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a)<=>(x-4)(x-7)(x-5)(x-6)=1680
<=>(x2-11x+28)(x2-11x+30)=1680
đặt a=x2-11x+28 khi đó ptr trở thành :
a(a+2)=1680
=>a2+2a=1680
=>a2+2a+1=1681
=>(a+1)2=1681
=>a+1=41 hoặc a+1=-41
=>a=40 hoặc a=-42
=>x2-11x+28=40 hoặc -42
TH1:x2-11x+28=40
=>x2-11x+121/4-9/4=40
=>(x-11/2)2-9/4=40
=>(x-11/2)2=169/4
đến đây tự làm tiếp nhé
câu b thì nhóm x+2 với x-5 và x+3 với x-6 ,nhân vào phá ngoặc và đặt (như câu a) thôi
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+2\right)\left(x+3\right)\left(x-5\right)\left(x-6\right)=180\)
\(\left[\left(x+2\right)\left(x-5\right)\right]\left[\left(x+3\right)\left(x-6\right)\right]=180\)
\(\left(x^2-3x-10\right)\left(x^2-3x-18\right)=180\)
Đặt \(x^2-3x-10=a\) ta có
\(a\left(a-8\right)=180\)
\(a^2-8a-180=0\)
\(\left(a-18\right)\left(a+10\right)=0=>\orbr{\begin{cases}a=18\\a=-10\end{cases}}\)
=> \(\orbr{\begin{cases}x^2-3x-10=18\\x^2-3x-10=-10\end{cases}}\) => \(\orbr{\begin{cases}x^2-3x-28=0\\x^2-3x=0\end{cases}}\)
Đến đây bn tự giải tiếp nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
[(x+2)(x-5)][(x+3)(x-6)]=180
(x2-3x-10)(x2-3x-18)=180
Đặt x2-3x-10=y
y(y-8)=180
y2-8y-180=0
y2+10y-18y-180=0
y(y+10)-18(y+10)=0
(y+10)(y-18)=0
y={-10;18}
TH1: x2-3x-10= -10
x(x-3)=0
x={0;3}
TH2: x2-3x-10=18
x2-3x-28=0
x2+4x-7x-28=0
x(x+4)-7(x+4)=0
(x-7)(x+4)=0
x={-4;7}
Vậy x={-4;0;3;7}
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x^2+5x\right)+10\left(x^2-5x\right)+24=0\)
\(\Leftrightarrow\left(x^2+5x\right)-10\left(x^2+5x\right)+24=0\)
\(\Leftrightarrow\left(x^2+5x\right)\left(1-10\right)+14=0\)
\(\Leftrightarrow\left(-9\right)\left(x^2+5x\right)+14=0\)
\(\Leftrightarrow-9\left(x^2+5x\right)=-14\)
\(\Leftrightarrow x^2+5x=\frac{14}{9}\)
\(\Leftrightarrow x=0,2938.....\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu trên làm (a) câu này làm (b)
b)
\(\left(x^2+x-2\right)\left(x^2+x-3\right)=12\)
đặt: \(x^2+x-2=\left(x+\frac{1}{2}\right)^2-\frac{9}{4}=t\)
\(t\left(t-1\right)=12\Leftrightarrow t^2-t+\frac{1}{4}=12+\frac{1}{4}=\frac{49}{4}\)
\(\left(t-\frac{1}{2}\right)^2=\left(\frac{7}{2}\right)^2\Rightarrow\left[\begin{matrix}t=\frac{1-7}{2}=-3\left(loai\right)\\t=\frac{1+7}{2}=4\end{matrix}\right.\)
\(t=4\Leftrightarrow\left(x+\frac{1}{2}\right)^2=4+\frac{9}{4}=\frac{25}{4}\Rightarrow\left[\begin{matrix}x=\frac{-1-5}{2}=-3\\x=\frac{-1+5}{2}=2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\text{a) }x\left(x+1\right)\left(x-1\right)\left(x+2\right)=24\\ \Leftrightarrow\left(x^2+x\right)\left(x^2-x+2x-2\right)=24\\ \Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)
Đặt \(x^2+x-1=t\)
