\(8x^3+12x^2+6x+1=\left(2x+1\right)^3\)

\(=\left(2\cdot24.5+1\right)^3=50^3=125000\)

\(B=x^3-3x^2+3x\)

\(=x^3-3x^21+3x1^2-1^3+1\)

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

thay x=11 vào P ta đc:

\(B=\left(11-1\right)^3+1=1001\)

Vậy B=1001

với x=11

Bài làm:

Ta có: Tại x = 11 thì giá trị của B là

\(B=x\left(x^2-3x+3\right)=11\left(11^2-3.11+3\right)\)

\(=11.91=1001\)

\(B=x^3-3x^2+3x\)

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

thay x = 11 vào biểu thức ta có

\(B=11\left(11^2-3.11+3\right)\)

\(B=11.91=1001\)

\(A=4\left(3^2+1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)

\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)

\(=\frac{1}{2}\left(3^4-1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)

\(=\frac{1}{2}\left(3^{128}-1\right)< B\)

\(A=4\left(3^2+1\right)\left(3^4+1\right)....\left(3^{64}+1\right)\)

\(\Rightarrow2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right).....\left(3^{64}+1\right)\)

\(=\left(3^4-1\right)\left(3^4+1\right).....\left(3^{64}+1\right)=\left(3^{64}-1\right)\left(3^{64}+1\right)=3^{128}-1=B\)

\(\Rightarrow A< B\)

\(A=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(=\frac{1}{2}\left(3^4-1\right)\left(3^4+1\right)....\left(3^{64}+1\right)\)

                          \(.........\)

\(=\frac{1}{2}\left(3^{168}-1\right)\)\(< \)\(3^{168}-1\)

\(\Rightarrow\)\(A< B\)

Tại sao 4 lại trở thành 2 vậy. Giải thích giúp mình nhé.

$a^4-8\\=(a^2-\sqrt 8)(a^2+\sqrt 8)$

a) \(a^2+9-6a\)

\(=\left(a-3\right)^2\)

b) \(x^2-x+\frac{1}{4}\)

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

c) \(-x^2+4x-x\)

\(=3x-x^2\)

\(=x\left(3-x\right)\)

Bài giải

\(a,\text{ }a^2+9-6a=a^2+2\cdot3a+3^2=\left(a-3\right)^2\)

\(b,\text{ }x^2-x+\frac{1}{4}=x^2-2\cdot\frac{1}{2}\cdot x+\left(\frac{1}{2}\right)^2=\left(x-\frac{1}{2}\right)^2\)

\(c,\text{ }-x^2+4x-x=3x-x^2=\left(\sqrt{3x}\right)^2-x^2=\left(\sqrt{3x}-x\right)\left(\sqrt{3x}+x\right)\)( Đề nói vận dụng hằng đẳng thức để rút gọn nên mình đưa về hiệu hai ình phương nha ! )

\(-y^2-13y+14=0\)

\(\Leftrightarrow y^2+\frac{2.y.13}{2}+\frac{169}{4}=\frac{225}{4}\)

\(\Leftrightarrow\left(y+\frac{13}{2}\right)^2=\left(\frac{15}{2}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}y+\frac{13}{2}=\frac{15}{2}\\y+\frac{13}{2}=-\frac{15}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}y=1\\y=-14\end{cases}}\)

xin lỗi , đề là tìm y

\(20162017.20162019=\left(20162018-1\right)\left(20162018+1\right)\)

\(=20162018^2-1< 20162018^2\)

\(\Rightarrow20162017.20162019< 20162018^2\)

Vậy...

Ta có:

\(20162017=20162018-1\)

\(20162019=20162018+1\)

\(\Rightarrow20162017.20162019\)

\(=\left(20162018-1\right).\left(20162018+1\right)\)

\(=20162018^2-1^2=20162018^2-1\)

\(\Rightarrow20162018^2-1< 20162018^2\)

Do đó \(20162017.20162018< 20162018^2\)

Chúc bạn học tốt!!!

\(\text{20162017 . 20162019}=\left(20162018-1\right)\left(20162018-1\right)\)

\(=20162018^2-1< 20162018^2\)

\(\Rightarrow20162017\cdot20162019< 20162018^2\)

Vậy .... \(20162017\cdot20162019< 20162018^2\)