\(ĐKXĐ:x\ne-1;x\ne-\frac{1}{2}\)

\(PT:\Leftrightarrow\frac{x^2-4x+1}{x+1}+1+\frac{x^2-5x+1}{2x+1}=0\)

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

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

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

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

\(x-1=0\Leftrightarrow x=1\)

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

\(3x+2=0\Leftrightarrow3x=-2\Leftrightarrow x=-\frac{2}{3}\)

\(\Rightarrow\hept{\begin{cases}x=1\\x=2\\x=-\frac{2}{3}\end{cases}}\)

\(\frac{x^2-4x+1}{x+1}+2=-\frac{x^2-5x+1}{2x+1}\)

\(\Leftrightarrow\left(x^2-4x+1\right)\left(x+1\right)+2\left(x+1\right)\left(2x+1\right)=-\left(x^2-5x+1\right)\left(x+1\right)\)

\(\Leftrightarrow2x^3-3x^2+4x+3=-x^3+4x^2+4x-1\)

\(\Leftrightarrow2x^3-3x^2+3+x^2-4x+1=0\)

\(\Leftrightarrow3x^2-7x^2+4=0\)

\(\Leftrightarrow\left(3x^2-4x-4\right)\left(x-1\right)=0\)

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

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

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

\(\Leftrightarrow\hept{\begin{cases}3x+2=0\\x-2=0\\x-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{2}{3}\\x=2\\x=1\end{cases}}\)

vậy:...

\(\frac{4x-1}{3}-\frac{2-x}{15}\le\frac{10x-3}{5}\)

\(\Rightarrow\frac{5\left(4x-1\right)}{15}-\frac{2-x}{15}-\frac{3\left(10x-3\right)}{15}\le0\)

\(\Rightarrow\frac{20x-5-2+x-30x+9}{15}\le0\)

\(\Rightarrow-9x+2\le0\)

\(\Rightarrow9x-2\ge0\)

\(\Rightarrow9x\ge2\)

\(\Rightarrow x\ge\frac{2}{9}\)

Lần này thì đúng rồi :

\(\frac{2x}{3}+\frac{2x-1}{6}=4-\frac{x}{3}\)

\(\Leftrightarrow\frac{2x}{3}+\frac{x}{3}-\frac{1}{6}=4-\frac{x}{3}\)

\(\Leftrightarrow\frac{2x}{3}+\frac{x}{3}+\frac{x}{3}=4+\frac{1}{6}\)

\(\Leftrightarrow\frac{4x}{3}=\frac{25}{6}\)

\(\Leftrightarrow x=\frac{25}{8}\)

Ta có:

\(\frac{x^2-5x+1}{2x+1}+2=\frac{x^2-5x+4x+1+2}{2x+1}\)

=\(\frac{x^2-x+3}{2x+1}=\frac{x^2-4x+1}{x+1}\)

=> (x² - x +3)(x+1)=(x² - 4x+1)(2x+1)

=>x³ +2x+3=2x³-7x²-2x+1

=>0=x³-7x²-4x-2

Đây là cách làm của mình :

\(\Leftrightarrow\frac{x^2-5x+1}{2x+1}+1+1=\frac{x^2-4x+1}{x+1}\) 

\(\Leftrightarrow\frac{x^2-5x+1}{2x+1}+1=\frac{x^2-4x+1}{x+1}-1\)

\(\Leftrightarrow\frac{x^2-3x+2}{2x+1}=\frac{x^2-5x}{x+1}\)

\(\Leftrightarrow\frac{\left(x-2\right)\left(x-1\right)}{2x+1}=\frac{x^2-5x}{x+1}\)

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

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

Bạn tự nhân phân phối vào nha :

\(\Leftrightarrow x^3-2x^2-x+2=2x^3-9x^2-5x\)

\(\Leftrightarrow x^3-7x^2-4x-2=0\)

Đến đây chỉ có nước bấm máy tính thôi chứ phân tích bình thường không ra được đâu

CASIO fx-570VN PLUS : Mode --> 5 --> 4 : giải pt bậc 3 một ẩn

Kết quả cho là x = 7.563793497...

Ta có  \(\frac{x+2}{13}+\frac{2x+45}{15}=\frac{3x+8}{37}+\frac{4x+69}{9}\)

\(\Leftrightarrow\left(\frac{x+2}{13}+1\right)+\left(\frac{2x+45}{15}-1\right)=\left(\frac{3x+8}{37}+1\right)+\left(\frac{4x+69}{9}-1\right)\)

\(\Leftrightarrow\frac{x+15}{13}+\frac{2\left(x+15\right)}{15}=\frac{3\left(x+15\right)}{37}+\frac{4\left(x+15\right)}{9}\)

\(\Leftrightarrow\left(x+15\right)\left(\frac{1}{13}+\frac{2}{15}-\frac{3}{37}-\frac{4}{9}\right)=0\Leftrightarrow x+15=0\)vì \(\left(\frac{1}{13}+\frac{2}{15}-\frac{3}{37}-\frac{4}{9}\right)\ne0\)

\(\Leftrightarrow x=-15\)

Vậy \(x=-15\)

giải pt: (x-20)+(x-19)+......+100+101=101

ĐK: \(\hept{\begin{cases}1-\frac{2}{x}\ge0\\2x-\frac{8}{x}\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{x-2}{x}\ge0\\\frac{2x^2-8}{x}\ge0\end{cases}}\)

<=> \(-2\le x< 0\) hoặc  \(x\ge2\)

TH1:  \(-2\le x< 0\)

Bất phương trình đúng

TH2: \(x\ge2\)(@@)

bất pt <=> \(2\sqrt{\frac{x-2}{x}}+\sqrt{\frac{2\left(x-2\right)\left(x+2\right)}{x}}\ge x\)

<=> \(\sqrt{\frac{x-2}{x}}\left(2+\sqrt{2\left(x+2\right)}\right)\ge x\)

<=> \(\sqrt{\frac{x-2}{x}}\left(\frac{2x}{\sqrt{2\left(x+2\right)}-2}\right)\ge x\)

<=> \(2\sqrt{\frac{x-2}{x}}+2\ge\sqrt{2\left(x+2\right)}\)

<=> \(4\left(1-\frac{2}{x}\right)+4+8\sqrt{1-\frac{2}{x}}\ge2x+4\)

<=> \(4\sqrt{1-\frac{2}{x}}\ge x-2+\frac{4}{x}\)

<=> \(16\left(1-\frac{2}{x}\right)\ge x^2+4+\frac{16}{x^2}-4x+8-\frac{16}{x}\)

<=> \(4\ge x^2+\frac{16}{x^2}-4x+\frac{16}{x}\)

<=> \(\left(x-\frac{4}{x}\right)^2-4\left(x-\frac{4}{x}\right)+4\le0\)

<=> \(\left(x-\frac{4}{x}+2\right)^2\le0\) vô nghiệm vì x > 2 => \(x-\frac{4}{x}+2>2\)

Vậy -2 \(\le\) x < 0