cho da thuc A= \(\left(xy^2\right)\left(x^5y^4\right)\left(x^9y^6\right)\left(x^{13}y^8\right)\left(x^{17}y^{10}\right)...\)Biet da thuc A co bac bang 3675 . bac cao nhat cua bien x la...
Thu gon don thuc sau (x;y la bien so) roi xac dinh: phan he so, phan bien, bac cua don thuc: \(-ax\left(xy^3\right)^2.\left(-by\right)\)
thu gọn : -a.(x. x^2).(y.y^6).(-b)= -a.x^3.y^7.(-b)
hệ số là :-a và -b
phần biến là :x và y
bậc :10
\(\left[3\left(x-y\right)^5-2\left(x-y\right)^4+3\left(x-y\right)^2\right]:5\left(x-y\right)^2\)
chia da thuc cho don thuc
\(\left(x^{10}-y^{10}\right):\left(x^4-x^3y+x^2y^2+xy^3+y^4\right)chung-minh-hai-da-thuc-chia-het-cho-nha\)
Cho x,y la cac so thuc duong. Tim gia tri nho nhat cua bieu thuc:
\(P=\frac{xy}{x^2+y^2}+\left(\frac{1}{x}+\frac{1}{y}\right)\sqrt{2\left(x^2+y^2\right)}\)
Hình như đề sai rùi bạn ơi !
Phải sửa xy/x^2+y^2 thành x^2+y^2/xy hoặc cái gì khác
Vì xy/x^2+y^2 chỉ có GTLN chứ ko có GTNN đâu
Mk nói có gì sai thì thông cảm nha !
đề không sai đâu bạn à. Đây là đề toán chuyên ở tỉnh mình mà
Theo B.C.S ta có \(\sqrt{2\left(x^2+y^2\right)}\)\(\ge\)(\(\sqrt{\left(x+y\right)^2}\)\(=x+y\)
\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}\right)\sqrt{2\left(x^2+y^2\right)}\ge\left(\frac{1}{x}+\frac{1}{y}\right)\left(x+y\right)=2+\frac{x^2+y^2}{xy}\)
\(\Leftrightarrow\)\(P\ge2+\frac{xy}{x^2+y^2}+\frac{x^2+y^2}{4xy}+\frac{3\left(x^2+y^2\right)}{4xy}\)
\(\Leftrightarrow\)\(P\ge2+2\sqrt{\frac{xy}{x^2+y^2}\times\frac{x^2+y^2}{4xy}}\)\(+\frac{3\times2xy}{4xy}\)
\(\Leftrightarrow\)\(P\ge2+1+\frac{3}{2}=\frac{9}{2}\)
Dấu bằng xảy ra \(\Leftrightarrow\)x=y
Cho da thuc :
f(x) \(2.\left(3x-1\right).\left(x+1\right)+\left(15x-10\right).\left(x-4\right)+9x-6\)
Phân tich da thuc thanh nhan tu
a)\(yz\left(y+z\right)+xz\left(z-x\right)-xy\left(x+y\right)\)
\(yz\left(y+z\right)+xz\left(z-x\right)-xy\left(x+y\right)\)
\(=-[xy(x+y)-yz(y+z)-zx(z-x)]\)
\(=-(y.[x(x+y)-z(y+z)]-zx(z-x))\)
\(=-[y.(x^2+xy-zy-z^2)-zx(z-x)]\)
\(=-[y.(x^2-z^2+xy-zy)-zx(z-x)]\)
\(=-(y.[(x+z)(x-z)+y.(x-z)]-zx(z-x))\)
\(=-[y.(x-z)(x+z+y)+zx(x-z)]\)
\(=[(x-z)[y(x+z+y)+zx]]\)
\(=-(x-z)(yx+yz+y2+zx)\)
\(=-(x-z)(yx+zx+yz+y2)\)
\(=-[(x-z)[x.(y+z)+y.(y+z)]]\)
\(=-(x-z)(y+z)(x+y)\)
a, biet x+y=0
tinh gia tri bieu thuc : M=\(x^4-xy^3+x^3y-y^4-1\)
b, biet xyz=2 va x+y+z=0
tinh gia tri bieu thuc : M= \(\left(x+y\right)\left(y+2\right)\left(x+2\right)\)
a/ \(M=x^4-xy^3+x^3y-y^4-1\)
\(\Leftrightarrow M=x^3\left(x+y\right)-y^3\left(x+y\right)-1\)
Mà \(x+y=0\)
\(\Leftrightarrow M=x^3.0-y^3.0-1\)
\(\Leftrightarrow M=-1\)
Vậy ...
a. Cho da thuc P(x) = mx^2 + 2mx - 3 co nghiem x = -1. Tim m
b. Cho da thuc P(x) = ax^2 + bx + c. Chung to rang \(P\left(-1\right)\cdot P\left(-2\right)\le0\) biet rang 5a - 3b + 2c = 0
BT:thu gon don thuc , tim he so va bac cua don thuc
a)-(\(\frac{-1}{2}\)\(xy^2z\))\(^2\) (\(4x^2yz\)\(^3\))
b)\(\left(\frac{-1}{3}x^2yz^3\right)^2.\left(-\frac{6}{7}xyz^2\right)\)
c)\(-3x^2.y^4.\left(\frac{-1}{3}y^4z^5x\right).\left(\frac{-1}{2}zyx^3\right)\)
d)\(\left(-\frac{2}{5}x^2y\right)^3.\left(-\frac{1}{3}xy^2\right)\)
e)\(\frac{3}{4}xy^3\left(-\frac{2}{3}x^2y^4\right)^2\)
g)\(\left(-\frac{3}{5}x^2y^3\right)^2\left(-\frac{1}{3}x^3y^2\right)^3\)
mn co gang giup mik vs mik dang can gap
a)\(-\left(\frac{-1}{2}xy^2z\right)^2\left(4x^2yz^3\right)\)
\(=-\left(\frac{1}{4}x^2y^4z^2\right)\left(4x^2yz^3\right)\)
\(=\left(\frac{-1}{4}.4\right)\left(x^2x^2\right)\left(y^4y\right)\left(z^2z^3\right)\)
\(=-x^4y^5z^5\) \(\Rightarrow\)Bậc là 14 Hệ số là -1
b)\(\left(\frac{-1}{3}x^2yz^3\right).\left(\frac{-6}{7}xyz^2\right)\)
\(=\left(\frac{-1}{3}.\frac{-6}{7}\right)\left(x^2x\right)\left(yy\right)\left(z^3z^2\right)\)
\(=\frac{2}{7}x^3y^2z^5\) \(\Rightarrow\)Bậc là 10 Hệ số là \(\frac{2}{7}\)
c)\(-3x^2.y^4.\left(\frac{-1}{3}y^4z^5x\right).\left(\frac{-1}{2}zyx^3\right)\)
\(=\left(-3.\frac{-1}{3}.\frac{-1}{3}\right)\left(x^2xx^3\right)\left(y^4y^4y\right)\left(z^5z\right)\)
\(=\frac{-1}{3}x^6y^9z^6\) \(\Rightarrow\)Bậc là 21 Hệ số là \(\frac{-1}{3}\)
d)\(\frac{3}{4}xy^3\left(\frac{-2}{3}x^2y^4\right)^2\)
\(=\frac{3}{4}xy^3\left(\frac{4}{9}x^4y^{16}\right)\)
\(=\left(\frac{3}{4}\cdot\frac{4}{9}\right)\left(xx^4\right)\left(y^3y^{16}\right)\)
\(=\frac{1}{3}x^5y^{19}\)