so sánh hai tổng \(\left(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{2020}\right)?\left(\frac{1}{20}+\frac{1}{21}+...+\hept{\begin{cases}1\\2019\end{cases}}\right)\)
giải hệ phương trình:
1) \(\hept{\begin{cases}2\left(x+y\right)+3\left(x+y\right)=4\\\left(x+y\right)+2\left(x-y\right)=5\end{cases}}\)
2)\(\hept{\begin{cases}\left(2x-3\right)\left(2y+4\right)=4x\left(y-3\right)+54\\\left(x+1\right)\left(3y-3\right)=3y\left(x+1\right)-12_{ }\end{cases}}\)
3) \(\hept{\begin{cases}\frac{2y-5x}{3}+5=\frac{y+27}{4}-2x\\\frac{x+1}{3}+y=\frac{6y-5x}{7}\end{cases}}\)
4)\(\hept{\begin{cases}\frac{1}{2}\left(x+2\right)\left(y+3\right)-\frac{1}{2}xy=50\\\frac{1}{2}xy-\frac{1}{2}\left(x-2\right)\left(y-2\right)=32\end{cases}}\)
5)\(\hept{\begin{cases}\left(x+20\right)\left(y-1\right)=xy\\\left(x-10\right)\left(y+1\right)=xy\end{cases}}\)
Những bài còn lại chỉ cần phân tích ra rồi rút gọn là được nha. Bạn tự làm nha!
Đặt \(\hept{\begin{cases}x+y=a\\x-y=b\end{cases}}\)\(\Rightarrow\)ta có hệ \(\hept{\begin{cases}2a+3b=4\\a+2b=5\end{cases}}\Rightarrow\hept{\begin{cases}a=-7\\b=6\end{cases}}\)Từ đó ta có \(\hept{\begin{cases}x+y=-7\\x-y=6\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{1}{2}\\y=-\frac{13}{2}\end{cases}}\)PS: Cái đề chỗ 3(x+y) phải thành 3(x-y) chứ
2) Từ hệ ta có \(\hept{\begin{cases}20x-6y=66\\-3x=-9\end{cases}}\Rightarrow\hept{\begin{cases}x=3\\y=-1\end{cases}}\)
Giải các hệ phương trình sau:
\(\hept{\begin{cases}\left(x-1\right)\left(2x+y\right)=0\\\left(y+1\right)\left(2y-x\right)=0\end{cases}}\)\(\hept{\begin{cases}x+y=\frac{21}{8}\\\frac{x}{y}+\frac{y}{x}=\frac{37}{6}\end{cases}}\)\(\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\\\frac{2}{xy}-\frac{1}{z^2}=4\end{cases}}\)\(\hept{\begin{cases}xy+x+y=71\\x^2y+xy^2=880\end{cases}}\)
\(\hept{\begin{cases}x\sqrt{y}+y\sqrt{x}=12\\x\sqrt{x}+y\sqrt{y}=28\end{cases}}\)
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
Cho \(\hept{\begin{cases}\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\\\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=12\end{cases}}\)tính \(\left(\frac{1}{a}-3\right)^{2020}+\left(\frac{1}{b}-3\right)^{2020}+\left(\frac{1}{c}-3\right)^{2020}\)
Ta có :\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=36\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=36\)
\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=12\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)
\(\Rightarrow\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}=\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}\)
=> \(\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}-\frac{2}{ab}-\frac{2}{bc}-\frac{2}{ca}=0\)
=> \(\left(\frac{1}{a^2}-\frac{2}{ab}+\frac{1}{b^2}\right)+\left(\frac{1}{b^2}-\frac{2}{bc}+\frac{1}{c^2}\right)+\left(\frac{1}{c^2}-\frac{2}{ac}+\frac{1}{a^2}\right)=0\)
=> \(\left(\frac{1}{a}-\frac{1}{b}\right)^2+\left(\frac{1}{b}-\frac{1}{c}\right)^2+\left(\frac{1}{c}-\frac{1}{a}\right)^2=0\)
=> \(\hept{\begin{cases}\frac{1}{a}-\frac{1}{b}=0\\\frac{1}{b}-\frac{1}{c}=0\\\frac{1}{c}-\frac{1}{a}=0\end{cases}}\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\)
Khi đó \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\Leftrightarrow3\frac{1}{a}=6\Rightarrow\frac{1}{a}=2\Leftrightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}=2\)
Khi đó Đặt P = \(\left(\frac{1}{a}-3\right)^{2020}+\left(\frac{1}{b}-3\right)^{2020}+\left(\frac{1}{c}-3\right)^{2020}\)
= (2 - 3)2020 + (2 - 3)2020 + (2 - 3)2020
= 1 + 1 + 1 = 3
Vậy P = 3
Cho A=\(\left(\hept{\begin{cases}1\\2^2\end{cases}}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)....\left(\frac{1}{100^2}-1\right)\)So sánh A với \(-\frac{1}{2}\)
giúp mình với ạ , mình đang cần gấp !!!
