a) Ta có: \(3x^2+2x-1=0\)

\(\Leftrightarrow3x^2+3x-x-1=0\)

\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)

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

Vậy: \(S=\left\{-1;\dfrac{1}{3}\right\}\)

b) Ta có: \(x^2-5x+6=0\)

\(\Leftrightarrow x^2-2x-3x+6=0\)

\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)

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

Vậy: S={2;3}

c) Ta có: \(x^2-3x+2=0\)

\(\Leftrightarrow x^2-x-2x+2=0\)

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

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

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

Vậy: S={1;2}

d) Ta có: \(2x^2-6x+1=0\)

\(\Leftrightarrow2\left(x^2-3x+\dfrac{1}{3}\right)=0\)

mà \(2\ne0\)

nên \(x^2-3x+\dfrac{1}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{23}{12}=0\)

\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=\dfrac{23}{12}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{\sqrt{69}}{6}\\x-\dfrac{3}{2}=\dfrac{-\sqrt{69}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9+\sqrt{69}}{6}\\x=\dfrac{9-\sqrt{69}}{6}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{9+\sqrt{69}}{6};\dfrac{9-\sqrt{69}}{6}\right\}\)

e) Ta có: \(4x^2-12x+5=0\)

\(\Leftrightarrow4x^2-10x-2x+5=0\)

\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{5}{2};\dfrac{1}{2}\right\}\)

cho vào máy tính là ra hết

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

\(\Leftrightarrow\left(6x^2+2x\right)\left(x+1+x-1\right)=0\)

\(\Leftrightarrow2x\left(3x+1\right).2x=0\)

\(\Leftrightarrow4x^2\left(3x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{3}\end{cases}}\)

Ta có: \(\left(x+1\right)\left(6x^2+2x\right)+\left(x-1\right)\left(6x^2+2x\right)=0\)

\(\Leftrightarrow\left(6x^2+2x\right)\left(x+1+x-1\right)=0\)

\(\Leftrightarrow\left(6x^2+2x\right).2x=0\)

\(\Leftrightarrow2x.\left(3x+1\right).2x=0\)

\(\Leftrightarrow4x^2\left(3x+1\right)=0\)

\(\Leftrightarrow4x^2=0\text{ hoặc }3x+1=0\)

\(\Leftrightarrow x=0\text{ hoặc }x=\frac{-1}{3}\)

\(\left(x+1\right)\left(6x^2+2x\right)+\left(x-1\right)\left(6x^2+2x\right)=0\)

\(\Leftrightarrow\left(6x^2+2x\right)\left(x+1+x-1\right)=0\)

\(\Leftrightarrow2x\left(3x+1\right).2x=0\)

\(\Leftrightarrow4x^2\left(3x+1\right)=0\)

\(\Rightarrow4x^2=0\)

\(\Rightarrow x=0\)

\(\Rightarrow3x+1=0\)

\(\Rightarrow3x=-1\)

\(x=-\frac{1}{3}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{3}\end{cases}}\)

Đây nhé! Tích giúp mình nha

Ta có: \(\dfrac{4x^4+3x^3}{-x^3}+\dfrac{15x^2+6x}{3x}=0\)

\(\Leftrightarrow-4x-3+5x+2=0\)

\(\Leftrightarrow x-1=0\)

hay x=1

pt quá vĩ đại =.= cx trên OLM lun 

câu a biến đổi to lắm

\(\Leftrightarrow-\left(12x\sqrt{6x-1}-2\sqrt{6x-1}-2x^3-9x^2+6x-8\right)=0\)rồi sao nx 

cái này ra nghiệm là 

\(2-\sqrt{2}\)và\(\sqrt{2}+2\)

bn ko bik lm hay sao, hay là bn chỉ đăng đề lên thôi

sao nhìu... z p , đăq từq câu 1 thôy nha p

Ôi trời sao lắm thế ít thôi bạn nên tách ra mà bạn cần gấp lắm à

đúng rồi pn. giúp mik đc bài nào cũng đc

c, |2\(x\) + 1| + |3\(x\) - 1| = 0

   vì |2\(x\) + 1| ≥ 0; |3\(x\) - 1| = 0

  ⇒ |2\(x\) + 1| + |3\(x\) - 1| = 0

   ⇔ \(\left\{{}\begin{matrix}2x+1=0\\3x-1=0\end{matrix}\right.\)

