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\(\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)=24x^2\) (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 \(\left[{}\begin{matrix}x=15\\x=2\end{matrix}\right.\)
Bạn ơi mik ra \(\dfrac{x^3+45x-54}{12\left(x-3\right)\left(x+3\right)}\) có đúng không bạn?
Mình rút chx hết bạn bạn gửi cách làm bạn qua mình tham khảo đc k ạ?
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:
\((a+1)^4+(a-1)^4=272\)
\(\Leftrightarrow a^4+4a^3+6a^2+4a+1+a^4-4a^3+6a^2-4a+1=272\)
\(\Leftrightarrow 2a^4+12a^2+2=272\)
\(\Leftrightarrow a^4+6a^2-135=0\)
\(\Leftrightarrow (a^2+3)^2-144=0\Leftrightarrow (a^2+3)^2-12^2=0\)
\(\Leftrightarrow (a^2+15)(a^2-9)=0\)
\(\Rightarrow a^2-9=0\Rightarrow a=\pm 3\)
\(\Rightarrow x=a+7=\left[\begin{matrix} 4\\ 10\end{matrix}\right.\)
(x-3)(x-5)(x-6)(x-10)-24x2
=(x-3)(x-10)(x-5)(x-6)-24x2
=(x2-13x+30)(x2-11x+30)-24x2
Đặt x2-12x+30=k
Khi đó ta có:
(k-x)(k+x)-24x2=k2-x2-24x2=k2-25x2
=(k-5x)(k+5x)
=(x2-12x+30-5x)(x2-12x+30+5x)
=(x2-17x+30)(x2-7x+30)
=(x2-2x-15x+30)(x2-7x+30)
=(x-2)(x-15)(x2-7x+30)
a) \(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{8}+\frac{2x-1}{12}\)
<=> \(\frac{x}{4}+\frac{5}{4}-\frac{2x}{3}+1=\frac{6x}{8}-\frac{1}{8}+\frac{2x}{12}-\frac{1}{12}\)
<=> \(-\frac{4}{3}x=-\frac{59}{24}\)
<=> \(x=\frac{59}{32}\)
Vậy S = { 59/32}
b) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
<=> \(\frac{x^2+14x+40}{12}-\frac{-x^2-2x+8}{4}=\frac{x^2+8x-20}{3}\)
<=> \(\left(\frac{x^2}{12}+\frac{x^2}{4}-\frac{x^2}{3}\right)+\left(\frac{14}{12}x+\frac{2}{4}x-\frac{8}{3}x\right)=-\frac{20}{8}+\frac{8}{4}-\frac{40}{12}\)
<=> \(-x=-8\)
<=> x = 8
Vậy S = { 8 }
\(\Leftrightarrow10\left(x^2+\dfrac{1}{x^2}+2\right)+5\left(x^2+\dfrac{1}{x^2}\right)^2-5\left(x^2+\dfrac{1}{x^2}\right)\left(x^2+\dfrac{1}{x^2}+2\right)=\left(x-5\right)^2-5\)
\(\Leftrightarrow10\left(x^2+\dfrac{1}{x^2}\right)+20+5\left(x^2+\dfrac{1}{x^2}\right)^2-5\left(x^2+\dfrac{1}{x^2}\right)^2-10\left(x^2+\dfrac{1}{x^2}\right)=\left(x-5\right)^2-5\)
\(\Leftrightarrow\left(x-5\right)^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)
a/ Đặt (x^2 - 5x) = a thì ta có
a^2 + 10a + 24 = 0
<=> (a + 4)(a + 6) = 0
Làm nốt
b/ (x - 4)(x - 5)(x - 6)(x - 7) = 1680
<=> (x - 4)(x - 7)(x - 5)(x - 6) = 1680
<=> (x^2 - 11x + 28)(x^2 - 11x + 30) = 1680
Đặt x^2 - 11x + 28 = a thì ta có
a(a + 2) = 1680
<=> (a - 40)(a + 42) = 0
Làm nốt
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 }