\(\left(2x^2+x-2015\right)^2+4.\left(x^2-5x-2016\right)^2=4.\left(2x^2+x-2015\right).\left(x^2-5x-2016\right)\) Giải phương trình trên..
Giải phương trình:
1,\(\left(x^2-x+1\right)^4+5x^4=6\left(x^2-x+1\right)^4\)
2,\(\frac{x+4}{x-1}+\frac{x-4}{x+1}=\frac{x-8}{x+2}+\frac{x+8}{x-2}+\frac{8}{3}\)
3,\(\left|x-2015\right|^{2015}+\left|x-2016\right|^{2016}=1\)
4,\(\frac{5}{2x-3}-\frac{1}{x+2}=\frac{5}{x-6}-\frac{7}{2x-1}\)
5,\(\left(x+2008\right)^4+\left(x+2009\right)^4=\frac{1}{8}\)
tớ ko bt lm abc , tớ lm d thôi nha , thứ lỗi
\(\frac{5}{2x-3}-\frac{1}{x+2}=\frac{5}{x-6}-\frac{7}{2x-1}\)
\(\frac{3x+13}{2x^2+x-6}=\frac{5}{x-6}+\frac{7}{1-2x}\)
\(\frac{3x+13}{\left(x+2\right)\left(2x-3\right)}=\frac{3x+37}{\left(x-6\right)\left(2x-1\right)}\)
\(\frac{10-9x}{-4x^3+32x^2-51x+18}=0\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{10}{9}\end{cases}}\)
Giải phương trình: \(\left(2x^2+x-2014\right)^2+4\left(x^2-5x-2013\right)^2=4\left(2x^2+x-2014\right)\left(x^2-5x-2013\right)\)
Đặt \(a=2x^2+x-2014\) , \(b=x^2-5x-2013\)
thì \(a^2+4b^2=4ab\Leftrightarrow a^2-4ab+4b^2=0\Leftrightarrow\left(a-2b\right)^2=0\)
Thay vào được \(\left[\left(2x^2+x-2014\right)-2\left(x^2-5x-2013\right)\right]^2=0\)
\(\Leftrightarrow11x+2012=0\Leftrightarrow x=-\frac{2012}{11}\)
\(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)^2\)\(=4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)\)
Giải phương trình
Đặt 2x^2 + x +2013 = a, x^2-5x+2012 = b
Ta có: a^2 + 4b^2 = 4ab
a^2 - 4ab + 4b^2 = 0
(a-2b)^2 = 0
Do đó: a = 2b
Hay: 2x^2 + x -2013 = 2(x^2 -5x -2012)
2x^2 + x -2013 = 2x^2 -10x -4024
x-2013 = -10x -4024
x+10x = -4024+2013
11x = -2011
x = -2011/11
Bạn hỏi nhiều câu hay đấy. Chúc bạn học tốt.
Bài Tập: Giải phương trình :
a) (x + 5)(2x - 3) = 0
b) \(\left(x^2-9\right)\left(4-x\right)=0\)
c) \(\left(2x+3\right)\left(4-5x\right)=0\)
d) \(2\left(x+3\right)\left(x-4\right)=0\)
e) \(\left(x^2-9\right)\left(4-x\right)=0\)
f) \(\left(2x+3\right)\left(x^2-16\right)=0\)
a: \(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{3}{2}\end{matrix}\right.\)
b: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=4\end{matrix}\right.\)
c: \(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\5x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{4}{5}\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x-4\right)=0\)
=>x+3=0 hoặc x-4=0
=>x=-3 hoặc x=4
e: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=4\end{matrix}\right.\)
f: \(\Leftrightarrow\left(2x+3\right)\left(x-4\right)\left(x+4\right)=0\)
hay \(x\in\left\{-\dfrac{3}{2};4;-4\right\}\)
a, \(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{3}{2}\end{matrix}\right.\)
b, \(\Leftrightarrow\left[{}\begin{matrix}x^2-9=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm3\\x=4\end{matrix}\right.\)
c, \(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\4-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{4}{5}\end{matrix}\right.\)
d, \(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)
e, tương tự d
f, \(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\x^2-16=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\pm4\end{matrix}\right.\)
\(\frac{2016\left(x+xy^2\right)\left(2x-y^2\right)\left(x^{8^{ }}-y^4\right)}{x^{2015^{ }}-y^{2015}}\)Với x=8 và y=4
