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

b: \(x^8+x^7+1\)

\(=x^8+x^7+x^6-x^6-x^5-x^4+x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\)

\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)

c: \(x^8+x^4+1\)

\(=\left(x^8+2x^4+1\right)-x^4\)

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

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

a)\(x^4+4\\ =\left(x^2\right)^2+4x^2+4-4x^2\\ =\left[\left(x^2\right)^2+4x^2+4\right]-\left(2x\right)^2\\ =\left(x^2+2\right)^2-\left(2x\right)^2\\ =\left(x^2+2+2x\right)\left(x^2+2-2x\right)\)

Ta có \(x1-\frac{1}{9}=x2-\frac{2}{8}=...=x9-\frac{9}{1}\)

\(=\frac{x1-1}{9}=\frac{x2-2}{8}=\frac{x3-3}{7}=...=\frac{x9-9}{1}\)

= \(\frac{x1-1+x2-2+x3-3+...+x9-9}{9+8+7+...+1}\)

\(=\frac{\left(x1+x2+x3+...+x9\right)-\left(1+2+3+...+9\right)}{9+8+7+....+1}\)

=\(\frac{90-45}{45}=\frac{45}{45}=1\)

=> \(\hept{\begin{cases}\begin{cases}x1=10\\x2=10\end{cases}\\.....\\x9=10\end{cases}}\)

Bạn cần viết đầy đủ đề: Bao gồm yêu cầu đề và công thức toán để được hỗ trợ tốt hơn.

\(D=\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^4-x^2+1\right)\left(x^8-x^4+1\right)\)

\(=\left[\left(x^2+1\right)^2-x^2\right]\left(x^4-x^2+1\right)\left(x^8-x^4+1\right)\)

\(=\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)\cdot\left(x^8-x^4+1\right)\)

\(=\left(x^8+2x^4+1-x^4\right)\left(x^8-x^4+1\right)\)

\(=\left(x^8+1\right)^2-x^8\)

\(=x^{16}+x^8+1\)

\(3 × 32 , 85 + 32 , 85 × a × ( a × 1 − a : 1 ) + 32 , 85 × 8 − 32 , 85 × 10\)

\(= 3 × 32 , 85 + 32 , 85 × a × ( a − a ) + 32 , 85 × 8 − 32 , 85 × 10\)

\(= 3 × 32 , 85 + 32 , 85 × a × 0 + 32 , 85 × 8 − 32 , 85 × 10\)

\(= 32 , 85 × ( 3 + a × 0 + 8 − 10 )\)

\(= 32 , 85 × ( 3 + 0 + 8 − 10 )\)

\(= 32 , 85 × 1=32,85\)

`#Ya`

a dù nhanh ;-;

X8 LÀ X MŨ 8 HAY X*8 VẬY BẠN

a) \(\sqrt{x^8}=256\)

\(\Leftrightarrow\sqrt{\left(x^4\right)^2}=256\)

\(\Leftrightarrow x^4=256\)

\(\Leftrightarrow x^4=\left(\pm4\right)^4\)

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

b) \(\sqrt{x^2-2x+1}=x-1\) (x≥1)

\(\Leftrightarrow\sqrt{\left(x-1\right)^2}=x-1\)

\(\Leftrightarrow\left|x-1\right|=x-1\)

Mà: \(x\ge1\Rightarrow x-1\ge0\)

\(\Leftrightarrow x-1=x-1\)

\(\Leftrightarrow0=0\) (luôn đúng)

Vậy pt thỏa mãn với mọi x đk x ≥ 1