7}{8}ab^2+\dfrac{3}{4}a^2b-\dfrac{3}{8}a^2b-\dfrac{1}{2}ab^2\)
\(=\left(\dfrac{1}{2}ab^2-\dfrac{7}{8}ab^2-\dfrac{1}{2}ab^2\right)+\left(\dfrac{3}{4}a^2b-\dfrac{3}{8}a^2b\right)\)
\(=-\dfrac{7}{8}ab^2+\dfrac{3}{8}a^2b\)
c) \(2a^2b-8b^2+5a^2b+5c^2-3b^2+4c^2\)
\(=\left(2a^2b+5a^2b\right)+\left(-8b^2-3b^2\right)+\left(5c^2+4c^2\right)\)
\(=7a^2b-11b^2+9c^2\)
a: \(=\left(15x^2y^3-12x^2y^3\right)+\left(7x^2-12x^2\right)+\left(-8x^3y^2+11x^3y^2\right)\)
\(=3x^2y^3-5x^2+3x^3y^2\)
bậc là 5
b: \(=\left(3x^5y-\dfrac{1}{2}x^5y\right)+\left(\dfrac{1}{3}xy^4+2xy^4\right)+\left(\dfrac{3}{4}x^2y^3-x^2y^3\right)\)
\(=\dfrac{5}{2}x^5y+\dfrac{7}{3}xy^4-\dfrac{1}{4}x^2y^3\)
Bậc là 6
c: \(=5xy-2xy+4xy-y^2+3x-2y\)
\(=-y^2+3x-2y+7xy\)
Bậc là 2
a: \(=-10bx^3y^2\)
b: \(\dfrac{-4}{5}ab^2c\cdot\left(-20\right)a^4bx=16a^5b^3c\cdot x\)
c: \(=8\cdot\dfrac{1}{4}\cdot a^3\cdot b^2c^4=2a^3b^2c^4\)
d: \(=2ab\cdot\dfrac{4}{3}a^2\cdot b^4\cdot7abc=\dfrac{56}{3}a^4b^6c\)
4. \(1^2+2^2+3^2+...+10^2+11^2=506\)
Ta có: \(2^2+4^2+6^2+...+20^2+22^2\)
\(=2^2.1^2+2^2.2^2+2^2.3^2+...+2^2.10^2+2^2.11^2\)
\(=2^2\left(1^2+2^2+3^2+...+10^2+11^2\right)\)
\(=2^2.506=2024\)
Vậy....
1.
Ta có: \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\)
\(\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}=\dfrac{a^2-b^2+2c^2}{4-9+32}=\dfrac{108}{27}=4\)
\(\Rightarrow a^2=16\)
\(\Rightarrow b^2=36\)
\(\Rightarrow c^2=64\)
\(\Rightarrow a=\pm4\) , \(b=\pm6\) , \(c=\pm8\)
\(\left(2x+y^3\right)^2=4x^2+4xy^3+y^6\)
\(\left(\dfrac{1}{2}x-y\right)^2=\dfrac{1}{2}x^2-xy+y^2\)
\(\left(xy+5\right)^2=xy^2+10xy+25\)
\(\left(2y^2-3\right)^2=4y^4-12y^2+9\)
Các câu sau làm tương tự nha,dựa vào HĐT đó
1. Thu gọn các đơn thức sau rồi tìm hệ số và bậc của nó :
a) \(\left(-2xy^3\right)\left(\dfrac{1}{3}xy\right)^2\)
\(=\left(-2.\dfrac{1}{9}\right)\left(x.x^2\right)\left(y^3.y^2\right)\)
\(=\dfrac{-2}{9}x^3y^5\)
Hệ số : \(\dfrac{-2}{9}\)
Bậc : 8
b) \(\left(-18x^2y^2\right)\left(\dfrac{1}{6}ax^2y^3\right)\)
\(=\left(-18.\dfrac{1}{6}a\right)\left(x^2.x^2\right)\left(y^2.y^3\right)\)
\(=-3ax^4y^5\)
Hệ số : \(-3a\)
Bậc : 9
c) \(3x^2yz\left(-xy\right)\left(\dfrac{-2}{3}xy^2z^3\right)\)
