Bài 1

a) \(x=x^5\)

\(x^5-x=0\)

\(x\left(x^4-1\right)=0\)

\(x=0\) hoặc \(x^4-1=0\)

* \(x^4-1=0\)

\(x^4=1\)

\(x=1\)

Vậy x = 0; x = 1

b) \(x^4=x^2\)

\(x^4-x^2=0\)

\(x^2\left(x^2-1\right)=0\)

\(x^2=0\) hoặc \(x^2-1=0\)

*) \(x^2=0\)

\(x=0\)

*) \(x^2-1=0\)

\(x^2=1\)

\(x=1\)

Vậy \(x=0\); \(x=1\)

c) \(\left(x-1\right)^3=x-1\)

\(\left(x-1\right)^3-\left(x-1\right)=0\)

\(\left(x-1\right)\left[\left(x-1\right)^2-1\right]=0\)

\(x-1=0\) hoặc \(\left(x-1\right)^2-1=0\)

*) \(x-1=0\)

\(x=1\)

*) \(\left(x-1\right)^2-1=0\)

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

\(x-1=1\) hoặc \(x-1=-1\)

**) \(x-1=1\)

\(x=2\)

**) \(x-1=-1\)

\(x=0\)

Vậy \(x=0\); \(x=1\); \(x=2\)

Bài 1:

a) \(4^{x+2}+4^x=68\)

\(\Rightarrow4^x\cdot\left(4^2+1\right)=68\)

\(\Rightarrow4^x\cdot17=68\)

\(\Rightarrow4^x=\dfrac{68}{17}\)

\(\Rightarrow4^x=4\)

\(\Rightarrow4^x=4^1\)

\(\Rightarrow x=1\)

b) \(5\cdot2^{x+4}-3\cdot2^x=308\)

\(\Rightarrow2^x\cdot\left(5\cdot2^4-3\right)=308\)

\(\Rightarrow2^x\cdot\left(5\cdot16-3\right)=308\)

\(\Rightarrow2^x\cdot77=308\)

\(\Rightarrow2^x=\dfrac{308}{77}\)

\(\Rightarrow2^x=4\)

\(\Rightarrow2^x=2^2\)

\(\Rightarrow x=2\)

c) \(4\cdot3^{x+1}+7\cdot3^x=513\)

\(\Rightarrow3^x\cdot\left(4\cdot3+7\right)=513\)

\(\Rightarrow3^x\cdot19=513\)

\(\Rightarrow3^x=\dfrac{513}{19}\)

\(\Rightarrow3^x=27\)

\(\Rightarrow3^x=3^3\)

\(\Rightarrow x=3\)

d) \(5^{x+4}-5^x=3120\)

\(\Rightarrow5^x\cdot\left(5^4-1\right)=3120\)

\(\Rightarrow5^x\cdot\left(625-1\right)=3120\)

\(\Rightarrow5^x\cdot624=3120\)

\(\Rightarrow5^x\cdot\dfrac{3120}{624}\)

\(\Rightarrow5^x=5\)

\(\Rightarrow5^x=5^1\)

\(\Rightarrow x=1\)

f) \(3\cdot4^{2x+1}-16^x=2816\)

\(\Rightarrow3\cdot4^{2x+1}-\left(4^2\right)^x=2816\)

\(\Rightarrow3\cdot4^{2x+1}-4^{2x}=2816\)

\(\Rightarrow4^{2x}\cdot\left(3\cdot4-1\right)=2816\)

\(\Rightarrow4^{2x}\cdot11=2816\)

\(\Rightarrow4^{2x}=\dfrac{2816}{11}\)

\(\Rightarrow4^{2x}=256\)

\(\Rightarrow\left(2^2\right)^{2x}=2^8\)

\(\Rightarrow2^{4x}=2^8\)

\(\Rightarrow4x=8\)

\(\Rightarrow x=2\)

Bài 2:

\(2^x+124=5^y\)

\(\Rightarrow5^y-2^x=124\)

\(\Rightarrow5^y-2^x=125-1\)

\(\Rightarrow\left\{{}\begin{matrix}5^y=125\\2^x=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}5^y=5^3\\2^x=2^0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y=3\\x=0\end{matrix}\right.\)

Vậy: .... 

a)<=>

A,=(x+y)(x-y)=x^2-y^2

x=(-1/2)^5:(1/2)^4=-1/2

x^2=1/4

y=8^2/(-2)^5=-2

y^2=4

A=1/4-4=-15/4

1)1/9 x 3^x = 2187:81=27

            3^x=27:1/9=243=3⁵

            =>x=5

\(\frac{1}{9}.3^4.3^x=3^7\)

\(\Leftrightarrow3^x=3^7:\frac{1}{9}:3^4=243\)

\(\Leftrightarrow3^x=3^5\)

\(\Leftrightarrow x=5\)

a) \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)

\(\Leftrightarrow\left(x^2+6x+9\right)-\left(x^2+4x-32\right)-1=0\)

\(\Leftrightarrow2x=-40\)

\(\Rightarrow x=-20\)

b) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)

\(\Leftrightarrow x^3+27-x^3+4x=15\)

\(\Leftrightarrow4x=-12\)

\(\Rightarrow x=-3\)

c) \(\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)

\(\Leftrightarrow\left(x^2-4x+4\right)-\left(x^2+6x+9\right)-\left(4x+4\right)=5\)

\(\Leftrightarrow-14x=14\)

\(\Rightarrow x=-1\)

d) \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)

\(\Leftrightarrow4x^2-9-\left(x^2-2x+1\right)-\left(3x^2-15x\right)=-44\)

\(\Leftrightarrow17x=-34\)

\(\Rightarrow x=-2\)

e) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)

\(\Leftrightarrow24x=24\)

\(\Rightarrow x=1\)

f) \(5x\left(x-3\right)^2-5\left(x-1\right)^3+15\left(x+2\right)\left(x-2\right)=5\)

\(\Leftrightarrow5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-60=5\)

\(\Leftrightarrow30x=60\)

\(\Rightarrow x=2\)

g) \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)-3x^2=42\)

\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1-3x^2=42\)

\(\Leftrightarrow26x=14\)

\(\Rightarrow x=\frac{7}{13}\)