1 a ) \(3x\left(5x^2-2x-1\right)\)
b ) \(\left(x^2+2xy-3\right)\left(-xy\right)\)
c ) \(\frac{1}{2}x^2y\left(2x^3-\frac{2}{5}xy^2-1\right)\)
Chỉ cần cho đáp án thôi
thực hiện phép tính
a.\(-2xy^2.\left(x^3y-2x^2y^2+5xy^3\right)\)
b.\(\left(-2x\right).\left(x^3-3x^2-x+1\right)\)
c.3x\(^2\left(2x^3-x+5\right)\)
d.\(\left(-10x^3+\frac{2}{5}y-\frac{1}{3}z\right).\left(-\frac{1}{2}xy\right)\)
e.\(\left(3x^2y-6xy+9x\right).\left(-\frac{4}{3}xy\right)\)
f.\(\left(4xy+3y-5x\right).x^2y\)
a) Ta có: \(-2xy^2\cdot\left(x^3y-2x^2y^2+5xy^3\right)\)
\(=-2x^4y^3+4x^3y^4-10x^2y^5\)
b) Ta có: \(\left(-2x\right)\cdot\left(x^3-3x^2-x+1\right)\)
\(=-2x^4+6x^3+2x^2-2x\)
c) Ta có: \(3x^2\left(2x^3-x+5\right)\)
\(=6x^5-3x^3+15x^2\)
d) Ta có: \(\left(-10x^3+\frac{2}{5}y-\frac{1}{3}z\right)\cdot\left(-\frac{1}{2}xy\right)\)
\(=5x^4y-\frac{1}{5}xy^2+\frac{1}{6}xyz\)
e) Ta có: \(\left(3x^2y-6xy+9x\right)\cdot\left(-\frac{4}{3}xy\right)\)
\(=-4x^3y^2+8x^2y^2-12x^2y\)
f) Ta có: \(\left(4xy+3y-5x\right)\cdot x^2y\)
\(=4x^3y^2+3x^2y^2-5x^3y\)
giải hệ phương trình
a) \(\left\{{}\begin{matrix}\sqrt{2x^2+2y^2}+\sqrt{\frac{4}{3}\left(x^2+xy+y^2\right)}=2\left(x+y\right)\\\sqrt{3x+1}+\sqrt{5x+4}=3xy-y+3\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\\\sqrt{x+2y+1}+2\sqrt[3]{12x+7y+8}=2xy+x+5\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}x^2+xy+x+3=0\\\left(x+1\right)^2+3\left(y+1\right)+2\left(xy-\sqrt{x^2y+2y}\right)=0\end{matrix}\right.\)
b)\(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\)
\(\Rightarrow\left(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}\right)^2=\left(3\left(x+y\right)\right)^2\)
\(\Leftrightarrow\sqrt{\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)}=x^2+7xy+y^2\)
\(\Rightarrow\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)=\left(x^2+7xy+y^2\right)^2\)
\(\Leftrightarrow9\left(x-y\right)^2\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)
\(\rightarrow\left(x;y\right)\in\left\{\left(0;0\right),\left(1;1\right)\right\}\)
caau a) binh phuong len ra no x=y tuong tu
c)
ĐK $y \geqslant 0$
Hệ đã cho tương đương với
$\left\{\begin{matrix} 2x^2+2xy+2x+6=0\\ (x+1)^2+3(y+1)+2xy=2\sqrt{y(x^2+2)} \end{matrix}\right.$
Trừ từng vế $2$ phương trình ta được
$x^2+2+2\sqrt{y(x^2+2)}-3y=0$
$\Leftrightarrow (\sqrt{x^2+2}-\sqrt{y})(\sqrt{x^2+2}+3\sqrt{y})=0$
