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a) \(\left\{{}\begin{matrix}2x+3y=5\\4x-5y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=10\\4x-5y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=5\\11y=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+3\cdot\dfrac{9}{11}=5\\y=\dfrac{9}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{27}{11}=5\\y=\dfrac{9}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=\dfrac{28}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{14}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)
Vậy: \(x=\dfrac{14}{11};y=\dfrac{9}{11}\)
a) \(\hept{\begin{cases}\left(x-1\right)\left(2x+y\right)=0\\\left(y+1\right)\left(2y-x\right)=0\end{cases}}\)
\(\cdot x=1\Rightarrow\hept{\begin{cases}0=0\\\left(y+1\right)\left(2y-1\right)=0\end{cases}}\Leftrightarrow\hept{\begin{cases}0=0\\y=-1;y=\frac{1}{2}\end{cases}}\)
\(\cdot y=-1\Rightarrow\hept{\begin{cases}\left(x-1\right)\left(2x-1\right)=0\\0=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1;x=\frac{1}{2}\\0=0\end{cases}}\)
\(\cdot x=2y\Rightarrow\hept{\begin{cases}\left(2y-1\right)5y=0\\0=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}y=0\Rightarrow x=0\\y=\frac{1}{2}\Rightarrow x=1\end{cases}}\)
\(y=-2x\Rightarrow\hept{\begin{cases}0=0\\\left(1-2x\right)5x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\Rightarrow y=-1\\x=0\Rightarrow y=0\end{cases}}\)
b) \(\hept{\begin{cases}x+y=\frac{21}{8}\\\frac{x}{y}+\frac{y}{x}=\frac{37}{6}\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{21}{8}-y\\\left(\frac{21}{8}-y\right)^2+y^2=\frac{37}{6}y\left(\frac{21}{8}-y\right)\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{21}{8}-y\\2y^2-\frac{21}{4}y+\frac{441}{64}=-\frac{37}{6}y^2+\frac{259}{16}y\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{21}{8}-y\\1568y^2-4116y+1323=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{8}\\y=\frac{9}{4}\end{cases}}hay\hept{\begin{cases}x=\frac{9}{4}\\y=\frac{3}{8}\end{cases}}\)
c) \(\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\\\frac{2}{xy}-\frac{1}{z^2}=4\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{1}{z^2}=\left(2-\frac{1}{x}-\frac{1}{y}\right)^2\\\frac{1}{z^2}=\frac{2}{xy}-4\end{cases}}}\)\(\Leftrightarrow\hept{\begin{cases}\left(2xy-x-y\right)^2=-4x^2y^2+2xy\\\frac{1}{z^2}=\frac{2}{xy}-4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}8x^2y^2-4x^2y-4xy^2+x^2+y^2-2xy+2xy=0\\\frac{1}{z^2}=\frac{2}{xy}-4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}4x^2y^2-4x^2y+x^2+4x^2y^2-4xy^2+y^2=0\\\frac{1}{z^2}=\frac{2}{xy}-4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}\left(2xy-x\right)^2+\left(2xy-y\right)^2=0\\\frac{1}{z^2}=\frac{2}{xy}-4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=y=\frac{1}{2}\\z=\frac{-1}{2}\end{cases}}\)
d) \(\hept{\begin{cases}xy+x+y=71\\x^2y+xy^2=880\end{cases}}\). Đặt \(\hept{\begin{cases}x+y=S\\xy=P\end{cases}}\), ta có: \(\hept{\begin{cases}S+P=71\\SP=880\end{cases}}\Leftrightarrow\hept{\begin{cases}S=71-P\\P\left(71-P\right)=880\end{cases}}\Leftrightarrow\hept{\begin{cases}S=71-P\\P^2-71P+880=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}S=16\\P=55\end{cases}}hay\hept{\begin{cases}S=55\\P=16\end{cases}}\)
\(\cdot\hept{\begin{cases}S=16\\P=55\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y=16\\xy=55\end{cases}}\Leftrightarrow\hept{\begin{cases}x=16-y\\y\left(16-y\right)=55\end{cases}}\Leftrightarrow\hept{\begin{cases}x=16-y\\y^2-16y+55=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=5\\y=11\end{cases}}hay\hept{\begin{cases}x=11\\y=5\end{cases}}\)
\(\cdot\hept{\begin{cases}S=55\\P=16\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y=55\\xy=16\end{cases}}\Leftrightarrow\hept{\begin{cases}x=55-y\\y\left(55-y\right)=16\end{cases}}\Leftrightarrow\hept{\begin{cases}x=55-y\\y^2-55y+16=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{55-3\sqrt{329}}{2}\\y=\frac{55+3\sqrt{329}}{2}\end{cases}}hay\hept{\begin{cases}x=\frac{55+3\sqrt{329}}{2}\\y=\frac{55-3\sqrt{329}}{2}\end{cases}}\)
