Câu hỏi của Đinh Diệp

Question

giải phương trình :

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

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

Sáng · Accepted Answer

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

$\Leftrightarrow x^2-11x+29=41$

$\Leftrightarrow x^2-11x-12=0$

$\Leftrightarrow x^2+x-12x-12=0$

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

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

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

$\Rightarrow\left\{x_1=-1;x_2=12\right\}$Câu trả lời: 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

$\Leftrightarrow x^2-11x+29=41$

$\Leftrightarrow x^2-11x-12=0$

$\Leftrightarrow x^2+x-12x-12=0$

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

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

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

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

Sáng · Answer

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: $t=-14\Leftrightarrow x^2-3x-14=-14$

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

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

$\Rightarrow\left\{x_1=-4;x_2=7;x_3=0;x_4=3\right\}$Câu trả lời: b, $\left(x+2\right)\left(x+3\right)\left(x-5\right)\left(x-6\right)=180$