a) (x+3)⁴+(x+5)⁴=16

<=>(x+3)⁴+(x+5)⁴=0⁴+2⁴

TH1: \(\left\{{}\begin{matrix}x+3=0\\x+5=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\x=-3\end{matrix}\right.\Leftrightarrow x=-3\)

TH2:\(\left\{{}\begin{matrix}x+3=2\\x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=-5\end{matrix}\right.\)(loại)

b)(x-2)⁴+(x-3)⁴=1=0⁴+1⁴

TH1: \(\left\{{}\begin{matrix}x-2=0\\x-3=1\end{matrix}\right.\)loại

TH2: \(\left\{{}\begin{matrix}x-2=1\\x-3=0\end{matrix}\right.\)=>x=3.

c)(x+1)⁴+(x-3)⁴=82=3⁴+(-1)⁴

làm tương tự => x=2.

d) làm tương tự câu b

Kết quả là X=2 đúng 100% mình làm rồi.

x = 2 đó mk đảm bảo 100% luôn

a) Sửa đề

\(\left(x+1\right)^4-\left(x-3\right)^4=82\)

Đặt x - 1 = a

\(\left(a+2\right)^4-\left(a-2\right)^4=82\)

\(\Rightarrow\left[\left(a+2\right)^2\right]^2-\left[\left(a-2\right)^2\right]^2=82\)

\(\Rightarrow\left(a^2+4a+4\right)^2-\left(a^2-4a+4\right)^2=82\)

\(\Rightarrow\left(a^2+4\right)^2+8a\left(a^2+4\right)+16a^2+\left(a^2+4\right)^2-8a\left(a^2+4\right)+16a^2=82\)

\(\Rightarrow\left(a^2+4\right)^2+16a^2=41\)

\(\Rightarrow a^4+8a^2+16+16a^2=41\)

\(\Rightarrow a^4+24a^2=25\)

\(\Rightarrow a^4+24a^2-25=0\)

\(\Rightarrow a^4-a^2+25a^2-25=0\)

\(\Rightarrow a^2\left(a^2-1\right)+25\left(a^2-1\right)=0\)

\(\Rightarrow\left(a^2-1\right)\left(a^2+25\right)=0\)

\(\Rightarrow\left(a-1\right)\left(a+1\right)\left(a^2+25\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a-1=0\\a+1=0\\a^2+25=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}a=1\\a=-1\\a^2=-25\end{matrix}\right.\)

Do a²= -25 không tồn tại

Vậy a = 1 ; a = -1

b) \(\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-2\right)=24\)

\(\Rightarrow\left[\left(x-1\right)\left(x-2\right)\right]\left[\left(x+1\right)\left(x+2\right)\right]=24\)

\(\Rightarrow\left(x^2-3x+2\right)\left(x^2+3x+2\right)=24\)

\(\Rightarrow\left(x^2+2\right)^2-\left(3x\right)^2=24\)

\(\Rightarrow x^4+4x^2+4-9x^2-24=0\)

\(\Rightarrow x^4-5x^2-20=0\)

\(\Rightarrow\left(x^2\right)^2-2.x^2\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{25}{4}-20=0\)

\(\Rightarrow\left(x^2-\dfrac{5}{2}\right)^2-\dfrac{105}{4}=0\)

\(\Rightarrow\left(x^2-\dfrac{5}{2}\right)^2=\dfrac{105}{4}\)

\(\Rightarrow\left(x^2-\dfrac{5}{2}\right)=\left(\dfrac{\sqrt{105}}{2}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x^2-\dfrac{5}{2}=\dfrac{\sqrt{105}}{2}\\x^2-\dfrac{5}{2}=-\dfrac{\sqrt{105}}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2=\dfrac{5+\sqrt{105}}{2}\\x^2=\dfrac{5-\sqrt{105}}{2}\end{matrix}\right.\)

...

Đặt \(x+2=t\)

\(\Rightarrow\left(x+1\right)^4+\left(x+3\right)^4=82\)

\(\Leftrightarrow\left(t-1\right)^4+\left(t+1\right)^4=82\)

\(\Leftrightarrow\left[\left(t-1\right)^2\right]^2+\left[\left(t+1\right)^2\right]^2=82\)

\(\Leftrightarrow\left(t^2-2t+1\right)^2+\left(t+2t+1\right)^2=82\)

\(\Leftrightarrow\left(t^2+1\right)^2-4t\left(t^2+1\right)+4t^2+\left(t^2+1\right)^2+4t\left(t^2+1\right)+4t^2=82\)

\(\Leftrightarrow\left(t^2+1\right)^2+4t^2=41\)

\(\Leftrightarrow t^4+6t^2+1=41\)

\(\Leftrightarrow t^4+6t^2-40t=0\)

\(\Leftrightarrow\left[\begin{matrix}t^2=-10\left(lo\text{ại}\right)\\t^2=4\Rightarrow\left[\begin{matrix}t=2\\t=-2\end{matrix}\right.\end{matrix}\right.\)

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

x=0 hoat 4 nha bn

chuc bn hoc tot

happy new year

Bài này có hai cách giải:
*Cách 1:
Đặt t = x + 3
=> x + 2 = t - 1; x + 4 = t + 1.
ta có pt: (t - 1)^4 + (t + 1)^4 = 82
<=>[(t -1)²]² + [(t + 1)²]² = 82
<=> (t² - 2t + 1)² + (t² + 2t + 1)² = 82
<=> (t²+1)² - 4t(t²+1) + 4t² + (t²+1)² + 4t(t²+1) + 4t² = 82
<=> (t² + 1)² + 4t² = 41
<=> t^4 + 6t² + 1 = 41
<=> (t²)² + 6t² - 40 = 0
<=> t² = -10 (loại) hoặc t² = 4
<=> t = 2 hoặc t = -2
với t = -2 => x = -5
với t = 2 => x = -1
vậy pt có hai nghiệm là : x = -1 hoặc x = -5
*Tổng quát:
(x+a)^4 + (x+b)^4 = c
đặt: t = x + (a+b)/2, sau khi chuyển qua ẩn phụ rồi khai triển chắc chắn sẽ ra pt trùng phương.
**Cách 2/ chú ý hai hằng đẳng thức:
a² + b² = (a - b)² + 2ab. và
a² + b² = (a + b)² - 2ab.
pt: (x + 2)^4 + (x + 4)^4 = 82
Đặt: t = (x + 2)(x + 4). ta có:
*(x+2)² + (x+4)² = [(x+2)-(x+4)]² + 2(x+2)(x+4) =
= (-2)² + 2t = 4 + 2t
*(x + 2)^4 + (x + 4)^4 = [(x + 2)²]² + [(x + 4)²]² =
= [(x+2)² + (x+4)²]² - 2(x+2)².(x+4)² =
= [4 + 2t]² - 2t²
= 16 + 16t + 4t² - 2t²
thay vào pt đã cho ta có:
16 + 16t + 2t² = 82
<=> t² + 8t - 33 = 0
<=> t = -11 hoặc t = 3
+Với t = -11:
(x + 2)(x + 4) = -11
<=> x² + 6x +19 = 0 => vn
+Với t = 3:
(x + 2)(x + 4) = 3
<=> x² + 6x + 5 = 0
<=> x = -1 hoặc x = -5