\(a,\) Ta có : \(\hept{\begin{cases}2^{10}=2^{10}\\3^{12}=3^{10}.3^2\end{cases}}\)

Vì \(3^{10}>2^{10}\Rightarrow2^{10}< 3^{10}.3^2\)

Hay \(2^{10}< 3^{12}\)

\(b,\)  Ta có : \(\hept{\begin{cases}33^{52}=\left(33^4\right)^{13}=1185921^{13}\\44^{39}=\left(44^3\right)^{13}=85184^{13}\end{cases}}\)

Vì \(1185921^{13}>85184^{13}\)

Do đó : \(33^{52}>44^{39}\)

6x+2x+x = 10

8x+x = 10

9x = 10

<=> x = 10/9

\(6x-2x+x=10\)

\(\Rightarrow x.\left(6-2+1\right)=10\)

\(\Rightarrow x.5=10\)

\(\Rightarrow x=10:5\)

\(\Rightarrow x=2\)

a) Do \(2x=3y=-2z\) nên \(\frac{2x}{1}=\frac{3y}{1}=\frac{4z}{-2}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta được:
\(\frac{2x}{1}=\frac{3y}{1}=\frac{4z}{-2}=\frac{2x-3y+4z}{1-1+\left(-2\right)}=\frac{48}{-2}=-24\)    ( do 2x - 3y + 4z = 48 )
Khi đó: 
\(\frac{2x}{1}=-24\)\(\Rightarrow2x=-24\)\(\Rightarrow x=\frac{-24}{2}=-12\)
\(\frac{3y}{1}=-24\)\(\Rightarrow3y=-24\)\(\Rightarrow y=\frac{-24}{3}=-8\)
\(\frac{4z}{-2}=-24\)\(\Rightarrow-2z=-24\)\(\Rightarrow z=\frac{-24}{-2}=12\)
Vậy x = -12 ; y = -8 ; z = 12

Vũ Quang Vinh: tks bạn nhiềuu

x chỉ có thể bằng 0

hoặc 1

\(\frac{a}{2}=\frac{b}{3}\Rightarrow\frac{a}{8}=\frac{b}{12};\frac{b}{4}=\frac{c}{5}\Rightarrow\frac{b}{12}=\frac{c}{15}\)

\(\Rightarrow\frac{a}{8}=\frac{b}{12}=\frac{c}{15}=\frac{a-b+c}{8-12+15}=\frac{33}{11}=3\)

\(\frac{a}{8}=3\Rightarrow a=3.8=24\)

\(\frac{b}{12}=3\Rightarrow b=12.3=36\)

\(\frac{c}{15}=3\Rightarrow c=45\)

\(\left(x-5\right).\left(x+\frac{1}{2}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+\frac{1}{2}=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=5\\x=-\frac{1}{2}\end{cases}}\)

Vậy x = 5 ; x = -1/2

\(\left(x-5\right)\cdot\left(x+\frac{1}{2}\right)=0\)

• x - 5 =0 => x = 5
• x + 1/2 = 0 => x = -1/2

bằng 10

\(A=x^5-5x^4+5x^3-5x^2+5x\)

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

\(=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x\)

\(=x\)

\(=4\)