giải phương trình \(x^2-x-\dfrac{1}{x}+\dfrac{1}{x^2}-10=0\)
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H
26 tháng 3 2021
Bài 4 :
24 phút = \(\dfrac{24}{60} = \dfrac{2}{5}\) giờ
Gọi thời gian dự định đi từ A đến B là x(giờ) ; x > 0
Suy ra quãng đường AB là 36x(km)
Khi vận tốc sau khi giảm là 36 -6 = 30(km/h)
Vì giảm vận tốc nên thời gian đi hết AB là x + \(\dfrac{2}{5}\)(giờ)
Ta có phương trình:
\(36x = 30(x + \dfrac{2}{5})\\ \Leftrightarrow x = 2\)
Vậy quãng đường AB dài 36.2 = 72(km)
HN
1
XO
28 tháng 8 2023
ĐKXĐ : \(x\notin\left\{0;-1;-2;-3;-4\right\}\)
Ta có \(\dfrac{1}{x}+\dfrac{1}{x+1}+\dfrac{1}{x+2}+\dfrac{1}{x+3}+\dfrac{1}{x+4}=0\)
\(\Leftrightarrow\dfrac{2x+4}{x.\left(x+4\right)}+\dfrac{2x+4}{\left(x+1\right).\left(x+3\right)}+\dfrac{1}{x+2}=0\)
\(\Leftrightarrow\dfrac{2x+4}{\left(x+2\right)^2-4}+\dfrac{2x+4}{\left(x+2\right)^2-1}+\dfrac{1}{x+2}=0\) (*)
Đặt x + 2 = a \(\left(a\ne0\right)\)
(*) \(\Leftrightarrow\dfrac{2a}{a^2-4}+\dfrac{2a}{a^2-1}+\dfrac{1}{a}=0\)
\(\Leftrightarrow\dfrac{2}{a-\dfrac{4}{a}}+\dfrac{2}{a-\dfrac{1}{a}}+\dfrac{1}{a}=0\) (**)
Đặt \(\dfrac{1}{a}=b\left(b\ne0\right)\) \(\Rightarrow ab=1\)
Ta được (**) \(\Leftrightarrow\dfrac{2}{a-4b}+\dfrac{2}{a-b}+b=0\)
\(\Leftrightarrow\dfrac{2b}{1-4b^2}+\dfrac{2b}{1-b^2}+b=0\)
\(\Leftrightarrow\dfrac{2}{1-4b^2}+\dfrac{2}{1-b^2}=-1\)
\(\Rightarrow4-10b^2=-4b^4+5b^2-1\)
\(\Leftrightarrow4b^4-15b^2+5=0\) (***)
Đặt b2 = t > 0
Ta có (***) <=> \(4t^2-15t+5=0\Leftrightarrow t=\dfrac{15\pm\sqrt{145}}{8}\) (tm)
\(\Leftrightarrow b=\pm\sqrt{\dfrac{15\pm\sqrt{145}}{8}}\)
mà x + 2 = a ; ab = 1
nên \(x=\pm\sqrt{\dfrac{8}{15\pm\sqrt{145}}}-2\)
Thử lại ta có phương trình có 4 nghiệm như trên
28 tháng 4 2023
\(\dfrac{1}{x^2+2x}+\dfrac{1}{x^2+6x+8}+\dfrac{1}{x^2+10x+24}+\dfrac{1}{x^2+14x+48}=\dfrac{4}{105}\)
\(\Leftrightarrow\dfrac{2}{x\left(x+2\right)}+\dfrac{2}{\left(x+2\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}+\dfrac{2}{\left(x+6\right)\left(x+8\right)}=\dfrac{8}{105}\)
\(\Leftrightarrow\left(\dfrac{1}{x}-\dfrac{1}{x+2}\right)+\left(\dfrac{1}{x+2}-\dfrac{1}{x+4}\right)+\left(\dfrac{1}{x+4}-\dfrac{1}{x+6}\right)+\left(\dfrac{1}{x+6}-\dfrac{1}{x+8}\right)=\dfrac{8}{105}\)
