\(\text{a) }\left(x^2+x\right)^2+4\left(x^2+x\right)=12\\ \Leftrightarrow\text{Đặt }x^2+x=y\\ \Leftrightarrow y^2+4y=12\\ \Leftrightarrow y^2+6y-2y-12=0\\ \Leftrightarrow\left(y^2+6y\right)-\left(2y+12\right)=0\\ \Leftrightarrow y\left(y+6\right)-2\left(y+6\right)=0\\ \Leftrightarrow\left(y+6\right)\left(y-2\right)=0\\ \Leftrightarrow\left(x^2+x+6\right)\left(x^2+x-2\right)=0\\ \Leftrightarrow\left(x^2+x+\dfrac{1}{4}+\dfrac{23}{4}\right)\left(x^2+2x-x-2\right)=0\\ \Leftrightarrow\left[\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{23}{4}\right]\left[\left(x^2+2x\right)-\left(x+2\right)\right]=0\\ \Leftrightarrow\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\right]\left[x\left(x+2\right)-\left(x+2\right)\right]=0\\ \Leftrightarrow\left(x-1\right)\left(x+2\right)=0\left(Vì\text{ }\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\ne0\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\\ \text{Vậy }S=\left\{1;-2\right\}\\ \)

\(\text{b) }6x^4-5x^3-38x^2-5x+6=0\\ \Leftrightarrow x^2\left(6x^2-5x-38-\dfrac{5}{x}+\dfrac{6}{x^2}\right)=0\\ \Leftrightarrow x^2\left[\left(6x^2+12+\dfrac{6}{x^2}\right)-\left(5x+\dfrac{5}{x}\right)-50\right]=0\\ \Leftrightarrow x^2\left[6\left(x^2+2+\dfrac{1}{x^2}\right)-5\left(x+\dfrac{1}{x}\right)-50\right]=0\\ \Leftrightarrow x^2\left[6\left(x+\dfrac{1}{x}\right)^2-5\left(x+\dfrac{1}{x}\right)-50\right]=0\\ \text{Đặt }x+\dfrac{1}{x}=y\\ \Leftrightarrow x^2\left(6y^2-5y-50\right)=0\\ \Leftrightarrow x^2\left(6y^2-20y+15y-50\right)=0\\ \Leftrightarrow x^2\left[\left(6y^2-20y\right)+\left(15y-50\right)\right]=0\\ \Leftrightarrow x^2\left[2y\left(3y-10\right)+5\left(3y-10\right)\right]=0\\ \Leftrightarrow x^2\left(2y+5\right)\left(3y-10\right)=0\\ \Leftrightarrow x^2\left(2x+\dfrac{2}{x}+5\right)\left(3x+\dfrac{3}{x}-10\right)=0\\ \Leftrightarrow\left(2x^2+2+5x\right)\left(3x^2+3-10x\right)=0\\ \Leftrightarrow\left(2x^2+4x+x+2\right)\left(3x^2-9x-x+3\right)=0\\ \Leftrightarrow\left[\left(2x^2+4x\right)+\left(x+2\right)\right]\left[\left(3x^2-9x\right)-\left(x-3\right)\right]=0\\ \Leftrightarrow\left[2x\left(x+2\right)+\left(x+2\right)\right]\left[3x\left(x-3\right)-\left(x-3\right)\right]=0\\ \Leftrightarrow\left(2x+1\right)\left(x+2\right)\left(3x-1\right)\left(x-3\right)=0\\ \)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\x+2=0\\3x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-1\\x=-2\\3x=1\\x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-2\\x=\dfrac{1}{3}\\x=3\end{matrix}\right.\\ \text{Vậy }S=\left\{-\dfrac{1}{2};-2;\dfrac{1}{3};3\right\}\)

casio fx 570vn

a: \(\Leftrightarrow x^2\left(x^2+x-12\right)=0\)

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

hay \(x\in\left\{0;-4;3\right\}\)

d: \(\left(x^2+5x\right)^2-2\left(x^2+5x\right)-24=0\)

\(\Leftrightarrow\left(x^2+5x-6\right)\left(x^2+5x+4\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(x-1\right)\left(x+1\right)\left(x+4\right)=0\)

hay \(x\in\left\{-6;1;-1;-4\right\}\)

f: \(x\left(x+1\right)\left(x-1\right)\left(x+2\right)=24\)

\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)

\(\Leftrightarrow\left(x^2+x\right)^2-2\left(x^2+x\right)-24=0\)

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

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

hay \(x\in\left\{-3;2\right\}\)

a) Ta có: \(2x^3+5x^2-3x=0\)

\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)

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

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

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

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

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

b) Ta có: \(2x^3+6x^2=x^2+3x\)

\(\Leftrightarrow2x^2\left(x+3\right)=x\left(x+3\right)\)

\(\Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\)

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

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

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

c) Ta có: \(x^2+\left(x+2\right)\left(11x-7\right)=4\)

\(\Leftrightarrow x^2+11x^2-7x+22x-14-4=0\)

\(\Leftrightarrow12x^2+15x-18=0\)

\(\Leftrightarrow12x^2+24x-9x-18=0\)

\(\Leftrightarrow12x\left(x+2\right)-9\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(12x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\12x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\12x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)

Vậy: \(S=\left\{-2;\dfrac{3}{4}\right\}\)

Trong đó có nhiều phương trình kiến thức cơ bản mà nhỉ? Ít nâng cao, bạn lọc ra câu nào k làm đc thôi chứ!

