\(\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)=24x^2\)

\(\Leftrightarrow x^4-24x^3+203x^2-720x+900=24x^4\)

\(\Leftrightarrow x^4-24x^3+203x^2-720x+900-24x^2=0\)

\(\Leftrightarrow x^4-24x^3+179x^3-720x+900=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-15\right)\left(x^2-7x+30\right)=0\)

có: \(x^2-7x+30\ne0\), nên:

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-15=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=15\end{cases}}\)

\(\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)=24x^2\)

\(\Leftrightarrow\left[\left(x-5\right)\left(x-6\right)\right]\cdot\left[\left(x-3\right)\left(x-10\right)\right]=24x^2\)

\(\Leftrightarrow\left(x^2-11x+30\right)\left(x^2-13x+30\right)-24x^2=0\)

Đặt: \(x^2-13x+30=t\)

Lúc này PT trở thành:

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

\(\Leftrightarrow t^2+2tx-24x^2=0\)

\(\Leftrightarrow t^2+6tx-4tx-24x^2=0\)

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

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

\(\Leftrightarrow\left(x^2-7x+30\right)\left(x^2-17x+30\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-7x+30=0\\x^2-17x+30=0\end{matrix}\right.\)

Ta có: \(x^2-7x+30=\left(x-\dfrac{7}{2}\right)^2+\dfrac{71}{4}>0\)(vô nghiệm)

=> \(x^2-17x+30=0\)

\(\Leftrightarrow\) \(\left(x-15\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-15=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=2\end{matrix}\right.\)

Vậy x = 2 hoặc x = 15

Đặt: \(\sqrt{x+9}=v;\sqrt{x+6}=u\)

Ta có: \(v+5u=5+vu\)

\(\Leftrightarrow v+5u-5-uv=0\)

\(\Leftrightarrow-v\left(u-1\right)+5\left(u-1\right)\)

\(\Leftrightarrow\left(5-v\right)\left(u-1\right)\)

\(\left\{{}\begin{matrix}5-v=0\Leftrightarrow5=\sqrt{x+9}\Leftrightarrow x=16\left(N\right)\\u-1=0\Leftrightarrow\sqrt{x+6}=1\Leftrightarrow x=-5\left(L\right)\end{matrix}\right.\)          ĐKXĐ:\(x>=-6\)

\(S=\left\{16\right\}\)

Đặt:\(\sqrt{x+9}=v;\sqrt{x+6}=u\)

Ta có: \(v+5u=5+vu\Leftrightarrow-v\left(u-1\right)+5\left(u-1\right)\Leftrightarrow\left(5-v\right)\left(u-1\right)\)

\(\left\{{}\begin{matrix}5-v=0\Leftrightarrow5=\sqrt{x+9}\Leftrightarrow x=16\left(N\right)\\u-1=0\Leftrightarrow\sqrt{x+6}=1\Leftrightarrow x=-5\left(N\right)\end{matrix}\right.ĐKXĐ:x>=-6\)

\(S=\left\{16,-5\right\}\)

Câu trên mình quên -5>-6

\(\left\{{}\begin{matrix}6\left(x+y\right)=8+2x-3y\\5\left(y-x\right)=5+3x+2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+6y=8+2x-3y\\5y-5x=5+3x+2y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}6x-2x+6y+3y=8\\-5x-3x+5y-2y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\-8x+3y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\-24x+9y=15\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}28x=-7\\4x+9y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{7}{28}=-\dfrac{1}{4}\\4.\left(-\dfrac{1}{4}\right)+9y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{4}\\y=1\end{matrix}\right.\\ Vậy:\left(x;y\right)=\left(-\dfrac{1}{4};1\right)\)

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

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