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Theo t/c dãy tỉ số bằng nhau, ta có:
\(\frac{x}{2}-\frac{y}{3}=\frac{y^2-x^2}{3^2-2^2}=\frac{20}{5}=4\)
\(=>\hept{\begin{cases}\frac{x}{2}=4\\\frac{y}{3}=4\end{cases}}=>\hept{\begin{cases}x=8\\y=12\end{cases}}\)
\(x^2\left(x^2-4\right)=3\left(x^2-4\right)\)
\(\Rightarrow x^2\left(x^2-4\right)-3\left(x^2-4\right)=0\)
\(\Rightarrow\left(x^2-4\right)\left(x^2-3\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x^2-4=0\\x^2-3=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}\left[\begin{array}{nghiempt}x=2\\x=-2\end{array}\right.\\\left[\begin{array}{nghiempt}x=\sqrt{3}\\x=-\sqrt{3}\end{array}\right.\end{array}\right.\)
Vậy x=2; x= - 2 ; x=\(\sqrt{3}\) ; x=\(-\sqrt{3}\)
Có : \(x^2\left(x^2-4\right)=3\left(x^2-4\right)\)
\(\Leftrightarrow x^2\left(x^2-4\right)-3\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(x^2-3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x^2-4=0\\x^2-3=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x^2=4\\x^2=3\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=2;x=-2\\x=\sqrt{3};x=-\sqrt{3}\end{array}\right.\)
Vậy \(x=-2;x=2;x=-\sqrt{3};x=\sqrt{3}\)
c) \(\dfrac{x+4}{20}=\dfrac{5}{x+4}\)
⇔\(\left(x+4\right)\left(x+4\right)=100\)
⇔\(\left(x+4\right)^2=10^2\)
⇔\(\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=6\\x=-14\end{matrix}\right.\)
\(c,ĐK:x\ne-4\\ PT\Leftrightarrow\left(x+4\right)^2=100\\ \Leftrightarrow\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(tm\right)\\x=-14\left(tm\right)\end{matrix}\right.\\ d,ĐK:x\ne-2;x\ne-3\\ PT\Leftrightarrow\left(x-1\right)\left(x+3\right)=\left(x-2\right)\left(x+2\right)\\ \Leftrightarrow x^2+2x-3=x^2-4\\ \Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\left(tm\right)\)
3|x-1|+2|x-4|=x+17
x=-1,
x=7
nha bạn chúc bạn học tốt nha
\(\frac{20-x}{x+7}=\frac{2}{5}\)
=> \(5\left(20-x\right)=2\left(x+7\right)\)
<=> 100 - 5x = 2x + 14
=> 2x + 5x = 100 - 14
=> 7x = 86
=> x = 86/7