Tại x = 16 => x +1 = 17

Thay vào A ta được:

A = x⁴ - (x+1)x³ + (x+1)x² - (x+1)x + 20

A= x⁴ -(x⁴ + x³)  + (x³ + x²)  -(x² + x) +20

A= x⁴ - x⁴ - x³ + x³ + x² - x² -x + 20

A= - x+20

Mà  x = 16

=> A= -16 + 20 = 4

Vậy A= 4 khi x =16

a) Ta có: \(\left(2^2\right)^3\cdot4^5\)

\(=2^6\cdot2^{10}\)

\(=2^{16}=65536\)

b) Ta có: \(\left[\left(-4\right)^2\right]^2\cdot6\)

\(=16^2\cdot6\)

\(=256\cdot6=1536\)

c) Ta có: \(\frac{16}{25}\cdot\left(\frac{4}{5}\right)^3\)

\(=\left(\frac{4}{5}\right)^2\cdot\left(\frac{4}{5}\right)^3\)

\(=\left(\frac{4}{5}\right)^5\)

\(=\frac{1024}{3125}\)

d) Ta có: \(\left(\frac{121}{64}\right)^2\cdot\left(-\frac{64}{11}\right)^2\)

\(=\frac{121^2}{64^2}\cdot\frac{64^2}{11^2}\)

\(=11^2=121\)

e) Ta có: \(\left[\left(-3\right)^3\right]^3\cdot271:125\)

\(=\left(-27\right)^3\cdot\frac{271}{125}\)

\(=\frac{-5334093}{125}\)

4³.2⁵.16=(2²)³.2⁵.2⁴=2⁶.2⁵.2⁴=2¹⁵

32.64.8²=2⁵.2⁶.2⁶=2¹⁷

=> A=2¹⁷:2¹⁵=2^17-15=2²=4

Đáp số: A=4

\(\frac{32.64.8^2}{4^3\cdot2^5\cdot16}=\frac{2^5.2^6.2^6}{2^6.2^5.2^3}\)

\(=\frac{2^{5+6+6}}{2^{6+5+3}}=\frac{2^{17}}{2^{14}}=2^{17-14}=2^3=8\)

Mình giải thích thêm nha 

\(8^2=\left(2^3\right)^2=2^6\)

\(4^3=\left(2^2\right)^3=2^6\)

bài này câu hỏi là gì bạn

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

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

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

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

Bài 5: 

1) Ta có: \(2x\left(x+1\right)-2x^2-2x\)

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

=0

2) Ta có: \(3x\left(x-2\right)-3\left(x^2-2x\right)+4\)

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

=4

3) Ta có: \(\left(x-1\right)\left(x-5\right)-x^2+6x-5\)

\(=x^2-6x+5-x^2+6x-5\)

=0

4) Ta có: \(\left(2x+1\right)\left(x-1\right)-2x^2+x-5\)

\(=2x^2-2x+x-1-2x^2+x-5\)

=-6

5) Ta có: \(\left(3x-2\right)\left(x-1\right)-3x^2+5x-4\)

\(=3x^2-3x-2x+2-3x^2+5x-4\)

=-2

6) Ta có: \(2x\left(x+1\right)-x\left(x+3\right)-x^2+x+5\)

\(=2x^2+2x-x^2-3x-x^2+x+5\)

=5

cảm ơn bạn 

\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)

\(\rightarrow A=\frac{3}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)

\(\rightarrow A=\frac{7}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)

\(\rightarrow A=\frac{15}{16}+\frac{1}{32}+\frac{1}{64}\)

\(\rightarrow A=\frac{31}{32}+\frac{1}{64}\)

\(\rightarrow A=\frac{63}{64}\)

\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\Rightarrow64A=32+16+8+4+2+1\Rightarrow64A=63\Rightarrow A=\frac{63}{64}\)