Đang rảnh, buồn ngủ nên giải cho tỉnh táo :D

Ta nhận thấy x=0 không phải là nghiệm của phương trình, vậy ta chia cả 2 vế của phương trình cho x² khác 0, ta được:

\(6x^2+5x-38+\dfrac{5}{x}+\dfrac{6}{x^2}=0\)

\(\Leftrightarrow6\left(x^2+\dfrac{1}{x}\right)+5\left(x+\dfrac{1}{x}\right)-38=0\)

Đặt \(x+\dfrac{1}{x}=y\Rightarrow x^2+\dfrac{1}{x^2}=y^2-2\)

Ta được: \(6\left(y^2-2\right)+5y-38=0\)

Do đó: y₁=2,5;y₂=-10/3

Với y=2,5\(\Rightarrow x+\dfrac{1}{x}=2,5\Rightarrow x_1=2;x_2=0,5\)

Với y=-10/3

\(\Rightarrow x+\dfrac{1}{x}=-\dfrac{10}{3}\Rightarrow x_3=-\dfrac{1}{3};x_4=-3\)

Vậy: \(S=\left\{2;0,5;-\dfrac{1}{3};-3\right\}\)

Bài a tự giải

Bài b thì biến đổi xong rồi đặt ẩn phụ \(y=x+\dfrac{1}{x}\)

Bài c:

Đặt x-1=y

Phương trình trở thành: \(\left(y+2\right)^4+\left(y-2\right)^4=82\)

Rút gọn ta được: \(2y^4+48y^2-50=0\)

Đặt y²=z ( \(z\ge0\) )

Phương trình này cho z₁=1, z₂=-25(Loại)

\(z=1\Rightarrow y^2=1\Rightarrow y=\pm1\)

\(\Rightarrow x_1=2;x_2=0\)

\(b,\)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)

\(\Rightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)

\(\Rightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)

\(\Rightarrow\left(x+9\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}\right)=\left(x+9\right)\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)\)

\(\Rightarrow\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}=\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\left(KTM\right)\)

\(\text{Giải}\)

\(b,\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)

\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)

\(\Leftrightarrow x+2009=0\Leftrightarrow x=-2009\)

Forever Miss You nếu (x-2009)=0 

thì \(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}=\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\text{ko}?\)

nếu làm cách đó xét 2 trường hợp :")

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

\(\Leftrightarrow x^2.\left(x^2-4x+4\right)=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\\left(x-2\right)^2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

\(b,\)\(x^2+5x+4=0\)

\(\Leftrightarrow x^2+x+4x+4=0\)

\(\Leftrightarrow x.\left(x+1\right)+4.\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right).\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+4=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-4\end{cases}}\)

\(c,\)\(9x-6x^2-3=0\)

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

\(\Leftrightarrow2x^2-3x+1=0\)

\(\Leftrightarrow2x^2-2x-x+1=0\)

\(\Leftrightarrow2x.\left(x-1\right)-\left(x-1\right)\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\2x=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)

\(d,\)\(2x^2+5x+2=0\)

\(\Leftrightarrow2x^2+4x+x+2=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\2x+1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\2x=-1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{2}\end{cases}}\)

@Lightning Farron

@soyeon_Tiểubàng giải

@Trần Việt Linh

c: \(\Leftrightarrow x-3=0\)

hay x=3

c: 

hay x=3

a) \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\)

\(\Leftrightarrow\)\(\frac{21\left(4x+3\right)-15\left(6x-2\right)}{105}=\frac{35\left(5x+4\right)+315}{105}\)

\(\Leftrightarrow21\left(4x+3\right)-15\left(6x-2\right)=35\left(5x+4\right)+315\)

\(\Leftrightarrow84x+63-90x+30=175x+140+315\)

\(\Leftrightarrow84x-90x-175x=140+315-63-30\)

\(\Leftrightarrow-181x=362\)

\(\Leftrightarrow x=-2\)

b)\(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right)\left(2x+3\right)}{8}+\frac{\left(x+4\right)^2}{6}=0\)

\(\Leftrightarrow\)\(\frac{8\left(x-2\right)^2-3\left(2x-3\right)\left(2x+3\right)+4\left(x+4\right)^2}{24}=0\)

\(\Leftrightarrow8\left(x^2-4x+4\right)-3\left(4x^2-9\right)+4\left(x^2+8x+16\right)=0\)

\(\Leftrightarrow8x^2-32x+32-12x^2+27+4x^2+32x+64=0\)

\(\Leftrightarrow8x^2-12x^2+4x^2-32x+32x=-64-27-32\)

\(\Leftrightarrow0x=-123\) (vô nghiệm)