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

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

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

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

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

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

\(\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)=24x^2\) (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 \(\left[{}\begin{matrix}x=15\\x=2\end{matrix}\right.\)

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

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:\(x>=-6\)

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

\(\sqrt{x+9}+5\sqrt{x+\text{6}}=5+\sqrt{\left(x+9\right)\left(x+\text{6}\right)}\Leftrightarrow\sqrt{x+9}+5=5+\sqrt{x+9}\Leftrightarrow\sqrt{x+9}-\sqrt{x+9}=0\Leftrightarrow x+9-x-9=0\Leftrightarrow0=0\)

Vậy x vô số nghiệm

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: =>|x-7|=3-2x

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(-2x+3\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(2x-3-x+7\right)\left(2x-3+x-7\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(x+4\right)\left(3x-10\right)=0\end{matrix}\right.\Leftrightarrow x=-4\)

b: =>|2x-3|=4x+9

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(4x+9-2x+3\right)\left(4x+9+2x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(2x+12\right)\left(6x+6\right)=0\end{matrix}\right.\Leftrightarrow x=-1\)

c: =>3x+5=2-5x hoặc 3x+5=5x-2

=>8x=-3 hoặc -2x=-7

=>x=-3/8 hoặc x=7/2

Ta có : |x - 2| + |x - 3| + |x - 4| + |x - 5| + |x - 6| -x + 7 = 0

=> |x - 2| + |x - 3| + |x - 4| + |x - 5| + |x - 6| = x - 7

ĐK \(x-7\ge0\Rightarrow x\ge7\)

Khi đó ta có x - 2 > 0 ; x - 3 > 0 ; ... x - 6 > 0

=> |x - 2| + |x - 3| + |x - 4| + |x - 5| + |x - 6| = x - 7

<=> x - 2 + x - 3 + x - 4 + x - 5 + x - 6 = x - 7

=> 5x - 20 = x - 7

=> 4x = 13

=> x = 4,25 (loại)

Vậy x \(\in\varnothing\)