à nhầm:x^5-5x^3+4x=0

mong các bạn giúp đỡ

Giải:

Ta có: x⁵ - 5x³ + 4x = 0 ⇔ x⁵ - x³ - 4x³ + 4x = 0

⇔ x³(x² - 1) - 4x(x² - 1) = 0

⇔ (x² - 1)(x³ - 4x) = 0

⇔ (x + 1)(x - 1)x(x² - 4) = 0

⇔ x(x + 1)(x - 1)(x + 2)(x - 2) = 0

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

⇔ x = {0;1;-1;2;-2}

Vập tập nghiệm của phương trình là S = {0;1;-1;2;-2}.

Chúc bạn học tốt!

\(\Leftrightarrow x\left(x^4-5x^3+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^4-5x^2+4=0\left(1\right)\end{matrix}\right.\)

Đặt x^2=y PT (1) trở thành \(y^2-5y+4=0\Leftrightarrow y\left(y-4\right)-\left(y-4\right)=0\)

\(\Leftrightarrow\left(y-1\right)\left(y-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}y=1\\y=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=+-1\\x=+-2\end{matrix}\right.\)

Vậy \(S=\left(0,+-1,+-2\right)\)

ĐKXĐ: \(x\ge-\frac{3}{2}\)

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

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

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

\(ĐK:\frac{2}{3}\ge x\ge\frac{5}{2}\)

\(PT\Leftrightarrow\left(4x^2-4x+1\right)+\left(2x-5\right)\sqrt{2+4x}-\left(2x+3\right)\sqrt{6-4x}+16=0\)

\(\Leftrightarrow\left(2x-1\right)^2+\left(2x-5\right)\sqrt{2+4x}-\left(2x+3\right)\sqrt{6-4x}+16=0\)

\(\Leftrightarrow\left(2x-1\right)^2+\left(2x-5\right)\left(\sqrt{2+4x}-2\right)-\left(2x+3\right)\left(\sqrt{6-4x}-2\right)=0\)

\(\Leftrightarrow\left(2x-1\right)^2+\left(2x-5\right)\frac{2+4x-4}{\sqrt{2+4x}+2}+\left(2x+3\right)\frac{6-4x-4}{\sqrt{6-4x}+2}=0\)

\(\Leftrightarrow\left(2x-1\right)^2+\left(2x-5\right)\frac{2\left(2x-1\right)}{\sqrt{2+4x}+2}+\left(2x+3\right)\frac{-2\left(2x-1\right)}{\sqrt{6-4x}+2}=0\)

\(\Leftrightarrow\left(2x-1\right)\left(2x-1+\left(2x-5\right)\frac{2}{\sqrt{2+4x}+2}+\left(2x+3\right)\frac{-2}{\sqrt{6-4x}+2}\right)=0\)

Theo ĐK ta chứng minh đc \(\left(2x-1+\left(2x-5\right)\frac{2}{\sqrt{2+4x}+2}+\left(2x+3\right)\frac{-2}{\sqrt{6-4x}+2}\right)>0\)

Do đó \(2x-1=0\Rightarrow x=\frac{1}{2}\left(TMĐKXĐ\right)\)

PT ⇒ \(2\left(x^2-4x+5\right)-3\sqrt{x^2-4x+5}=22\)

Đặt \(\sqrt{x^2-4x+5}=y>0\), ta có:

\(2y^2-3y-22=0\) \(\Rightarrow y=\frac{3\pm\sqrt{185}}{4}\)

Số xấu quá, ko muốn giải nữa :D

Có vẻ phương trình có 4 nghiệm

a: \(\dfrac{2x-1}{3}-\dfrac{5x+2}{7}=x+13\)

\(\Leftrightarrow21\left(x+13\right)=7\left(2x-1\right)-3\left(5x+2\right)\)

\(\Leftrightarrow21x+273=14x-7-15x-6=-x-13\)

=>22x=-286

hay x=-13

b: \(\dfrac{2x-3}{3}-\dfrac{x-3}{6}=\dfrac{4x+3}{5}-17\)

\(\Leftrightarrow10\left(2x-3\right)-5\left(x-3\right)=6\left(4x+3\right)-510\)

\(\Leftrightarrow20x-30-5x+15=24x+18-510\)

\(\Leftrightarrow15x-15=24x-492\)

=>-9x=-477

hay x=53