\(\dfrac{x+1}{2021}+\dfrac{x+2}{2020}=\dfrac{x+3}{2019}+\dfrac{x+4}{2018}\)

=>\(\dfrac{x+1}{2021}+1+\dfrac{x+2}{2020}+1=\dfrac{x+3}{2019}+1+\dfrac{x+4}{2018}+1\)

=>\(\dfrac{x+2022}{2021}+\dfrac{x+2022}{2020}=\dfrac{x+2022}{2019}+\dfrac{x+2022}{2018}\)

=> (x+2022)(\(\dfrac{1}{2021}+\dfrac{1}{2020}-\dfrac{1}{2019}-\dfrac{1}{2018}\))=0

=>x+2022=0

=> x=-2022

\(x-4\sqrt{x-2}+1=0\)(Đk x>2)

⇔\(x-2-4\sqrt{x-2}+4-1=0\)

\(\Leftrightarrow\left(\sqrt{x-2}-2\right)^2-1=0\)

\(\Leftrightarrow\left(\sqrt{x-2}-3\right)\left(\sqrt{x-2}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}-3=0\\\sqrt{x-2}-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=3\\\sqrt{x-2}=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=9\\x-2=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=11\\x=3\end{matrix}\right.\)(thảo đk)

Vậy\(\left[{}\begin{matrix}x=11\\x=3\end{matrix}\right.\)là nghiệm của pt

ĐKXĐ: x≥2

x+1=\(4\sqrt{x-2}\) bình phương 2 vế ta đc:\(\left(x+1\right)^2=16\cdot\left(x-2\right)< =>x^2+2x+1=16x-32< =>x^2-14x+33=0\)

giải phương trình này ta đc:x1=11(nhận); x2=3(nhận)

vậy phương trình có 2 nghiệm: x1=11;x2=3

Câu 1 : 

a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)

\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)

\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)

Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)

tương tự 

\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)

\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)

\(< =>95-24x+40=6-4x-15x+5\)

\(< =>-24x+135=-19x+11\)

\(< =>5x=135-11=124\)

\(< =>x=\frac{124}{5}\)

\(\frac{\left(x-2\right).3}{2}+3+\frac{\left(x-3\right).5}{3}+5+\frac{\left(x-5\right).2}{5}+2=10\)

\(< =>\frac{\left(x-2\right).3.15}{30}+\frac{\left(x-3\right).5.10}{30}+\frac{\left(x-5\right).2.6}{30}=10-2-3-5\)

\(< =>\frac{\left(x-2\right).45+\left(x-3\right).50+\left(x-5\right).12}{30}=0\)

\(< =>45x-90+50x-150+12x-60=0\)

\(< =>107x-300=0< =>x=\frac{300}{107}\)

1: \(2^x=64\)

=>\(x=log_264=6\)

2: \(2^x\cdot3^x\cdot5^x=7\)

=>\(\left(2\cdot3\cdot5\right)^x=7\)

=>\(30^x=7\)

=>\(x=log_{30}7\)

3: \(4^x+2\cdot2^x-3=0\)

=>\(\left(2^x\right)^2+2\cdot2^x-3=0\)

=>\(\left(2^x\right)^2+3\cdot2^x-2^x-3=0\)

=>\(\left(2^x+3\right)\left(2^x-1\right)=0\)

=>\(2^x-1=0\)

=>\(2^x=1\)

=>x=0

4: \(9^x-4\cdot3^x+3=0\)

=>\(\left(3^x\right)^2-4\cdot3^x+3=0\)

Đặt \(a=3^x\left(a>0\right)\)

Phương trình sẽ trở thành:

\(a^2-4a+3=0\)

=>(a-1)(a-3)=0

=>\(\left[{}\begin{matrix}a-1=0\\a-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=1\left(nhận\right)\\a=3\left(nhận\right)\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}3^x=1\\3^x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)

5: \(3^{2\left(x+1\right)}+3^{x+1}=6\)

=>\(\left[3^{x+1}\right]^2+3^{x+1}-6=0\)

=>\(\left(3^{x+1}\right)^2+3\cdot3^{x+1}-2\cdot3^{x+1}-6=0\)

=>\(3^{x+1}\left(3^{x+1}+3\right)-2\left(3^{x+1}+3\right)=0\)

=>\(\left(3^{x+1}+3\right)\left(3^{x+1}-2\right)=0\)

=>\(3^{x+1}-2=0\)

=>\(3^{x+1}=2\)

=>\(x+1=log_32\)

=>\(x=-1+log_32\)

6: \(\left(2-\sqrt{3}\right)^x+\left(2+\sqrt{3}\right)^x=2\)
=>\(\left(\dfrac{1}{2+\sqrt{3}}\right)^x+\left(2+\sqrt{3}\right)^x=2\) 

