\(\dfrac{x+1}{x-1}\)-\(\dfrac{1}{x+1}\)=\(\dfrac{x^2+2}{x^2-1}\)
giúp mik vs ah
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
AH
Akai Haruma
Giáo viên
20 tháng 6 2023
Dấu ngoặc và cuối là sai nhé bạn. Phải là ngoặc vuông (x=0 hoặc x=-8) mới đúng, vì x không thể nhận 2 giá trị khác nhau cùng lúc.
20 tháng 6 2023
=>8(x+1/x)^2+4[(x+1/x)^2-2]^2-4[(x+1/x)^2-2](x+1/x)^2=(x+4)^2
Đặt x+1/x=a(a>=2)
=>8a^2+4[a^2-2]^2-4[a^2-2]*a^2=(x+4)^2
=>8a^2+4a^4-16a^2+16-4a^4+8a^2=(x+4)^2
=>(x+4)^2=16
=>x+4=4 hoặc x+4=-4
=>x=-8;x=0
20 tháng 12 2022
a: \(=\dfrac{x^2-x+x+1+2x}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{x-1}\)
b: \(=\dfrac{x^2+2x-4x-2x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{x+2}\)
c: \(=\dfrac{2x^2-3x-9-x^2+3x+x^2+6x+9}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{2x^2+6x}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x}{x-3}\)
25 tháng 9 2023
`a, 2/3 +3/4 = (8+9)/12=17/12.`
`1 1/3+4/5 = 4/3 + 4/5 = (20+12)/15=32/15`.
`=> x=2.`
`b, 5/6-1/4=(20-6)/24=7/12`.
`2 1/3-2/5= 7/3-2/5 = (35-6)/15=29/15`.
`=> x=1`.
HX
2
NH
22 tháng 3 2023
\(\dfrac{5}{x+2}-\dfrac{x-1}{x-2}=\dfrac{12}{x^2-4}+1\left(x\ne-2;x\ne2\right)\)
\(< =>\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
suy ra
`5x-10-(x^2 +2x-x-2)=12+x^2 -4`
`<=>5x-10-x^2 -2x+x+2-12-x^2 +4=0`
`<=>-x^2 -x^2 +5x-2x+x-10+2+4=0`
`<=>-x^2 +4x-4=0`
`<=>x^2 -4x+4=0`
`<=>(x-2)^2 =0`
`<=>x-2=0`
`<=>x=2(ktmđk)`
vậy phương trình vô nghiệm
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 3 2023
ĐKXĐ: \(x\ne\pm2\)
\(\dfrac{5\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow5\left(x-2\right)-\left(x-1\right)\left(x+2\right)=12+\left(x-2\right)\left(x+2\right)\)
\(\Leftrightarrow5x-10-\left(x^2+x-2\right)=12+x^2-4\)
\(\Leftrightarrow-x^2+4x-8=x^2+8\)
\(\Leftrightarrow2x^2-4x+16=0\)
\(\Leftrightarrow2\left(x-1\right)^2+14=0\)
Do \(\left\{{}\begin{matrix}2\left(x-1\right)^2\ge0\\14>0\end{matrix}\right.\) ;\(\forall x\)
\(\Rightarrow2\left(x-1\right)^2+14>0\)
Vậy phương trình đã cho vô nghiệm
NH
15 tháng 12 2022
a)
\(\left|x-2\right|-\dfrac{3}{5}=\dfrac{1}{2}\\ \left|x-2\right|=\dfrac{1}{2}+\dfrac{3}{5}\\ \left|x-2\right|=\dfrac{11}{10}\\ =>\left[{}\begin{matrix}x-2=\dfrac{11}{10}\\x-2=-\dfrac{11}{10}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{31}{10}\\x=\dfrac{9}{10}\end{matrix}\right.\)
b)
\(\left(x-\dfrac{7}{3}\right):\dfrac{-1}{3}=0,4\\ x-\dfrac{7}{3}=0,4\cdot\dfrac{-1}{3}\\ x-\dfrac{7}{3}=-\dfrac{2}{15}\\ x=-\dfrac{2}{15}+\dfrac{7}{3}\\ x=\dfrac{11}{5}\)
c)
\(\left|x-3\right|=5\\ =>\left[{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\left[{}\begin{matrix}x=5+3\\x=-5+3\end{matrix}\right.\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
d)
\(\left(2x+3\right)^2=25\\ =>\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
e)
\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\dfrac{1}{4}:x=-\dfrac{7}{20}\)
\(x=\dfrac{1}{4}:\dfrac{-7}{20}\\ x=-\dfrac{5}{7}\)
f)
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\\ =>x-\dfrac{1}{2}=\dfrac{1}{3}\\ x=\dfrac{1}{3}+\dfrac{1}{2}\\ x=\dfrac{5}{6}\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 4 2021
Em coi lại đề bài, \(8\left(x+\dfrac{1}{x}\right)\) hay \(8\left(x+\dfrac{1}{x}\right)^2\) nhỉ?
H9
HT.Phong (9A5)
CTVHS
15 tháng 3 2023
a) \(2x-6=0\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=\dfrac{6}{2}=3\)
b) \(x^2-4x=0\)
\(\Leftrightarrow x\left(x-4\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(\dfrac{x+1}{x-1}-\dfrac{1}{x+1}=\dfrac{x^2+2}{x^2-1}\left(ĐKXĐ:x\ne1;x\ne-1\right)\)
\(\Leftrightarrow\dfrac{x+1}{x-1}-\dfrac{1}{x+1}=\dfrac{x^2+2}{\left(x+1\right)\left(x-1\right)}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{x^2+2}{\left(x+1\right)\left(x-1\right)}\)
\(\Leftrightarrow\left(x+1\right)\left(x+1\right)-\left(x-1\right)=x^2+2\)
\(\Leftrightarrow x^2+2x+1-x+1=x^2+2\)
\(\Leftrightarrow x^2+2x+1-x+1-x^2-2=0\)
\(\Leftrightarrow x=0\left(tm\right)\)
Vậy phương trình có nghiệm là x=0
mong có thể giúp bạn