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10) \(9x^2+4y^2=20xy\)
\(\Leftrightarrow\left(3x-2y\right)^2=8xy\)
\(\Rightarrow\left(3x-2y\right)=\sqrt{8xy}\)
--- \(9x^2+4y^2=20xy\)
\(\Leftrightarrow\left(3x+2y\right)^2=32xy\)
\(\Rightarrow\left(3x+2y\right)=\sqrt{32xy}\)
\(A=\frac{3x-2y}{3x+2y}=\frac{\sqrt{8xy}}{\sqrt{32xy}}=\frac{1}{2}=0,5\)
5) \(x^3+8-\left(x+2\right)\left(x^2+3x+3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+4\right)-\left(x+2\right)\left(x^2+3x+3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(-5x+1\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x+2=0\Leftrightarrow x=-2\\-5x+1=0\Leftrightarrow x=0,2\end{matrix}\right.\)
Tổng các nghiệm là: -2+0,2=-1,8
K=\(\frac{x^2-3x+2x-6}{x^2-4}\)
=\(\frac{x\left(x-3\right)+2\left(x-3\right)}{x^2-4}\)
=\(\frac{\left(x-3\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}\)
=\(\frac{x-3}{x-2}\)
ta có K=3
\(\Rightarrow3=\frac{x-3}{x-2}\)
\(\Leftrightarrow3x-6=x-3\)
\(\Leftrightarrow2x=3\)
\(\Rightarrow x=\frac{3}{2}\)=1.5
vây x=1,5
(Nhập kết quả dưới dạng số thập phân gọn nhất).
\(\dfrac{x^9-1}{x^9+1}=7=\dfrac{7}{1}\Rightarrow\dfrac{x^9-1}{7}=\dfrac{x^9+1}{1}=\dfrac{-2}{6}=\dfrac{-1}{3}\)
\(\Rightarrow\dfrac{x^9-1}{7}=\dfrac{-1}{3}\Rightarrow x^9=1-\dfrac{7}{3}=\dfrac{-4}{3}\)
\(\Rightarrow x^{18}=\left(x^9\right)^2=\left(\dfrac{-4}{3}\right)^2=\dfrac{16}{9}\)
\(A=\dfrac{x^{18}-1}{x^{18}+1}=\dfrac{\dfrac{16}{9}-1}{\dfrac{16}{9}+1}=\dfrac{7}{25}\)
0,5
\(9x^2+4y^2=20xy\)
\(\Leftrightarrow\left(3x\right)^2+\left(2y\right)^2-20xy=0\)
\(\Leftrightarrow\left(3x\right)^2+\left(2y\right)^2-12xy-8xy=0\)
\(\Leftrightarrow\left(3x-2y\right)^2-8xy=0\)
\(\Leftrightarrow\left(3x-2y\right)^2=8xy\)
\(\Leftrightarrow3x-2y=\sqrt{8xy}\)(1)
- \(9x^2+4y^2=20xy\)
\(\Leftrightarrow\left(3x\right)^2+\left(2y\right)^2-20xy=0\)
\(\Leftrightarrow\left(3x\right)^2+\left(2y\right)^2+12xy-32xy=0\)
\(\Leftrightarrow\left(3x+2y\right)^2-32xy=0\)
\(\Leftrightarrow\left(3x+2y\right)^2=32xy\)
\(\Leftrightarrow3x+2y=\sqrt{32xy}\)(2)
Thay (1) và (2) vào A, ta có:
\(A=\dfrac{3x-2y}{3x+2y}=\dfrac{\sqrt{8xy}}{\sqrt{32xy}}=0,5\)
\(9x^2+4y^2=20xy\)
=>\(\left(9x^2-18xy\right)-\left(2xy-4y^2\right)=0\)
=>\(9x\left(x-2y\right)-2y\left(x-2y\right)=0\)
=>\(\left(x-2y\right)\left(9x-2y\right)=0\)
=>\(x-2y=0\) (vì 2y<3x<0 nên 9x\(\ne2y\) )
=>x=2y
Thay x=2y vào A ta có
\(A=\dfrac{6y-2y}{6y+2y}=\dfrac{4y}{8y}=0,5\)
Vậy A=0,5
Quá dài
\(9x^2+4y^2=20xy\Rightarrow\left\{{}\begin{matrix}\left[\left(3x\right)^2-2.\left(3x\right).\left(2y\right)+\left(2y\right)^2\right]=8xy\left(1\right)\\\left[\left(3x\right)^2+2.\left(3x\right).\left(2y\right)+\left(2y\right)^2\right]=32xy\left(2\right)\end{matrix}\right.\)Lấy (1) chia cho (2)
\(\left|\dfrac{3x-2y}{3x+2y}\right|=\left|A\right|=\dfrac{8}{32}=\dfrac{1}{2}\)
\(2y< 3x< 0\Rightarrow\left\{{}\begin{matrix}3x-2y>0\\3x+2y< 0\end{matrix}\right.\)
\(A< 0\Rightarrow A=-\dfrac{1}{2}\)