TINH
B=(1+3)nhân(1+3^2)nhân(1+3^4)nhân(1+3^8)trừ 3^16phan2
Tim x
a) (x-5)^2-(x+3)^2=1
b) (2x-1)^2-(2x-3)^2=4
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1/
\(B=\frac{\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-3^{16}}{4}\)
\(=\frac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)-3^{16}}{4}\)
\(=\frac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)-3^{16}}{4}\)
\(=\frac{\left(3^8-1\right)\left(3^8+1\right)-3^{16}}{4}\)
\(=\frac{3^{16}-1-3^{16}}{4}=\frac{-1}{4}\)
2/
a, (x-5)2-(x+3)2=1
<=>(x-5+x+3)(x-5-x-3)=1
<=>-16.(x-1)=1
<=>x-1=-1/16
<=>x=15/16
b, (2x-1)2-(2x-3)2=4
<=>(2x-1+2x-3)(2x-1-2x+3)=4
<=>-8(x-1)=4
<=>x-1=-1/2
<=>x=1/2
a) \(\frac{8}{9}\cdot x-\frac{2}{3}=\frac{1}{3}\cdot x+1\frac{1}{3}\)
=> \(\frac{8x}{9}-\frac{2}{3}=\frac{x}{3}+\frac{4}{3}\)
=> \(\frac{8x}{9}-\frac{6}{9}=\frac{x+4}{3}\)
=> \(\frac{8x-6}{9}=\frac{x+4}{3}\)
=> \(3\left(8x-6\right)=9\left(x+4\right)\)
=> \(24x-18=9x+36\)
=> \(24x-18-9x=36\)
=> \(24x-9x=54\)
=> \(15x=54\)
=> \(5x=18\)
=> \(x=\frac{18}{5}\)
Vậy x = \(\frac{18}{5}\)
b) \(\left(x-\frac{1}{2}\right)\left(\frac{3}{2}-2x\right)=0\)
=> \(\orbr{\begin{cases}x-\frac{1}{2}=0\\\frac{3}{2}-2x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\2x=\frac{3}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{3}{2}:2=\frac{3}{4}\end{cases}}\)
Vậy \(x\in\left\{\frac{1}{2};\frac{3}{4}\right\}\)
X x \(\dfrac{3}{4}\)+ X x\(\dfrac{1}{5}\)+ X x \(\dfrac{1}{20}\)+ X= 1000
\(\left(2x+1\right)\left(4x^2-2x+1\right)\)
\(=\left(2x\right)^3+1\)
\(=8x^3+1\)
⇒ Chọn C