\(C=\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{997.999}\)

\(\Leftrightarrow C=\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{997.999}\right)\)

\(\Leftrightarrow C=\frac{5}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{997}-\frac{1}{999}\right)\)

\(\Leftrightarrow C=\frac{5}{2}\left(1-\frac{1}{999}\right)=\frac{5}{2}.\frac{998}{999}=\frac{2495}{999}=2\frac{497}{999}\)

\(A=\frac{2}{4}+\frac{2}{28}+\frac{2}{70}+\frac{2}{130}+\frac{2}{208}\)

\(\Leftrightarrow A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+\frac{2}{10.13}+\frac{2}{13.16}\)

\(\Leftrightarrow A=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}\right)\)

\(\Leftrightarrow A=\frac{2}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\right)\)

\(\Leftrightarrow A=\frac{2}{3}\left(1-\frac{1}{16}\right)=\frac{2}{3}.\frac{15}{16}=\frac{5}{8}\)

C = 5/1x3 + 5/3x5 + 5/5x7 + ... + 5/997x999

C = 5 - 5/3 + 5/3 - 5/5 + 5/5 - 5/7 + ... + 5/997 - 5/999

C = 5 - 5/999

C = bạn tự tính nhé !

A = 2/4 + 2/28 + 2/70 + 2/130 + 2/208

A = 2/1x4 + 2/4x7 + 2/7x10 + 2/10x13 + 2/13x16

A = 2 - 2/4 + 2/4 - 2/7 + 2/7 - 2/10 + 2/10 - 2/13 + 2/13 - 2/16

A = 2 - 2/16

A = bạn tự tính nhé !

\(\frac{2C}{5}=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+..+\frac{2}{97.99}\)

\(\frac{2C}{5}=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{99-97}{97.99}\)

\(\frac{2C}{5}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

\(\frac{2C}{5}=1-\frac{1}{99}=\frac{98}{99}\Rightarrow C=\frac{98.5}{99.2}=\frac{245}{99}\)

\(\frac{3A}{2}=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}\)

\(\frac{3A}{2}=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+\frac{13-10}{10.13}+\frac{16-13}{13.16}\)

\(\frac{3A}{2}=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\)

\(\frac{3A}{2}=1-\frac{1}{16}=\frac{15}{16}\Rightarrow A=\frac{15.2}{16.3}=\frac{5}{8}\)

a)x^m+4+x^m+3-x-1

=(x^m+4-x)+(x^m+3-1)

=x(x^m+3-1)+(x^m+3-1)

=(x+1)(x^m+3-1)

Với x=-2 ta có:... bn tự thay

b)x⁶-x⁴+2x³+2x²=x⁶-2x⁵+2x⁴+2x⁵-4x⁴+4x³+x⁴-2x³+2x²

=x⁴(x²-2x+2)+2x³(x²-2x+2)+x²(x²-2x+2)

=(x⁴+2x³+x²)(x²-2x+2)

=[x²(x²+2x+1)](x²-2x+2)

=x²(x+1)²(x²-2x+2)

Với x=-2 bn tự thay nhé h mk bận

a: =152,3+7,7+2021,19-2021,19

=160

b: =7/15*3/14*20/13

\(=\dfrac{7}{14}\cdot\dfrac{3}{15}\cdot\dfrac{20}{13}=\dfrac{1}{2}\cdot\dfrac{1}{5}\cdot\dfrac{20}{13}=\dfrac{2}{13}\)

c: \(=\dfrac{7}{4}\left(\dfrac{13}{12}-\dfrac{10}{12}\right)+\dfrac{5}{6}=\dfrac{7}{16}+\dfrac{5}{6}=\dfrac{61}{48}\)

Ta có : A = x² - 4x + 1 

=> A = x² - 2.x.2 + 4 - 3 

=> A = (x - 2)² - 3 

Mà : (x - 2)² \(\ge0\forall x\in R\)

Nên :   (x - 2)² - 3 \(\ge-3\forall x\in R\)

Vậy GTNN của A là -3 khi x = 2 

\(B=4x^2+4x+11=\left(2x\right)^2+2.2x.1+1+10=\left(2x+1\right)^2+10\)

