Giải phương trình:
a) \(\left|x-2018\right|^{2019}+\left|x-2019\right|^{2018}=1\)
b)\(\frac{2x}{x^2-x+1}-\frac{x}{x^2+x+1}=\frac{5}{3}\)
Giải phương trình:\(\frac{\left(2018-x\right)^2+\left(2018-x\right)\left(x-2019\right)+\left(x-2019\right)^2}{\left(2018-x\right)^2-\left(2018-x\right)\left(x-2019\right)+\left(x-2019\right)^2}=\frac{19}{49}\)
Đặt \(\left\{{}\begin{matrix}2018-x=a\\x-2019=b\end{matrix}\right.\) \(\Rightarrow a+b=-1\Rightarrow b=-1-a\)
\(\frac{a^2+ab+b^2}{a^2-ab+b^2}=\frac{19}{49}\Leftrightarrow49\left(a^2+ab+b^2\right)=19\left(a^2-ab+b^2\right)\)
\(\Leftrightarrow15a^2+34ab+15b^2=0\)
\(\Leftrightarrow\left(5a+3b\right)\left(3a+5b\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5a=-3b\\3a=-5b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}5a=-3\left(-1-a\right)\\3a=-5\left(-1-a\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2a=3\\2a=-5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}a=\frac{3}{2}\\a=-\frac{5}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2018-x=\frac{3}{2}\\2018-x=-\frac{5}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{4033}{2}\\x=\frac{4041}{2}\end{matrix}\right.\)
Tìm MinS= \(\frac{1}{\left(x-2018\right)^2}+\frac{1}{\left(x-2019\right)^2}+\frac{1}{\left(x-2018\right)\left(x-2019\right)}\)
Cho số thực x thỏa mãn \(\frac{9x}{2x^2+x+3}\)-\(\frac{x}{2x^2-x+3}\)=\(\frac{5}{4}\)
Hãy tính giá trị biểu thức A=\(\frac{\left(2x-3\right)\left(x-1\right)^{2018}}{2019}\)-\(\frac{2x^2-5x+3}{\left(x-2018\right)^2}\)+ 2019
Giải phương trình: \(\frac{\left(2017-x\right)^2+\left(2017-x\right)\left(x-2018\right)+\left(x-2018^2\right)}{\left(2017-x\right)^2-\left(2107-x\right)\left(x-2018\right)+\left(x-2018\right)^2}=\frac{13}{37}\)
Đây là đề thi hoc sinh giỏi lớp 9 cấp tỉnh Phú yên năm 2018-2019
Dễ thấy \(x=2017\)không là nghiệm của phương trình.
Ta có:
\(\frac{1+\frac{x-2018}{2017-x}+\left(\frac{x-2018}{2017-x}\right)^2}{1-\frac{x-2018}{2017-x}+\left(\frac{x-2018}{2017-x}\right)}=\frac{13}{37}\)
Đặt \(\frac{x-2018}{2017-x}=a\)
\(\Rightarrow\frac{1+a+a^2}{1-a+a^2}=\frac{13}{37}\)
\(\Leftrightarrow24a^2+50a+24=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=-\frac{3}{4}\\a=-\frac{4}{3}\end{cases}}\)
Giải các phương trình sau:
a) \(\left(\frac{x-2}{x-1}\right)^2-5\left(\frac{x+2}{x+1}\right)^2+4\left(\frac{x^2-4}{x^2-1}\right)=1\)
b) \(\left(\frac{x-1}{x}\right)^2+\left(\frac{x-1}{x-2}\right)^2=\frac{40}{9}\)
c) \(x.\frac{4-x}{x+2}.\left(\frac{8-2x}{x+2}\right)=3\)
d) \(\frac{1}{3x-2020}+\frac{1}{4x-2018}+\frac{1}{5x-2017}=\frac{1}{12x-2019}\)
Giải hệ phương trình:
\(\left\{{}\begin{matrix}\frac{x+2019}{x+2018}+\frac{2}{y-2019}=6\\\frac{5}{x+2016}-\frac{1}{y-2019}=3\end{matrix}\right.\)
ĐKXĐ:....
