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1) \(\left|x-2\right|+2=x\)
\(\Leftrightarrow\left|x-2\right|=x-2\)
\(\Leftrightarrow x-2\ge0\Leftrightarrow x\ge2\)
2) \(x^2+5x+4=0\)
\(\Leftrightarrow x^2+4x+x+4=0\)
\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-4\end{cases}}\)
3) \(8\sqrt{x}=x^2\)
Bình phương hai vế, ta được: \(64x=x^4\)
\(\Leftrightarrow x^4-64x=0\)
\(\Leftrightarrow x\left(x^3-64\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^3-64=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
4) \(\frac{x+29}{31}-\frac{x+27}{33}=\frac{x+17}{43}-\frac{x+15}{45}\)
\(\Leftrightarrow\frac{x+29}{31}-\frac{x+27}{33}-\frac{x+17}{43}+\frac{x+15}{45}=0\)
\(\Leftrightarrow\frac{x+29}{31}+1-\frac{x+27}{33}-1-\frac{x+17}{43}-1+\frac{x+15}{45}+1=0\)
\(\Leftrightarrow\frac{x+60}{31}+\frac{x+60}{45}-\frac{x+60}{33}-\frac{x+60}{43}=0\)
\(\Leftrightarrow\left(x+60\right)\left(\frac{1}{31}+\frac{1}{45}-\frac{1}{33}-\frac{1}{43}\right)=0\)
\(\Leftrightarrow x+60=0\Leftrightarrow x=-60\)
5)\(\left|x-1\right|+3x=1\)
\(\Leftrightarrow\left|x-1\right|=1-3x\)(1)
* Nếu \(x\ge1\)thì \(\left(1\right)\Leftrightarrow x-1=1-3x\Leftrightarrow4x=2\Leftrightarrow x=\frac{1}{2}\left(L\right)\)
* Nếu \(x< 1\)thì \(\left(1\right)\Leftrightarrow1-x=1-3x\Leftrightarrow2x=0\Leftrightarrow x=0\left(TM\right)\)
Vậy x = 0
\(\frac{x+32}{11}+\frac{x+33}{12}=\frac{x+34}{13}+\frac{x+35}{14}\)
\(\Leftrightarrow\left(\frac{x+32}{11}-1\right)+\left(\frac{x+33}{12}-1\right)=\left(\frac{x+34}{13}-1\right)+\left(\frac{x+35}{14}-1\right)\)
\(\Leftrightarrow\frac{x-21}{11}+\frac{x-21}{12}=\frac{x-21}{13}+\frac{x-21}{14}\)
\(\Leftrightarrow\left(x-21\right)\left(\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Vì \(\left(\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\ne0\)
\(\Rightarrow x-21=0\Rightarrow x=21\)
\(\frac{x+32}{11}+\frac{x+33}{12}=\frac{x+34}{13}+\frac{x+35}{14}\)
\(\Leftrightarrow\left(\frac{x+32}{11}-1\right)+\left(\frac{x+33}{12}-1\right)=\left(\frac{x+34}{13}-1\right)+\left(\frac{x+35}{14}-1\right)\)( trừ cả hai vế cho 2 )
\(\Leftrightarrow\frac{x-21}{11}+\frac{x-21}{12}=\frac{x-21}{13}+\frac{x-21}{14}\)
\(\Leftrightarrow\frac{x-21}{11}+\frac{x-21}{12}-\frac{x-21}{13}-\frac{x-21}{14}=0\)
\(\Leftrightarrow\left(x-21\right)\left(\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Mà \(\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\)
\(\Rightarrow x-21=0\)
\(\Leftrightarrow x=21\)
Vậy \(x=21\)
\(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)
<=>\(\frac{x+1}{35}+1+\frac{x+3}{33}+1=\frac{x+5}{31}+1+\frac{x+7}{29}+1\)
<=>\(\frac{x+1+35}{35}+\frac{x+3+33}{33}=\frac{x+5+31}{31}+\frac{x+7+29}{29}\)
<=>\(\frac{x+36}{35}+\frac{x+36}{33}=\frac{x+36}{31}+\frac{x+36}{29}\)
<=>\(\frac{x+36}{35}+\frac{x+36}{33}-\frac{x+36}{31}-\frac{x+36}{29}=0\)
<=>\(\left(x+36\right)\left(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\right)=0\)
Vì \(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\ne0\Rightarrow x+36=0\Rightarrow x=-36\)
\(\frac{x-11}{95}+\frac{x-13}{93}=\frac{x-15}{91}+\frac{x-17}{89}\) => \(\frac{x-11}{95}-1+\frac{x-13}{93}-1=\frac{x-15}{91}-1+\frac{x-17}{89}-1\)
=>\(\frac{x-106}{95}+\frac{x-106}{93}=\frac{x-106}{91}+\frac{x-106}{89}\)
=>\(\left(\frac{1}{95}+\frac{1}{93}\right)\left(x-106\right)-\left(\frac{1}{91}+\frac{1}{89}\right)\left(x-106\right)=0\)
<=>\(\left[\left(\frac{1}{95}+\frac{1}{93}\right)-\left(\frac{1}{91}+\frac{1}{89}\right)\right]\left(x-106\right)=0\).Vì\(\frac{1}{95}< \frac{1}{91};\frac{1}{93}< \frac{1}{89}\) nên\(\frac{1}{95}+\frac{1}{93}< \frac{1}{91}+\frac{1}{89}\)
=>\(\left(\frac{1}{95}+\frac{1}{93}\right)-\left(\frac{1}{91}+\frac{1}{89}\right)< 0\) hay khác 0.Vậy x - 106 = 0, tìm được x = 106
Cộng 3 vào cả 2 vế và chuyển vế xong đặt nhân tử x+50 ra ngoài ta được :
(x+50)(1/39+1/37+1/35-1/33-1/31-1/29)=0
vì (1/39+1/37+1/35-1/33-1/31-1/29) khác 0
=> x+50=0
=> x=-50
Nhớ k mình nhé
Chúc bạn học tốt
x = -50 nha