Đặt \(x^2-4x+5=a\) (\(a\ge1\))

\(\frac{21}{a}-a-1=0\)

\(\Leftrightarrow-a^2-a+21=0\)

Nghiệm xấu, bạn coi lại dề

Đặt \(x^2-4x+5=a\)

\(\frac{5}{a}-a+4=0\)

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

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

\(\left(x-2\right)\left(4x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\frac{5}{4}\end{matrix}\right.\\ \Rightarrow S=\left\{-\frac{5}{4};2\right\}\)

x-2=0 hoặc 4x+5=0

x=2 hoặc x=\(\frac{-5}{4}\)

( x - 2 ) ( 4x + 5 ) = 0

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

➜\(\left[{}\begin{matrix}x=2\\4x=-5\end{matrix}\right.\)

➜\(\left[{}\begin{matrix}x=2\\x=\frac{-5}{4}\end{matrix}\right.\)

Vậy S∈\(\left\{2;\frac{-5}{4}\right\}\)

\(ĐK:\frac{2}{3}\ge x\ge\frac{5}{2}\)

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

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

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

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

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

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

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

\(\Leftrightarrow\dfrac{3\left(x+7\right)}{15}+\dfrac{5\left(4x+5\right)}{15}\ge0\)

\(\Leftrightarrow3\left(x+7\right)+5\left(4x+5\right)\ge0\)

\(\Leftrightarrow23x+46\ge0\)

\(\Leftrightarrow23x\ge-46\)

\(\Leftrightarrow x\ge-2\)

Lời giải:

$\frac{x+7}{5}+\frac{4x+5}{3}\geq 0$

$\Leftrightarrow \frac{x}{5}+\frac{4x}{3}+\frac{7}{5}+\frac{5}{3}\geq 0$

$\Leftrightarrow \frac{23}{15}x+\frac{46}{15}\geq 0$

$\Leftrightarrow 23x+46\geq 0$

$\Leftrightarrow 23x\geq -46$

$\Leftrightarrow x\geq -2$

\(3x^3+6x^2-12x+8=0\)

\(\Leftrightarrow4x^3=x^3-6x^2+12x-8\)

\(\Leftrightarrow4x^3=\left(x-2\right)^3\)

\(\Rightarrow\sqrt[3]{4}.x=x-2\)

\(\Rightarrow x=\dfrac{2}{1-\sqrt[3]{4}}\)

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

Thêm 5 vào hai vế suy ra:

\(\left(x^2-4x+5\right)+\frac{10}{x^2-4x+5}=7\)

Đặt \(t=x^2-4x+5=\left(x^2-4x+4\right)+1=\left(x-2\right)^2+1\ge1\). PT trở thành:

\(t+\frac{10}{t}=7\Leftrightarrow\frac{t^2+10}{t}=7\Leftrightarrow t^2-7t+10=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=5\\t=2\end{matrix}\right.\left(C\right)\). Với t = 5 suy ra \(x^2-4x+5=5\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

Với t = 2 suy ra \(x^2-4x+5=2\Leftrightarrow x^2-4x+3=0\Leftrightarrow x^2-x-3x+3=0\)

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

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

Vậy tập hợp nghiệm của PT là S = (0;1;3;4)