\(D=\dfrac{15}{3\left|2x+1\right|+5}\)

Ta có:

\(\left\{{}\begin{matrix}15>0\\3\left|2x+1\right|\ge5\forall x\end{matrix}\right.\)Nên:

\(\Rightarrow D=\dfrac{15}{3\left|2x-1\right|+5}\le3\left(=\dfrac{15}{5}\right)\forall x\) 

Dấu "=" xảy ra:

\(\dfrac{15}{3\left|2x+1\right|+5}=3\)

\(\Rightarrow3\left|2x+1\right|+5=5\)

\(\Rightarrow3\left|2x+1\right|=0\)

\(\Rightarrow\left|2x+1\right|=0\)

\(\Rightarrow2x+1=0\)

\(\Rightarrow2x=-1\)

\(\Rightarrow x=-\dfrac{1}{2}\)

Vậy: \(D_{max}=3\) khi \(x=-\dfrac{1}{2}\)

D = \(\dfrac{15}{3.\left|2x-1\right|+5}\)  vì |2\(x\) - 1| ≥ 0 ∀ \(x\) ⇒3.|2\(x-1\)| + 5 ≥ 5 ∀ \(x\)

⇒D = \(\dfrac{15}{3.\left|2x-1\right|+5}\) ≤ \(\dfrac{15}{5}\) = 3 dấu bằng xảy ra khi 2\(x\) - 1 =0 ⇒ \(x=\dfrac{1}{2}\)

Kết luận D_min = 3 ⇔ \(x\) = \(\dfrac{1}{2}\)

a)

Vì tam giác ABC cân tại A (gt)

=> AB = AC (TC tam giác cân)

Xét tam giác ABM và tam giác ACM có:

AB = AC (CMT)

AM chung

BM = CM (AM là đường trung tuyến)

=> tam giác ABM = tam giác ACM (c - c - c)

a. \(\dfrac{1}{2.4}\) + \(\dfrac{1}{4.6}\) + \(\dfrac{1}{6.8}\) + ...... + \(\dfrac{1}{20.22}\)

= 1/2 ( 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + ..... + 1/20 - 1/22)

=1/2 ( 1/2 - 1/22)

= 1/2 . 5/11

= 5/22

b. 1+ 2/3 + 2/6 + 2/10 +...+ 2/45

=>1/2.(1+2/3+2/6+....+2/45)=1/2+2/6+2/12+...+2/90

=1/2+2/2.3+2/3.4+...+2/9.10
=2.(1/4+3-2/2.3+4-3/3.4+...+10-9/9.10)

=2. ( 1/4+1/2-1/3+1/3-1/4+.....+1/9-1/10)
= 2.( 1/4-1/10)=2.3/20=3/10

=> vì 1/2.*=3/10

=> *=3/10:1/2=3/5

tick mình nhé

B = 1 + \(\dfrac{2}{3}\) + \(\dfrac{2}{6}\) +\(\dfrac{2}{10}\) + \(\dfrac{2}{15}\)+...+ \(\dfrac{2}{45}\)

B = 1 + 2.(\(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\)+...+ \(\dfrac{1}{45}\))

B = 1 + \(\dfrac{4}{2}\).(\(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) +\(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + ...+ \(\dfrac{1}{45}\))

B = 1 + 4.( \(\dfrac{1}{6}\) +\(\dfrac{1}{12}\)+ \(\dfrac{1}{20}\)+ \(\dfrac{1}{30}\)+...+ \(\dfrac{1}{90}\))

B = 1 + 4.(\(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\)+ \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\)+...+\(\dfrac{1}{9.10}\))

B = 1 + 4 .( \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\)+...+ \(\dfrac{1}{9}\) - \(\dfrac{1}{10}\))

B = 1 + 4.( \(\dfrac{1}{2}\) - \(\dfrac{1}{10}\))

B = 1 + 4. \(\dfrac{2}{5}\)

B = \(\dfrac{13}{5}\)

