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Do \(\left|x\right|\ge2;\left|y\right|\ge2\Rightarrow xy\ne0\)
Ta luôn có \(\left\{{}\begin{matrix}\frac{1}{x}\le\frac{1}{\left|x\right|}\le\frac{1}{2}\\\frac{1}{y}\le\frac{1}{\left|y\right|}\le\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\frac{1}{x}+\frac{1}{y}\le\frac{1}{2}+\frac{1}{2}=1\)
\(\frac{xy}{x+y}=\frac{2003}{2004}\Leftrightarrow\frac{x+y}{xy}=\frac{2004}{2003}\Leftrightarrow\frac{1}{x}+\frac{1}{y}=\frac{2004}{2003}\)
Ta có \(\frac{2004}{2003}>1\) mà \(\frac{1}{x}+\frac{1}{y}\le1\Rightarrow VT< VP\Rightarrow\) phương trình vô nghiệm
\(b,\)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Rightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)
\(\Rightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)
\(\Rightarrow\left(x+9\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}\right)=\left(x+9\right)\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)\)
\(\Rightarrow\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}=\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\left(KTM\right)\)
\(\text{Giải}\)
\(b,\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2009=0\Leftrightarrow x=-2009\)
Ta có: \(\frac{x^2}{yz}+\frac{y^2}{xz}+\frac{z^2}{xy}\Rightarrow x^3+y^3+z^3-3xyz=0\Rightarrow\orbr{\begin{cases}x=y=z\\x+y+z=0\end{cases}}\)
Vì nghiệm của phương trình là bộ ba số khác O nên các số a,b,c là ba số khác nhau và khác O
+) Nếu: \(\frac{a}{b-c}=\frac{b}{c-a}=\frac{c}{a-b}=k\ne0\Rightarrow a=k\left(b-c\right);b=k\left(c-a\right);c=k\left(a-b\right)\)
\(\Rightarrow a+b+c=0\Rightarrow a+b=-c\)
Từ: \(\frac{a}{b-c}=\frac{b}{c-a}\Rightarrow\frac{a}{b+a+b}=\frac{b}{-a-b-a}\Rightarrow\left(a+b\right)^2+a^2+b^2=0\)
\(\Rightarrow a=b=0\Rightarrow a=b=c=0\)(loại)
+) Nếu: \(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\Rightarrow\frac{a}{b-c}=\frac{b}{a-c}+\frac{c}{b-a}=\frac{b\left(b-a\right)+c\left(a-c\right)}{\left(c-a\right)\left(a-b\right)}\)
\(\Rightarrow\frac{a}{\left(b-c\right)^2}=\frac{b^2-ba+ca-c^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left(1\right)\)
Tương tự ta có: \(\frac{b}{\left(c-a\right)^2}=\frac{c^2-cb+ab-a^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left(2\right);\frac{c}{\left(a-b\right)^2}=\frac{a^2-ac+bc-b^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left(3\right)\)
Từ (1),(2) và (3) \(\Rightarrow\frac{a}{\left(b-c^2\right)}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)
Đặt \(m=\frac{a}{\left(b-c\right)^2};n=\frac{b}{\left(c-a\right)^2};p=\frac{c}{\left(a-b\right)^2}\Rightarrow m+n+p=0\)
\(\Rightarrow m^3+n^3+p^3=3mnp\Rightarrow\frac{m^2}{np}+\frac{n^2}{mp}+\frac{p^2}{mn}=3\left(ĐPCM\right)\)
câu 2 :
\(\Leftrightarrow\)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}-\frac{x+4}{2005}-\frac{x+5}{2004}-\frac{x+6}{2003}\)=0
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x-2009}{2003}\)=0
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\)
\(\Rightarrow x+2009=0\)
\(\Rightarrow x=-2009\)
