Giải các phương trình:
a)\(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2}-25\)
b) x2+ 6x+9= 144
c) \(\frac{2-x}{2016}-1=\frac{1-x}{2017}-\frac{x}{2018}\)
( mink đag cần gấp)
Giải phương trình:
a) \(2\sqrt{x^2-4}-3=6\sqrt{x-2}-\sqrt{x+2}\)
b) \(\frac{\sqrt{x-2016}-1}{x-2016}+\frac{\sqrt{y-2017}-1}{y-2017}+\frac{\sqrt{z-2018}-1}{z-2018}=\frac{3}{4}\)
c) \(\sqrt{3+\sqrt{3+x}}=x\)
d) \(\sqrt{6x^2+1}=\sqrt{2x-3}+x^2\)
e) \(\sqrt{x^2+3x+5}+\sqrt{x^2-2x+5}=5\sqrt{x}\)
f) \(\sqrt{x^2+3x}+2\sqrt{x+2}=2x+\sqrt{x+\frac{6}{x}+5}\)
a/ ĐKXĐ: \(x\ge2\)
\(\Leftrightarrow2\sqrt{\left(x-2\right)\left(x+2\right)}-6\sqrt{x-2}+\sqrt{x+2}-3=0\)
\(\Leftrightarrow2\sqrt{x-2}\left(\sqrt{x+2}-3\right)+\sqrt{x+2}-3=0\)
\(\Leftrightarrow\left(2\sqrt{x-2}+1\right)\left(\sqrt{x+2}-3\right)=0\)
\(\Leftrightarrow\sqrt{x+2}-3=0\Rightarrow x=11\)
b/ ĐKXĐ: ....
Đặt \(\left\{{}\begin{matrix}\sqrt{x-2016}=a>0\\\sqrt{y-2017}=b>0\\\sqrt{z-2018}=a>0\end{matrix}\right.\)
\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{4}-\frac{a-1}{a^2}+\frac{1}{4}-\frac{b-1}{b^2}+\frac{1}{4}-\frac{c-1}{c^2}=0\)
\(\Leftrightarrow\frac{\left(a-2\right)^2}{a^2}+\frac{\left(b-2\right)^2}{b^2}+\frac{\left(c-2\right)^2}{c^2}=0\)
\(\Leftrightarrow a=b=c=2\Rightarrow\left\{{}\begin{matrix}x=2020\\y=2021\\z=2022\end{matrix}\right.\)
a/ ĐK: \(x\ge0\)
\(\Leftrightarrow\sqrt{3+x}=x^2-3\)
Đặt \(\sqrt{3+x}=a>0\Rightarrow3=a^2-x\) pt trở thành:
\(a=x^2-\left(a^2-x\right)\)
\(\Leftrightarrow x^2-a^2+x-a=0\)
\(\Leftrightarrow\left(x-a\right)\left(x+a+1\right)=0\)
\(\Leftrightarrow x=a\) (do \(x\ge0;a>0\))
\(\Leftrightarrow\sqrt{3+x}=x\Leftrightarrow x^2-x-3=0\)
d/ ĐKXĐ: ...
