a: =>1+3x-6=-x+3

=>3x-5=-x+3

=>4x=8

=>x=2(loại)

b: \(\Leftrightarrow\dfrac{3\left(x-3\right)+2\left(x-2\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\dfrac{x-1}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)

=>3x-9+2x-4=x-1

=>5x-13=x-1

=>4x=12

=>x=3(loại)

c: =>x^2-2x+4+x^3+8=12

=>x^3+x^2-2x=0

=>x(x^2+x-2)=0

=>x(x+2)(x-1)=0

=>x=0 hoặc x=1

b: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)

\(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=0\)

\(\Leftrightarrow\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24=0\)

\(\Leftrightarrow x^2+7x+6=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)

a) (x+2)³−x+1=(x−1)(x+1)

=>(x+2)(x²-2x+4)-x+1=(x²-1²)

=>x³-2x²+4x+2x²-4x+8-x+1=(x²-1)

=>x³-2x²+4x+2x²-4x+8+1-x²+1=0

=>x³-x²+10=0

=>x³-x²=-10

=>x²(x-1)=-10

=>....

Sửa lại câu c) đặt \(\sqrt{x}+1=\)t \(\Rightarrow\left[2\left(t+\dfrac{1}{2}\right)\right]\left(t-3\right)\)=7⇒\(\left\{{}\begin{matrix}t=3\\t=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=4\\x=\dfrac{9}{4}\end{matrix}\right.\)

a) \(\left(\sqrt{4-3x}\right)^2=8^2\)\(\Leftrightarrow4-3x=64\Rightarrow x=-20\)

b) \(\sqrt{4x-8}+1=12\sqrt{\dfrac{x-2}{9}}\Leftrightarrow2\sqrt{x-2}+1\)\(=\left(12\sqrt{\left(x-2\right).\dfrac{1}{9}}\right)\)

\(\Leftrightarrow2t+1=12.\dfrac{1}{3}t\) (Đặt t = \(\sqrt{x-2}\))

\(\Rightarrow t=\dfrac{1}{2}\) \(\Rightarrow\sqrt{x-2}=\dfrac{1}{2}\)\(\Rightarrow x=\dfrac{9}{4}\)

c) pt\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x}+1=7\\\sqrt{x}-2=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=9\\x=4\end{matrix}\right.\)

a) Ta có: \(\sqrt{4-3x}=8\)

\(\Leftrightarrow4-3x=64\)

\(\Leftrightarrow3x=4-64=-60\)

hay x=-20

b) Ta có: \(\sqrt{4x-8}-12\cdot\sqrt{\dfrac{x-2}{9}}=-1\)

\(\Leftrightarrow2\cdot\sqrt{x-2}-12\cdot\dfrac{\sqrt{x-2}}{3}=-1\)

\(\Leftrightarrow-2\cdot\sqrt{x-2}=-1\)

\(\Leftrightarrow\sqrt{x-2}=\dfrac{1}{2}\)

\(\Leftrightarrow x-2=\dfrac{1}{4}\)

hay \(x=\dfrac{9}{4}\)

b) Ta có: \(x^3+4x+5=0\)

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

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

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

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

mà \(x^2-x+5>0\forall x\)

nên x+1=0

hay x=-1

Vậy: S={-1}

a)x²-(x+3)(3x+1)=9

⇔(x-3)(x+3)-(x+3)(3x+1)=0

⇔x+3=0 hoặc 3x+1=0 

1.x+3=0 ⇔x=-3

2.3x+1=0⇔x=-1/3

phương trình có 2 nghiệm x=-3 và x=-1/3

\(a,\left|2x\right|=x-6\)

\(\Leftrightarrow2x=x-6\)

\(\Leftrightarrow2x-x=-6\)

\(\Leftrightarrow x=-6\)

____________________

\(b,\left|-3x\right|=x-8\)

\(\Leftrightarrow3x=x-8\)

\(\Leftrightarrow3x-x=-8\)

\(\Leftrightarrow2x=-8\)

\(\Leftrightarrow x=-4\)

____________________

\(c,\left|4x\right|=2x+12\)

\(\Leftrightarrow4x=2x+12\)

\(\Leftrightarrow4x-2x=12\)

\(\Leftrightarrow2x=12\)

\(\Leftrightarrow x=6\)

____________________

\(d,\left|-5x\right|-16=3x\)

\(\Leftrightarrow5x-16=3x\)

\(\Leftrightarrow5x-3x=16\)

\(\Leftrightarrow2x=16\)

\(\Leftrightarrow x=8\)

b: Đặt \(x^2+5x+4=a\)

\(\Leftrightarrow a=5\sqrt{a+24}\)

\(\Leftrightarrow a^2=25a+600\)

\(\Leftrightarrow a^2-25a-600=0\)

\(\Leftrightarrow\left(a-40\right)\left(a+15\right)=0\)

\(\Leftrightarrow a=-15\)

hay S=∅

a) \(x^4-x^2+\dfrac{1}{4}-\dfrac{225}{4}=0\\ \left(x^2-\dfrac{1}{2}\right)^2-\dfrac{15}{2}^2=0\\ \left(x+7\right)\left(x-8\right)=0\\ \left[{}\begin{matrix}x=8\\x=-7\end{matrix}\right.\)

Vậy x = 8 hoặc x = -7

a: Ta có: \(x^4-x^2-56=0\)

\(\Leftrightarrow x^4-8x^2+7x^2-56=0\)

\(\Leftrightarrow\left(x^2-8\right)\left(x^2+7\right)=0\)

\(\Leftrightarrow x^2-8=0\)

hay \(x\in\left\{2\sqrt{2};-2\sqrt{2}\right\}\)