\(\Leftrightarrow\left(t+1\right)\left(t-1\right)=24\\ \Leftrightarrow t^2-1-24=0\\ \Leftrightarrow t^2-25=0\\ \Leftrightarrow\left(t+5\right)\left(t-5\right)=0\\ \Leftrightarrow\left(x^2+x-1+5\right)\left(x^2+x-1-5\right)=0\\ \Leftrightarrow\left(x^2+x+4\right)\left(x^2+x-6\right)=0\\ \Leftrightarrow\left(x^2+x+\dfrac{1}{4}+\dfrac{15}{4}\right)\left(x^2+3x-2x-6\right)=0\\ \Leftrightarrow\left[\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{15}{4}\right]\left[\left(x^2+3x\right)-\left(2x+6\right)\right]=0\\ \Leftrightarrow\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}\right]\left[x\left(x+3\right)-2\left(x+3\right)\right]=0\\ \Leftrightarrow\left(x-2\right)\left(x+3\right)=0\left(\text{Vì }\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}\ne0\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy tập nghiệm phương trình là \(S=\left\{2;-3\right\}\)
\(\text{b) }\left(x-4\right)\left(x-5\right)\left(x-6\right)\left(x-7\right)=1680\\ \Leftrightarrow\left(x^2-4x-7x+28\right)\left(x^2-5x-6x+30\right)=1680\\ \Leftrightarrow\left(x^2-11x+28\right)\left(x^2-11x+30\right)=1680\)
Đặt \(x^2-11x+29=t\)
\(\Leftrightarrow\left(t-1\right)\left(t+1\right)=1680\\ \Leftrightarrow t^2-1-1680=0\\ \Leftrightarrow t^2-1681=0\\ \Leftrightarrow\left(t+41\right)\left(t-41\right)=0\\ \Leftrightarrow\left(x^2-11x+29+41\right)\left(x^2-11x+29-41\right)=0\\ \Leftrightarrow\left(x^2-11x+70\right)\left(x^2-11x-12\right)=0\\ \Leftrightarrow\left(x^2-11x+\dfrac{121}{4}+\dfrac{159}{4}\right)\left(x^2-12x+x-12\right)=0\\ \Leftrightarrow\left[\left(x^2-11x+\dfrac{121}{4}\right)+\dfrac{159}{4}\right]\left[\left(x^2-12x\right)+\left(x-12\right)\right]=0\\ \Leftrightarrow\left[\left(x-\dfrac{11}{2}\right)^2+\dfrac{159}{4}\right]\left[x\left(x-12\right)+\left(x-12\right)\right]=0\\ \Leftrightarrow\left(x+1\right)\left(x-12\right)=0\left(\text{Vì }\left(x-\dfrac{11}{2}\right)^2+\dfrac{159}{4}\ne0\right)\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=12\end{matrix}\right.\)
Vậy tập nghiệm phương trình là \(S=\left\{-1;12\right\}\)
\(\text{c) }\left(x+2\right)\left(x+3\right)\left(x-5\right)\left(x-6\right)=180\\ \Leftrightarrow\left(x^2+2x-5x-10\right)\left(x^2+3x-6x-18\right)=180\\ \Leftrightarrow\left(x^2-3x-10\right)\left(x^2-3x-18\right)=180\) Đặt \(x^2-3x-14=t\) \(\Leftrightarrow\left(t+4\right)\left(t-4\right)=180\\ \Leftrightarrow t^2-16-180=0\\ \Leftrightarrow t^2-196=0\\ \Leftrightarrow\left(t+14\right)\left(t-14\right)=0\\ \Leftrightarrow\left(x^2-3x-14+14\right)\left(x^2-3x-14-14\right)=0\\ \Leftrightarrow\left(x^2-3x\right)\left(x^2-3x-28\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x^2-7x+4x-28\right)=0\\ \Leftrightarrow x\left(x-3\right)\left[x\left(x-7\right)+4\left(x-7\right)\right]=0\\ \Leftrightarrow x\left(x-3\right)\left(x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\\x+4=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-4\\x=7\end{matrix}\right.\) Vậy tập nghiệm phương trình là \(S=\left\{0;3;-4;7\right\}\)![](https://rs.olm.vn/images/avt/0.png?1311)
a.
\(6x^2-3x=7x^2\)
\(\Leftrightarrow-x^2-3x=0\)
\(\Leftrightarrow x^2+3x=0\)
\(\Leftrightarrow x\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
a.
