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: \(x>0\)

\(\Leftrightarrow\sqrt{x\left(x+3\right)}+2\sqrt{x+2}=2x+\sqrt{\frac{\left(x+2\right)\left(x+3\right)}{x}}\)

\(\Leftrightarrow\sqrt{\frac{\left(x+2\right)\left(x+3\right)}{x}}-2\sqrt{x+2}+2x-2\sqrt{x\left(x+3\right)}=0\)

\(\Leftrightarrow\sqrt{\frac{x+2}{x}}\left(\sqrt{x+3}-2\sqrt{x}\right)-2\sqrt{x}\left(\sqrt{x+3}-2\sqrt{x}\right)=0\)

\(\Leftrightarrow\left(\sqrt{\frac{x+2}{x}}-2\sqrt{x}\right)\left(\sqrt{x+3}-2\sqrt{x}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{x+3}{x}=4x\\x+3=4x\end{matrix}\right.\)

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:..............

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) \(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}=\frac{x+1}{8}+\frac{x+1}{9}\)

\(\Leftrightarrow\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}-\frac{x+1}{8}-\frac{x+1}{9}=0\)

\(\Leftrightarrow\left(x+1\right).\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\right)=0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=0-1\)

\(\Rightarrow x=-1\)

Vậy \(x=-1.\)

Mình chỉ làm câu a) thôi nhé.

Chúc bạn học tốt!

Lê Thị Thục HiềnTrần Thanh PhươngVũ Minh Tuấn?Amanda?Nguyễn Việt LâmHISINOMA KINIMADONguyễn Huy TúAkai Haruma

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

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 !

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 \(S=\left\{-1;2\right\}\)