Ta có A = 2018.2020 + 2019.2021
= (2020 - 2).2020 + 2019.(2019 + 2)
= 20202 - 2.2020 + 20192 + 2.2019
= 20202 + 20192 - 2(2020 - 2019) = 20202 + 20192 - 2 = B
=> A = B
b) Ta có B = 964 - 1= (932)2 - 12
= (932 + 1)(932 - 1) = (932 + 1)(916 + 1)(916 - 1) = (932 + 1)(916 + 1)(98 + 1)(98 - 1)
= (932 + 1)(916 + 1)(98 + 1)(94 + 1)(94 - 1)
= (932 + 1)(916 + 1)(98 + 1)(94 + 1)(92 + 1)(92 - 1)
(932 + 1)(916 + 1)(98 + 1)(94 + 1)(92 + 1).80
mà A = (932 + 1)(916 + 1)(98 + 1)(94 + 1)(92 + 1).10
=> A < B
c) Ta có A = \(\frac{x-y}{x+y}=\frac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2}=\frac{x^2-y^2}{x^2+2xy+y^2}< \frac{x^2-y^2}{x^2+xy+y^2}=B\)
=> A < B
d) \(A=\frac{\left(x+y\right)^3}{x^2-y^2}=\frac{\left(x+y\right)^3}{\left(x+y\right)\left(x-y\right)}=\frac{\left(x+y\right)^2}{x-y}=\frac{x^2+2xy+y^2}{x-y}< \frac{x^2-xy+y^2}{x-y}=B\)
=> A < B
\(a,\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)\(\Leftrightarrow\frac{x^2+3x+2+x^2-3x+2}{x^2-4}=\frac{2\left(x^2+2\right)}{x^2-4}\)
\(\Leftrightarrow2\left(x^2+2\right)=2\left(x^2+2\right)\)(luôn đúng)
Vậy pt có vô số nghiệm
\(b,\Leftrightarrow\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\)
\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left(2x+3-x+5\right)=0\)\(\Leftrightarrow\left(\frac{-4x+10}{2-7x}\right)\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-4x+10=0\\x+8=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-8\end{cases}}\)
Mấy câu rút gọn bạn quy đồng nha
a/ Đặt \(\hept{\begin{cases}\frac{x+1}{x-2}=a\\\frac{x+1}{x-4}=b\end{cases}}\) thì có
\(a^2+b-\frac{12b^2}{a^2}=0\)
\(\Leftrightarrow\left(a^2-3b\right)\left(a^2+4b\right)=0\)
b/ \(2x^2+3xy-2y^2=7\)
\(\Leftrightarrow\left(2x-y\right)\left(x+2y\right)=7\)
a)VP lẻ => VT lẻ =>x2-y2=2k+1 (k\(\in\)Z) (số lẻ)
\(\Rightarrow10y+9=\left(2k+1\right)^2\Rightarrow y=\frac{2\left(k+2\right)\left(k-1\right)}{5}\in Z^+\)
\(\Rightarrow\orbr{\begin{cases}\left(k+2\right)⋮5\Rightarrow k=5t-2\Rightarrow y=2t\left(5t-3\right)\left(1\right)\\\left(k-1\right)⋮5\Rightarrow k=5t+1\Rightarrow y=2t\left(5t+3\right)\left(2\right)\end{cases}}\left(t\in Z^+\right)\)
Mà \(\hept{\begin{cases}\left(10t^2-6t\right)^2< \left(10t^2-6t\right)^2+10t-3< \left(10t^2-6t+1\right)^2\left(\text{khi}\text{ t }\ge1\right)\\\left(10t^2-6t-1\right)^2< \left(10t^2-6t\right)^2+10t-3< \left(10t^2-6t\right)^2\left(\text{khi t}\le-1\right)\\\left(10t^2-6t\right)^2+10t-3=-3< 0\left(\text{khi t}=0\right)\end{cases}}\)
Suy ra pt vô nghiệm
Mà \(\left(10t^2+6t\right)^2< \left(10t^2+6t\right)^2+10t+3< \left(10t^2+6t+1\right)^2\left(\text{khi t}\ge1\right)\) (*)
\(\left(10t^2+6t-1\right)^2< \left(10t^2+6t\right)^2+10t+3< \left(10t^2+6t\right)^2\left(\text{khi t}< -1\right)\)(*)
\(\left(10t^2+6t\right)^2+10t+3=3^2\left(\text{khi t}=-1\right)\)(*)
\(1^2< \left(10t^2+6t\right)^2+10t+3=3< 2^2\left(\text{khi t}=0\right)\)(*)
Suy ra \(t=-1;y=4;x=\pm3\) (thỏa mãn)
Vậy....
P/s:Ngoặc nhọn 4 dòng có dấu (*) vào
Xin lỗi bạn mình chưa học lớp 8
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