pt \(\Rightarrow\)\(x\left(x+1\right)\left(x+4\right)\left(x+3\right)+1=3\left[\left(x+2\right)^2\left(x+5\right)\left(x-1\right)+2\right]\)

\(\Leftrightarrow\left(x^2+4x\right)\left(x^2+4x+3\right)+1=3\left[\left(x^2+4x+4\right)\left(x^2+4x-5\right)+2\right]\)

đến dây bn đặt  \(x^2+4x=a\)

pt \(\Leftrightarrow a\left(a+3\right)+1=3\left[\left(a+4\right)\left(a-5\right)+2\right]\)

đén đay bn làm nốt nhé

Đống nhất hệ số đưa và dạng 2 pt bậc 2 nhân vs nhau :v
1 có nghiệm 
2 vô nghiệm 
:)

Theo như đã nhìn 

Ta thấy 2 điều

1. Đây là 1 bài toán

2. Sau khi xài máy tính tính , nó = 0,7320508076 

Câu hỏi của Bùi Thị Vân - Toán lớp 9 | Học trực tuyến

\(\left(x+2\right)^2=9\left(x^2-4x+4\right)\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x+2=3x-6\\x+2=6-3x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-8=0\\4x-4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=1\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{4;1\right\}\)

\(\Leftrightarrow x^2+4x+4=9x^2-36x+36\)

\(\Leftrightarrow-8x^2+40x-32=0\)

\(\Leftrightarrow-8\left(x^2-5x+4\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=1\end{cases}}\)

Vậy ...

\(\left(x+2\right)^2=9\left(x^2-4x+4\right)\))

\(\Leftrightarrow x^2+4x+4=9x^2-36x+36\)

\(\Leftrightarrow x^2-4x+4-9x^2+36x-36=0\)

\(\Leftrightarrow-8x^2+40x-32=0\)

<=>\(-8x^2+32x+8x-32=0\)

<=>\(\left(-8x^2+8x\right)+\left(32x-32\right)=0\)

\(\Leftrightarrow8x\left(x-1\right)+32\left(x-1\right)=0\)

\(\orbr{\begin{cases}8x+32=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=1\end{cases}}}\)

b: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)

\(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=0\)

\(\Leftrightarrow\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24=0\)

\(\Leftrightarrow x^2+7x+6=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)

a: =>|x-7|=3-2x

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(-2x+3\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(2x-3-x+7\right)\left(2x-3+x-7\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(x+4\right)\left(3x-10\right)=0\end{matrix}\right.\Leftrightarrow x=-4\)

b: =>|2x-3|=4x+9

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(4x+9-2x+3\right)\left(4x+9+2x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(2x+12\right)\left(6x+6\right)=0\end{matrix}\right.\Leftrightarrow x=-1\)

c: =>3x+5=2-5x hoặc 3x+5=5x-2

=>8x=-3 hoặc -2x=-7

=>x=-3/8 hoặc x=7/2

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

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

\(\Leftrightarrow2x\left(x-1\right)=\left(1-x\right)\left(x+3\right)\)

\(\Leftrightarrow2x\left(x-1\right)+\left(x-1\right)\left(x+3\right)=0\)

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

\(\Rightarrow x=\pm1\)

Giúp tớ mấy câu còn lại đi các cậu, tớ cần gấp lắm ạ ;;-;;

\(x^4-4x^3+3x^2+4x-4=0\)

\(x^4-4x^3+4x^2-x^2+4x-4=0\)

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

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

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

\(Th1:\left(x-2\right)^2=0^2\Leftrightarrow x-2=0\Leftrightarrow x=2\)

\(Th2:x^2-1=0\Leftrightarrow x^2=1\Leftrightarrow x=\pm\sqrt{1}\)

\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)

\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)

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

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

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

\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)

\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)

\(5.\left|3x+2\right|=4.\Leftrightarrow\left[{}\begin{matrix}3x+2=4.\\3x+2=-4.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}.\\x=-2.\end{matrix}\right.\)

\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)

\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)

\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)

\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)

\(6,\left|2x-5\right|=\left|-x+2\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=-x+2\\2x-5=x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)

\(\Leftrightarrow x^2-4x+5+3\sqrt{x^2-4x+5}-2=0\)

Đặt \(\sqrt{x^2-4x+5}=t>0\)

\(\Rightarrow t^2+3t-2=0\Rightarrow\left[{}\begin{matrix}t=\dfrac{-3+\sqrt{17}}{2}\\t=\dfrac{-3-\sqrt{17}}{2}\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow x^2-4x+5=\dfrac{13-3\sqrt{17}}{2}\)

\(\Leftrightarrow x^2-4x+\dfrac{-3+3\sqrt{17}}{2}=0\)

\(x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=4^2-2\left(\dfrac{-3+3\sqrt{17}}{2}\right)=19-3\sqrt{17}\)