\(\left(4x+1\right)\left(-2+\dfrac{1}{3}\right)=0\\ \Rightarrow\left(4x+1\right)\left(-\dfrac{6}{3}+\dfrac{1}{3}\right)=0\\ \Rightarrow\left(4x+1\right)\left(-\dfrac{5}{3}\right)=0\\ \Rightarrow4x+1=0:\left(-\dfrac{5}{3}\right)\\ \Rightarrow4x+1=0\\ \Rightarrow4x=0-1\\ \Rightarrow4x=-1\\ \Rightarrow x=-\dfrac{1}{3}\)

\(\dfrac{1}{7}-\dfrac{3}{5}x\text{=}\dfrac{3}{5}\)

\(\dfrac{3}{5}x\text{=}\dfrac{1}{7}-\dfrac{3}{5}\)

\(\dfrac{3}{5}x\text{=}\dfrac{-16}{35}\)

\(x\text{=}\dfrac{-16}{35}:\dfrac{3}{5}\)

\(x\text{=}\dfrac{-16}{21}\)

Đổi 5 tấn = 5000kg; 1 hec-ta= 10000m²

Lượng thóc thu được 1 sào bắc bộ:

(360:10000) x 5000= 180(kg)

Lượng gạo thu được trên 1 sào bắc bộ:

180 x 4/5 = 144(kg)

Đ.số: 144kg gạo

a) Vì \(-45< -16\) nên \(\left(-\dfrac{45}{17}\right)^{15}< \left(\dfrac{-16}{17}\right)^{15}\)

b) Vì \(21< 23\) nên \(\left(-\dfrac{8}{9}\right)^{21}< \left(-\dfrac{8}{9}\right)^{23}\)

c) \(27^{40}=3^{3^{40}}=3^{120}\)

\(64^{60}=8^{2^{60}}=8^{120}\)

Vì \(3< 8\) nên \(3^{120}< 8^{120}\) hay \(27^{40}< 64^{60}\)

con ai kooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo

5a.

$\frac{1}{2}-\sqrt{x}=0$

$\Rightarrow \sqrt{x}=\frac{1}{2}$
$\Rightarrow x=(\frac{1}{2})^2=\frac{1}{4}$

5b.

$\frac{5}{11}\sqrt{x}-\frac{1}{6}=\frac{1}{3}$

$\Rightarrow \frac{5}{11}\sqrt{x}=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}$

$\Rightarrow \sqrt{x}=\frac{1}{2}: \frac{5}{11}=\frac{11}{10}$

$\Rightarrow x=(\frac{11}{10})^2=\frac{121}{100}$

5c.

$-\frac{4}{3}\sqrt{x}+\frac{8}{5}=\frac{1}{3}+\frac{2}{3}=1$

$\Rightarrow -\frac{4}{3}\sqrt{x}=1-\frac{8}{5}=\frac{-3}{5}$

$\Rightarrow \frac{4}{3}\sqrt{x}=\frac{3}{5}$

$\Rightarrow \sqrt{x}=\frac{3}{5}: \frac{4}{3}=\frac{9}{20}$
$\Rightarrow x=(\frac{9}{20})^2=\frac{81}{400}$

5d.

$x-6\sqrt{x}=0$

$\Rightarrow \sqrt{x}(\sqrt{x}-6)=0$

$\Rightarrow \sqrt{x}=0$ hoặc $\sqrt{x}-6=0$

$\Rightarrow \sqrt{x}=0$ hoặc $\sqrt{x}=6$

$\Rightarrow x=0$ hoặc $x=36$

5e.

$1-3x^2=7$

$3x^2=1-7=-6$

$x^2=-2<0$ (vô lý)

Do đđ không tồn tại $x$ thỏa mãn đề.

5f.

$7x^2-4=1$

$7x^2=1+4=5$

$x^2=\frac{5}{7}=(\sqrt{\frac{5}{7}})^2=(-\sqrt{\frac{5}{7}})^2$

$\Rightarrow x=\pm \sqrt{\frac{5}{7}}$

C = 1/(9.10) - 1/(8.9) - 1/(7.8) - ... - 1/(2.3) - 1/(1.2)

= 1/9 - 1/10 - 1/8 + 1/9 - 1/7 + 1/8 - ... - 1/2 + 1/3 - 1 + 1/2

= 1/9 - 1/10 + 1/9 - 1

= 2/9 - 11/10

= -79/90