\(=\left(3.\dfrac{-2}{3}\right).\left(x^2.-x.x\right)\left(y.y.y^2\right).z^3\)
\(=-2x^4y^4x^3\)
Hệ số : -2
Bậc : 11
d) \(\left(-3x^2y\right)^2xz^2.\dfrac{1}{2}xy^3\)
\(=\left(-3.\dfrac{1}{2}\right)\left(x^4.x.x\right)\left(y^2.y^3\right).z^2\)
\(=\dfrac{-3}{2}x^6y^5z^2\)
Hệ số : \(\dfrac{-3}{2}\)
Bậc : 13
e) \(-3x^2yz\left(-5xy^3z^2\right)\)
\(=\left(-3.-5\right)\left(x^2.x\right)\left(y.y^3\right)\left(z.z^2\right)\)
\(=-15x^3y^4z^3\)
Hệ số : -15
Biến : 10
I . Trắc Nghiệm
1B . 2D . 3C . 5A
II . Tự luận
2,a,Ta có: A+(x\(^2\)y-2xy\(^2\)+5xy+1)=-2x\(^2\)y+xy\(^2\)-xy-1
\(\Leftrightarrow\) A=(-2x\(^2\)y+xy\(^2\)-xy-1) - (x\(^2\)y-2xy\(^2\)+5xy+1)
=-2x\(^2\)y+xy\(^2\)-xy-1 - x\(^2\)y+2xy\(^2\)-5xy-1
=(-2x\(^2\)y - x\(^2\)y) + (xy\(^2\)+ 2xy\(^2\)) + (-xy - 5xy ) + (-1 - 1)
= -3x\(^2\)y + 3xy\(^2\) - 6xy - 2
b, thay x=1,y=2 vào đa thức A
Ta có A= -3x\(^2\)y + 3xy\(^2\) - 6xy - 2
= -3 . 1\(^2\) . 2 + 3 .1 . 2\(^2\) - 6 . 1 . 2 -2
= -6 + 12 - 12 - 2
= -8
3,Sắp xếp
f(x) =9-x\(^5\)+4x-2x\(^3\)+x\(^2\)-7x\(^4\)
=9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x
g(x) = x\(^5\)-9+2x\(^2\)+7x\(^4\)+2x\(^3\)-3x
=-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x
b,f(x) + g(x)=(9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x) + (-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x)
=9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x
=(9-9)+(-x\(^5\)+x\(^5\))+(-7x\(^4\)+7x\(^4\))+(-2x\(^3\)+2x\(^3\))+(x\(^2\)+2x\(^2\))+(4x-3x)
= 3x\(^2\) + x
g(x)-f(x)=(-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x) - (9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x)
=-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x-9+x\(^5\)+7x\(^4\)+2x \(^3\)-x\(^2\)-4x
=(-9-9)+(x\(^5\)+x\(^5\))+(7x\(^4\)+7x\(^4\))+(2x\(^3\)+2x\(^3\))+(2x\(^2\)-x\(^2\))+(3x-4x)
= -18 + 2x\(^5\) + 14x\(^4\) + 4x\(^3\) + x\(^2\) - x
a: \(A=\left(5xy-2xy+1.3xy\right)+3x-2y-3.5y^2\)
\(=4.3xy+3x-2y-3.5y^2\)
b: \(B=\left(\dfrac{1}{2}ab^2-\dfrac{1}{2}ab^2-\dfrac{7}{8}ab^2\right)+\left(\dfrac{3}{4}a^2b-\dfrac{3}{8}a^2b\right)\)
\(=-\dfrac{7}{8}ab^2+\dfrac{3}{8}a^2b\)
c: \(C=\left(2a^2b+5a^2b\right)+\left(-8b^2-3b^2\right)+\left(5c^2+4c^2\right)\)
\(=7a^2b-11b^2+9c^2\)