\(\sqrt{6x^2+1}=\sqrt{2x-3}+x^2\)
\(\Leftrightarrow\sqrt{2x-3}-1+x^2+1-\sqrt{6x^2+1}\)
\(\Leftrightarrow\frac{2\left(x-2\right)}{\sqrt{2x-3}+1}+\frac{x^4+2x^2+1-6x^2-1}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}=0\)
\(\Leftrightarrow\frac{2\left(x-2\right)}{\sqrt{2x-3}+1}+\frac{x^2\left(x+2\right)\left(x-2\right)}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{2}{\sqrt{2x-3}+1}+\frac{x^2\left(x+2\right)}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}\right)=0\)
\(\Leftrightarrow x=2\) (phần trong ngoặc luôn dương với mọi \(x\ge\frac{3}{2}\))
e/ ĐKXĐ: \(x\ge0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+3x+5}=a>0\\\sqrt{x^2-2x+5}=b>0\\\sqrt{x}=c\ge0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=5c^2\)
Ta được hệ: \(\left\{{}\begin{matrix}a^2-b^2=5c^2\\a+b=5c\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)\left(a+b\right)=5c^2\\a+b=5c\end{matrix}\right.\)
\(\Rightarrow5c\left(a-b\right)=5c^2\)
\(\Leftrightarrow\left[{}\begin{matrix}c=0\\a-b=c\end{matrix}\right.\)
f/ ĐKXĐ: \(x>0\)
\(\Leftrightarrow\sqrt{x\left(x+3\right)}+2\sqrt{x+2}=2x+\sqrt{\frac{\left(x+2\right)\left(x+3\right)}{x}}\)
\(\Leftrightarrow\sqrt{\frac{\left(x+2\right)\left(x+3\right)}{x}}-2\sqrt{x+2}+2x-2\sqrt{x\left(x+3\right)}=0\)
\(\Leftrightarrow\sqrt{\frac{x+2}{x}}\left(\sqrt{x+3}-2\sqrt{x}\right)-2\sqrt{x}\left(\sqrt{x+3}-2\sqrt{x}\right)=0\)
\(\Leftrightarrow\left(\sqrt{\frac{x+2}{x}}-2\sqrt{x}\right)\left(\sqrt{x+3}-2\sqrt{x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{x+3}{x}=4x\\x+3=4x\end{matrix}\right.\)
Giải phương trình :
a) \(\frac{x+1}{2020}+\frac{x+2}{2019}+\frac{x+3}{2018}=\frac{x+4}{2017}+\frac{x+5}{2016}+\frac{x+6}{2015}\)
b) \(\frac{2-x}{2004}-1=\frac{1-x}{2005}-\frac{x}{2006}\)
a, Làm
\(\frac{x+1}{2020}+\frac{x+2}{2019}+\frac{x+3}{2018}=\frac{x+4}{2017}+\frac{x+5}{2016}+\frac{x+6}{2015}\)
<=>\(\frac{x+2021}{2020}+\frac{x+2021}{2019}+\frac{x+2021}{2018}=\frac{x+2021}{2017}+\frac{x+2021}{2016}+\frac{x+2021}{2015}\)
<=>\(\left(x+2021\right)\left(\frac{1}{2020}+\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)=0\)
<=> x+2021=0
<=> x=-2021
Kl:......................
b, Làmmmmm
\(\frac{2-x}{2004}-1=\frac{1-x}{2005}-\frac{x}{2006}\)
<=> \(\frac{2006-x}{2004}=\frac{2006-x}{2005}+\frac{2006-x}{2006}\)
<=> \(\left(2006-x\right)\left(\frac{1}{2004}-\frac{1}{2005}-\frac{1}{2006}\right)=0< =>2006-x=0\)
<=> x=2006
Kl:..............
Giải phương trình
a , \(\frac{3x+2}{x-1}+\frac{2x-4}{x+2}=5\)
b , \(\frac{4-3x}{1-x^2}-\frac{2}{x+1}=0\)
c , \(\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{x^2-4x+3}\)
d , \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
e ,\(\frac{x}{2016}+\frac{x+1}{2019}+\frac{x+2}{2018}+\frac{x+3}{2019}+\frac{x+4}{2020}=5\)
Cần gấp giúp nhanh hộ mình với ạ !!!!
d, \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
\(\Leftrightarrow\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)
\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)
\(\Leftrightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)
\(\Leftrightarrow x+10=0\) (Vì \(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\) ≠ 0)
\(\Leftrightarrow x=-10\)
Vậy x = -10 là nghiệm của phương trình.
e, Đề sai.
Sửa đề: \(\frac{x}{2016}+\frac{x+1}{2017}+\frac{x+2}{2018}+\frac{x+3}{2019}+\frac{x+4}{2020}=5\)
\(\Leftrightarrow\frac{x}{2016}-1+\frac{x+1}{2017}-1+\frac{x+2}{2018}-1+\frac{x+3}{2019}-1+\frac{x+4}{2020}-1=0\)
\(\Leftrightarrow\frac{x-2016}{2016}+\frac{x-2016}{2017}+\frac{x-2016}{2018}+\frac{x-2016}{2019}+\frac{x-2016}{2020}=0\)
\(\Leftrightarrow\left(x-2016\right)\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}+\frac{1}{2020}\right)=0\)
\(\Leftrightarrow x-2016=0\) (Vì \(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}+\frac{1}{2020}\ne0\)
\(\Leftrightarrow x=2016\)
Vậy x = 2016 là nghiệm của phương trình.
