5^x+5^(x+2)=650

=>5^x+5^x. 5^2=650

=>5^x.(1+5^2)=650

=>5^x.26=650

=>5^x=650:26

=>5^x=25

=>5^x=5^2

=>x=2

Vậy x=2

\(5^{x+2}+5^x=650\)

\(\Leftrightarrow5^x.5^2+5^x=650\)

\(\Leftrightarrow5^x\left(5^2+1\right)=650\)

\(\Leftrightarrow5^x=25=5^2\)

\(\Leftrightarrow x=2\)

a) Đặt  \(A=4x-x^2-5\)

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

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

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

Mà  \(\left(x-2\right)^2\ge0\forall x\)

\(\Rightarrow-A\ge1\)

\(\Leftrightarrow A\le-1< 0\left(đpcm\right)\)

b) Đặt  \(B=x^2-2x+5\)

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

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

Mà  \(\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow B\ge4>0\left(đpcm\right)\)

a)4x-x²-5 = -(x²-4x+4)-1= -(x-2)^2 -1 < 0 với mọi x (đpcm)

b) x² -2x+5= (x²-2x+1)+4=(x-1)^2 +4 >0  với mọi x (đpcm)

cảm ơn bn nhiều

a: =>6x-3x^2-5=4-3x^2-2

=>6x-5=2

=>6x=7

=>x=7/6

b: =>20x+5-12x^2-3x=6x^2-10x+3x-5

=>-12x^2+17x+5-6x^2+7x+5=0

=>-18x^2+24x+10=0

=>x=5/3 hoặc x=-1/3

Ta có \(\frac{-5}{25}=\frac{-4}{5-x}\)

=> (5- x).(-5) = -4.25

=> 5 - x = 20

=> x = -15

Vậy x = -15

\(\frac{-5}{25}=\frac{-4}{5}-x\)

\(=>x=\frac{-4}{5}+\frac{-5}{25}\)

\(=>x=-1\)

3x(2-x)-5=1-(3x²+2)

<=>6x-3x²-5=-3x²-2

<=>6x=3

<=>x=1/2

a) \(\left(x^2-9\right)\cdot\left(4^x-16\right)=0\)

\(\Rightarrow x^2-9=0\)hoặc \(4^x-16=0\)

               \(x^2=9\)                     \(4^x=16\)

               \(x^2=\left(\pm3\right)^2\)         \(4^x=4^2\)                

\(\Rightarrow x=\pm3\)hoặc \(x=2\)

b) \(5^x+5^{x+2}=650\)

\(\Rightarrow5^x+5^x\cdot25=650\)

\(\Rightarrow5^x\cdot\left(1+25\right)=650\)

\(\Rightarrow5^x\cdot26=650\)

\(\Rightarrow5^x=650\div26=25\)

\(\Rightarrow5^x=5^2\)

\(\Rightarrow x=2\)Vậy \(x=2\)

c) \(2^{x+2}-2^x=96\)

\(2^x\cdot4-2^x=96\)

\(2^x\cdot\left(4-1\right)=96\)

\(2^x\cdot3=96\)

\(2^x=96\div3=32\)

\(2^x=2^5\)Vậy \(x=5\)