\(A=1+4+...+4^{23}\)

\(\Rightarrow4A=4+4^2+...+4^{24}\)

\(\Rightarrow3A=4^{24}-1\)

\(\Rightarrow3A+1=4^{24}=64^8>62^7\)

Vậy...

a, \(B=\dfrac{10^{12}+1}{10^{12}+1}=1\)

+) Xét \(n>12\Rightarrow A>1=B\)

+) Xét \(n< 12\Rightarrow A< B=1\)

Vậy...

b, \(\overline{abc}-\overline{deg}⋮7\)

\(\Rightarrow\left\{{}\begin{matrix}\overline{abc}⋮7\\\overline{deg}⋮7\end{matrix}\right.\)

Ta có: \(\overline{abcdeg}=1000\overline{abc}+\overline{deg}⋮7\) ( do \(\left(1000;7\right)=1\) )

\(\Rightarrowđpcm\)

nick moi cua LeChi Cong ne

b, \(\dfrac{\sqrt{2}+\sqrt{3}}{2+\sqrt{6}}=\dfrac{\sqrt{2}+\sqrt{3}}{\sqrt{2}\left(\sqrt{2}+\sqrt{3}\right)}=\dfrac{1}{\sqrt{2}}\)

d, sai đề?

\(\left\{{}\begin{matrix}\left|5-x\right|\ge0\\\left|x-2\right|\ge0\end{matrix}\right.\Rightarrow\left|5-x\right|+\left|x-2\right|\ge0\)

\(\Rightarrow-3x\ge0\Rightarrow x\le0\)

\(\Rightarrow5-x+2-x=-3x\)

\(\Rightarrow x=-7\)

Vậy x = -7

ĐKXĐ: \(x\ne1\), x > 0

\(H=\dfrac{1}{\sqrt{x-1}-\sqrt{x}}+\dfrac{1}{\sqrt{x-1}+\sqrt{x}}+\dfrac{\sqrt{x^3}-x}{\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x-1}+\sqrt{x}+\sqrt{x-1}-\sqrt{x}}{\left(\sqrt{x-1}-\sqrt{x}\right)\left(\sqrt{x-1}+\sqrt{x}\right)}+\dfrac{x\left(\sqrt{x}-1\right)}{\sqrt{x-1}}\)

\(=\dfrac{2\sqrt{x-1}}{x-1-x}+x\)

\(=x-2\sqrt{x-1}\)

1,

Để \(\overline{^{\circledast}471}⋮3\Rightarrow12+^{\circledast}⋮3\)

Và \(\overline{^{\circledast}471}⋮9\)

\(\Rightarrow^{\circledast}\in\left\{3;9\right\}\) ( * khác 0 do * là số đầu )

Vậy...

2, tương tự

3, \(\overline{135xy}⋮5,9\Rightarrow y=5\)

Thay y = 5\(\Rightarrow\overline{135x5}⋮9\Rightarrow14+x⋮9\Rightarrow x=4\)

Vậy x = 4, y = 5

Phân số chỉ số học sinh giỏi là:

\(\dfrac{5}{4+5}=\dfrac{5}{9}\) ( số học sinh cả lớp )

Phân số chỉ 5 em học sinh trung bình là:

\(1-\left(\dfrac{5}{9}+\dfrac{1}{3}\right)=\dfrac{1}{9}\) ( số học sinh cả lớp )

Số học sinh cả lớp là:

\(5:\dfrac{1}{9}=45\) ( học sinh )

Số học sinh giỏi là:

\(45.\dfrac{5}{9}=25\) ( học sinh )

Số học sinh khá là:

\(45.\dfrac{1}{3}=15\) ( học sinh )

Vậy...

Ta có: \(\left(\dfrac{4}{9}\right)^{15}=\left(\dfrac{2}{3}\right)^{30}\)

\(\left(\dfrac{8}{27}\right)^{13}>\left(\dfrac{8}{27}\right)^{10}=\left(\dfrac{2}{3}\right)^{30}\)

\(\Rightarrow\left(\dfrac{8}{27}\right)^{13}>\left(\dfrac{4}{9}\right)^{15}\)

a, \(9-x^2+2xy-y^2\)

\(=9-\left(x-y\right)^2\)

\(=\left(3-x+y\right)\left(3+x-y\right)\)

b, \(x^4-x^2+4x-4\)

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

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

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

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

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

c, \(x^3-2x^2y+xy^2\)

\(=x^3-x^2y-x^2y+xy^2\)

\(=x^2\left(x-y\right)-xy\left(x-y\right)\)

\(=\left(x^2-xy\right)\left(x-y\right)\)

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

d, \(1-x^2-2xz-z^2\)

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

\(=\left(1-x-z\right)\left(1+x+z\right)\)