Question

[Đinh Diệp]

giải phương trình :

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

b)\(\left(x+2\right)\left(x+3\right)\left(x-5\right)\left(x-6\right)=180\)

Answer 1

[Sáng]

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

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-12=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-1\end{matrix}\right.\)

\(\Rightarrow\left\{x_1=-1;x_2=12\right\}\)

Answer 2

[Sáng]

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

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

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

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