=>\(\dfrac{1}{\left(2+\sqrt{3}\right)^x}+\left(2+\sqrt{3}\right)^x=2\)

Đặt \(b=\left(2+\sqrt{3}\right)^x\left(b>0\right)\)

Phương trình sẽ trở thành:

\(\dfrac{1}{b}+b=2\)

=>\(b^2+1=2b\)

=>\(b^2-2b+1=0\)

=>(b-1)²=0

=>b-1=0

=>b=1

=>\(\left(2+\sqrt{3}\right)^x=1\)

=>x=0

7: ĐKXĐ: \(x^2+3x>0\)

=>x(x+3)>0

=>\(\left[{}\begin{matrix}x>0\\x< -3\end{matrix}\right.\)
\(log_4\left(x^2+3x\right)=1\)

=>\(x^2+3x=4^1=4\)

=>\(x^2+3x-4=0\)

=>(x+4)(x-1)=0

=>\(\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)

\(x+\sqrt{9-x^2}-x\sqrt{9-x^2}=3\left(-3\le x\le3\right)\)

\(\Leftrightarrow\sqrt{9-x^2}-x\sqrt{9-x^2}=3-x\\ \Leftrightarrow9-x^2+x^2\left(9-x^2\right)-2x\sqrt{\left(9-x^2\right)^2}=9-6x+x^2\\ \Leftrightarrow9+8x^2-x^4-2x\left(9-x^2\right)=x^2-6x+9\\ \Leftrightarrow-x^4+2x^3+7x^2-12x=0\\ \Leftrightarrow-x\left(x^3-2x^2-7x+12\right)=0\Leftrightarrow-x\left(x^3-3x^2+x^2-3x-4x+12\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x^2+x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(N\right)\\x=3\left(N\right)\\x^2+x-4=0\left(1\right)\end{matrix}\right.\)

 \(\Delta\left(1\right)=1-4\left(-4\right)=17>0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1-\sqrt{17}}{2}\left(N\right)\\x=\dfrac{-1+\sqrt{17}}{2}\left(N\right)\end{matrix}\right.\)

Vậy \(S=\left\{0;3;\dfrac{-1-\sqrt{17}}{2};\dfrac{-1+\sqrt{17}}{2}\right\}\)

Tick ✔

\(\left(x+4\right)\left(x^2-4x+16\right)-x\left(x-4\right)^2=8\left(x-3\right)\left(x+3\right)\)3)

\(\Leftrightarrow x^3+4^3-x\left(x-4\right)^2=8\left(x^2-3^2\right)\)

\(\Leftrightarrow x^3+64-x\left(x^2-8x+16\right)=8x^2-72\)

\(\Leftrightarrow x^3+64-x^3+8x^2-16x-8x^2-72=0\)

\(\Leftrightarrow-16x-8=0\)

\(\Leftrightarrow-8\left(2x-1\right)=0 \)

\(\Rightarrow2x-1=0\)

\(\Leftrightarrow2x=1\)

\(\Leftrightarrow x=\frac{1}{2}\)

Vậy   \(x=\frac{1}{2}\)

Bài 1:

1. \(x-8=3-2\left(x+4\right)\)

\(x-8=3-2x-8\)

\(3x=3\Rightarrow x=1\)

2. \(2\left(x+3\right)-3\left(x-1\right)=2\)

\(2x+6-3x+3=2\)

\(-x+9=2\Rightarrow x=7\)

3. \(4\left(x-5\right)-\left(3x-1\right)=x-19\)

\(4x-20-3x+1=x-19\)

\(0x=0\Rightarrow x=0\)

4. \(7-\left(x-2\right)=5\left(2x-3\right)\)

\(7-x+2=10x-15\)

\(-11x=-24\Rightarrow x=\frac{24}{11}\)

5. \(32-4\left(0,5y-5\right)=3y+2\)

\(32-2y+20=3y+2\)

\(-5y=-50\Rightarrow y=10\)

6. \(3\left(x-1\right)-x=2x-3\)

\(3x-3-x=2x-3\)

\(0x=0\Rightarrow x=0\)

Bài 2:

1. \(\frac{2-x}{3}=\frac{3-2x}{5}\)

\(\frac{\left(2-x\right)5}{15}-\frac{\left(3-2x\right)3}{15}=0\)

\(\frac{10-5x-9+6x}{15}=0\)

\(x+1=0\Rightarrow x=-1\)

2. \(\frac{3-4x}{4}=\frac{x+2}{5}\)

\(\frac{5\left(3-4x\right)}{20}-\frac{4\left(x+2\right)}{20}=0\)

\(\frac{15-20x-4x-8}{20}=0\)

\(7-24x=0\)

\(24x=7\Rightarrow x=\frac{7}{24}\)