$\Leftrightarrow x^2+2=y$
Bài 1: Thu gọn
a) \(\frac{1}{5}x^4y^3-3x^4y^3\)
b) \(5x^2y^5-\frac{1}{4}x^2y^5\)
c) \(\frac{1}{7}x^2y^3.\left(-\frac{14}{3}xy^2\right)-\frac{1}{2}xy.\left(x^2y^{\text{4}}\right)\)
d) \(\left(3xy\right)^2.\left(-\frac{1}{2}x^3y^2\right)\)
e) \(-\frac{1}{4}xy^2+\frac{2}{5}x^2y+\frac{1}{2}xy^2-x^2y\)
f) \(\frac{1}{2}x^4y.\left(-\frac{2}{3}x^3y^2\right)-\frac{1}{3}x^7y^3\)
g) \(\frac{1}{2}x^2y.\left(-10x^3yz^2\right).\frac{1}{4}x^5y^3z\)
h) \(4.\left(-\frac{1}{2}x\right)^2-\frac{3}{2}x.\left(-x\right)+\frac{1}{3}x^2\)
i) \(1\frac{2}{3}x^3y.\left(\frac{-1}{2}xy^2\right)^2-\frac{5}{4}.\frac{8}{15}x^3y.\left(-\frac{1}{2}xy^2\right)^2\)
k) \(-\frac{3}{2}xy^2.\left(\frac{3}{4}x^2y\right)^2-\frac{3}{5}xy.\left(-\frac{1}{3}x^4y^3\right)+\left(-x^2y\right)^2.\left(xy\right)^2\)
n) \(-2\frac{1}{5}xy.\left(-5x\right)^2+\frac{3}{4}y.\frac{2}{3}\left(-x^3\right)-\frac{1}{9}.\left(-x\right)^3.\frac{1}{3}y\)
m) \(\left(-\frac{1}{3}xy^2\right)^2.\left(3x^2y\right)^3.\left(-\frac{5}{2}xy^2z^3\right)^{^2}\)
p) \(-2y.\left|2\right|x^4y^5.\left|-\frac{3}{4}\right|x^3y^2z\)
Bài 1:
a) \(\frac{1}{5}x^4y^3-3x^4y^3\)
= \(\left(\frac{1}{5}-3\right)x^4y^3\)
= \(-\frac{14}{5}x^4y^3.\)
b) \(5x^2y^5-\frac{1}{4}x^2y^5\)
= \(\left(5-\frac{1}{4}\right)x^2y^5\)
= \(\frac{19}{4}x^2y^5.\)
Mình chỉ làm 2 câu thôi nhé, bạn đăng nhiều quá.
Chúc bạn học tốt!
1.Rút gọn các đơn thức sau và chỉ bra hệ số và phần biến
a)\(-2x^2y.\left(-xy^2\right)\)
b)\(\frac{1}{4}\left(x^2y^3\right)^2.\left(-2xy\right)\)
2.Tính các tích sau rồi tìm bậc của công thức thu được
a)\(\left(-7x^2yz\right).\frac{3}{7}xy^2z^3\)
b)\(-\frac{2}{3}xy^2z.\left(-3x^2y\right)^2\)
c)\(x^2yz.\left(2xy\right)^2z\)
d)\(-\frac{1}{3}x^2y.\left(-x^3yz\right)\)
3.Thực hiện phép nhân các đơn thức sau rồi tìm bậc đơn thức nhận được
a)\(4x^2y.\left(-5xy^4\right)\)
b)\(\frac{-1}{2}x^3y.\left(-xy\right)\)
c)\(\left(-2x^3y\right).3xy^4\)
d)\(\frac{-4}{5}x^3y.\left(-xy\right)\)
e)\(\frac{2}{3}xyz.\left(-6x^2y\right).\left(-xy^2z\right)\)
f)\(\left(-2x^2y\right).\left(\frac{-1}{2}\right)^2.\left(x^2y^3\right)^2\)
Làm tính nhân :
a) \(3x\left(5x^2-2x-1\right)\)
b) \(\left(x^2+2xy-3\right)\left(-xy\right)\)
c) \(\dfrac{1}{2}x^2y\left(2x^3-\dfrac{2}{5}xy^2-1\right)\)
a) \(3x\left(5x^2-2x-1\right)\)
\(=15x^3-6x^2-3x\)
b) \(\left(x^2+2xy-3\right)\left(-xy\right)\)
