a) x : 4 = 8

x = 8 × 4

x = 32

b) 4 × x = 12 + 8

4 × x = 20

x = 20 : 4

x = 5

c) 30 : x = 14 + 16

30 : x = 30

x = 30 : 30

x = 1

=>(x-6)^4+(x-8)^4=16

Đặt a=x-7

=>(a-1)^4+(a+1)^4=16

=>a^4+4a^3+6a^2+4a+1+a^4-4a^3+6a^2-4a+1=16

=>2a^4+12a^2-14=0

=>a^4+6a^2-7=0

=>(a^2+7)(a^2-1)=0

=>a^2=1

=>a=1 hoặc a=-1

=>x-7=1 hoặc x-7=-1

=>x=6 hoặc x=8

=>(x-6)^4+(x-8)^4=16

Đặt a=x-7

=>(a-1)^4+(a+1)^4=16

=>a^4+4a^3+6a^2+4a+1+a^4-4a^3+6a^2-4a+1=16

=>2a^4+12a^2-14=0

=>a^4+6a^2-7=0

=>(a^2+7)(a^2-1)=0

=>a^2=1

=>a=1 hoặc a=-1

=>x-7=1 hoặc x-7=-1

=>x=6 hoặc x=8

Trả lời

\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=\frac{23}{16}\)

\(\Leftrightarrow\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=\frac{23}{16}\)

\(\Leftrightarrow4x+\left(\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}\right)=\frac{23}{16}\)

\(\Leftrightarrow4x+\frac{8+4+2+1}{16}=\frac{23}{16}\)

\(\Leftrightarrow4x+\frac{15}{16}=\frac{23}{16}\)

\(\Leftrightarrow4x=\frac{23}{16}-\frac{15}{16}\)

\(\Leftrightarrow4x=\frac{8}{16}\)

\(\Leftrightarrow4x=\frac{1}{2}\)

\(\Leftrightarrow x=\frac{1}{2}:4\)

\(\Leftrightarrow x=\frac{1}{8}\)

Vậy x=\(\frac{1}{8}\)

\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=\frac{23}{16}\)

\(\Leftrightarrow\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=\frac{23}{16}\)

\(\Leftrightarrow4x+\left(\frac{8+4+2+1}{16}\right)=\frac{23}{16}\)

\(\Leftrightarrow4x+\frac{15}{16}=\frac{23}{16}\)

\(\Leftrightarrow4x=\frac{23}{16}-\frac{15}{16}\)

\(\Leftrightarrow4x=\frac{1}{2}\)

\(\Leftrightarrow x=\frac{1}{2}:4\)

\(\Leftrightarrow x=\frac{1}{8}\)

\(\frac{8}{3}:x=\frac{16}{9}:\frac{8}{3}\)

\(\frac{8}{3}:x=\frac{16}{9}.\frac{3}{8}\)

\(\frac{8}{3}:x=\frac{2}{3}\)

       \(x=\frac{8}{3}:\frac{2}{3}\)

      \(x=\frac{8}{3}.\frac{3}{2}\)

    \(x=4\)

Vậy x = 4

Đặt \(t=x-7\)

Thay t vào phương trình ban đầu ta có: 

\(\left(t+1\right)^4+\left(t-1\right)^4=16\)

\(\left(t^4+4t^3+6t^2+4t+1\right)-\left(t^4-4t^3+6t^2-4t+1\right)=16\)

\(8t^3+8t=16\)

\(t^3+t-2=0\)

\(t=1\)

=> \(x-7=1\)

=> x = 8

Vậy x = 8 là giá trị cần tìm 

16 – x = 21 – − 8 16 – x = 21 + 8 16 − x = 30 − x = 30 − 16 − x = 14       x = − 14.

x = 13