câu 1: giải pt
a) \(x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}=4\)
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AM
6 tháng 2 2022
\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{6}\\\dfrac{\dfrac{2}{3}}{x}+\dfrac{\dfrac{2}{3}}{y}+\dfrac{\dfrac{8}{9}}{y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{6}\\\dfrac{\dfrac{2}{3}}{x}+\dfrac{\dfrac{14}{9}}{y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{6}\left(1\right)\\\dfrac{2}{3x}+\dfrac{14}{9y}=1\left(2\right)\end{matrix}\right.\)
Nhân cả hai vế (1) cho \(\dfrac{2}{3}\) ta có: \(\left\{{}\begin{matrix}\dfrac{2}{3x}+\dfrac{2}{3y}=\dfrac{5.2}{6.3}\\\dfrac{2}{3x}+\dfrac{14}{9y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{3x}+\dfrac{2}{3y}=\dfrac{10}{18}\left(3\right)\\\dfrac{2}{3x}+\dfrac{14}{9y}=1\left(4\right)\end{matrix}\right.\)
Lấy (4) trừ (3) ta có:
\(\dfrac{14}{9y}-\dfrac{2}{3y}=1-\dfrac{10}{18}\)\(\Leftrightarrow\dfrac{8}{9y}=\dfrac{4}{9}\)\(\Leftrightarrow y=2\Rightarrow x=\dfrac{1}{\dfrac{5}{6}-\dfrac{1}{2}}=3\)
NN
1
23 tháng 1 2022
ĐK: \(x\ne0\) ; \(y\ne0\)
Hệ phương trình tương đương với:
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+\dfrac{1}{x}\right)+\left(y+\dfrac{1}{y}\right)=4\\\left(x+\dfrac{1}{x}\right)^2+\left(y+\dfrac{1}{y}\right)^2=8\end{matrix}\right.\)
Đặt \(S=\left(x+\dfrac{1}{x}\right)+\left(y+\dfrac{1}{y}\right)\)
\(P=\left(x+\dfrac{1}{x}\right)\left(y+\dfrac{1}{y}\right)\)
Mà \(S^2\ge4P\)
Ta có: \(\left\{{}\begin{matrix}S=4\\S^2-2P=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}S=4\\P=4\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}\left(x+\dfrac{1}{x}\right)+\left(y+\dfrac{1}{y}\right)=4\\\left(x+\dfrac{1}{x}\right)\left(y+\dfrac{1}{y}\right)=4\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}x+\dfrac{1}{x}=2\\y+\dfrac{1}{y}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 3 2021
ĐKXĐ: \(x\ne\left\{-3;-2;-1;0\right\}\)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}=\dfrac{x}{x\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}=\dfrac{x}{x\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{x}{x\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{3}{x\left(x+3\right)}=\dfrac{x}{x\left(x+3\right)}\)
\(\Leftrightarrow x=3\)
AH
Akai Haruma
Giáo viên
16 tháng 12 2021
Lời giải:
Đặt $\frac{1}{x-y+2}=a;\frac{1}{x+y-1}=b$ thì HPT trở thành cơ bản:
\(\left\{\begin{matrix}
14a-10b=9\\
3a+2b=4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
14a-10b=9\\
15a+10b=20\end{matrix}\right.\)
$\Rightarrow (14a-10b)+(15a+10b)=9+20$
$\Leftrightarrow 29a=29\Leftrightarrow a=1$.
$b=\frac{4-3a}{2}=\frac{1}{2}$
Vậy: \(\left\{\begin{matrix} \frac{1}{x-y+2}=1\\ \frac{1}{x+y-1}=\frac{1}{2}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x-y+2=1\\ x+y-1=2\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x-y=-1\\ x+y=3\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=1\\ y=2\end{matrix}\right.\)
Y
2 tháng 2 2019
\(x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}=4\)
\(\Leftrightarrow x^2-2+\dfrac{1}{x^2}+y^2-2+\dfrac{1}{y^2}=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-\dfrac{1}{x}\right)^2=0\\\left(y-\dfrac{1}{y}\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{x}\\y=\dfrac{1}{y}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=1\\y^2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\pm1\\y=\pm1\end{matrix}\right.\)
6 tháng 10 2021
a)Ta có:
\(\left(x-3,5\right)^2+\left(y-\dfrac{1}{10}\right)^4\le0\)
\(\Rightarrow x-3,5=y-\dfrac{1}{10}=0\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\y=\dfrac{1}{10}=0,1\end{matrix}\right.\)
b) Ta có:
\(\left(5x+1\right)^2=\dfrac{36}{49}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=\dfrac{-6}{7}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{35}\\x=\dfrac{-13}{35}\end{matrix}\right.\)
6 tháng 10 2021
b: ta có: \(\left(5x+1\right)^2=\dfrac{36}{49}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{-1}{7}\\5x=\dfrac{-13}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{35}\\x=\dfrac{-13}{35}\end{matrix}\right.\)
AN
1
\(\Leftrightarrow x^2-2+\dfrac{1}{x^2}+y^2-2+\dfrac{1}{y^2}=4-2-2\)
\(\Leftrightarrow\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2=0\)
Với mọi x, y ta luôn có \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{x}\right)^2\ge0\\\left(y-\dfrac{1}{y}\right)^2\ge0\end{matrix}\right.\)
=> \(\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2\ge0\)
mà \(\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\\y-\dfrac{1}{y}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x^2-1}{x}=0\\\dfrac{y^2-1}{y}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1;x=-1\\y=1;y=-1\end{matrix}\right.\)
Vậy....
mk giải luôn đó nha
Giải:
Áp dụng BĐT AM-GM cho hai số dương, ta có:
\(x^2+\dfrac{1}{x^2}\ge2\sqrt{x^2.\dfrac{1}{x^2}}=2\)
\(y^2+\dfrac{1}{y^2}\ge2\sqrt{y^2.\dfrac{1}{y^2}}=2\)
\(\Leftrightarrow x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}\ge4\)
Dấu "=" xảy ra khi:
\(x=y=\pm1\)
Vậy ...