a\(8\left(x+\dfrac{1}{x}\right)^{2^{ }}+4\left(x^{2^{ }}+\dfrac{1}{x^2}\right)-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)=\left(x+4\right)^2\)giải các phương trình\(\dfrac{x+4}{2x^2-5x+2}+\dfrac{x+1}{2x^2-7x+3}=\dfrac{2x+5}{2x^2-7x+3}\)
b)
ĐKXĐ: \(x\notin\left\{2;3;\dfrac{1}{2}\right\}\)
Ta có: \(\dfrac{x+4}{2x^2-5x+2}+\dfrac{x+1}{2x^2-7x+3}=\dfrac{2x+5}{2x^2-7x+3}\)
\(\Leftrightarrow\dfrac{x+4}{\left(x-2\right)\left(2x-1\right)}+\dfrac{x+1}{\left(x-3\right)\left(2x-1\right)}=\dfrac{2x+5}{\left(2x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{\left(x+4\right)\left(x-3\right)}{\left(x-2\right)\left(2x-1\right)\left(x-3\right)}+\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)\left(2x-1\right)}=\dfrac{\left(2x+5\right)\left(x-2\right)}{\left(2x-1\right)\left(x-3\right)\left(x-2\right)}\)
Suy ra: \(x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)
\(\Leftrightarrow2x^2-14=2x^2+x-10\)
\(\Leftrightarrow2x^2-14-2x^2-x+10=0\)
\(\Leftrightarrow-x-4=0\)
\(\Leftrightarrow-x=4\)
hay x=-4(nhận)
Vậy: S={-4}
Giải phương trình sau:
\(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)^2\)
\(=4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)\)
\(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)^2\)
\(=4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)\)
Đặt \(\hept{\begin{cases}2x^2+x-2013=a\\x^2-5x-2012=b\end{cases}}\) thì ta có :
\(a^2+4b^2=4ab\Rightarrow a^2+b^2-4ab=0\)
\(\Rightarrow\left(a-2b\right)^2=0\Rightarrow a-2b\Rightarrow a=2b\)
Tức là :
\(2x^2+x-2013=2\left(x^2-5x-2012\right)\)
\(\Leftrightarrow2x^2+x-2013=2x^2-10x-4024\)
\(\Leftrightarrow11x+2011=0\Leftrightarrow11x=-2011\Rightarrow x=-\frac{2011}{11}\)
Chúc bạn học tốt !!!
Giải các bất phương trình sau :
\(a.4\left(x-3\right)^2-\left(2x-1\right)^2\ge12\)
\(b.\left(x-4\right)\left(x+4\right)\ge\left(x+3\right)^2+5\)
c. \(\left(3x-1\right)^2-9\left(x+2\right)\left(x-2\right)< 5x\)
\(a,4\left(x-3\right)^2-\left(2x-1\right)^2\ge12\)
\(\Leftrightarrow4x^2-24x+36-4x^2-4x+1\ge12\)
\(\Leftrightarrow-28x+37\ge12\)
\(\Leftrightarrow-28x\ge12-37\)
\(\Leftrightarrow-28x\ge-25\)
\(\Leftrightarrow x\le\dfrac{25}{28}\)
Vậy \(S=\left\{x\left|x\le\dfrac{25}{28}\right|\right\}\)
b, \(\left(x-4\right)\left(x+4\right)\ge\left(x+3\right)^2+5\)
\(\Leftrightarrow x^2-16\ge x^2+6x+9+5\)
\(\Leftrightarrow x^2-x^2-6x\ge9+5+16\)
\(\Leftrightarrow-6x\ge30\)
\(\Leftrightarrow x\le-5\)
Vậy \(S=\left\{x\left|x\le-5\right|\right\}\)
\(c,\left(3x-1\right)^2-9\left(x+2\right)\left(x-2\right)< 5x\)
\(\Leftrightarrow9x^2-6x-1-9x^2+36< 5x\)
\(\Leftrightarrow9x^2-9x^2-6x-5x+36+1< 0\)
\(\Leftrightarrow-11x+37< 0\)
\(\Leftrightarrow-11x< -37\)
\(\Leftrightarrow x>\dfrac{37}{11}\)
vậy \(S=\left\{x\left|x>\dfrac{37}{11}\right|\right\}\)
Giải phương trình sau: \(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)^2=4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)\)
Sửa tí nha kết quả cuối sai dâu phải là \(x=\dfrac{-2011}{11}\)
\(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)^2=4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)\\ \Leftrightarrow\left(2x^2+x-2013\right)^2-4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)+4\left(x^2-5x-2012\right)^2=0\\ \Leftrightarrow\left[2x^2+x-2013-2\left(x^2-5x-2012\right)\right]^2=0\\ \Leftrightarrow\left(11x+2011\right)^2=0\\ \Leftrightarrow11x+2011=0\\ \Leftrightarrow x=\dfrac{2011}{11}\)