\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+8}=\dfrac{8}{105}\)
\(\Leftrightarrow\dfrac{8}{x\left(x+8\right)}=\dfrac{8}{105}\)
\(\Leftrightarrow x\left(x+8\right)=105\)
\(\Leftrightarrow x^2+8x-105=0\)
\(\Leftrightarrow x^2-7x+15x-105=0\)
\(\Leftrightarrow x\left(x-7\right)+15\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-15\end{matrix}\right.\)
Thử lại ta có nghiệm của phương trình trên là \(x=7\text{v}à\text{x}=15\)
XL
2
26 tháng 2 2022
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)-\left(x-2\right)\left(x+1\right)+14=0\)
\(\Leftrightarrow x^2-4x+3-\left(x^2-x-2\right)+14=0\)
\(\Leftrightarrow x^2-4x+17-x^2+x+2=0\)
=>-3x+19=0
hay x=19/3(nhận)
26 tháng 2 2022
ĐKXĐ:\(\left\{{}\begin{matrix}x\ne-1\\x\ne3\end{matrix}\right.\)
\(\dfrac{x-1}{x+1}-\dfrac{x-2}{x-3}+\dfrac{14}{x^2-2x-3}=0\\ \Leftrightarrow\dfrac{\left(x-3\right)\left(x-1\right)}{\left(x-3\right)\left(x+1\right)}-\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x+1\right)\left(x-3\right)}+\dfrac{14}{\left(x+1\right)\left(x-3\right)}=0\\ \Leftrightarrow\dfrac{\left(x-3\right)\left(x-1\right)-\left(x+1\right)\left(x-2\right)+14}{\left(x+1\right)\left(x-3\right)}=0\)
\(\Rightarrow\left(x^2-4x+3\right)-\left(x^2-x-2\right)+14=0\\ \Leftrightarrow x^2-4x+3-x^2+x+2+14=0\)
\(\Leftrightarrow-3x+19=0\\ \Leftrightarrow x=\dfrac{19}{3}\left(tm\right)\)
Vậy pt có tập nghiệm \(S=\left\{\dfrac{19}{3}\right\}\)
12 tháng 11 2023
a:
ĐKXĐ: \(x\notin\left\{\dfrac{3}{2};1\right\}\)
\(y=\dfrac{\left(x-2\right)^2}{\left(2x-3\right)\left(x-1\right)}=\dfrac{x^2-4x+4}{2x^2-2x-3x+3}\)
=>\(y=\dfrac{x^2-4x+4}{2x^2-5x+3}\)
=>\(y'=\dfrac{\left(x^2-4x+4\right)'\left(2x^2-5x+3\right)-\left(x^2-4x+4\right)\left(2x^2-5x+3\right)'}{\left(2x^2-5x+3\right)^2}\)
=>\(y'=\dfrac{\left(2x-4\right)\left(2x^2-5x+3\right)-\left(2x-5\right)\left(x^2-4x+4\right)}{\left(2x^2-5x+3\right)^2}\)
=>\(y'=\dfrac{4x^3-10x^2+6x-8x^2+20x-12-2x^3+8x^2-8x+5x^2-20x+20}{\left(2x^2-5x+3\right)^2}\)
=>\(y'=\dfrac{2x^3-5x^2-2x+8}{\left(2x^2-5x+3\right)^2}\)
b:
ĐKXĐ: x<>-3
\(y=\left(x+3\right)+\dfrac{4}{x+3}\)
=>\(y'=\left(x+3+\dfrac{4}{x+3}\right)'=1+\left(\dfrac{4}{x+3}\right)'\)
\(=1+\dfrac{4'\left(x+3\right)-4\left(x+3\right)'}{\left(x+3\right)^2}\)
