Bài 4: Phương trình tích

\(a,x^4-16x^2+32x-16=0\)

\(\Leftrightarrow\left(x^4-16\right)-16x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x^4+4\right)\left(x-2\right)\left(x+2\right)-16x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3+2x^2-12x+8\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3-2x^2+4x^2-8x-4x+8\right)=0\)\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x-2\right)+4x\left(x-2\right)-4\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-2\right)\left(x^2+4x-4\right)=0\)

\(\Leftrightarrow\left(x-2\right)^2\left[\left(x+2\right)^2-8\right]=0\Rightarrow\left[{}\begin{matrix}\left(x-2\right)^2=0\\\left(x+2\right)^2-8=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-2=0\\\left(x+2\right)^2=8\Rightarrow\left[{}\begin{matrix}x+2=\sqrt{8}\\x+2=-\sqrt{8}\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{8}-2\\x=-\sqrt{8}-2\end{matrix}\right.\)

câu nào dễ xơi trước

g) \(x^3+3x^2-2x-6=0\Leftrightarrow x^2\left(x+3\right)-2\left(x+3\right)=0\)

\(\Leftrightarrow\left(x^2-2\right)\left(x+3\right)=0\Leftrightarrow\left\{{}\begin{matrix}x=\pm\sqrt{2}\\x=-3\end{matrix}\right.\)

kl: ...........

Ta có: \(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x+4\right)^2\)

Ta có : \(x^2-4-5\left(x-2\right)^2=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)-5\left(x-2\right)^2=0\)

\(\Leftrightarrow\left(x-2\right)\left[x+2-5\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2-5x+10\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(12-4x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\12-4x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\-4x=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

Vậy tập nghiệm của phương trình là \(S=\left\{2;3\right\}\)

Do mẫu số lớn hơn 0

nên khi quy đồng bất phương trình không đổi chiều

hay nói dễ hiểu là bạn dùng dấu tương đương được

a ) ( x + 1 ) ( x + 2 ) ( x + 3 ) = 0

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\\x=-3\end{matrix}\right.\)

Vậy tập nghiệm của phương trình là \(S=\left\{-1;-2;-3\right\}\)

b ) \(\left(x-1\right)^2-16=0\)

\(\Leftrightarrow\left(x-1\right)^2-4^2=0\)

\(\Leftrightarrow\left(x-1+4\right)\left(x-1-4\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)

Vậy tập nghiệm của phương trình là S = {-3;5}

c ) \(\left(2x-1\right)^2-\left(x+3\right)^2=0\)

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

\(\Leftrightarrow\left(x-4\right)\left(3x+2\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)

Vậy ......

a, \(\left[{}\begin{matrix}x+1=0\\x+2=0\\x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-2\\x=-3\end{matrix}\right.\)

b, \(\left(x-1\right)^2-16=0\Leftrightarrow\left(x-1\right)^2-4^2=0\Leftrightarrow\left(x-5\right)\left(x+3\right)=0\Rightarrow\left[{}\begin{matrix}x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

c,\(\Rightarrow\left(x-2\right)\left(3x+4\right)\Rightarrow\left[{}\begin{matrix}x-2=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-4}{3}\end{matrix}\right.\)

\(\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{\left(x-1\right)\left(x-2\right)}+...+\dfrac{1}{\left(x-3\right)\left(x-4\right)}=\dfrac{1}{x-4}+2\)

Đk: \(x\ne0;x\ne1;x\ne2;x\ne3;x\ne4\)

\(\Leftrightarrow\dfrac{1}{\left(x-4\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-2\right)}+...+\dfrac{1}{\left(x-1\right)x}=\dfrac{1}{x-4}+2\)

\(\Leftrightarrow\dfrac{1}{x-4}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-2}+...+\dfrac{1}{x-1}-\dfrac{1}{x}=\dfrac{1}{x-4}+2\)

\(\Leftrightarrow\dfrac{1}{x-4}-\dfrac{1}{x}=\dfrac{1}{x-4}+2\)

\(\Leftrightarrow-\dfrac{1}{x}=2\)\(\Leftrightarrow-1=2x\Leftrightarrow x=-\dfrac{1}{2}\)

(3x-1) (2x-3) (2x-3) (x-5) = 0

=> 3x-1=0 hoặc 2x-3=0 hoặc x-5=0

Giải các PT ta có x=\(\dfrac{1}{3}\); x=\(\dfrac{3}{2}\); x=5

Suy ra 3x-1=0 hoặc 2x-3=0 hoặc x-5=0 suy ra x=1/3 hoặc x=3/2 hoặc x=5

Ta có:

\(x^3+5x^2+3x-9=0\)

\(\Leftrightarrow x^3+3x^2+2x^2+6x-3x-9=0\)

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

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

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

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

Vậy PT có nghiệm là \(\left\{1;-3\right\}\)

\(\left\{{}\begin{matrix}3x^2+y^2+2x-2y-1=0\left(1\right)\\2x\left(x+y\right)=2\left(2\right)\end{matrix}\right.\)

cộng vào nhau

Lấy (1) trừ (2)

\(x^2+y^2-2xy+2x-2y-1=-2\)

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

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

\(x-y+1=0\) thế ngược lại ra x,y