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cảm ơn 
a) \(\left(x-3\right)\left(x-2\right)< 0\)
Ta có : \(x-2>x-3\)
\(\Rightarrow\left\{{}\begin{matrix}x-3< 0\\x-2>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 3\\x>2\end{matrix}\right.\Rightarrow2< x< 3\)
Vậy \(2< x< 3\)
b) \(3x+x^2=0\)
\(x\left(3+x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\3+x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
Vậy \(x\in\left\{-3;0\right\}\)
a/ \(\left(4x^2y^3\right)\left(x^ny^7\right)=4x^5y^{10}\)
\(\Leftrightarrow4x^{2+n}y^{3+7}=4x^5y^{10}\)
\(\Rightarrow2+n=5\Rightarrow n=3\)
Vậy \(n=3\)
b/ \(\left(-7x^4y^m\right)\left(-5x^ny^4\right)=35x^9y^{15}\)
\(\Leftrightarrow35x^{4+n}y^{m+4}=35x^9y^{15}\)
\(\Rightarrow\left[{}\begin{matrix}4+n=9\\m+4=15\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}n=5\\m=11\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}m=11\\n=5\end{matrix}\right.\)
a) \(\left(4x^2\times y^3\right)\left(x^n\times y^7\right)=4x^5y^{10}\)
\(\Rightarrow4\times\left(x^2\times x^n\right)\times\left(y^3\times y^7\right)=4x^5y^{10}\)
\(\Rightarrow4x^{2+x}y^{10}=4x^5y^{10}\)
\(\Rightarrow x^{2+n}=x^5\)
\(\Rightarrow2+n=5\)
\(\Rightarrow n=5-2\)
\(\Rightarrow n=3\)
Vậy \(n=3\).
b) \(\left(-7x^4y^m\right)\left(-5x^ny^4\right)=35x^9y^{15}\)
\(\Rightarrow\left[\left(-7\right)\times\left(-5\right)\right]\times\left(x^4\times x^n\right)\times\left(y^m\times y^4\right)=35x^9y^{15}\)
\(\Rightarrow35x^{4+n}y^{m+4}=35x^9y^{15}\)
\(\Rightarrow\left\{{}\begin{matrix}x^{4+n}=x^9\\y^{m+4}=y^{15}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}4+n=9\\m+4=15\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n=9-4\\m=15-4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n=5\\m=9\end{matrix}\right.\)
Vậy \(m=9\) và \(n=5\).
\(3x^2y^4\)-\(5xy^3\)-\(\dfrac{3}{2}x^2y^4\)+\(3xy^3\)+\(2xy^3\)+1=1,5\(x^2y^4\)+1>0
mk ko chép đề mà tách luôn nha
M = x2x2 + x2x2 + x2y2 + x2y2 + x2y2 + y2y2 + y2
= ( x2x2 + x2y2 ) + ( x2x2 + x2y2 ) + ( x2y2 + y2y2 ) + y2
= x2( x2 + y2 ) + x2( x2 + y2 ) + y2( x2 + y2 ) + y2
= ( x2 + y2 ) (x2 + x2 + y2 ) + y2
= 1( x2 + 1) + y2
= x2 + y2 +1 = 2
a, (6x2+9xy-y2) - ( 5x2-2xy)=M
=> M= (6x2+9xy-y2) - ( 5x2-2xy)
=> M= 6x2+9xy-y2 - 5x2+2xy
=> M=(6x2- 5x2)+(9xy+2xy)-y2
=>M= 1x2 + 11xy - y2
Vậy M= 1x2 + 11xy - y2
b, N= (3xy-4y2) - (x2-7xy+8y2)
=> N= 3xy-4y2 - x2+7xy-8y2
=> N= (3xy+7xy)-(4y2+8y2)-x2
=> N= 10xy - 12y2 -x2
Vậy N= 10xy - 12y2 -x2
Chúc bạn học tốt nha!!!
a, M+(5x2-2xy)=6x2+9xy-y2
\(\Rightarrow\)M= (6x2+9xy-y2)-(5x2-2xy)
= 6x2+9xy-y2-5x2+2xy
= (6x2-5x2)+(9xy+2xy)-y2
= x2+11xy-y2
Vậy đa thức M= x2+11xy-y2
b, (3xy-4y2)-N=x2-7xy+8y2
\(\Rightarrow\)N= (3xy-4y2)-(x2-7xy+8y2)
= 3xy-4y2-x2+7xy-8y2
= (3xy+7xy)+(-4y2-8y2)-x2
= 10xy-12y2-x2
Vậy đa thức N=10xy-12y2-x2
\(A=\dfrac{4^2}{1.3}+\dfrac{4^2}{3.5}+\dfrac{4^2}{5.8}+...+\dfrac{4^2}{45.47}.\dfrac{1-3-5-...-49}{8}\)
\(A=4\left(\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.8}+...+\dfrac{4}{45.47}\right).\dfrac{1-3-5-...-49}{8}\)\(A=4\left[2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{45}-\dfrac{1}{47}\right)\right].\dfrac{1-3-5-...-49}{8}\)\(A=8\left(1-\dfrac{1}{47}\right).\dfrac{1-3-5-...-49}{8}\)
\(A=8\left(1-\dfrac{1}{47}\right).\dfrac{-623}{8}\)
\(A=\dfrac{368}{47}.\dfrac{-623}{8}=\dfrac{-28658}{47}\)
\(6x^{n+2}-4x^n+3x^{n+2}-5x^n+x^{n+2}-x^n=0\)
\(\left(6x^{n+2}+3x^{n+2}+x^{n+2}\right)+\left(-4x^n-5x^n-x^n\right)=0\)
\(x^{n+2}\left(6+3+1\right)+x^n\left(-4-5-1\right)=0\)
\(10x^{n+2}+\left(-10\right)x^n=0\)
\(10x^n.x^2+\left(-10\right)x^n=0\)
\(x^n\left(10x^2-10\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^n=0\\10x^2-10=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\left(n\in N\circledast\right)\\10x^2=10\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\\left\{{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\end{matrix}\right.\)
Vậy...
thanks nhìu nha