Lấy pt trên cộng 2 lần pt dưới ta được:
\(\frac{x+2019}{x+2018}+\frac{10}{x+2016}=12\)
Số to quá, đặt \(x+2016=a\Rightarrow\frac{a+3}{a+2}+\frac{10}{a}=12\)
\(\Leftrightarrow12a\left(a+2\right)=a\left(a+3\right)+10\left(a+2\right)\)
\(\Leftrightarrow12a^2+24a-a^2-3a-10a-20=0\)
\(\Leftrightarrow11a^2+11a-20=0\)
Nghiệm rất xấu, bạn tự giải tiếp
Tìm x,y,z biết:
a) \(2019-|x-2019|=x\)
b) \(\left(2x-1\right)^{2018}+\left(y-\frac{2}{5}\right)^{2018}+|x+y-z|=0\)
a)\(2019-\left|x-2019\right|=x\)
\(\Rightarrow2019-x=\left|x-2019\right|\)
=>\(\left|x-2019\right|=-\left(x-2019\right)\)
=>\(x-2019\le0\)
=>\(x\le2019\)
b) Vì \(\left(2x-1\right)^{2018}\ge0\forall x\)
\(\left(y-\frac{2}{5}\right)^{2018}\ge0\forall y\)
\(\left|x+y-z\right|\ge0\forall x,y,z\)
=> \(\left(2x-1\right)^{2018}+\left(y-\frac{2}{5}\right)^{2018}\)\(+\left|x+y-z\right|\ge0\forall x,y,z\)
mà \(\left(2x-1\right)^{2018}+\left(y-\frac{2}{5}\right)^{2018}\)\(+\left|x+y-z\right|=0\)
\(\Leftrightarrow\hept{\begin{cases}2x-1=0\\y-\frac{2}{5}=0\\x+y-z=0\end{cases}}\)=>\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{9}{10}\end{cases}}\)
a, Ta có:
\(\left|x-2019\right|=\orbr{\begin{cases}x-2019\ge0\Rightarrow x\ge2019\\-x+2019< 0\Rightarrow x< 2019\end{cases}}\)
Xét x<2019 thì |x-2019|=-x+2019
Khi đó: 2019-(-x+2019)=x
\(\Leftrightarrow\)-x+2019=2019-x
\(\Leftrightarrow\)-x+2019+x=2019
\(\Leftrightarrow\)0x+2019=2019
\(\Leftrightarrow\)0x=0 (thỏa mãn)
Xét 2019\(\le\)x thì |x-2019|=x-2019
Khi đó 2019-(x-2019)=x
\(\Leftrightarrow\)2019-x+2019=x
\(\Leftrightarrow\)4038-x=x
\(\Leftrightarrow\)4038=2x
\(\Leftrightarrow\)x=2019(thỏa mãn)
Vậy .......................................................!!!
ta có
\(\frac{\left(2018-x\right)^2+\left(2018-x\right)\left(x-2019\right)+\left(x-2019\right)^2}{\left(2018-x\right)^2-\left(2018-x\right)\left(x-2019\right)-\left(x-2019\right)^2}=\frac{19}{49}\) ( điều kiện : x khác : 2018;2019 )
đặt a = x - 2019 ( a khác 0 )
ta có hệ thức :
\(\frac{\left(a+1\right)^2-\left(a+1\right)a+a^2}{\left(a+1\right)^2+\left(a+1\right)a+a^2}=\frac{19}{49}\\ \Leftrightarrow\frac{a^2+a+1}{3a^2+3a+1}=\frac{19}{49}\)
\(\Leftrightarrow49\left(a^2+a+1\right)=19\left(3a^2+3a+1\right)\)
\(\Leftrightarrow49a^2+49a+49=57a^2+57a+19\)
\(\Leftrightarrow8a^2+8a-30=0\\ \left(2a+1\right)^2-4^2=0\\ \Leftrightarrow\left(2a+3\right)\left(2a+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=\frac{3}{2}\\a=-\frac{5}{2}\end{matrix}\right.\)( thỏa mãn điều kiện )
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{4041}{2}\\x=\frac{4033}{2}\end{matrix}\right.\)( thỏa mãn điều kiện )
vậy \(x\in\left\{\frac{4041}{2};\frac{4033}{2}\right\}\)
1/Cho a,b,c thỏa mãn \(\frac{2}{\left(x^2+1\right)\left(x-1\right)}=\frac{ax+b}{x^2+1}+\frac{c}{x-1}\)
Tính giá trị biểu thức M=\(\frac{a^{2017}+b^{2018}+c^{2019}}{a^{2017}b^{2018}c^{2019}}\)
2/Cho x,y,z≠0 và x+y+z=2008
Tính giá trị biểu thức P=\(\frac{x^3}{\left(x-y\right)\left(x-z\right)}+\frac{y^3}{\left(y-x\right)\left(y-z\right)}+\frac{z^3}{\left(z-y\right)\left(z-x\right)}\)