a,\(\hept{\begin{cases}3\left(x+1\right)+2\left(x+2y\right)=4\\4\left(x+1\right)-\left(x+2y\right)=9\end{cases}}\)
b, \(\hept{\begin{cases}x+\frac{1}{y}=\frac{-1}{2}\\2x-\frac{3}{y}=\frac{-7}{2}\end{cases}}\)
c,\(\hept{\begin{cases}\frac{x+2}{x+1}+\frac{2}{y-2}=6\\\frac{5}{x+1}-\frac{1}{y-2}=3\end{cases}}\)
c) Ta có: \(\left\{{}\begin{matrix}\dfrac{x+2}{x+1}+\dfrac{2}{y-2}=6\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x+1}+\dfrac{10}{y-2}=25\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{y-2}=22\\\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-2=\dfrac{1}{2}\\\dfrac{1}{x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=1\\y-2=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{5}{2}\end{matrix}\right.\)
Tìm x,y , biết :
a,\(\hept{\begin{cases}x\left(x+y\right)=\frac{1}{48}\\y\left(x+y\right)=\frac{1}{24}\end{cases}}\)
b,\(\hept{\begin{cases}x\left(x-y\right)=\frac{3}{10}\\y\left(x+y\right)=-\frac{3}{10}\end{cases}}\)
Giải phương trình
\(\hept{\begin{cases}\frac{1}{x}+2\left(x+y\right)=3\\3x\left(x+y\right)-x=2\end{cases}}\)\(\hept{\begin{cases}\frac{x+y}{x-y}+\frac{2x}{y+1}=3\\\frac{x+y}{2\left(x-y\right)}-\frac{3x}{y+1}=\frac{-1}{2}\end{cases}}\)\(\hept{\begin{cases}2x+3y=xy+5\\\frac{1}{x}+\frac{1}{y+1}=1\end{cases}}\)1.
\(ĐK:x\ne0\)
HPT
\(\Leftrightarrow\hept{\begin{cases}2x\left(x+y\right)-3x+1=0\\3x\left(x+y\right)-x-2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}3x\left(x+y\right)-\frac{9}{2}x+\frac{3}{2}=0\left(1\right)\\3x\left(x+y\right)-x-2=0\left(2\right)\end{cases}}\)
\(\left(1\right)-\left(2\right)\Leftrightarrow\frac{7}{2}x=\frac{7}{2}\)
\(\Leftrightarrow x=1\left(3\right)\)
\(\left(1\right),\left(3\right)\Rightarrow3\left(1+y\right)-3=0\)
\(\Leftrightarrow y=0\)
Vay nghiem cua HPT la \(\left(1;0\right)\)
giúp mình với ạ , mình đang cần gấp !!!
a,\(\hept{\begin{cases}3\left(x+1\right)+2\left(x+2y\right)=4\\4\left(x+1\right)-\left(x+2y\right)=9\end{cases}}\)
b, \(\hept{\begin{cases}x+\frac{1}{y}=\frac{-1}{2}\\2x-\frac{3}{y}=\frac{-7}{2}\end{cases}}\)
c,\(\hept{\begin{cases}\frac{x+2}{x+1}+\frac{2}{y-2}=6\\\frac{5}{x+1}-\frac{1}{y-2}=3\end{cases}}\)
a) \(\hept{\begin{cases}3\left(x+1\right)+2\left(x+2y\right)=4\\4\left(x+1\right)-\left(x+2y\right)=9\end{cases}}\Leftrightarrow\hept{\begin{cases}3\left(x+1\right)+2\left(x+2y\right)=4\\8\left(x+1\right)-2\left(x+2y\right)=18\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}11\left(x+1\right)=22\\3\left(x+1\right)+2\left(x+2y\right)=4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\4y+8=4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-1\end{cases}}\)
b) ĐK : y khác 0
\(\hept{\begin{cases}x+\frac{1}{y}=-\frac{1}{2}\\2x-\frac{3}{y}=-\frac{7}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}3x+\frac{3}{y}=-\frac{3}{2}\\2x-\frac{3}{y}=-\frac{7}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}5x=-5\\3x+\frac{3}{y}=-\frac{3}{2}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=-1\\-3+\frac{3}{y}=-\frac{3}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\\frac{3}{y}=\frac{3}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=2\left(tm\right)\end{cases}}\)
c) ĐK : x khác -1 ; y khác 2
\(\hept{\begin{cases}\frac{x+2}{x+1}+\frac{2}{y-2}=6\\\frac{5}{x+1}-\frac{1}{y-2}=3\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{1}{x+1}+\frac{2}{y-2}=5\\\frac{5}{x+1}-\frac{1}{y-2}=3\end{cases}}\). Đặt \(\hept{\begin{cases}\frac{1}{x+1}=a\\\frac{1}{y-2}=b\end{cases}\left(a,b\ne0\right)}\)
\(\Leftrightarrow\hept{\begin{cases}a+2b=6\\5a-b=3\end{cases}}\Leftrightarrow\hept{\begin{cases}a+2b=5\\10a-2b=6\end{cases}}\Leftrightarrow\hept{\begin{cases}11a=11\\a+2b=5\end{cases}}\Leftrightarrow\hept{\begin{cases}a=1\\b=2\end{cases}\left(tm\right)}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{x+1}=1\\\frac{1}{y-2}=2\end{cases}}\Rightarrow\hept{\begin{cases}x+1=1\\y-2=\frac{1}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\y=\frac{5}{2}\end{cases}\left(tm\right)}\)
1)\(\hept{\begin{cases}\left(x+y\right)\left(1+\frac{1}{xy}\right)=4\\xy+\frac{1}{xy}+\frac{\left(x^2+y^2\right)}{xy}=4\end{cases}}\)
2)\(\hept{\begin{cases}4xy+4\left(x^2+y^2\right)+\frac{3}{\left(x+y\right)^2}=7\\2x+\frac{1}{x+y}=3\end{cases}}\)
sử dụng bất đẳng thức đối với pt2 he 1
pt 2<=>\(xy+\frac{1}{xy}+\frac{x}{y}+\frac{y}{x}=4\)
áp dụng bdt cô si ta dễ dàng chứng minh được VT>=4. dau = xay ra <=>x=y=1
nhưng x,y có không âm đâu mà được phép áp dụng cosi
khong su dung co si thi su dung bunhiacopxi