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\(3x^2y^4\)-\(5xy^3\)-\(\dfrac{3}{2}x^2y^4\)+\(3xy^3\)+\(2xy^3\)+1=1,5\(x^2y^4\)+1>0
Ta có:
A =2100-299+298-297+.....+22-21
=>2A=2101-2100+299-298+.....+23-22
=>2A+A=(2101-2100+299-298+.....+23-22) + (2100-299+298-297+....+22-21)
=>3A=2101-2
=>A=\(\frac{2^{101}-2}{3}\)
Vậy A=\(\frac{2^{101}-2}{3}\).
\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(\Rightarrow2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
\(\Rightarrow2A+A=\left(2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\right)+\left(2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\right)\)
\(\Rightarrow3A=2^{101}-2\)
\(\Rightarrow A=\frac{2^{101}-2}{3}\)
!)
=> x(x - 1)=0
=> \(\left[\begin{array}{nghiempt}x=1\\x-1=0\end{array}\right.\)
=>\(\left[\begin{array}{nghiempt}x=0\\x=1\end{array}\right.\)
Vậy đa thức có nghiệm là x=0 ; x=1
1) \(x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-1=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=1\end{array}\right.\)
b) \(x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-2=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=2\end{array}\right.\)
c)\(x^2-3x=0\)
\(\Leftrightarrow x\left(x-3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-3=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=3\end{array}\right.\)
d)\(3x^2-4x=0\)
\(\Leftrightarrow x\left(3x-4\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\3x-4=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{4}{3}\end{array}\right.\)
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 \(n=5\).
\(7\left(x-2004\right)^2=23-y^2\)
\(\Rightarrow7\left(x-2004\right)^2+y^2=23\left(1\right)\)
Vì \(y^2\ge0\) nên \(\left(x-2004\right)^2\le\frac{23}{7}\) suy ra \(\left[\begin{matrix}\left(x-2004\right)^2=0\\\left(x-2004\right)^2=1\end{matrix}\right.\)
*)Xét \(\left(x-2004\right)^2=0\) thay vào \((1)\) ta có: \(y^2=23\) (loại)
*)Xét \((x-2004)^2=1\) thay vào \((1)\) ta có \(y^2=16\)
Từ đó ta tìm được \(\left[\begin{matrix}\left\{\begin{matrix}x=2005\\y=4\end{matrix}\right.\\\left\{\begin{matrix}x=2003\\y=4\end{matrix}\right.\end{matrix}\right.\)
Sửa đề: \(C=\left(x^2y^3+x^3y^2-x^2-y^2+5\right)-\left(x^2y^3+x^3y^2+2y^2-1\right)\)
\(C=x^2y^3+x^3y^2-x^2-y^2+5-x^2y^3-x^3y^2-2y^2+1\)
\(=-3y^2-x^2+6\le6\)
Dấu '=' xảy ra khi x=y=0
Câu 1:
Nếu \(x\ge-2\)thì .......\(\Leftrightarrow\)x+2+3x=3012........x=752,5(t/m)
Nếu x<-2 thì........\(\Leftrightarrow\)-x-2+3x=3012.......x=1507(ko t/m)
\(\left\{{}\begin{matrix}P\left(x\right)=x+x^2-x^3+2x^3+2=x^3+x^2+x+2\\Q\left(x\right)=1+3x-x^2-4x+x^3=x^3-x^2-x+1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}P\left(x\right)+Q\left(x\right)=2x^3+3\\P\left(x\right)-Q\left(x\right)=2x^2+2x+1\end{matrix}\right.\)
Dễ thấy với các sô mũ m chăn tích \(x^m.y^m=1\)
Với số mũ n lẻ thì tích \(x^n.y^n=1-1\)
\(=>A=\left(-1+1\right)+\left(-1+1\right)+.....+\left(-1+1\right)+\left(-1\right)\)
=> A= - 1