Nghiệm nhỏ nhất của phương trình \(\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}=\dfrac{3}{4}\)
Nghiệm nhỏ nhất của phương trình \(\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}=\dfrac{3}{4}\)
\(\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}=\frac{3}{4}\)
\(\frac{1}{x-1}-\frac{1}{x+2}=\frac{3}{4}\)
tự tính nhé bạn
Nghiệm lớn nhất của phương trình: \(\dfrac{1}{\left(x-1\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-7\right)}=\dfrac{3}{16}\)
giải phương trình
\(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)
ĐKXĐ: \(x\notin\left\{-1;-2;-3;-4\right\}\)
Ta có: \(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+4}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{x+4}{\left(x+1\right)\left(x+4\right)}-\dfrac{x+1}{\left(x+1\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{x+4-x-1}{\left(x+1\right)\left(x+4\right)}=\dfrac{x^2+5x+4}{6\left(x+1\right)\left(x+4\right)}\)
\(\Leftrightarrow\dfrac{18}{6\left(x+1\right)\left(x+4\right)}=\dfrac{x^2+5x+4}{6\left(x+1\right)\left(x+4\right)}\)
Suy ra: \(x^2+5x+4=18\)
\(\Leftrightarrow x^2+5x-14=0\)
\(\Leftrightarrow x^2+7x-2x-14=0\)
\(\Leftrightarrow x\left(x+7\right)-2\left(x+7\right)=0\)
\(\Leftrightarrow\left(x+7\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)
Vậy: S={-7;2}
ĐKXĐ: $x \neq -1;-2;-3;-4$
$pt⇔\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}=\dfrac{1}{6}$
$⇔\dfrac{1}{x+1}-\dfrac{1}{x+4}=\dfrac{1}{6}$
$⇔\dfrac{3}{(x+1)(x+4)}=\dfrac{1}{6}$
$⇔x^2+5x+4=18$
$⇔x^2+5x-14=0$
$⇔(x-2)(x+7)=0$
$⇔$ \(\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\)(t/m)
Vậy...
Giải các bất phương trình sau rồi biểu diễn tập nghiệm của chúng trên trục số:
1) \(\left(x+3\right)^2-3\left(2x-1\right)>x\left(x-4\right)\)
2) \(1+\dfrac{x+1}{3}>\dfrac{2x-1}{6}-2\)
3) \(x-\dfrac{2x-7}{4}< \dfrac{2x}{3}-\dfrac{2x+3}{2}-1\)
4) \(\dfrac{2x+1}{x-3}\le2\)
5) \(\dfrac{12-3x}{2x+6}>3\)
6) \(x^2+3x-4\le0\)
7) \(\dfrac{5}{5x-1}< \dfrac{-3}{5-3x}\)
8) \(\left(2x-1\right)\left(3-2x\right)\left(1-x\right)>0\)
1: \(\Leftrightarrow x^2+6x+9-6x+3>x^2-4x\)
=>-4x<12
hay x>-3
2: \(\Leftrightarrow6+2x+2>2x-1-12\)
=>8>-13(đúng)
4: \(\dfrac{2x+1}{x-3}\le2\)
\(\Leftrightarrow\dfrac{2x+1-2x+6}{x-3}< =0\)
=>x-3<0
hay x<3
6: =>(x+4)(x-1)<=0
=>-4<=x<=1
Nghiệm lớn nhất của phương trình: \(\dfrac{1}{\left(x-1\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-7\right)}=\dfrac{3}{16}\)
Ta có: \(\dfrac{1}{\left(x-1\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-7\right)}=\dfrac{3}{16}\) \(\Rightarrow\dfrac{1}{2}\left(\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-5}+\dfrac{1}{x+5}-\dfrac{1}{x-7}\right)=\dfrac{3}{16}\)
\(\Rightarrow\) \(\dfrac{1}{2}\left(\dfrac{1}{x-1}-\dfrac{1}{x-7}\right)=\dfrac{3}{16}\)
\(\Rightarrow\) \(\dfrac{6}{x^2-8x+7}=\dfrac{3}{8}\)
\(\Rightarrow\) \(x^2-8x+7=16\)
\(\Rightarrow\) \(x^2-8x-9=0\)
\(\Rightarrow\) \(\left(x-9\right)\left(x+1\right)=0\)
\(\Rightarrow\) \(\left\{{}\begin{matrix}x=9\\x=-1\end{matrix}\right.\)
Vậy: Nghiệm lớn nhất của phương trình là: \(x=9\).
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}
tìm n của phương trình \(x^2-\dfrac{2n-2x}{4}-2x+5n=x^3-9x^2+10\)
có nghiệm bằng \(\dfrac{1}{3}\)của phương trình \(\left(x+1\right)\left(x+3\right)=x\left(x-3\right)+24\)
Giải phương trình:
\(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)=\left(x+4\right)^2\)
đkxđ: x khác 0
\(\Leftrightarrow8.\left(x+\dfrac{1}{x}\right)\left(x+\dfrac{1}{x}\right)-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)+4\left(x^2+\dfrac{1}{x^2}\right)^2=x^2+8x+16\)
\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left[\left(8.x+\dfrac{1}{x}\right)-4\left(x^2+\dfrac{1}{x^2}\right)\right]+4\left(x^4+2+\dfrac{1}{x^2}\right)-x^2-8x-16=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left[\left(\dfrac{8x^2+1}{x}-4x^2-\dfrac{4}{x^2}\right)\right]+4x^4+8+\dfrac{4}{x^2}-x^2-8x-16=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left(\dfrac{x\left(8x^2+1\right)}{x^2}-\dfrac{4x^2.x^2}{x^2}-\dfrac{4}{x^2}\right)+......=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left(\dfrac{8x^3+x-4x^4-4}{x^2}\right)+...=0\)
\(\Leftrightarrow\dfrac{x^2}{x}.-\dfrac{4x^4+8x^3+x-4}{x^2}+.....=0\)
\(\Leftrightarrow-\dfrac{4x^6+8x^5+x^3-4x^2}{x^3}+\dfrac{4x^4+8+4x^2}{1}-\dfrac{x^2-8x-16}{1}=0\)
\(\Leftrightarrow......+\dfrac{x^3.\left(4x^4+8+4x^2\right)}{x^3}-\dfrac{x^3\left(x^2-8x-16\right)}{x^3}=0\)
\(\Leftrightarrow-4x^6+8x^5+x^3-4x^2+4x^7+8x^3+4x^5-x^5+8x^4+16x^3=0\)
\(\Leftrightarrow4x^7-4x^6+12x^5+8x^4+25x^3-4x^2=0\)
=> x=0 ( loại , ko tm)
Vậy pt vô nghiệm
Tìm giá trị nhỏ nhất :
\(D=\left|x-\dfrac{1}{2}\right|+\left|x-\dfrac{1}{3}\right|+\left|x-\dfrac{1}{4}\right|+...+\left|x-\dfrac{1}{2022}\right|\)