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

   \(\Rightarrow\) \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

       \(-\dfrac{1}{2}\) < \(\dfrac{1}{3}\) 

Vậy \(x\) \(\in\) \(\varnothing\)

a, Nếu 4.|3\(x\) - 1| = |6\(x\) - 2| + |-1,5|

             4.|3\(x\) -1| - 2.|3\(x\) - 1|  = 1,5

           Nếu 3\(x\) - 1 ≥ 0 ⇒ \(x\) ≥ \(\dfrac{1}{3}\)

Ta có: 4.(3\(x\) - 1) - 2.(3\(x\) - 1) = 1,5

           12\(x\) - 4 - 6\(x\) + 2 = 1,5

            6\(x\) - 2  = 1,5

            6\(x\)        = 1,5 + 2

            6\(x\)       = 3,5

               \(x\)      = 3,5: 6

                \(x\)    = \(\dfrac{7}{12}\)

Nếu 3\(x\) - 1 < 0 ⇒ \(x\) < \(\dfrac{1}{3}\)

Ta có: - 4.(3\(x\) - 1) = - (6\(x\) - 2) + 1,5

           -12\(x\) + 4 + 6\(x\) - 2 = 1,5

             -6\(x\) + 2 = 1,5

              6\(x\)         = 2- 1,5

              6\(x\)          = 0,5

                 \(x\)         = 0,5 : 6

                 \(x\)        = \(\dfrac{1}{12}\)

Vậy \(x\) \(\in\) {\(\dfrac{1}{12}\); \(\dfrac{7}{12}\)}

b, 2024.|2\(x\) - 1| = 2025.|1 - 2\(x\)| - |-2|

    2025.|1 - 2\(x\)| - 2024.|1 - 2\(x\)| = |-2|

             |1 - 2\(x\)| = 2 

             Nếu 1 - 2\(x\) ≥ 0 ⇒ \(x\) ≥ \(\dfrac{1}{2}\)

       với \(x\)    ≥ \(\dfrac{1}{2}\)  ta có: 1 - 2\(x\) = 2 ⇒ 2\(x\) = -1 ⇒ \(x\) = - \(\dfrac{1}{2}\) (1)

      Nếu \(1-2x\) < 0 ⇒ 2\(x\) ≤ 1 ⇒ \(x\) < \(\dfrac{1}{2}\) 

Với \(x\) < \(\dfrac{1}{2}\) ta có: -1 + 2\(x\) = 2 ⇒ 2\(x\) = 3 ⇒ \(x\) = \(\dfrac{3}{2}\) (2)

Kết hợp(1) và (2) ta có: \(x\) \(\in\) { - \(\dfrac{1}{2}\); \(\dfrac{3}{2}\)}

\(1,\dfrac{3x-1}{4}+\dfrac{6x-2}{8}=\dfrac{1-3x}{6}\\ \Leftrightarrow1,\dfrac{3x-1}{4}+\dfrac{3x-1}{4}=\dfrac{1-3x}{6}\\ \Leftrightarrow2.\dfrac{3x-1}{4}=\dfrac{1-3x}{6}\\ \Leftrightarrow\dfrac{3x-1}{4}=\dfrac{1-3x}{12}\\ \Leftrightarrow12\left(3x-1\right)=4\left(1-3x\right)\\ \Leftrightarrow3\left(3x-1\right)=1-3x\\ \Leftrightarrow9x-3-1+3x=0\\ \Leftrightarrow12x-4=0\\ \Leftrightarrow x=\dfrac{1}{3}\)

\(2,\left(2x-1\right)^2+\left(x-3\right)\left(2x-1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(2x-1+x-3\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(3x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{4}{3}\end{matrix}\right.\)