2^x+1-2^x=32

2^xx2^-2x^x1=32

2^xx(2-1)=32

2^xx1=32

2^x=32:1

2^x=32

=>2^x=2⁵

=>x=5

HT

2^{x + 1} - 2^x = 32

2^x . 2 - 2^x . 1 = 32

2^x . ( 2 - 1 ) = 32

2^x . 1 = 32

2^x = 32

2^x = 2⁵

=> x = 5

câu 1: =15

câu 2:=-98

câu 3:   54-(-16)-(-13)+27

         =     70   -    14

         =           56

\(A=\left|x-10\right|+2021\ge2021\)

Dấu = xảy ra khi x = 10

\(A=\left|x-10\right|+2021\ge2021\forall x\)

Dấu '=' xảy ra khi x=10

\(\Leftrightarrow\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{1}{3}-\dfrac{1}{x+2}=\dfrac{1}{5}:\dfrac{3}{2}=\dfrac{2}{15}\)

\(\Leftrightarrow\dfrac{1}{x+2}=\dfrac{1}{5}\)

=>x+2=5

hay x=3

a) \(6xy+4x-9y-7=0\)

  \(\Leftrightarrow2x.\left(3y+2\right)-9y-6-1=0\)

\(\Leftrightarrow2x.\left(3y+x\right)-3.\left(3y+2\right)=1\)

\(\Leftrightarrow\left(2x-3\right).\left(3y+2\right)=1\)

Mà \(x,y\in Z\Rightarrow2x-3;3y+2\in Z\)

Tự làm típ

\(A=x^3+y^3+xy\)

\(A=\left(x+y\right)\left(x^2-xy+y^2\right)+xy\)

\(A=x^2-xy+y^2+xy\)( vì \(x+y=1\))

\(A=x^2+y^2\)

Áp dụng bất đẳng thức Bunhiakovxky ta có :

\(\left(1^2+1^2\right)\left(x^2+y^2\right)\ge\left(x\cdot1+y\cdot1\right)^2=\left(x+y\right)^2=1\)

\(\Leftrightarrow2\left(x^2+y^2\right)\ge1\)

\(\Leftrightarrow x^2+y^2\ge\frac{1}{2}\)

Hay \(x^3+y^3+xy\ge\frac{1}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=\frac{1}{2}\)

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1,x=6

3,x=-9