Giải phương trình:
(\(\dfrac{x}{x+1}\))2 + (\(\dfrac{x}{x-1}\))2 = 90
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27 tháng 1 2022
\(\Leftrightarrow\dfrac{4\cdot90\cdot\left(x+5\right)-4\cdot90\cdot x}{4x\left(x+5\right)}=\dfrac{x\left(x+5\right)}{4x\left(x+5\right)}\)
\(\Leftrightarrow x^2+5x-1800=0\)
\(\text{Δ}=5^2-4\cdot1\cdot\left(-1800\right)=7225>0\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-5-85}{2}=\dfrac{-90}{2}=-45\left(nhận\right)\\x_2=\dfrac{-5+85}{2}=40\left(nhận\right)\end{matrix}\right.\)
LN
3
27 tháng 1 2022
đk : x khác 1 ; -1
<=> \(-x\left(x+1\right)+x^2+2=2\left(x-1\right)\)
\(\Leftrightarrow-x+2=2x-2\Leftrightarrow x=\dfrac{4}{3}\)(tm)
27 tháng 1 2022
\(\Leftrightarrow-x\left(x+1\right)+x^2+2=2x-2\)
\(\Leftrightarrow-x^2-x+x^2+2-2x+2=0\)
=>-3x+4=0
hay x=4/3(nhận)
PN
1
25 tháng 2 2023
=>(x^2+1)^2+x^2/x*(x^2+1)=5/2
=>\(\dfrac{\left(x^2+1\right)^2+x^2}{x\left(x^2+1\right)}=\dfrac{5}{2}\)
=>\(2\left(x^4+2x^2+1+x^2\right)=5\left(x^3+x\right)\)
=>2x^4+6x^2+2-5x^3-5x=0
=>2x^4-5x^3+6x^2-5x+2=0
=>2x^4-2x^3-3x^3+3x^2+3x^2-3x-2x+2=0
=>(x-1)(2x^3-3x^2+3x-2)=0
=>(x-1)(2x^3-2x^2-x^2+x+2x-2)=0
=>(x-1)^2*(2x^2-x+2)=0
=>x-1=0
=>x=1
5N
1
24 tháng 1
Sửa đề: \(\dfrac{2x-1}{x+2}+\dfrac{3x+2}{x^2+2x}=\dfrac{x+1}{x}\)
ĐKXĐ: \(x\notin\left\{0;-2\right\}\)
\(\dfrac{2x-1}{x+2}+\dfrac{3x+2}{x^2+2x}=\dfrac{x+1}{x}\)
=>\(\dfrac{2x-1}{x+2}+\dfrac{3x+2}{x\left(x+2\right)}=\dfrac{x+1}{x}\)
=>\(x\left(2x-1\right)+3x+2=\left(x+1\right)\left(x+2\right)\)
=>\(2x^2-x+3x+2=x^2+3x+2\)
=>\(2x^2+2x-x^2-3x=0\)
=>\(x^2-x=0\)
=>x(x-1)=0
=>\(\left[{}\begin{matrix}x=0\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
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 \(x=7\text{v}à\text{x}=15\)
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
2
25 tháng 3 2021
ĐKXĐ: \(x\ne\pm2\)
\(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x+2\right)\left(x-2\right)}+\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\\ \Leftrightarrow\left(x+1\right)\left(x+2\right)-5\left(x-2\right)=12+\left(x+2\right)\left(x-2\right)\\ \Leftrightarrow x^2+x+2x+2-5x+10=12+x^2-4\\ \Leftrightarrow-2x=-4\\ \Leftrightarrow x=2\left(ktm\right)\)
Vậy \(S\in\left\{\varnothing\right\}\)
PD
27 tháng 3 2021
ĐKXĐ: \(\begin{cases}x-2\ne 0\\x+2\ne 0\end{cases}\leftrightarrow x\ne 2\\x\ne -2\end{cases}\)
\(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)
\(\leftrightarrow \dfrac{(x+1)(x+2)}{(x-2)(x+2)}-\dfrac{5(x-2)}{(x+2)(x-2)}=\dfrac{12}{(x-2)(x+2)}+\dfrac{(x-2)(x+2)}{(x-2)(x+2)}\)
\(\to x^2+3x+2-5x+10=12+x^2-4\)
\(\leftrightarrow x^2-2x-x^2=12-12-4\)
\(\leftrightarrow -2x=-4\)
\(\leftrightarrow x=2(\rm KTM)\)
Vậy pt đã cho vô nghiệm \(S=\varnothing\)
`(x/(x+1))^2+(x/(x-1))^2=90(x ne -1,1)`
`<=>x^2/(x+1)^2+x^2/(x-1)^2=90`
`<=>x^2(x-1)^2+x^2(x-1)^2=90(x^2-1)^2`
`<=>x^2(2x^2+2)=90(x^4-2x^2+1)`
`<=>2x^4+2x^2=90x^4-180x^2+90`
`<=>88x^4-182x^2+90=0`
`<=>88x^4-110x^2-72x^2+90=0`
`<=>22x^2(4x^2-5)-18(4x^2-5)=0`
`<=>(4x^2-5)(22x^2-18)=0`
`<=>(4x^2-5)(11x^2-9)=0`
`<=>` $\left[ \begin{array}{l}4x^2=5\\11x^2=9\end{array} \right.$
`<=>` $\left[ \begin{array}{l}x=\sqrt{\dfrac{5}{4}}\\x=-\sqrt{\dfrac{5}{4}}\\x=\sqrt{\dfrac{9}{11}}\\x=-\sqrt{\dfrac{9}{11}}\end{array} \right.$
Vậy `S={\sqrt{9/11},-\sqrt{9/11},\sqrt{5/4},-\sqrt{5/4}}`
\(\left(\dfrac{x}{x+1}\right)^2+\left(\dfrac{x}{x-1}\right)^2=90\)
\(\Leftrightarrow\dfrac{x^2}{\left(x+1\right)^2}+\dfrac{x^2}{\left(x-1\right)^2}=90\)
\(\Leftrightarrow\dfrac{x^2\left(x-1\right)^2}{\left(x+1\right)^2\left(x-1\right)^2}+\dfrac{x^2\left(x+1\right)^2}{\left(x+1\right)^2\left(x-1\right)^2}=90\)
\(\Leftrightarrow\dfrac{x^2\left(x-1\right)^2+x^2\left(x+1\right)^2-90\left(x-1\right)^2\left(x+1\right)^2}{\left(x+1\right)^2\left(x-1\right)^2}=0\)
\(\Rightarrow x^2\left(x^2-2x+1\right)+x^2\left(x^2+2x+1\right)-90\left(x^2-1\right)^2=0\)
\(\Leftrightarrow x^4-2x^3+x^2+x^4+2x^3+x^2-90x^4+90x^2-90=0\)
\(\Leftrightarrow-88x^4+92x^2-90=0\)