\(=-x^3y-2x^2y^2+3xy\)
c) \(\dfrac{1}{2}x^2y\left(2x^3-\dfrac{2}{5}xy^2-1\right)\)
\(=x^5y-\dfrac{1}{5}x^3y^3-\dfrac{1}{2}x^2y\)
a) 3x(5x2-2x-1) = 15x3 - 6x2 - 3x
b) (x2+2xy-3)(-xy) = -x3y - 2x2y2 + 3xy
c) \(\dfrac{1}{2}\)x2y\((\)2x3 - \(\dfrac{2}{5}\)xy2 - 1) = x5y - \(\dfrac{1}{5}\)x3y3 - \(\dfrac{1}{2}\)x2y
2) Giải phương trình
a) \(\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)
b) \(\left(2x+3\right).\left(\frac{3x+8}{2-7x}+1\right)=\left(x-5\right).\left(\frac{3x+8}{2-7x}+1\right)\)
3) Rút gọn
a) \(\frac{2x-1}{x^3+1}+\frac{2x}{x^2-x+1}+\frac{-x}{x+1}+2\)
b) \(\frac{x+1}{2x-2}+\frac{x^2+3}{2-2x^2}+\frac{1}{1-x}-1,5\)
c) \(\left(\frac{x^2}{x^3-4x}-\frac{6}{3x-6}+\frac{1}{x+2}\right).\frac{x+2}{6}\)
d) \(\left(\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}\right):\frac{x^2-2xy+y^2}{x^2y-xy^2}\)
e) \([\frac{1}{\left(2x-y\right)^2}+\frac{2}{4x^2-y^2}-\frac{1}{\left(2x+y\right)^2}].\frac{x^2+4xy+y^2}{16x}\)
Mn giúp mik vs mik đang cần gấp
\(a,\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)\(\Leftrightarrow\frac{x^2+3x+2+x^2-3x+2}{x^2-4}=\frac{2\left(x^2+2\right)}{x^2-4}\)
\(\Leftrightarrow2\left(x^2+2\right)=2\left(x^2+2\right)\)(luôn đúng)
Vậy pt có vô số nghiệm
\(b,\Leftrightarrow\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\)
\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left(2x+3-x+5\right)=0\)\(\Leftrightarrow\left(\frac{-4x+10}{2-7x}\right)\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-4x+10=0\\x+8=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-8\end{cases}}\)
Mấy câu rút gọn bạn quy đồng nha
Giải PT và HPT:
1)\(\left\{{}\begin{matrix}xy+x+y=3\\\frac{1}{x^2+2x}+\frac{1}{y^2+2y}=\frac{2}{3}\end{matrix}\right.\)
2)\(\left(\sqrt{x+4}-2\right)\left(\sqrt{4-x}+2\right)=2x\)
3)\(\left\{{}\begin{matrix}xy\left(x+y\right)=2\\9xy\left(3x-y\right)+6=26x^3-2y^3\end{matrix}\right.\)
4)\(\left\{{}\begin{matrix}x^2-2xy+x-2y+3=0\\y^2-x^2+2xy+2x-2=0\end{matrix}\right.\)
Tìm đa thức M , biết :
a) \(M-\left(\frac{1}{2}x^2y-5xy^2+x^3-y^3\right)=\frac{3}{4}xy^2-2x^2y+\)\(2y^3-\frac{1}{3}x^3\)
b)\(\left(-\frac{1}{3}x^3y^3+5x^2y^2-\frac{5}{2}xy\right)-M=xy-\frac{1}{6}x^3y^3-3x^2y^2\)
c)\(\left(\frac{2}{7}xy^4-5x^5+7x^2y^3-3\right)+M=0\)
Làm tính nhân:
a,3x(5x2-2x-1)
b,(x2+2xy-3)(-xy)
c,\(\frac{1}{2}x^2y\left(2x^3-\frac{2}{5}xy^2-1\right)\)