a)\(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}=\frac{x+1}{8}+\frac{x+1}{9}\)
b)\(\frac{x+6}{2015}+\frac{x+5}{2016}+\frac{x+4}{2017}=\frac{x+3}{2018}+\frac{x+2}{2019}+\frac{x+1}{2010}\)
c)\(\frac{x+6}{2016}+\frac{x+7}{2017}+\frac{x+8}{2018}=\frac{x+9}{2019}+\frac{x+10}{2020}+1\)
d)\(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\)
ai xong nhanh nhất và đúng em xin gửi 2 SP ạ
a) \(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}=\frac{x+1}{8}+\frac{x+1}{9}\)
\(\Leftrightarrow\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}-\frac{x+1}{8}-\frac{x+1}{9}=0\)
\(\Leftrightarrow\left(x+1\right).\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=0-1\)
\(\Rightarrow x=-1\)
Vậy \(x=-1.\)
Mình chỉ làm câu a) thôi nhé.
Chúc bạn học tốt!
a)\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
b)\(\frac{x+2015}{5}+\frac{x+2016}{6}=\frac{x+2017}{7}+\frac{x+2018}{8}\)
a) \(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\frac{x+2015}{5}+\frac{5}{5}+\frac{x+2016}{4}+\frac{4}{4}=\frac{x+2017}{3}+\frac{3}{3}+\frac{x+2018}{2}+\frac{2}{2}\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2002}{2}\)
\(\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right).\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)
\(\Leftrightarrow x=-2020\)
Vậy : \(x=-2020\)
Chúc bạn học tốt !!
a) \(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\\ \left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\\ \frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2020}{2}\\ \frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\\ \left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\\ \Rightarrow x+2020=0\\ \Rightarrow x=-2020\)
Vậy x = -2020
b) \(\frac{x+2015}{5}+\frac{x+2016}{6}=\frac{x+2017}{7}+\frac{x+2018}{8}\\ \left(\frac{x+2015}{5}-1\right)+\left(\frac{x+2016}{6}-1\right)=\left(\frac{x+2017}{7}-1\right)+\left(\frac{x+2018}{8}-1\right)\\ \frac{x+2010}{5}+\frac{x+2010}{6}=\frac{x+2010}{7}+\frac{x+2010}{8}\\ \frac{x+2010}{5}+\frac{x+2010}{6}-\frac{x+2010}{7}-\frac{x+2010}{8}=0\\ \left(x+2010\right)\left(\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\right)=0\\ \Rightarrow x+2010=0\\ \Rightarrow x=-2010\)
Vậy x = -2010
Bài 2: Giải các phương trình sau:
a \(\frac{3x+2}{2}-\frac{3x+1}{6}=\frac{5}{3}+2x\)
b \(\left(x+1\right)\left(x+2\right)=\left(2-x\right)\left(x+2\right)\)
c \(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\)
d\(x^2+6x+9=114\)
Bài 2: Giải các phương trình sau:
a) \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\) b) (x2 – 25) + (x – 5)(2x – 11) = 0
c) (x2 – 6x + 9) – 4 = 0 d) \(\frac{x+3}{x+1}+\frac{x-5}{x}=2\)
giải các phương trình
a)\(\frac{2}{x-14}-\frac{5}{x-13}=\frac{2}{x-9}-\frac{5}{x-11}\)
b) \(\frac{12x+1}{6x-2}-\frac{9x-5}{3x+1}=\frac{108x-36x^2-9}{4\left(9x^2-1\right)}\)
a)\(-ĐKXĐ:\hept{\begin{cases}x-14\ne0;x-13\ne0\\x-9\ne0\\x-11\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne14;x\ne13\\x\ne9\\x\ne11\end{cases}}\)
