NX
0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
NM
Nguyễn Minh Quang
Giáo viên
5 tháng 3 2022
ta có :
\(\left|x+1\right|+\left|x-1\right|=1+\left|\left(x-1\right)\left(x+1\right)\right|\)
\(\Leftrightarrow\left|x-1\right|\left|x+1\right|-\left|x-1\right|-\left|x+1\right|+1=0\)
\(\Leftrightarrow\left(\left|x-1\right|-1\right)\left(\left|x+1\right|-1\right)=0\Leftrightarrow\orbr{\begin{cases}\left|x-1\right|=1\\\left|x+1\right|=1\end{cases}}\)
\(\Leftrightarrow x\in\left\{-2,0,2\right\}\)
N
1
1 tháng 2 2021
ĐKXĐ: \(x\notin\left\{0;-1\right\}\)
Ta có: \(\dfrac{1}{x^2}+\dfrac{1}{\left(x+1\right)^2}=15\)
\(\Leftrightarrow\left(\dfrac{1}{x}\right)^2+\left(\dfrac{1}{x+1}\right)^2=15\)
\(\Leftrightarrow\left(\dfrac{1}{x}\right)^2+\left(\dfrac{1}{x+1}\right)^2-\dfrac{2}{x\left(x+1\right)}+\dfrac{2}{x\left(x+1\right)}=15\)
\(\Leftrightarrow\left(\dfrac{1}{x}-\dfrac{1}{x+1}\right)^2+\dfrac{2}{x\left(x+1\right)}=15\)
\(\Leftrightarrow\left(\dfrac{x+1}{x\left(x+1\right)}-\dfrac{x}{x\left(x+1\right)}\right)^2+\dfrac{2}{x\left(x+1\right)}=15\)
\(\Leftrightarrow\left(\dfrac{1}{x\left(x+1\right)}\right)^2+\dfrac{2}{x\left(x+1\right)}=15\)
\(\Leftrightarrow\dfrac{1}{x^2\cdot\left(x+1\right)^2}+\dfrac{2}{x\left(x+1\right)}-15=0\)(1)
Đặt \(\dfrac{1}{x\left(x+1\right)}=a\)(Điều kiện: \(x\notin\left\{0;-1\right\}\)
(1)\(\Leftrightarrow a^2+2a-15=0\)
\(\Leftrightarrow a^2+5a-3a-15=0\)
\(\Leftrightarrow a\left(a+5\right)-3\left(a+5\right)=0\)
\(\Leftrightarrow\left(a+5\right)\left(a-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+5=0\\a-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=-5\\a=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{x\left(x+1\right)}=-5\\\dfrac{1}{x\left(x+1\right)}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\left(x+1\right)=-\dfrac{1}{5}\\x\left(x+1\right)=\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+\dfrac{1}{5}=0\\x^2+x-\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{20}=0\\x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{7}{12}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{20}\\\left(x+\dfrac{1}{2}\right)^2=\dfrac{7}{12}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{\sqrt{5}}{10}\\x+\dfrac{1}{2}=-\dfrac{\sqrt{5}}{10}\\x+\dfrac{1}{2}=\dfrac{\sqrt{21}}{6}\\x+\dfrac{1}{2}=-\dfrac{\sqrt{21}}{6}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5+\sqrt{5}}{10}\left(nhận\right)\\x=\dfrac{-5-\sqrt{5}}{10}\left(nhận\right)\\x=\dfrac{-3+\sqrt{21}}{6}\left(nhận\right)\\x=\dfrac{-3-\sqrt{21}}{6}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{-5+\sqrt{5}}{10};\dfrac{-5-\sqrt{5}}{10};\dfrac{-3+\sqrt{21}}{6};\dfrac{-3-\sqrt{21}}{6}\right\}\)
D
0
MT
1
MB
3
8 tháng 2 2021
1x+1−5x−2=15(x+1)(2−x)1x+1−5x−2=15(x+1)(2−x)(1)
ĐKXĐ : x≠1;x≠2x≠1;x≠2
(1)⇔1x+1+52−x=15(x+1)(2−x)⇔1x+1+52−x=15(x+1)(2−x)
⇔2−x+5x+5=15⇔2−x+5x+5=15
⇔4x+7=15⇔4x+7=15
⇔4x=8⇔4x=8
8 tháng 2 2021
ĐKXĐ : \(\left\{{}\begin{matrix}x\ne\pm1\\x\ne2\end{matrix}\right.\)
PT \(\Leftrightarrow\dfrac{\left(x-2\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)\left(x-2\right)}-\dfrac{5\left(x+1\right)\left(x-1\right)}{\left(x-2\right)\left(x+1\right)\left(x-1\right)}=\dfrac{-15\left(x+1\right)}{\left(x+1\right)\left(x-1\right)\left(x-2\right)}\)\(\Leftrightarrow\left(x-2\right)\left(x-1\right)-5\left(x+1\right)\left(x-1\right)=-15\left(x+1\right)\)
\(\Leftrightarrow x^2-2x-x+2-5x^2+5=-15x-15\)
\(\Leftrightarrow x^2-2x-x+2-5x^2+5+15x+15=0\)
\(\Leftrightarrow-4x^2+12x+22=0\)
\(\Leftrightarrow x=\dfrac{3\pm\sqrt{31}}{2}\) ( TM )
Vậy ...
NY
1
MT
4 tháng 4 2016
Đặt t=(x+1)2
-----------------------------ra phương trình bậc 2 giải bằng cách đưa về PT tích ==" theo t rồi tìm x
28 tháng 6 2023
a: =>x-1=0 hoặc 3x-1=0
=>x=1 hoặc x=1/3
b: ĐKXĐ: x<>2; x<>-1
PT =>x-2-5(x+1)=15
=>x-2-5x-5=15
=>-4x-7=15
=>-4x=22
=>x=-11/2(nhận)
c: ĐKXĐ: x<>2; x<>-2
PT =>(x-1)(x-2)-x(x+2)=5x-2
=>x^2-3x+2-x^2-2x=5x-2
=>-5x+2=5x-2
=>-10x=-4
=>x=2/5(nhận)
20 tháng 9 2020
1) \(\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{4x+15}{9-x^2}\)
ĐKXĐ : \(x\ne\pm3\)
\(\Leftrightarrow\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{-4x-15}{x^2-9}\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow\frac{x^2-4x+3}{\left(x-3\right)\left(x+3\right)}-\frac{x^2+3x}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow\frac{x^2-4x+3-x^2-3x}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow-7x+3=-4x-15\)
\(\Leftrightarrow-7x+4x=-15-3\)
\(\Leftrightarrow-3x=-18\)
\(\Leftrightarrow x=6\)( tmđk )
Vậy x = 6 là nghiệm của phương trình
2) 2x + 3 < 6 - ( 3 - 4x )
<=> 2x + 3 < 6 - 3 + 4x
<=> 2x - 4x < 6 - 3 - 3
<=> -2x < 0
<=> x > 0
Vậy nghiệm của bất phương trình là x > 0