=>\(y'=1+\dfrac{-4}{\left(x+3\right)^2}=\dfrac{\left(x+3\right)^2-4}{\left(x+3\right)^2}\)
y'=0
=>\(\left(x+3\right)^2-4=0\)
=>\(\left(x+3+2\right)\left(x+3-2\right)=0\)
=>(x+5)(x+1)=0
=>x=-5 hoặc x=-1
c:
ĐKXĐ: x<>-2
\(y=\dfrac{\left(5x-1\right)\left(x+1\right)}{x+2}\)
=>\(y=\dfrac{5x^2+5x-x-1}{x+2}=\dfrac{5x^2+4x-1}{x+2}\)
=>\(y'=\dfrac{\left(5x^2+4x-1\right)'\left(x+2\right)-\left(5x^2+4x-1\right)\left(x+2\right)'}{\left(x+2\right)^2}\)
=>\(y'=\dfrac{\left(5x+4\right)\left(x+2\right)-\left(5x^2+4x-1\right)}{\left(x+2\right)^2}\)
=>\(y'=\dfrac{5x^2+10x+4x+8-5x^2-4x+1}{\left(x+2\right)^2}\)
=>\(y'=\dfrac{10x+9}{\left(x+2\right)^2}\)
\(y'\left(-1\right)=\dfrac{10\cdot\left(-1\right)+9}{\left(-1+2\right)^2}=\dfrac{-1}{1}=-1\)
d:
ĐKXĐ: x<>2
\(y=x-2+\dfrac{9}{x-2}\)
=>\(y'=\left(x-2+\dfrac{9}{x-2}\right)'=1+\left(\dfrac{9}{x-2}\right)'\)
\(=1+\dfrac{9'\left(x-2\right)-9\left(x-2\right)'}{\left(x-2\right)^2}\)
=>\(y'=1+\dfrac{-9}{\left(x-2\right)^2}=\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}\)
y'=0
=>\(\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}=0\)
=>\(\left(x-2\right)^2-9=0\)
=>(x-2-3)(x-2+3)=0
=>(x-5)(x+1)=0
=>x=5 hoặc x=-1
I
1
14 tháng 12 2023
ĐKXĐ: x<>-1
\(\dfrac{x^2}{\left(x+1\right)^2}+\dfrac{x}{x+1}-2=0\)
\(\Leftrightarrow\left(\dfrac{x}{x+1}\right)^2+\left(\dfrac{x}{x+1}\right)-2=0\)
=>\(\left(\dfrac{x}{x+1}\right)^2+2\left(\dfrac{x}{x+1}\right)-\dfrac{x}{x+1}-2=0\)
=>\(\dfrac{x}{x+1}\left(\dfrac{x}{x+1}+2\right)-\left(\dfrac{x}{x+1}+2\right)=0\)
=>\(\left(\dfrac{x}{x+1}+2\right)\left(\dfrac{x}{x+1}-1\right)=0\)
=>\(\dfrac{x+2x+2}{x+1}\cdot\dfrac{x-x-1}{x+1}=0\)
=>\(\dfrac{3x+2}{x+1}\cdot\dfrac{-1}{x+1}=0\)
=>3x+2=0
=>x=-2/3(nhận)
\(x^2-x-\dfrac{1}{x}+\dfrac{1}{x^2}-10=0\)
\(\Rightarrow\left(x^2+\dfrac{1}{x^2}\right)-\left(x+\dfrac{1}{x}\right)-10=0\)
Đặt: \(x+\dfrac{1}{x}=t\) ta có: \(\left(x+\dfrac{1}{x}\right)^2=t^2\Leftrightarrow x^2+2+\dfrac{1}{x^2}=t^2\Leftrightarrow x^2+\dfrac{1}{x^2}=t^2-2\)
\(\Rightarrow t^2-2-t-10=0\)
\(\Rightarrow t^2-t-12=0\)
\(\Rightarrow t^2-4t+3t-12=0\)
\(\Rightarrow t\left(t-4\right)+3\left(t-4\right)=0\)
\(\Rightarrow\left(t+3\right)\left(t-4\right)=0\Leftrightarrow\left[{}\begin{matrix}t=-3\\t=4\end{matrix}\right.\)
Thay vào rồi giải tiếp nha bạn