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i) Ta có: \(\frac{x}{3}=\frac{y}{6}.\)
=> \(\frac{x}{3}=\frac{y}{6}\) và \(x+y=90.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{x}{3}=\frac{y}{6}=\frac{x+y}{3+6}=\frac{90}{9}=10.\)
\(\left\{{}\begin{matrix}\frac{x}{3}=10\Rightarrow x=10.3=30\\\frac{y}{6}=10\Rightarrow y=10.6=60\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(30;60\right).\)
ii) Ta có: \(\frac{x}{3}=\frac{y}{6}.\)
=> \(\frac{4x}{12}=\frac{y}{6}\) và \(4x-y=42.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{4x}{12}=\frac{y}{6}=\frac{4x-y}{12-6}=\frac{42}{6}=7.\)
\(\left\{{}\begin{matrix}\frac{x}{3}=7\Rightarrow x=7.3=21\\\frac{y}{6}=7\Rightarrow y=7.6=42\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(21;42\right).\)
Chúc bạn học tốt!
1, \(\left(xy\right)^2-\frac{1}{2}x^2y^2+3xy^2.\left(-\frac{1}{3}x\right)\)
\(=x^2y^2-\frac{1}{2}x^2y^2-x^2y^2\)
\(=-\frac{1}{2}x^2y^2\)
2, \(4.\left(-\frac{1}{2}x\right)^2-\frac{3}{2}x.\left(-x\right)+\frac{1}{3}x^2\)
\(=x^2+\frac{3}{2}x^2+\frac{1}{3}x^2\)
\(=\frac{17}{6}x^2\)
3, \(-4.\left(2x\right)^2y^3+\frac{1}{2}xy.\left(-2xy^2\right)+\frac{1}{4}x^2y^3\)
\(=-16x^2y^3-x^2y^3+\frac{1}{4}x^2y^3\)
\(=-\frac{67}{4}x^2y^3\)
4, \(\frac{1}{3}x^4y-\frac{5}{3}x^3.\left(\frac{5}{2}xy\right)+\frac{3}{4}x^4y\)
\(=\frac{1}{3}x^4y-\frac{25}{6}x^4y+\frac{3}{5}x^4y\)
\(=-\frac{97}{30}x^4y\)
5, \(\left(-2x^3y^4\right)^2-5x^2y.\left(\frac{3}{4}x^4y^7\right)-\frac{2}{3}x^6y^8\)
\(=4x^6y^8-\frac{15}{4}x^6y^8-\frac{2}{3}x^6y^8\)
\(=-\frac{5}{12}x^6y^8\)
Bài 1 :
a/ \(x^2-7x+6=0\)
\(\Leftrightarrow x^2-6x-x+6=0\)
\(\Leftrightarrow x\left(x-6\right)-\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)
Vậy....
b/ \(x^2-10x+9=0\)
\(\Leftrightarrow x^2-9x-x+9=0\)
\(\Leftrightarrow x\left(x-9\right)-\left(x-9\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-9=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)
Vậy...
c/ \(x^2+9x+8=0\)
\(\Leftrightarrow x^2+8x+x+8=0\)
\(\Leftrightarrow\left(x+8\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=-1\end{matrix}\right.\)
Vậy ...
d/ \(x^2-11x+10=0\)
\(\Leftrightarrow x^2-11x+10=0\)
\(\Leftrightarrow x^2-x-10x+10=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-10=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=10\end{matrix}\right.\)
Vậy...
Bài 2 :
Ta có :
\(\frac{2x-y}{x+y}=\frac{2}{3}\)
\(\Leftrightarrow3\left(2x-y\right)=2\left(x+y\right)\)
\(\Leftrightarrow6x-3y=2x+2y\)
\(\Leftrightarrow6x-2x=2y+3y\)
\(\Leftrightarrow4x=5y\)
\(\Leftrightarrow\frac{x}{y}=\frac{5}{4}\)
Vậy....
Bài 3 : không hiểu đề lắm ???!!!!
Bài 4 :
Ta có :
\(\frac{x}{y^2}=2\Leftrightarrow x=2y^2\left(1\right)\)
Thay (1) ta có :
\(\frac{x}{y}=16\)
\(\Leftrightarrow\frac{2y^2}{y}=16\)
\(\Leftrightarrow2y=16\)
\(\Leftrightarrow y=8\Leftrightarrow x=128\)
Vậy...
1.
\(-3x^5y^4+3x^2y^3-7x^2y^3+5x^5y^4\)
\(=(-3x^5y^4+5x^5y^4)+(3x^2y^3-7x^2y^3)\)
\(=2x^5y^4-4x^2y^3\)
2.
\(\frac{1}{2}x^4y-\frac{3}{2}x^3y^4+\frac{5}{3}x^4y-x^3y^4\)
\(=(\frac{1}{2}x^4y+\frac{5}{3}x^4y)-(\frac{3}{2}x^3y^4+x^3y^4)\)
\(=\frac{13}{6}x^4y-\frac{5}{2}x^3y^4\)
3.
\(5x-7xy^2+3x-\frac{1}{2}xy^2\)
\(=(5x+3x)-(7xy^2+\frac{1}{2}xy^2)\)
\(=8x-\frac{15}{2}xy^2\)
4.
\(\frac{-1}{5}x^4y^3+\frac{3}{4}x^2y-\frac{1}{2}x^2y+x^4y^3\)
\(=(\frac{-1}{5}x^4y^3+x^4y^3)+(\frac{3}{4}x^2y-\frac{1}{2}x^2y)\)
\(=\frac{4}{5}x^4y^3+\frac{1}{4}x^2y\)
5.
\(\frac{7}{4}x^5y^7-\frac{3}{2}x^2y^6+\frac{1}{5}x^5y^7+\frac{2}{3}x^2y^6\)
\(=(\frac{7}{4}x^5y^7+\frac{1}{5}x^5y^7)+(-\frac{3}{2}x^2y^6+\frac{2}{3}x^2y^6)\)
\(=\frac{39}{20}x^5y^7-\frac{5}{6}x^2y^6\)
6.
\(\frac{1}{3}x^2y^5(-\frac{3}{5}x^3y)+x^5y^6=(\frac{1}{3}.\frac{-3}{5})(x^2.x^3)(y^5.y)+x^5y^6\)
\(=\frac{-1}{5}x^5y^6+x^5y^6=\frac{4}{5}x^5y^6\)
A=\(x^3.\left(\frac{-5}{4}x^2y\right)\)=\(x^5\).\(\left(\frac{-5}{4}\right)y\)
-Bậc là: 6
-Hệ số:\(\frac{-5}{4}\)
B=\(\left(\frac{-3}{4}x^5y^4\right).\left(xy^2\right).\left(\frac{-8}{9}\right)\)\(x^2y^5\)
=\(\frac{2}{3}.x^8.y^{11}\)
-Bậc là: 19
-Hệ số:\(\frac{2}{3}\)
C=\(\frac{1}{6}x\left(2y^3\right)^2.\left(-9x^5y\right)\)
=\(\frac{1}{6}x\left(4.y^6\right).\left(-9x^5y\right)\)
=-6.\(x^6\).\(y^7\)
-Bậc là: 13
-Hệ số: -6