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\(x^2-y^2=1\)
Ta có : \(\left(\frac{x}{5}\right)^2=\left(\frac{y}{4}\right)^2\)
\(=>\frac{x^2}{25}=\frac{y^2}{16}\)
A/d dãy ............
\(\frac{x^2-y^2}{25-16}=\frac{1}{9}=>\frac{x}{5}=\frac{y}{4}=\frac{1}{3}\)
\(=>\frac{x}{5}=\frac{1}{3}=>x=\frac{5}{3}\)
\(=>\frac{y}{4}=\frac{1}{3}=>x=\frac{4}{3}\)
\(\frac{x}{5}=\frac{y}{4}\)nên \(\frac{x^2}{25}=\frac{y^2}{16}=\frac{x^2-y^2}{25-16}=\frac{1}{9}\)=> \(\frac{x}{5}=\sqrt{\frac{1}{9}};-\sqrt{\frac{1}{9}}=\frac{1}{3};\frac{-1}{3}\)
=> x = \(\frac{1}{3}.5;\frac{-1}{3}.5=\frac{5}{3};\frac{-5}{3}\)
Kho..................wa.....................troi.....................thi......................lanh.................ret.......................ai........................tich..........................ung.....................ho........................minh.....................cho....................do....................lanh
Ta có: \(\left\{{}\begin{matrix}\left(3x-\dfrac{1}{5}\right)^{200}\ge0\\\left(\dfrac{2}{5}y+\dfrac{4}{7}\right)^{1968}\ge0\end{matrix}\right.\Leftrightarrow\left(3x-\dfrac{1}{5}\right)^{200}+\left(\dfrac{2}{5}y+\dfrac{4}{7}\right)^{1968}\ge0\)
Mà \(\left(3x-\dfrac{1}{5}\right)^{200}+\left(\dfrac{2}{5}y+\dfrac{4}{7}\right)^{1968}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(3x-\dfrac{1}{5}\right)^{200}=0\\\left(\dfrac{2}{5}y+\dfrac{4}{7}\right)^{1968}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=\dfrac{1}{5}\\\dfrac{2}{5}y=\dfrac{-4}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{15}\\y=\dfrac{-10}{7}\end{matrix}\right.\)
Vậy \(x=\dfrac{1}{15},y=\dfrac{-10}{7}\)
Bài này có mũ 200 hay 200 hay 20000 thì nó cũng chẳng ảnh hưởng gì nhé =))
Ta có : $(3x-\dfrac{1}{5})^{200})+(\dfrac{2}{5}y+\dfrac{4}{7})^1968=0$
Vì $3x-\dfrac{1}{5})^{200}\geq 0$
$\dfrac{2}{5}y+\dfrac{4}{7}^1968\geq 0$
\(=>\left\{{}\begin{matrix}\left(3x-\dfrac{1}{5}\right)^{200}=0\\\left(\dfrac{2}{5}y+\dfrac{4}{7}\right)^{1968}=0\end{matrix}\right.\)
\(=>\left\{{}\begin{matrix}\left(3x-\dfrac{1}{5}\right)=0\\\left(\dfrac{2}{5}y+\dfrac{4}{7}\right)^{ }=0\end{matrix}\right.\)
\(=>\left\{{}\begin{matrix}3x=\dfrac{1}{5}\\\dfrac{2}{5}y=\dfrac{-4}{7}\end{matrix}\right.\)
\(=>\left\{{}\begin{matrix}x=\dfrac{1}{15}\\y=\dfrac{-10}{7}\end{matrix}\right.\)
Bài 6:
b: \(x^2+165=y^2\)
TH1: x=2
\(y^2=x^2+165\)
=>\(y^2=2^2+165=4+165=169=13^2\)
=>y=13(nhận)
TH2: x lẻ
=>\(x^2\) lẻ
=>\(x^2+165\) chẵn và \(x^2+165>165\)
=>\(y^2\) chẵn và \(y^2>165\)
=>y là số chẵn lớn hơn 2
=>y không là số nguyên tố
=>Loại
Bài 5:
\(m=\left\lbrack\log_{10}2^{200}\right\rbrack+1=\left\lbrack200\cdot\log_{10}2\right\rbrack+1\)
=>\(m=\left\lbrack200\cdot0,30103\right\rbrack+1\)
=>m=60+1=61
=>\(2^{200}\) có 61 chữ số
=>m=61
\(n=\left\lbrack\log_{10}5^{200}\right\rbrack+1\)
\(=\left\lbrack200\cdot\log_{10}5\right\rbrack+1\)
=>n≃140
=>\(5^{200}\) có 140 chữ số
=>n=140
n+m=60+141=201