Toán

Xét (O) có

ΔABC nội tiếp

AB là đường kính

=>ΔABC vuông tại C

=>BC vuông góc AC

Xét ΔKAB vuông tại A có AC là đường cao

nên BC*BK=BA^2=4*R^2

\(\dfrac{7}{x}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+\dfrac{4}{13.17}+...+\dfrac{4}{41.45}=\dfrac{29}{45}\)

\(\Rightarrow\dfrac{7}{x}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{17}+...+\dfrac{1}{37}-\dfrac{1}{41}+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{29}{45}\)

\(\Rightarrow\dfrac{7}{x}+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{29}{45}\)

\(\Rightarrow\dfrac{7}{x}+\dfrac{8}{45}=\dfrac{29}{45}\)

\(\Rightarrow\dfrac{7}{x}=\dfrac{29}{45}-\dfrac{8}{45}=\dfrac{7}{15}\)

\(\Leftrightarrow x=15\)

=>\(\dfrac{7}{x}+\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}\right)=\dfrac{29}{45}\)

=>7/x+1/5-1/45=29/45

=>7/x+8/45=29/45

=>7/x=21/45

=>7/x=7/15

=>x=15

\(\dfrac{7}{x}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}=\dfrac{29}{45}\)

\(\Rightarrow\dfrac{7}{x}+\left(\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}\right)=\dfrac{29}{45}\)

\(\Rightarrow\dfrac{7}{x}+\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}\right)=\dfrac{29}{45}\)

\(\Rightarrow\dfrac{7}{x}+\left(\dfrac{1}{5}-\dfrac{1}{45}\right)=\dfrac{29}{45}\)

\(\Rightarrow\dfrac{7}{x}+\dfrac{8}{45}=\dfrac{29}{45}\)

\(\Rightarrow\dfrac{7}{x}=\dfrac{29}{45}-\dfrac{8}{45}\)

\(\Rightarrow\dfrac{7}{x}=\dfrac{21}{45}\)

\(\Rightarrow\dfrac{7}{x}=\dfrac{7}{15}\)

\(\Rightarrow x=15\)

Lớp 6A góp được 200*1/5=40 quyển

Lớp 6B góp được 40*150%=60 quyển

Lớp 6C góp được:

200-40-60=100 quyển

a) \(\dfrac{-5}{9}-\dfrac{4}{15}+\dfrac{2}{9}-\dfrac{11}{15}\)

\(=\left(\dfrac{-5}{9}+\dfrac{2}{9}\right)-\left(\dfrac{4}{15}+\dfrac{11}{15}\right)\)

\(=\dfrac{-3}{9}-\dfrac{15}{15}\)

\(=-\dfrac{1}{3}-1\)

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

b) \(\dfrac{-8}{13}-\dfrac{7}{16}+\dfrac{21}{13}-\dfrac{1}{6}\)

\(=\left(\dfrac{-8}{13}+\dfrac{21}{13}\right)-\dfrac{7}{16}-\dfrac{1}{6}\)

\(=\dfrac{13}{13}-\dfrac{29}{48}\)

\(=1-\dfrac{29}{48}\)

\(=\dfrac{19}{48}\)

c) \(\dfrac{-15}{25}-\dfrac{14}{21}+\dfrac{8}{5}-\dfrac{5}{3}\)

\(=\dfrac{-3}{5}-\dfrac{2}{3}+\dfrac{8}{5}-\dfrac{5}{3}\)

\(=\left(\dfrac{-3}{5}+\dfrac{8}{5}\right)-\left(\dfrac{2}{3}+\dfrac{5}{3}\right)\)

\(=\dfrac{5}{5}-\dfrac{7}{3}\)

\(=1-\dfrac{7}{3}\)

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

a: =-5/9+2/9-11/15-4/15

=-3/9-1

=-12/9=-4/3

b: =-8/13+21/13-7/16-1/6

=1-1/6-7/16

=5/6-7/16

=19/48

c: =-15/25+8/5-2/3-5/3

=-7/3+1

=-4/3

a) \(-\dfrac{5}{9}-\dfrac{4}{15}+\dfrac{2}{9}-\dfrac{11}{15}\)

\(=\left(-\dfrac{5}{9}+\dfrac{2}{9}\right)+\left(-\dfrac{4}{15}-\dfrac{11}{15}\right)\)

\(=-\dfrac{1}{3}-1=-\dfrac{4}{3}\)

b) \(-\dfrac{8}{13}-\dfrac{7}{16}+\dfrac{21}{13}-\dfrac{1}{16}\)(sửa đề)

\(=\left(-\dfrac{8}{13}+\dfrac{21}{13}\right)+\left(-\dfrac{7}{16}-\dfrac{1}{16}\right)\)

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

c) \(-\dfrac{15}{25}-\dfrac{14}{21}+\dfrac{8}{5}-\dfrac{5}{3}\)