\(6x^2-3x=7x^2\)
\(\Leftrightarrow6x^2-7x^2-3x=0\)
\(\Leftrightarrow-x^2-3x=0\)
\(\Leftrightarrow-x\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
vậy,...
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\left(x-4\right)\left(x-5\right)\left(x-6\right)\left(x-7\right)=1680\)
\(\Leftrightarrow\left[\left(x-4\right)\left(x-7\right)\right]\left[\left(x-5\right)\left(x-6\right)\right]=1680\)
\(\Leftrightarrow\left(x^2-11x+28\right)\left(x^2-11x+30\right)=1680\)
Gọi \(k=x^2-11x+29\)
\(\Rightarrow\left(k-1\right)\left(k+1\right)=1680\)
\(\Rightarrow k^2-1=1680\Rightarrow k^2=1681\)
\(\Rightarrow k=\sqrt{1681}=\pm41\)
* TH1: k = -41
\(\Leftrightarrow x^2-11x+29=-41\)
\(\Leftrightarrow x^2-11x+70=0\)
\(\Leftrightarrow x^2-2.\dfrac{11}{2}x+\dfrac{121}{4}-\dfrac{121}{4}+70=0\)
\(\Leftrightarrow\left(x-\dfrac{11}{2}\right)^2+\dfrac{159}{4}=0\Leftrightarrow\left(x-\dfrac{11}{2}\right)^2=\dfrac{-159}{4}\left(vôli\right)\)
Vì \(\left(x-\dfrac{11}{2}\right)^2\ge0\forall x\) mà \(\dfrac{-159}{4}< 0\Rightarrow\left(x-\dfrac{11}{2}\right)^2=\dfrac{-159}{4}\left(loại\right)\)
* TH2: k = 41
\(\Leftrightarrow x^2-11x+29=41\)
\(\Leftrightarrow x^2-11x-12=0\)
\(\Leftrightarrow x^2+x-12x-12=0\)
\(\Leftrightarrow x\left(x+1\right)-12\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-12\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-12=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-1\end{matrix}\right.\)
\(\Rightarrow\left\{x_1=-1;x_2=12\right\}\)
b, \(\left(x+2\right)\left(x+3\right)\left(x-5\right)\left(x-6\right)=180\)
\(\Leftrightarrow\left[\left(x+2\right)\left(x-5\right)\right]\left[\left(x+3\right)\left(x-6\right)\right]=180\)
\(\Leftrightarrow\left(x^2-3x-10\right)\left(x^2-3x-18\right)=180\)
Đặt \(k=x^2-3x-14\)
Ta có pt: \(\left(k-4\right)\left(k+4\right)=180\)
\(\Leftrightarrow k^2-16=180\Leftrightarrow k^2=196\)
\(\Leftrightarrow k=\sqrt{196}=\pm14\)
* TH1: \(t=14\Leftrightarrow x^2-3x-14=14\)
\(\Leftrightarrow x^2-3x-28=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=7\end{matrix}\right.\)
* TH2: \(t=-14\Leftrightarrow x^2-3x-14=-14\)
\(\Leftrightarrow x^2-3x=0\Leftrightarrow x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
\(\Rightarrow\left\{x_1=-4;x_2=7;x_3=0;x_4=3\right\}\)
( x + 2)( x + 3)( x - 5)( x - 6) = 180
<=> ( x + 2)( x - 5)( x + 3)( x - 6) = 180
<=> ( x2 - 3x - 10 )( x2 - 3x - 18 ) = 180
Đặt : x2 - 3x - 14 = a , ta có :
( a + 4)( a - 4) = 180
<=> a2 - 16 - 180 = 0
<=> a2 - 196 = 0
<=> ( a - 14)( a + 14 ) = 0
<=> a = 14 hoặc a = -14
* Với , a = 14 , ta có :
x2 - 3x - 14 = 14
<=> x2 - 3x - 28 = 0
<=> x2 - 7x + 4x - 28 = 0
<=> x( x - 7) + 4( x - 7) = 0
<=> ( x + 4)( x - 7) = 0
<=> x = -4 hoặc : x = 7
* Với : a = -14 , ta có :
x2 - 3x - 14 = -14
<=> x( x - 3) = 0
<=> x = 0 hoặc : x = 3
Vậy,...