Vì \(\left(2x+1\right)^2\ge0\Rightarrow B=\left(2x+1\right)^2+10\ge10\)

Dấu "=" xảy ra khi (2x+1)²=0 <=> 2x+1=0 <=> x=-1/2

Vậy gtnn của B là 10 khi x=-1/2
---

\(C=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)=\left(x^2+5x-6\right)\left(x^2+5x+6\right)=\left(x^2+5x\right)^2-36\ge-36\)

Dấu "=" xảy ra khi x=0 hoặc x=-5

\(D=5-8x-x^2=5+16-16-8x-x^2=21-\left(16+8x+x^2\right)=21-\left(x+4\right)^2\le21\)

Dấu "=" xảy ra khi x=-4

---

\(E=4x-x^2+1=1+4-x^2+4x-4=5-\left(x^2-4x+4\right)=5-\left(x-2\right)^2\le5\)

Dấu "=" xảy ra khi x=2

Trình bày thì tương tự phần B mình đã trình bày

Vì 13 chia hết cho 13;

133 chia hết cho 13;

17.135 chia hết cho 13;

12 không chia hết cho 13.

Do đó B = 13 + 133 + 177.135 – 12 không chia hết cho 13.

Vì 13 chia hết cho 13;

133 chia hết cho 13;

17.135 chia hết cho 13;

12 không chia hết cho 13.

Do đó B = 13 + 133 + 177.135 – 12 không chia hết cho 13

15.

\(\Delta'=m^2+m-2>0\Leftrightarrow\left[{}\begin{matrix}m>1\\m< -2\end{matrix}\right.\)

Đáp án B

16.

\(\dfrac{\pi}{2}< a< \pi\Rightarrow\dfrac{\pi}{4}< \dfrac{a}{2}< \dfrac{\pi}{2}\Rightarrow\dfrac{\sqrt{2}}{2}< sin\dfrac{a}{2}< 1\Rightarrow\dfrac{1}{2}< sin^2\dfrac{a}{2}< 1\)

\(sina=\dfrac{3}{5}\Leftrightarrow sin^2a=\dfrac{9}{25}\Leftrightarrow4sin^2\dfrac{a}{2}.cos^2\dfrac{a}{2}=\dfrac{9}{25}\)

\(\Leftrightarrow sin^2\dfrac{a}{2}\left(1-sin^2\dfrac{a}{2}\right)=\dfrac{9}{100}\Leftrightarrow sin^4\dfrac{a}{2}-sin^2\dfrac{a}{2}+\dfrac{9}{100}=0\)

\(\Rightarrow\left[{}\begin{matrix}sin^2\dfrac{a}{2}=\dfrac{1}{10}< \dfrac{1}{2}\left(loại\right)\\sin^2\dfrac{a}{2}=\dfrac{9}{10}\end{matrix}\right.\)

\(\Rightarrow sin\dfrac{a}{2}=\dfrac{3\sqrt{10}}{10}\)

17.

Áp dụng công thức trung tuyến:

\(AM=\dfrac{\sqrt{2\left(AB^2+AC^2\right)-BC^2}}{2}=\dfrac{\sqrt{201}}{2}\)

18.

\(\Leftrightarrow x^2+2x+4>m^2+2m\) ; \(\forall x\in\left[-2;1\right]\)

\(\Leftrightarrow m^2+2m< \min\limits_{\left[-2;1\right]}\left(x^2+2x+4\right)\)

Xét \(f\left(x\right)=x^2+2x+4\) trên \(\left[-2;1\right]\)

\(-\dfrac{b}{2a}=-1\in\left[-2;1\right]\) ; \(f\left(-2\right)=4\) ; \(f\left(-1\right)=3\) ; \(f\left(1\right)=7\)

\(\Rightarrow\min\limits_{\left[-2;1\right]}\left(x^2+2x+4\right)=f\left(1\right)=3\)

\(\Rightarrow m^2+2m< 3\Leftrightarrow m^2+2m-3< 0\)

\(\Rightarrow-3< m< 1\Rightarrow m=\left\{-2;-1;0\right\}\)

Đáp án C

19. Hình đa giác là bát giác đều như hình vẽ

\(S=8S_{OAB}=8.\dfrac{1}{2}.IB.OA=4.y_B.x_A=4.\dfrac{\sqrt{2}}{2}.1=2\sqrt{2}\)

Cả 4 đáp án đều không chính xác?

