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

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

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

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

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

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

Giải:

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}5x+2=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{5}\\x=4\end{matrix}\right.\)

Vậy ...

b) \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)

\(\Leftrightarrow48x^2-20x-12x+5+3x-7-48x^2+112x=81\)

\(\Leftrightarrow83x-2=81\)

\(\Leftrightarrow83x=83\)

\(\Leftrightarrow x=1\)

Vậy ...

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

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

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

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

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

\(\Rightarrow x=\pm1\)

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

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

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

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

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

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

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

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

Ta có: \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)

\(\Leftrightarrow48x^2-12x-20x+5+3x-48x^2-7+112x=81\)

\(\Leftrightarrow83x-2=81\)

\(\Leftrightarrow83x=81+2=83\)

\(\Leftrightarrow x=1\)

(12x – 5)(4x – 1) + (3x – 7)(1 – 16x) = 81.

48x² – 12x – 20x + 5 + 3x – 48x² – 7 + 112x = 81.

83x – 2 = 81.

83x = 83.

x = 1.

Ta co:(12x-5)(4x-1) +(3x-7)(1-16x)=81

⇔48x²-12x -20x +5 +3x-48x² -7 ++112x=81

⇔83x-2=81

⇔83x=81 +2=83

⇔x = 1

Chuc ban hoc tot❓❗⛇

câu a, b, c dễ mà. Bạn áp dụng 7 hằng đẳng thúc là làm đc thoii!!

vd: a) \(\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)\)

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

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

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

\(\Rightarrow\left(3x+2\right)\left(x+1\right)\left(2x-1\right)=0\) (bạn phá ngoặc ra rồi tính là ra bước này)

\(\Leftrightarrow3x+2=0\) hoặc \(x+1=0\) hoặc \(2x-1=0\) ( đến đây bạn chia làm 3 trường hợp r tự tính nhé)

Chúc bạn học tốt!!

d/

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

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

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

e/

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

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

\(\Leftrightarrow\left(x-1\right)\left(x^2+x-6\right)=0\)

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

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

h.

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

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

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

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

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

g/

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm1\\x=\pm2\end{matrix}\right.\)

f/

\(x^4-9-4x\left(x^2-3\right)=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=\pm\sqrt{3}\\x=1\\x=3\end{matrix}\right.\)