e) \(\hept{\begin{cases}x\sqrt{y}+y\sqrt{x}=12\\x\sqrt{x}+y\sqrt{y}=28\end{cases}}\). Đặt \(\hept{\begin{cases}S=\sqrt{x}+\sqrt{y}\\P=\sqrt{xy}\end{cases}}\), ta có \(\hept{\begin{cases}SP=12\\P\left(S^2-2P\right)=28\end{cases}}\Leftrightarrow\hept{\begin{cases}S=\frac{12}{P}\\P\left(\frac{144}{P^2}-2P\right)=28\end{cases}}\Leftrightarrow\hept{\begin{cases}S=\frac{12}{P}\\2P^4+28P^2-144P=0\end{cases}}\)
Tự làm tiếp nhá! Đuối lắm luôn
đkxđ: \(\left\{{}\begin{matrix}x\ne0\\y\ne0\end{matrix}\right.\)
pt đầu \(\Leftrightarrow x+\dfrac{2}{x}+y+\dfrac{1}{y}=6\) (3)
pt thứ 2 \(\Leftrightarrow x^2+\dfrac{4}{x^2}+y^2+\dfrac{1}{y^2}=14\) \(\Leftrightarrow\left(x^2+2.x.\dfrac{2}{x}+\dfrac{4}{x^2}\right)+\left(y^2+2y.\dfrac{1}{y}+\dfrac{1}{y^2}\right)=20\)
\(\Leftrightarrow\left(x+\dfrac{2}{x}\right)^2+\left(y+\dfrac{1}{y}\right)^2=20\) (4)
Đặt \(\left\{{}\begin{matrix}x+\dfrac{2}{x}=u\left(\left|u\right|\ge2\sqrt{2}\right)\\y+\dfrac{1}{y}=v\left(\left|v\right|\ge2\right)\end{matrix}\right.\) thì từ (3) và (4) suy ra \(\left\{{}\begin{matrix}u+v=6\\u^2+v^2=20\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}v=6-u\\u^2+\left(6-u\right)^2=20\end{matrix}\right.\)
\(u^2+\left(6-u\right)^2=20\) \(\Leftrightarrow u^2+36-12u+u^2=20\) \(\Leftrightarrow2u^2-12u+16=0\) \(\Leftrightarrow u^2-6u+8=0\) \(\Leftrightarrow\left(u-2\right)\left(u-4\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}u=2\left(loại\right)\\u=4\left(nhận\right)\end{matrix}\right.\).
\(\Rightarrow v=6-u=2\), suy ra \(\left\{{}\begin{matrix}x+\dfrac{2}{x}=4\\y+\dfrac{1}{y}=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\pm\sqrt{2}\\y=1\end{matrix}\right.\) (nhận).
Vậy hpt đã cho có các nghiệm \(\left(x;y\right)\in\left\{\left(2-\sqrt{2};1\right);\left(2+\sqrt{2};1\right)\right\}\)
\(y\left(x+1\right)^2=-x^2+2018x-1\)
\(\Leftrightarrow y=\dfrac{-x^2+2018x-1}{\left(x+1\right)^2}=-1+\dfrac{2020x}{\left(x+1\right)^2}\)
\(\Rightarrow\dfrac{2020x}{\left(x+1\right)^2}\in Z\)
Mà x và \(x\left(x+2x\right)+1\) nguyên tố cùng nhau
\(\Rightarrow2020⋮\left(x+1\right)^2\)
Ta có 2020 chia hết cho đúng 2 số chính phương là 1 và 4
\(\Rightarrow\left[{}\begin{matrix}\left(x+1\right)^2=1\\\left(x+1\right)^2=4\end{matrix}\right.\) \(\Rightarrow x=\left\{0;1\right\}\) \(\Rightarrow y\)
b.
Từ pt đầu:
\(x^2+xy-2y^2+2\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+2y\right)+2\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+2y+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-2y-2\end{matrix}\right.\)
Thế xuống dưới ...
\(\Rightarrow\left\{{}\begin{matrix}x+y=71-xy\\xy\left(x+y\right)=880\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+y=71-xy\\xy\left(71-xy\right)=880\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=71-xy\\x^2y^2-71xy+880=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+y=71-xy\\\left(xy-16\right)\left(xy-55\right)=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=71-xy\\\left[{}\begin{matrix}xy=16\\xy=55\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}xy=55\\x+y=16\end{matrix}\right.\\\left\{{}\begin{matrix}xy=16\\x+y=55\end{matrix}\right.\end{matrix}\right.\)
+ TH1: \(\left\{{}\begin{matrix}xy=55\\x+y=16\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=\frac{55}{x}\\x+\frac{55}{x}=16\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=\frac{55}{x}\\x^2-16x+55=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y=\frac{55}{x}\\\left(x-5\right)\left(x-11\right)=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=5\\y=11\end{matrix}\right.\\\left\{{}\begin{matrix}x=11\\y=5\end{matrix}\right.\end{matrix}\right.\)
TH còn lại lm tương tự
lấy vế trên trừ dưới bạn có 2 kết quả
thế từng kết quả vào là ra