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

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

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

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

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

có: \(x^2-7x+30\ne0\), nên:

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

\(\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)=24x^2\) (1)

\(\Leftrightarrow\left(x^2-5x-3x+15\right)\left(x-6\right)\left(x-10\right)=24x^2\)

\(\Leftrightarrow\left(x^2-8x+15\right)\left(x-6\right)\left(x-10\right)=24x^2\)

\(\Leftrightarrow\left(x^3-6x^2-8x^2+48x+15x-90\right)\left(x-10\right)=24x^2\)

\(\Leftrightarrow\left(x^3-14x^2+63x-90\right)\left(x-10\right)=24x^2\)

\(\Leftrightarrow x^4-10x^3-14x^3+140x^2+63x^2-630x-90x+900=24x^2\)

\(\Leftrightarrow x^4-2x^3-22x^3+44x^2+135x^2-270x-450x+900=0\)

\(\Leftrightarrow x^3\left(x-2\right)-22x^2\left(x-2\right)+135x\left(x-2\right)-450\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3-22x^2+135x-450\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3-15x^2-7x^2+105x+30x-450\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-15=0\\x^2-7x+30=0\end{matrix}\right.\)

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

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

Vậy tập nghiệm phương trình (1) là \(S=\left\{2;15\right\}\)

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

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

Nhận thấy x=0 không là nghiệm của PT. Chia cả hai vế của phương trình cho \(x^2\) ta được:

PT\(\Leftrightarrow\)\(\left(x-13+\dfrac{30}{x}\right)\left(x-11+\dfrac{30}{x}\right)=24\)

Đặt \(x+\dfrac{30}{x}=t\) (1)

PT\(\Leftrightarrow\)\(\left(t-13\right)\left(t-11\right)=24\)

Tìm được \(\left[{}\begin{matrix}t=17\\t=7\end{matrix}\right.\)

Thay vào (1):\(\left[{}\begin{matrix}x^2-17x+30=0\\x^2-7x+30=0\end{matrix}\right.\)

Tìm được \(\left[{}\begin{matrix}x=15\\x=2\end{matrix}\right.\)

\(9.\left(x+5\right).\left(x+6\right).\left(x+7\right)=24.x\)

\(\Leftrightarrow\left(9.x+45\right).\left(x+6\right).\left(x+7\right)=24.x\)

\(\Leftrightarrow\left(9.x^2+54.x+45.x+270\right).\left(x+7\right)=24.x\)

\(\Leftrightarrow\left(9.x^2+99.x+270\right).\left(x+7\right)=24.x\)

\(\Leftrightarrow9.x^3+63.x^2+99.x^2+693.x+270.x+1890=24.x\)

\(\Leftrightarrow9.x^3+162.x^2+963.x+1890=24.x\)

\(\Leftrightarrow9.x^3+162.x^2+963.x+1890-24.x=0\)

\(\Leftrightarrow9.x^3+162.x^2+939.x+1890=0\)

\(\Leftrightarrow3.\left(3.x^3+54.x^2+313+630\right)=0\)

\(\Leftrightarrow3.\left(3.x^3+27.x^2+27.x^2+243.x+70.x+630\right)=0\)

\(\Leftrightarrow3.\left(3.x^2.\left(x+9\right)+27.x.\left(x+9\right)+70.\left(x+9\right)\right)=0\)

\(\Leftrightarrow3.\left(x+9\right).\left(3.x^2+27.x+70\right)=0\)

\(\Leftrightarrow\left(x+9\right).\left(3.x^2+27.x+70\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+9=0\\3.x^2+27.x+70=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=-9\\x\notinℝ\end{cases}}\)

Vậy x = -9

\(9\left(x+5\right)\left(x+6\right)\left(x+7\right)=24x\)

\(\Leftrightarrow3\left(x+5\right)\left(x+6\right)\left(x+7\right)=8x\)

\(\Leftrightarrow3x^3+54x^2+321x+630=8x\)

\(\Leftrightarrow3x^3+54x^2+313x+630=0\)

\(\Leftrightarrow\left(x+9\right)\left(3x^2+27x+70\right)=0\)

\(\Leftrightarrow x+9=0\)

\(\Leftrightarrow x=9\)

Mà: \(3x^2+27x+70=3\left(x+\frac{9}{2}\right)^2+\frac{37}{4}>0\)

Vậy ..............