- Ta có : \(\frac{2}{x-14}-\frac{5}{x-13}=\frac{2}{x-9}-\frac{5}{x-11}\)
\(\Leftrightarrow\frac{2}{x-14}-\frac{5}{x-13}-\frac{2}{x-9}+\frac{5}{x-11}=0\)
\(\Leftrightarrow\left(\frac{2}{x-14}-\frac{2}{x-9}\right)-\left(\frac{5}{x-13}-\frac{5}{x-11}\right)=0\)
\(\Leftrightarrow2\left(\frac{1}{x-14}-\frac{1}{x-9}\right)-5\left(\frac{1}{x-13}-\frac{1}{x-11}\right)=0\)\(\Leftrightarrow2.\frac{\left(x-9\right)-\left(x-14\right)}{\left(x-9\right)\left(x-14\right)}-5.\frac{\left(x-11\right)-\left(x-13\right)}{\left(x-11\right)\left(x-13\right)}=0\)
\(\Leftrightarrow2.\frac{5}{\left(x-9\right)\left(x-14\right)}-5.\frac{2}{\left(x-11\right)\left(x-13\right)}=0\)
\(\Leftrightarrow\frac{10}{\left(x-9\right)\left(x-14\right)}-\frac{10}{\left(x-11\right)\left(x-13\right)}=0\)
\(\Leftrightarrow10\left[\frac{1}{\left(x-9\right)\left(x-14\right)}-\frac{1}{\left(x-11\right)\left(x-13\right)}\right]=0\)
\(\Leftrightarrow\frac{\left(x-11\right)\left(x-13\right)}{\left(x-9\right)\left(x-14\right)\left(x-11\right)\left(x-13\right)}-\frac{\left(x-9\right)\left(x-14\right)}{\left(x-9\right)\left(x-14\right)\left(x-11\right)\left(x-13\right)}=\) \(0\)
\(\Leftrightarrow\left(x-11\right)\left(x-13\right)-\left(x-9\right)\left(x-14\right)=0\)
\(\Leftrightarrow x^2-24x+143-x^2+23x-126=0\)
\(\Leftrightarrow-x+17=0\Leftrightarrow-x=-17\Leftrightarrow x=17\)
Vậy pt có tập nghiệm S = { 17 }
P/s: Mk làm hơi lòng vòng, bn thông cảm nhé !
Giải phương trình:
a,\(\frac{2-x}{2016}\)-1=\(\frac{1-x}{2017}\)-\(\frac{x}{2018}\)
b,\(\frac{x-19}{1999}\)+\(\frac{x-23}{1995}\)+\(\frac{x+82}{700}\)=5
c,x^3-3*x^2+4=0
a) \(\frac{2-x}{2016}-1=\frac{1-x}{2017}-\frac{x}{2018}\)
\(\Leftrightarrow\frac{2-x}{2016}+1=\frac{1-2}{2017}+1-\frac{x}{2018}+1\)
\(\Leftrightarrow\frac{2018-x}{2016}=\frac{2018-x}{2017}+\frac{2018-x}{2018}\)
\(\Leftrightarrow\frac{2018-x}{2016}-\frac{2018-x}{2017}-\frac{2018-x}{2018}=0\)
\(\Leftrightarrow\left(2018-x\right)\left(\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\right)=0\)
\(\Leftrightarrow2018-x=0\) ( vì \(\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\ne0\))
\(\Leftrightarrow x=2018\)
Vậy nghiệm của pt x=2018
b)\(\frac{x-19}{1999}+\frac{x-23}{1995}+\frac{x+82}{700}=5\)
\(\Leftrightarrow\left(\frac{x-19}{1999}-1\right)+\left(\frac{x-23}{1995}+-1\right)+\left(\frac{x+82}{700}-3\right)=0\)
\(\Leftrightarrow\frac{x-2018}{1999}+\frac{x-2018}{1995}+\frac{x-2018}{700}=0\)
\(\Leftrightarrow\left(x-2018\right)\left(\frac{1}{1999}+\frac{1}{1995}+\frac{1}{700}\right)=0\)
\(\Leftrightarrow x-2018=0\)( vì \(\frac{1}{1999}+\frac{1}{1995}+\frac{1}{700}\ne0\))
\(\Leftrightarrow x=2018\)
Vậy nghiệm của pt x=2018
c) \(x^3-3x^2+4=0\)
\(\Leftrightarrow x^3+x^2-4x^2+4=0\)
\(\Leftrightarrow x^2\left(x+1\right)-4\left(x^2-1\right)=0\)
\(\Leftrightarrow x^2\left(x+1\right)-4\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\\left(x-2\right)^2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}}\)
Vậy tập hợp nghiệm \(S=\left\{-1;2\right\}\)