\(\Rightarrow\left(x-4\right)\left(2x+x-4\right)=0\\ \Rightarrow\left(x-4\right)\left(3x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{4}{3}\end{matrix}\right.\)

S=1+2+2^2+2^3+...+2^2002

2S=2+2^2+2^3+2^4+...+2^2003

2S-S=(2+2^2+2^3+2^4+...+2^2003)-(1+2+2^2+2^3+...+2^2002)

2S-S=(2-2)+(2^2-2^2)+(2^3-2^3)+...+(2^2002-2^2002)+(2^2003-1)

S=0+0+0+...+0+(2^2003-1)

S=2^2003-1 (rút gọn)

tick nha

S=1+2+2^2+2^3+...+2^2002

=>2S=2+2²+2³+2⁴+...+2²⁰¹³

=>2S-S=2+2²+2³+2⁴+...+2²⁰¹³-1-2-2²-2³-...-2²⁰¹²

=>S=2²⁰¹³-1

\(a,=\left(x+2-3x\right)\left(x+2+3x\right)=4\left(1-x\right)\left(2x+1\right)\\ b,=25-\left(x+y\right)^2=\left(5-x-y\right)\left(5+x+y\right)\)

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

S=3⁰+3²+3⁴+3⁶+...+3²⁰⁰

6S=3²+3⁴+3⁶+...+3²⁰²

6S-S=(3²+3⁴+3⁶+...+3²⁰²)-(1+3²+3⁴+...+3²⁰⁰)

5S=1+(3²-3²)+(3⁴-3⁴)+...+(3²⁰⁰-3²⁰⁰)+3²⁰²

S=(3²⁰⁰+1):5\(\frac{ }{ }\)

\(\text{C=1+2-3-4+5+6-7-8+9+...+2002-2003-2004+2005+2006}\)

\(\text{C=1+(2-3-4+5)+(6-7-8+9)+...+(2002-2003-2004+2005)+2006}\)

\(\text{C=1+0+0+...+0+2006}\)

\(\text{C=1+2006}\)

\(C=2007\)

HỌC TỐT!!!

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                                       Bài giải

\(S=(1+2-3-4)+(5+6-7-8)+......+(2001+2002-2003-2004)+(2005+2006)\)

\(S=(-4)+(-4)+.......+(-4)+(2005+2006)\) ( Có tất cả ( 2004 - 1 + 1 ) : 4 = 501 số hạng - 4 )
\(S= -2004+(2005+2006)\)
\(S=-2004+4011\)
\(S=2007\)

a: =>\(4\cdot3^x\cdot\dfrac{1}{3}+2\cdot3^x\cdot9=4\cdot3^6+2\cdot3^9\)
=>3^x(4*1/3+2*9)=3^6(4+2*3^3)

=>3^x*58/3=3^6*58

=>3^x/3^6=3

=>x-6=1

=>x=7

b: =>\(2^x\cdot\left(\dfrac{1}{5}+\dfrac{1}{3}\cdot2\right)=2^7\left(\dfrac{1}{5}+\dfrac{1}{3}\cdot2\right)\)

=>2^x=2^7

=>x=7