Đặt pt là (1)

Ta có :

(1) <=> \(\left[\left(x-3\right)\left(x-10\right)\right]\left[\left(x-5\right)\left(x-6\right)\right]-24x^2=0\)

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

Đặt \(x^2-12x+30=t\) (*)

Phương trình trở thành \(\left(t-x\right)\left(t+x\right)-24x^2=0\)

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

\(\Leftrightarrow t^2-25x^2=0\)

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

Thay (*) vào ta có :

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

Để ý thấy \(x^2-7x+30\ne0\)

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

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

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

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

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

Vậy S={1 ; 15 }

a)

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

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

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

Đặt \(x^2-11x+30=a\). PT trở thành:
\((a-2x)a=24x^2\)

\(\Leftrightarrow a^2-2ax-24x^2=0\)

\(\Leftrightarrow a^2-6ax+4ax-24x^2=0\)

\(\Leftrightarrow a(a-6x)+4x(a-6x)=0\)

\(\Leftrightarrow (a+4x)(a-6x)=0\)

\(\Rightarrow \left[\begin{matrix} a+4x=0\\ a-6x=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x^2-7x+30=0\\ x^2-17x+30=0\end{matrix}\right.\)

\(\Rightarrow \left[\begin{matrix} (x-3,5)^2+17,75=0(\text{vô lý})\\ (x-15)(x-2)=0\end{matrix}\right.\)

\(\Rightarrow x=15\) hoặc $x=2$

b)

Đặt \(x-7=a\). PT trở thành:

\((a+1)^4+(a-1)^4=272\)

\(\Leftrightarrow a^4+4a^3+6a^2+4a+1+a^4-4a^3+6a^2-4a+1=272\)

\(\Leftrightarrow 2a^4+12a^2+2=272\)

\(\Leftrightarrow a^4+6a^2-135=0\)

\(\Leftrightarrow (a^2+3)^2-144=0\Leftrightarrow (a^2+3)^2-12^2=0\)

\(\Leftrightarrow (a^2+15)(a^2-9)=0\)

\(\Rightarrow a^2-9=0\Rightarrow a=\pm 3\)

\(\Rightarrow x=a+7=\left[\begin{matrix} 4\\ 10\end{matrix}\right.\)

c)

\(x^4-3x^3+2x^2-9x+9=0\)

Ta để ý tổng các hệ số bằng $0$ nên có một nghiệm bằng $1$

Vậy ta thực hiện tách hợp lý:

\(\Leftrightarrow (x^4-x^3)-(2x^3-2x^2)-(9x-9)=0\)

\(\Leftrightarrow x^3(x-1)-2x^2(x-1)-9(x-1)=0\)

\(\Leftrightarrow (x-1)(x^3-2x^2-9)=0\)

\(\Leftrightarrow (x-1)[(x^3-3x^2)+x^2-9]=0\)

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

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

Dễ thấy \(x^2+x+3=(x+\frac{1}{2})^2+\frac{11}{4}\geq 0+\frac{11}{4}>0\) với mọi $x$

Do đó: \((x-1)(x-3)=0\Rightarrow \left[\begin{matrix} x=1\\ x=3\end{matrix}\right.\)

a) \(8 - \left( {x - 15} \right) = 2.\left( {3 - 2x} \right)\) 

\(8 - x + 15 = 6 - 4x\)

\( - x + 4x = 6 - 8 - 15\)

\(3x =  - 17\)

\(x = \left( { - 17} \right):3\)

\(x = \dfrac{{ - 17}}{3}\)

Vậy nghiệm của phương trình là \(x = \dfrac{{ - 17}}{3}\).

b) \( - 6\left( {1,5 - 2u} \right) = 3\left( { - 15 + 2u} \right)\)

\( - 9 + 12u =  - 45 + 6u\)

\(12u - 6u =  - 45 + 9\)

\(u = \left( { - 36} \right):6\)

\(6u =  - 36\)

\(u =  - 6\)

Vậy nghiệm của phương trình là \(u =  - 6\).

c) \({\left( {x + 3} \right)^2} - x\left( {x + 4} \right) = 13\)

\(\left( {{x^2} + 6x + 9} \right) - \left( {{x^2} + 4x} \right) = 13\)

\({x^2} + 6x + 9 - {x^2} - 4x = 13\)

\(\left( {{x^2} - {x^2}} \right) + \left( {6x - 4x} \right) = 13 - 9\)

\(2x = 4\)

\(x = 4:2\)

\(x = 2\)

Vậy nghiệm của phương trình là \(x = 2\).

d) \(\left( {y + 5} \right)\left( {y - 5} \right) - {\left( {y - 2} \right)^2} = 5\)

\(\left( {{y^2} - 25} \right) - \left( {{y^2} - 4y + 4} \right) = 5\)

\({y^2} - 25 - {y^2} + 4y - 4 = 5\)

\(\left( {{y^2} - {y^2}} \right) + 4y = 5 + 4 + 25\)

\(4y = 34\)

\(y = 34:4\)

\(y = \dfrac{{17}}{2}\)

Vậy nghiệm của phương trình là \(y = \dfrac{{17}}{2}\).