\(\left(2x-5\right)^4+\left(2x-3\right)^4=16\)
\(\Leftrightarrow\left[\left(2x-5\right)-\left(2x-3\right)\right]^4+2\left(2x-5\right)^2\left(2x-3\right)^2=16\)
\(\Leftrightarrow\left(-2\right)^4+2\left(2x-5\right)^2\left(2x-3\right)^2=16\)
\(\Leftrightarrow16+2\left(2x-5\right)^2\left(2x-3\right)^2=16\)
\(\Leftrightarrow2\left(2x-5\right)^2\left(2x-3\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-5\right)^2=0\\\left(2x-3\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\)
- Vậy \(S=\left\{\dfrac{5}{2};\dfrac{3}{2}\right\}\)
Áp dụng A^4 +B ^4 = (A^2+B^2) ( A^2 - B^2) = (A^2 + B^2) (A-B ) (A+B)
=> [(2x-5)^2 + (2x-3)^2] (2x-5 +2x -3)(2x - 5 - 2x +3) =16
=> [ 4x^2 - 20x +25 + 4x^2 - 12x +9 ] 4( x -2) (-2) =16
=> [ 8x^2 - 32x + 34 ] (x-2) = -2
=> 8x^3 - 16x^2 - 32 x^2 + 64x +34x -68 = -2
=> x= 3/2
Bạn xem lại cách mình khai triển hằng nhá với bấm lại máy PT cuối nha!
Có 3 kết quả x nhưng có 1 x là đúng thôi
- Bài trước mình nhầm nhé :)) . Để mình làm lại.
- Ta có hằng đẳng thức: \(\left(A+B\right)^4=A^4+4A^3B+6A^2B^2+4AB^3+B^4\).
\(\Rightarrow A^4+B^4=\left(A+B\right)^4-2AB\left(2A^2+3AB+2B^2\right)\)
- Quay lại bài toán:
\(\left(2x-5\right)^4+\left(2x-3\right)^4=16\)
\(\Leftrightarrow\left[\left(2x-5\right)-\left(2x-3\right)\right]^4+2\left(2x-5\right)\left(2x-3\right)\left[2\left(2x-5\right)^2+3\left(2x-5\right)\left(2x-3\right)+2\left(2x-3\right)^2\right]=16\)
\(\Leftrightarrow\left[-2\right]^4+2\left(2x-5\right)\left(2x-3\right)\left[2\left(2x-5\right)^2+3\left(2x-5\right)\left(2x-3\right)+2\left(2x-3\right)^2\right]=16\)
\(\Leftrightarrow2\left(2x-5\right)\left(2x-3\right)\left[2\left(2x-5\right)^2+3\left(2x-5\right)\left(2x-3\right)+2\left(2x-3\right)^2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-3=0\\2\left(2x-5\right)^2+3\left(2x-5\right)\left(2x-3\right)+2\left(2x-3\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{3}{2}\\2\left(2x-5\right)^2+3\left(2x-5\right)\left(2x-3\right)+2\left(2x-3\right)^2=0\left(1\right)\end{matrix}\right.\)
- Giải phương trình (1).
- Đặt \(\left\{{}\begin{matrix}2x-5=a\\2x-3=b\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2a^2+3ab+2b^2=0\)
\(\Leftrightarrow a^2+\dfrac{3}{2}ab+b^2=0\)
\(\Leftrightarrow\left(a^2+\dfrac{3}{2}ab+\dfrac{9}{16}b^2\right)+\dfrac{7}{16}b^2=0\)
\(\Leftrightarrow\left(a+\dfrac{3}{4}b\right)^2+\dfrac{7}{16}b^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a+\dfrac{3}{4}b\right)^2=0\\\dfrac{7}{16}b^2=0\end{matrix}\right.\Leftrightarrow a=b=0\)
\(\Rightarrow2x-5=2x-3\Leftrightarrow0x=2\left(PTVN\right)\)
- Vậy \(S=\left\{\dfrac{3}{2};\dfrac{5}{2}\right\}\)
