đầu bài là tìm x ạ

\(a,< =>2\left(x-4\right)^2+4x\left(x-4\right)=0< =>\left(x-4\right)\left(2x-8+4x\right)=0\)\(< =>\left(x-4\right)\left(6x-8\right)=0< =>\left[{}\begin{matrix}x=4\\x=\dfrac{4}{3}\end{matrix}\right.\)

b,\(< =>4x\left(x-2\right)=0< =>\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

c,\(< =>3x\left(x+2\right)=0< =>\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

d,\(< =>4x^2\left(2+x\right)=0< =>\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

d) \(2x^2+5x-7=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\) \(\left(a+b+c=1\right)\)

\(a,x\left(x+9\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\\ b,\Rightarrow x\left(x^2+4x+4\right)=0\\ \Rightarrow x\left(x+2\right)^2=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\ c,\Rightarrow\left(x-5-4\right)\left(x-5+4\right)=0\\ \Rightarrow\left(x-9\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\x=9\end{matrix}\right.\\ d,\Rightarrow3\left(x+2\right)-x\left(x+2\right)=0\\ \Rightarrow\left(x+2\right)\left(3-x\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\\ e,\Rightarrow x^3+6x^2+12x+8-x^3-6x^2=4\\ \Rightarrow12x=-4\Rightarrow x=-\dfrac{1}{3}\\ g,\Rightarrow\left(x+2\right)\left(x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)

Bài 1: 

b: \(3x-6=x^2-16\)

\(\Leftrightarrow x^2-3x-10=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

*vn:vô nghiệm.

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

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

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

\(\Leftrightarrow x=\pm\sqrt{2}\)

-Vậy \(S=\left\{\pm\sqrt{2}\right\}\).

b. \(16x^2-8x+5=0\)

\(\Leftrightarrow16x^2-8x+1+4=0\)

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

-Vậy S=∅.

c. \(2x^3-x^2-8x+4=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\pm2\end{matrix}\right.\)

-Vậy \(S=\left\{\dfrac{1}{2};\pm2\right\}\).

d. \(3x^3+6x^2-75x-150=0\)

\(\Leftrightarrow3x^2\left(x+2\right)-75\left(x+2\right)=0\)

\(\Leftrightarrow3\left(x+2\right)\left(x^2-25\right)=0\)

\(\Leftrightarrow3\left(x+2\right)\left(x+5\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\pm5\end{matrix}\right.\)

-Vậy \(S=\left\{-2;\pm5\right\}\)

a, \(2x\left(x-3\right)-15+5x=0\\ \Rightarrow2x\left(x-3\right)-\left(15-5x\right)=0\\ \Rightarrow2x\left(x-3\right)-5\left(3-x\right)=0\\ \Rightarrow\left(2x+5\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=3\end{matrix}\right.\)

b, \(x^3-7x=0\\ \Rightarrow x\left(x^2-7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm7\end{matrix}\right.\)

c, \(\left(2x-3\right)^2-\left(x+5\right)^2=0\\ \Rightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\\ \Rightarrow\left(x-8\right)\left(3x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)

Xem lại đề câu d 

e: ta có: \(4x^2+4x-6=2\)

\(\Leftrightarrow4x^2+4x-8=0\)

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

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

f: Ta có: \(2x^2+7x+3=0\)

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

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

\(a,\Leftrightarrow x\left(2x-7\right)+2\left(2x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{7}{2}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-2\left(2x-1\right)^2=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-4x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(-2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Rối quá bn ơi

Bn ko dùng dấu enter để xuống dòng à 

Mk sẽ giúp bn 3 câu đầu thôi còn 3 câu sau thì..... chịu vì ko biết làm sao

a) \(9\left(2x+1\right)^2-4\left(x+1\right)^2=0\)

(=)\(\left[9\left(2x+1\right)-4\left(x+1\right)\right]\left[9\left(2x+1\right)+4\left(x+1\right)\right]=0\)

(=)\(\left(18x+9-4x-4\right)\left(18x+9+4x+4\right)=0\)

(=)\(\left(14x+5\right)\left(22x+12\right)=0\)

(=)\(\left(14x+5\right)=0\) hoặc \(\left(22x+12\right)=0\)

1) \(14x+5=0\Rightarrow x=-\dfrac{5}{14}\)

2) \(22x+12=0\Rightarrow x=-\dfrac{6}{11}\)

Vậy........

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

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

(=) \(\left(x+2\right)^2=0\)

(=) \(x+2=0\Rightarrow x=-2\)

Vậy........

c) \(\left(x-1\right)\left(x^2-9\right)+x+3=0\)

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

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

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

(=) \(x+3=0\) hoặc \(x-2=0\) hoặc \(x-1=0\)

1)\(x+3=0\Rightarrow x=-3\)

2)\(x-2=0\Rightarrow x=2\)

3) \(x-1=0\Rightarrow x=1\)

Vậy.........