(x-4)(x-5)(x-6)(x-7)=1680
<=>(x-4)(x-7))(x-5)(x-6)=1680
<=>(x²-11x+28)(x²-11x+30)=1680
<=>(x²-11x+29-1)(x²-11x+29+1)=1680
Đặt x²-11x+29= t ta có phương trình
(t-1)(t+1)=1680
t²-1=1680
t²=1680+1=1681
\(t=\sqrt{1681}\)
\(t=+-41\)

Thay t=x²-11x+29 ta được
x²-11x+29=41=>x²-11x-12=0=>(x-12)(x+1)=0=>x=12;x= -1
hoặc
x²-11x+29=-41=>x²-11x+70=0=>(x²-11x+\(\frac{121}{4}\))+\(\frac{159}{4}\)=0 (loại)
vậy x=12;x=-1
 

(x - 4)(x - 5)(x - 6)(x - 7) = 1680

<=>(x-4)(x-7)(x-5)(x-6)-1680=0

<=>(x²-11x+28)(x²-11x+30)-1680=0

<=>(x²-11x+28)(x²-11x+28+2)-1680=0

<=>(x²-11x+28)²+2(x²-11x+28)+1-1681=0

<=>(x²-11x+29)²-41²=0

<=>(x²-11x+70)(x²-11x-12)=0

<=>x²-11x+70=0 hoặc x²-11x-12=0

Phần còn lại cho bạn mình đi ngủ

<=>(x-7)(x-6)(x-5)(x-4)-1680=(x-12)(x+1)(x^2-11x+70)

TH1:=>x-12=0

=>x=12

TH2:x+1=0

=>x=-1

Ta có : (-11)^2-4(1.70)=-159

=>D<0=> phương trình ko có nghiệm thực

=>x=-1 hoặc 12

nhớ tick nhé

Ta có : (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

<=>(x^2-11x+28)(x^2-11x+28+2)=1680

Đặt t = x^2-11x+28,ta được: t(t+2)=1680

<=> t^2 + 2t =1680

<=>t^2 + 2t -1680 = 0

<=> t^2 + 42t - 40t -1680 = 0

<=> t(t+42)-40(t+42)=0

<=>(t+42)(t-40)=0

<=>  t+42 =0 hoặc t-40=0

<=> t= - 42 hoặc t=40

Vì t=x^2 -11x +28 nên :

x^2 -11x +28 =40 (chuyển 40 qua vế trái rồi tách -11 x = x-12x ,phân tích nhân tử rồi giải ,ta được x = -1 và 12 )

x^2-11x + 28 = -42 (PT này vô nghiem,mình bấm máy tính nó ghi như thế )

Vậy : S={-1;12}

* Hì,mình học toán không giỏi nên không biết bài này đúng hay sai nữa,rảnh không có gì làm nên làm đại thôi =)

Bài b) (x-4)(x-7)(x-6)(x-5)=1680

=> (x²-11x+28)(x²-11x+30)=1680

Đặt t=x²-11x+28

=> t(t+2)=1680

=>t²+2t-1680=0

=> t²+2t+1-1681=0

=> (t+1)²-41²=0

=> (t-40)(t+42)=0

=> t=40 hoặc t=-42

Bạn thế vào như câu a) để giải nhé !!!

a.X=-3

b.X=-1

a.X=-3

b.x=-1

x³ + 2x² + 2x +1 = 0

(=) x³ + x² +x² + x + x + 1 = 0

(=) x².(x+1) + x.(x+1) + (x+1) = 0

(=) (x² + x +1 ).(x+1) = 0

(=) \(\orbr{\begin{cases}x+1=0\\x^2+x+1=0\left(lo\text{ại}\right)\end{cases}}\)(=) x=-1

Vậy phương trình có nghiệm là x=-1

Ta có: \(\left(x-4\right)\left(x-5\right)\left(x-6\right)\left(x-7\right)=1680\)

\(\Leftrightarrow\left(x-4\right)\left(x-7\right)\left(x-5\right)\left(x-6\right)=1680\)

\(\Leftrightarrow\left(x^2-11x+28\right)\left(x^2-11x+30\right)=1680\)

Gọi: \(x^2-11x+29=a\)

\(\Rightarrow\left(a-1\right)\left(a+1\right)=1680\)

\(\Leftrightarrow a^2-1=1680\)

\(\Leftrightarrow a^2=1681\)

\(\Leftrightarrow a=\pm41\)

* Nếu \(a=-41\)

\(\Leftrightarrow x^2-11x+29=-41\)

\(\Leftrightarrow x^2-11x+70=0\)

\(\Leftrightarrow x^2-2.\dfrac{11}{2}x+\dfrac{121}{4}-\dfrac{121}{4}+70=0\)

\(\Leftrightarrow\left(x-\dfrac{11}{2}\right)^2+\dfrac{159}{4}=0\) ( vô nghiệm )

*Nếu \(a=41\)

\(\Leftrightarrow x^2-11x+29=41\)

\(\Leftrightarrow x^2-11x-12=0\)

\(\Leftrightarrow x^2+x-12x-12=0\)

\(\Leftrightarrow x\left(x+1\right)-12\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=12\end{matrix}\right.\)

Vây: Tập nghiệm của phương trình là: \(S=\left\{-1;12\right\}\)

_Chúc bạn học tốt_

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: \(t=-14\Leftrightarrow x^2-3x-14=-14\)

\(\Leftrightarrow x^2-3x=0\Leftrightarrow x\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

\(\Rightarrow\left\{x_1=-4;x_2=7;x_3=0;x_4=3\right\}\)

a/ \(x^3+1+2x^2+2x=0\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2+x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x^2+x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\\\left(x+\frac{1}{2}\right)^2+\frac{3}{4}=0\left(vn\right)\end{matrix}\right.\)

b/ \(\left(x-4\right)\left(x-7\right)\left(x-5\right)\left(x-6\right)-1680=0\)

\(\Leftrightarrow\left(x^2-11x+28\right)\left(x^2-11x+30\right)-1680=0\)

Đặt \(x^2-11x+28=a\Rightarrow x^2-11x+30=a+2\)

Pt trở thành:

\(a\left(a+2\right)-1680=0\Leftrightarrow a^2-2a-1680=0\) \(\Rightarrow\left[{}\begin{matrix}a=42\\a=-40\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2-11x+28=42\\x^2-11x+28=-40\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-11x-14=0\\x^2-11x+68=0\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{11+\sqrt{177}}{2}\\x=\frac{11-\sqrt{177}}{2}\end{matrix}\right.\)

A\(=\left(x-4\right)\left(x-7\right)\left(x-5\right)\left(x-6\right)-1680\)

\(=\left(x^2-11x+28\right)\left(x^2-11x+30\right)-1680\)

Đặt \(\left(x^2-11x+28\right)=t\)

A\(=t\left(t+2\right)-1680=\left(t+1\right)^2-41^2=\left(t-40\right)\left(t+42\right)\)

Thay \(\left(x^2-11x+28\right)=t\)

A\(=\left(x^2-11x-12\right)\left(x^2-11x+70\right)=\left(x-12\right)\left(x+1\right)\left(x^2-11x+70\right)\)