\(=-\dfrac{3}{5}-\dfrac{2}{3}+\dfrac{8}{5}-\dfrac{5}{3}\)

\(=\left(-\dfrac{3}{5}+\dfrac{8}{5}\right)-\left(\dfrac{2}{3}+\dfrac{5}{3}\right)\)

\(=1-\dfrac{7}{3}=\dfrac{-4}{3}\)

\(B=\dfrac{1}{6.10}+\dfrac{1}{10.14}+\dfrac{1}{14.18}+...+\dfrac{1}{402.406}\)

\(4B=\dfrac{4}{6.10}+\dfrac{4}{10.14}+\dfrac{4}{14.18}...+\dfrac{4}{402.406}\)

\(=\dfrac{1}{6}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{18}+...+\dfrac{1}{398}-\dfrac{1}{402}+\dfrac{1}{402}-\dfrac{1}{406}\)

\(=\dfrac{1}{6}-\dfrac{1}{406}=\dfrac{100}{609}\)

\(\Rightarrow B=\dfrac{25}{609}\)

\(B=\dfrac{1}{4}\cdot\left(\dfrac{4}{6\cdot10}+\dfrac{4}{10\cdot14}+...+\dfrac{4}{402\cdot406}\right)\)

=1/4(1/6-1/10+1/10-1/14+...+1/402-1/406)

=1/4*(1/6-1/406)

=1/4*400/(6*406)

=100/(6*406)=25/609

`TXĐ: R\\{-m}`

Ta có: `y'=[2m-1]/[(x+m)^2]`

Để hàm số đồng biến trên `(-oo;-2)=>{(2m-1 >= 0),(x < -2; x ne -m):}`

                  `<=>{(m >= 1/2),(-m >= -2):}<=>{(m >= 1/2),(m <= 2):}` 

       `->\bb D`

\(A=\dfrac{-1}{199}-\dfrac{1}{199\cdot198}-\dfrac{1}{198\cdot197}-...-\dfrac{1}{3\cdot2}-\dfrac{1}{2}\)

\(A=\dfrac{-1}{199}-\left(\dfrac{1}{198}-\dfrac{1}{199}\right)-\left(\dfrac{1}{197}-\dfrac{1}{198}\right)-...-\left(\dfrac{1}{2}-\dfrac{1}{3}\right)-\dfrac{1}{2}\)

\(A=\dfrac{-1}{199}-\dfrac{1}{198}+\dfrac{1}{199}-\dfrac{1}{197}+\dfrac{1}{198}-...-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{2}\)

\(A=-\dfrac{1}{2}-\dfrac{1}{2}\)

\(A=\dfrac{-1-1}{2}\)

\(A=\dfrac{-2}{2}\)

\(A=-1\)

\(=\dfrac{-1}{199}-\left(\dfrac{1}{2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{198\cdot199}\right)\)

\(=\dfrac{-1}{199}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{198}-\dfrac{1}{199}\right)\)

=-1/199-(1-1/199)

=-1/199-1+1/199=-1

\(A=\dfrac{-1}{199}-\dfrac{1}{199.198}-\dfrac{1}{198.197}-\dfrac{1}{197.196}-...-\dfrac{1}{3.2}-\dfrac{1}{2}\)

\(=-\dfrac{1}{199}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{197.198}+\dfrac{1}{198.199}\right)\)

\(=-\dfrac{1}{199}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-...-\dfrac{1}{197}+\dfrac{1}{197}-\dfrac{1}{198}+\dfrac{1}{198}-\dfrac{1}{199}\right)\)

\(=-\dfrac{1}{199}-\left(1-\dfrac{1}{199}\right)\)

\(=-\dfrac{1}{199}-\dfrac{198}{199}=-\dfrac{199}{199}=-1\)

Xét ΔABC vuông tại A áp dụng định lý Py-ta-go ta có:

\(BC^2=AB^2+AC^2\)

\(\Rightarrow BC^2=3^2+4^2\)

\(\Rightarrow BC^2=9+16\)

\(\Rightarrow BC^2=25\)

\(\Rightarrow BC=\sqrt{25}\)

\(\Rightarrow BC=5\left(cm\right)\)

`a,x^2 + 4x + 4`

`=x^2 + 2 . x . 2 + 2^2`

`=(x+2)^2`

`b,16a^{2} - 16ab + 4b^{2}`

`=(4a)^{2} - 2 . 4a . 2b + (2b)^{2}`

`=(4a-2b)^{2}`

a)\(x^2+4x+4\)

\(=x^2+2\cdot x\cdot2+2^2\)

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

b) \(16a^2-16ab+4b^2\)

\(=\left(4a\right)^2-2\cdot4a\cdot2b+\left(2b\right)^2\)

\(=\left(4a-2b\right)^2\)