Bài 4:

a. Ta có: $-18a+15b=3(-6a+5b)\vdots 3$

b. Vì $-18a+15b$ chia hết cho $3$ với $a,b$ nguyên, mà $-2015\not\vdots 3$ nên không tồn tại hai số $a,b$ nguyên thỏa mãn $-18a+15b=-2015$

Bài 5:

a.

$23\vdots x-2$

$\Rightarrow x-2\in$ Ư$(23)$

$\Rightarrow x-2\in\left\{\pm 1;\pm 23\right\}$

$\Rightarrow x\in\left\{3; 1; 25; -21\right\}$

b.

$2x+1\in$ Ư$(-12)$, mà $2x+1$ lẻ nên:

$2x+1\in\left\{\pm 1; \pm 3\right\}$

$\Rightarrow x\in\left\{0; -1; 1; -2\right\}$

c.

$x-1\vdots x+2$

$(x+2)-3\vdots x+2$

$3\vdots x+2$

$\Rightarrow x+2\in$ Ư$(3)$

$\Rightarrow x+2\in\left\{\pm 1;\pm 3\right\}$

$\Rightarrow x\in\left\{-1; -3; -5; 1\right\}$

\(21_{10}=10101_2\)

\(\dfrac{5^4.20^4}{25^5.4^5}=\dfrac{5^4.5^44^4}{5^4.5.5^4.5.4^4.4}=\dfrac{5^8.4^4}{5^4.5^4.5.5.4^4.4.4.}=\dfrac{5^8.4^4}{5^8.5^2.4^4.4}\)\(=\dfrac{1}{5^2.4}=\dfrac{1}{100}=0,01\)

c)

\(\dfrac{5^4.20^4}{25^5.4^5}\)

\(=\dfrac{5^4.\left(5.4\right)^4}{\left(5^2\right)^5.\left(2^2\right)^5}\)

\(=\dfrac{5^4.5^4.4^4}{5^{10}.2^{10}}\)

\(=\dfrac{5^8.\left(2^2\right)^4}{5^{10}.2^{10}}\)

\(=\dfrac{5^8.2^8}{5^{10}.2^{10}}\)

\(=\dfrac{1}{5^2.2^2}\)

\(=\dfrac{1}{25.4}\)

\(=\dfrac{1}{100}\)

\(a,\left(\dfrac{3}{7}+\dfrac{1}{2}\right)^2=\left(\dfrac{6}{14}+\dfrac{7}{14}\right)^2=\left(\dfrac{13}{14}\right)^2=\dfrac{169}{196}\)

\(b,\left(\dfrac{3}{4}-\dfrac{4}{5}\right)^2=\left(\dfrac{15}{20}-\dfrac{16}{20}\right)^2=\left(\dfrac{-1}{20}\right)^2=\dfrac{1}{400}\)

\(c,\dfrac{5^4\cdot20^4}{25^5\cdot4^5}=\dfrac{\left(5\cdot20\right)^4}{\left(25\cdot4\right)^5}=\dfrac{\left(100\right)^4}{\left(100\right)^5}=\dfrac{100000000}{10000000000}=\dfrac{1}{100}\)

\(d,\dfrac{\left(\dfrac{2}{3}\right)^3\cdot\left(\dfrac{-3}{4}\right)^2\cdot\left(-1\right)^{2003}}{\left(\dfrac{2}{5}\right)^2\cdot\left(-\dfrac{5}{12}\right)^3}=\dfrac{\dfrac{8}{27}\cdot\dfrac{9}{16}\cdot\left(-1\right)}{\dfrac{4}{25}\cdot\left(-\dfrac{125}{1728}\right)}\)

\(=\dfrac{\dfrac{72}{432}\cdot\left(-1\right)}{\left(-\dfrac{500}{43200}\right)}=\dfrac{\dfrac{1}{6}\cdot\left(-1\right)}{\left(-\dfrac{5}{432}\right)}=\dfrac{-\dfrac{1}{6}}{-\dfrac{5}{432}}=\dfrac{72}{5}\)

328.8

4.8⋅3.5+0.6⋅65⋅8

328,8

b) \(B=\dfrac{\sqrt{10}\left(\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}-\sqrt{5}}+\dfrac{6\left(2+\sqrt{10}\right)}{4-10}+\sqrt{\left(3\sqrt{7}+2\right)^2}\)

\(=\sqrt{10}-2-\sqrt{10}+3\sqrt{7}+2=3\sqrt{7}\)

- Lê Hữu Trác

Lê Hữu Trác

Lê Hữu Trác

1³+2³+3³+...+100³

=1+2+1.2.3+3+2.3.4+100+99.100.101

=(1+2+3+...+100)+(1.2.3+2.3.4+...+99.100.101)

=5050+101989800

=101994850

NHỚ T.I.C.K và KB với mk nha

1.A
2.A
3.B
4.C
5.B
6.C
7.A
8.A
9.B
10.A
11.B
12.A
13.C
14.B
15.B
16.A
17.A
18.A
19.A
20.C

Cửa hàng đã bán:

420 x 25% = 105(kg)

Lượng gạo cửa hàng còn lại sau bán:

420 - 105= 315 (kg)

\(Q=x-2-2\sqrt{x-2}+4\)

\(=\left(\sqrt{x-2}-1\right)^2+3>=3\)

Dấu = xảy ra khi x=3

\(Q=x+2-2\sqrt{x-2} \)

\(Q=x-2-2\sqrt{x-2}+4 \)

\(Q=(\sqrt{x-2}-1)^2+3\)

Vì \((\sqrt{x-2}-1)^2\)\(\ge\)\(0\)

\(\Leftrightarrow\)\((\sqrt{x-2}-1)^2+3\)\(\ge3\)

Vậy GTNN Q=3