a,dk x>0

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

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

\(\Rightarrow\dfrac{x+2}{\sqrt{2x^2+x+1}-\sqrt{x^2-x+1}}=3\)

\(\Rightarrow\sqrt{2x^2+x+1}-\sqrt{x^2-x+1}=\dfrac{x+2}{3}\)

kh vs dé bài ta có hệ \(\left\{{}\begin{matrix}\sqrt{2x^2+x+1}+\sqrt{x^2-x+1}=3x\\\sqrt{2x^2+x+1}-\sqrt{x^2-x+1}=\dfrac{x+2}{3}\end{matrix}\right.\)

cộng vs nhau ta có

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

\(\Leftrightarrow3\sqrt{2x^2+x+1}=5x+1\)

giải ra ta có x=1(tm) x=-8/7 (l)

b, dk tu xd nhé

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

\(\Leftrightarrow2x\left(\dfrac{1}{\sqrt{x^2+x+1}+\sqrt{x^2-x+1}}-1\right)=0\)

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

ns \(\sqrt{x^2+x+1}+\sqrt{x^2-x+1}>1\)

\(\Rightarrow x=0\left(tm\right)\)

\(6-2\left(x-1\right)=4\)

\(\Rightarrow2\left(x-1\right)=6-4\)

\(\Rightarrow2\left(x-1\right)=2\)

\(\Rightarrow x-1=1\)

\(\Rightarrow x=1+1=2\)

________________

\(2\cdot\left(x-2\right)+1=7\)

\(\Rightarrow2\cdot\left(x-2\right)=7-1\)

\(\Rightarrow2\cdot\left(x-2\right)=6\)

\(\Rightarrow x-2=3\)

\(\Rightarrow x=3+2=5\)

_______________

\(\left(2\cdot x-3\right)+4=9\)

\(\Rightarrow2\cdot x-3=5\)

\(\Rightarrow2\cdot x=3+5\)

\(\Rightarrow2\cdot x=8\)

\(\Rightarrow x=\dfrac{8}{2}=4\)

________________

\(\left(3\cdot x-2\right)-1=3\)

\(\Rightarrow3\cdot x-2=3+1\)

\(\Rightarrow3\cdot x-2=4\)

\(\Rightarrow3\cdot x=6\)

\(\Rightarrow x=\dfrac{6}{3}=2\)

a: =>2(x-1)=2

=>x-1=1

=>x=2

b: =>2(x-2)=6

=>x-2=3

=>x=5

c; =>2x-3=5

=>2x=8

=>x=4

d: =>3x-2=4

=>3x=6

=>x=2

e: =>2(6-x)=4

=>6-x=2

=>x=4

f: =>x-2=5

=>x=7

g: =>10-2x=4

=>2x=6

=>x=3

h: =>2x+4=3

=>2x=-1

=>x=-1/2

j: =>x+2=12

=>x=10

l: =>2x+3=3

=>2x=0

=>x=0

\(6-2\left(x-1\right)=4\\ 2\left(x-1\right)=6-4=2\\ x-1=\dfrac{2}{2}=1\\ x=1+1=2\)

\(2\left(6-x\right)+1=5\\ 2\left(6-x\right)=5-1=4\\ 6-x=\dfrac{4}{2}=2\\ x=6-2=4\)

\(\left(4+2x\right)-1=8\\ 3\left(4+2x\right)=8+1=9\\ 4+2x=\dfrac{9}{3}=3\\ 2x=3-4=-1\\ x=-\dfrac{1}{2}\)

\(2\left(x-2\right)+1=7\\ 2\left(x-2\right)=7-1=6\\ x-2=\dfrac{6}{2}=3\\ x=3+2=5\)

\(20+\left(x-2\right)=25\\ x-2=25-20=5\\ x=5+2=7\)

\(\left(x+2\right)-5=7\\ x+2=7+5=12\\ x=12-2=10\)

\(\left(2x-3\right)+4=9\\ 2x-3=9-4=5\\ 2x=5+3=8\\ x=\dfrac{8}{2}=4\)

\(\left(10-\dfrac{2}{x}\right)-1=3\\ 10-\dfrac{2}{x}=3+1=4\\ \dfrac{2}{x}=10-4=6\\ x=\dfrac{2}{6}=\dfrac{1}{3}\)

\(\left(3+2x\right)-1=3\\ 3+2x=3+1=4\\ 2x=4-3=1\\ x=\dfrac{1}{2}\)

\(\left(3x-2\right)-1=3\\ 3x-2=3+1=4\\ 3x=4+2=6\\ x=\dfrac{6}{3}=2\)

tham khảo:

99/100

a ) \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=17\)

\(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\)

\(\Leftrightarrow\left(x^3-x^3\right)+\left(3x^2-3x^2\right)+\left(6x+3x\right)+\left(8-1\right)=17\)

\(\Leftrightarrow9x+7=17\)

\(\Leftrightarrow9x=10\)

\(\Leftrightarrow x=\dfrac{10}{9}\)

Vậy nghiệm của p/t là : \(\dfrac{10}{9}\)

b ) \(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)

\(\Leftrightarrow x\left(x^2-25\right)-\left(x^3+8\right)=3\)

\(\Leftrightarrow x^3-25x-x^3-8=3\)

\(\Leftrightarrow-25x-8=3\)

\(\Leftrightarrow-25x=11\)

\(\Leftrightarrow x=-\dfrac{11}{25}\)

Vậy nghiệm của p/t là : \(-\dfrac{11}{25}\)

=>\(\dfrac{1}{x}-\dfrac{1}{x+1}+.........+\dfrac{1}{x+2}-\dfrac{1}{x+3}-\dfrac{1}{x}=\dfrac{1}{2010}\)

(gạch đi những phân số trùng nhau )

=> \(\dfrac{-1}{x+3}=\dfrac{1}{2010}\)

=> -1=1 ; x+3 = 2010

Vì cả mẫu và tử đều phải là số âm mới có thể ra 1 phân số dương vì thế

=> x = -2013

=> \(\dfrac{-1}{-2013}=\dfrac{1}{2010}\)

diễn giải ra nhé

X x \(\dfrac{3}{4}\)+ X x\(\dfrac{1}{5}\)+ X x \(\dfrac{1}{20}\)+ X= 1000

a) \(\frac{1}{x-1}\)+\(\frac{2}{x+1}\)=\(\frac{x}{x^2-1}\) (ĐKXĐ:x≠1;x≠-1)

⇔\(\frac{x+1}{\left(x-1\right)\left(x+1\right)}\)+\(\frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)=\(\frac{x}{\left(x-1\right)\left(x+1\right)}\)

⇒x+1+2x-2=x

⇔2x-1=0

⇔x=\(\frac{1}{2}\) (TMĐKXĐ)

Vậy tập nghiệm của phương trình đã cho là:S={\(\frac{1}{2}\)}

những cách làm câu còn lại chẳng khác gì cách làm của câu này, bạn tự làm được mà!