20.

\(M\in\Delta\Rightarrow a+b+1=0\Rightarrow b=-a-1\Rightarrow M\left(a;-a-1\right)\)

\(\left\{{}\begin{matrix}\overrightarrow{AM}=\left(a+1;-a-4\right)\\\overrightarrow{BM}=\left(a-1;-a-2\right)\end{matrix}\right.\)

\(AM+BM=\sqrt{\left(a+1\right)^2+\left(-a-4\right)^2}+\sqrt{\left(-a-2\right)^2+\left(a-1\right)^2}\)

\(AM+BM\ge\sqrt{\left(a+1-a-2\right)^2+\left(-a-4+a-1\right)^2}=\sqrt{26}\)

Dấu "=" xảy ra khi:

\(\left(a+1\right)\left(a-1\right)=\left(-a-4\right)\left(-a-2\right)\Leftrightarrow a=-\dfrac{3}{2}\Rightarrow b=\dfrac{1}{2}\)

\(\Rightarrow ab=-\dfrac{3}{4}\)

a, \(5-\left(\frac{a}{b}+\frac{1}{2}\right)=2\frac{1}{3}\)   =>  \(\frac{a}{b}+\frac{1}{2}=5-2\frac{1}{3}\) =>  \(\frac{a}{b}+\frac{1}{2}=\frac{8}{3}\)  => \(\frac{a}{b}=\frac{8}{3}-\frac{1}{2}\) =>  \(\frac{a}{b}=\frac{13}{6}\)

b, \((\frac{3}{4}+2\frac{1}{2}):\frac{3}{5-3}=\left(\frac{3}{4}+\frac{5}{4}\right):\frac{3}{5}-1=\frac{9}{4}:\frac{-2}{5}=\frac{-45}{8}\)

a, 5-(\(\frac{a}{b}\)+\(\frac{1}{2}\))=2\(\frac{1}{3}\)

<=>5-\(\frac{a}{b}-\frac{1}{2}\)=\(\frac{7}{3}\)

<=>\(\frac{a}{b}=5-\frac{1}{2}-\frac{7}{3}\)

<=>\(\frac{a}{b}=\frac{13}{6}\)

b,(\(\frac{3}{4}\)+2\(\frac{1}{2}\)):\(\frac{3}{5}\)-3

=(\(\frac{3}{4}\)+\(\frac{5}{2}\)).\(\frac{5}{3}\)-3

=\(\frac{23}{4}\).\(\frac{5}{3}\)-3

=\(\frac{115}{12}\)-3

=\(\frac{115-36}{12}\)

=\(\frac{79}{12}\)

a) \(5-\left(\frac{a}{b}+\frac{1}{2}\right)=2\frac{1}{3}\)

\(\Rightarrow5-\left(\frac{a}{b}+\frac{1}{2}\right)=\frac{7}{3}\)

\(\Rightarrow\frac{a}{b}+\frac{1}{2}=5-\frac{7}{3}\)

\(\Rightarrow\frac{a}{b}+\frac{1}{2}=\frac{8}{3}\)

\(\Rightarrow\frac{a}{b}=\frac{8}{3}-\frac{1}{2}\)

\(\Rightarrow\frac{a}{b}=\frac{13}{6}\)

Vậy \(\frac{a}{b}=\frac{13}{6}\)

b) \(\left(\frac{3}{4}+2\frac{1}{2}\right):\frac{3}{5}-3\)

\(=\left(\frac{3}{4}+\frac{5}{2}\right):\frac{3}{5}-3\)

\(=\frac{13}{4}:\frac{3}{5}-3\)

\(=\frac{65}{12}-3\)

\(=\frac{29}{12}\)

Học tốt #

a) 13/12 

b)  7/8

c)  81/4

d) 20/21