A B C D 150 37,5 Gọi D như hình vẽ ta có \(\Delta ABD\) chính là tam giác chung đỉnh A với \(\Delta ABC\) cà cũng chính là đường cao của \(\Delta ABD\)

Chiều cao của \(\Delta ABC\) hay \(\Delta ABD\) là :

   \(37,5.2:5=15\left(cm\right)\)

Cạnh BC là :

   \(150.2:5=20\left(cm\right)\)

    Đáp số : \(20cm\)

hay

A C B x 3cm 5cm
a. Trong ba điểm A, B, C điểm nào nằm giữa hai điểm còn lại?
Trên tia Ax, ta có: AC < AB (vì 3cm < 5cm)
=> Điểm C nằm giữa A và B
b. Tính độ dài đoạn thẳng BC
Ta có: Điểm C nằm giữa A và B
=> AC + BC = AB
Hay 3 + BC = 5
=> BC = 5 - 3 = 2(cm)

a;\(\dfrac{-6}{11}\) : \(\dfrac{12}{55}\) = \(\dfrac{-5}{2}\)

b;\(\dfrac{7}{12}\) + \(\dfrac{5}{72}\) - \(\dfrac{11}{36}\) = \(\dfrac{47}{72}\) - \(\dfrac{11}{36}\) = \(\dfrac{25}{72}\)

c;\(\dfrac{13}{10}\) : \(\dfrac{-5}{13}\) = \(\dfrac{-169}{50}\)

d; {\(\dfrac{5}{12}\) + \(\dfrac{5}{11}\) } : { \(\dfrac{5}{3}\) -\(\dfrac{7}{11}\) } = \(\dfrac{115}{132}\) : \(\dfrac{34}{33}\) = \(\dfrac{115}{136}\)

lưu ý mk ko chép đầu bài

mình cần gấp lắm đến chiều mai là phải nộp rùi

giúp mình nha thanks cá bạn trước ko có tâm trạng mà cười nữa

Bài 1:

a)

\(\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Leftrightarrow\dfrac{x-1}{9}=\dfrac{24}{9}\\ \Leftrightarrow x-1=24\\ x=24+1\\ x=25\)

b)

\(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{8}\\ \dfrac{3x}{7}+1=\dfrac{-1}{8}\cdot\left(-4\right)\\ \dfrac{3x}{7}+1=\dfrac{1}{2}\\ \dfrac{3x}{7}=\dfrac{1}{2}-1\\ \dfrac{3x}{7}=\dfrac{-1}{2}\\ 3x=\dfrac{-1}{2}\cdot7\\ 3x=\dfrac{-7}{2}\\ x=\dfrac{-7}{2}:3\\ x=\dfrac{-7}{6}\)

c)

\(x+\dfrac{7}{12}=\dfrac{17}{18}-\dfrac{1}{9}\\ x+\dfrac{7}{12}=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{12}\\ x=\dfrac{1}{4}\)

d)

\(0,5x-\dfrac{2}{3}x=\dfrac{7}{12}\\ \dfrac{1}{2}x-\dfrac{2}{3}x=\dfrac{7}{12}\\ x\cdot\left(\dfrac{1}{2}-\dfrac{2}{3}\right)=\dfrac{7}{12}\\ \dfrac{-1}{6}x=\dfrac{7}{12}\\ x=\dfrac{7}{12}:\dfrac{-1}{6}\\ x=\dfrac{-7}{2}\)

e)

\(\dfrac{29}{30}-\left(\dfrac{13}{23}+x\right)=\dfrac{7}{46}\\ \dfrac{29}{30}-\dfrac{13}{23}-x=\dfrac{7}{46}\\ \dfrac{277}{690}-x=\dfrac{7}{46}\\ x=\dfrac{277}{690}-\dfrac{7}{46}\\ x=\dfrac{86}{345}\)

f)

\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\\ \left(x-\dfrac{1}{12}\right):\dfrac{23}{12}=\dfrac{7}{46}\\ x-\dfrac{1}{12}=\dfrac{7}{46}\cdot\dfrac{23}{12}\\ x-\dfrac{1}{12}=\dfrac{7}{24}\\ x=\dfrac{7}{24}+\dfrac{1}{12}\\ x=\dfrac{3}{8}\)

g)

\(\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{13}{15}-\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{1}{6}\\ \dfrac{13}{21}+x=\dfrac{1}{6}:\dfrac{7}{12}\\ \dfrac{13}{21}+x=\dfrac{2}{7}\\ x=\dfrac{2}{7}-\dfrac{13}{21}\\ x=\dfrac{-1}{3}\)

h)

\(2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}+\dfrac{3}{2}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}:2\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{8}\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\end{matrix}\right.\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\ \dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{29}{24}\\ x=\dfrac{29}{24}:\dfrac{1}{2}\\ x=\dfrac{29}{12}\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\\ \dfrac{1}{2}x=\dfrac{-7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{-13}{24}\\ x=\dfrac{-13}{24}:\dfrac{1}{2}\\ x=\dfrac{-13}{12}\)

i)

\(3\cdot\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=0-\dfrac{1}{9}\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}:3\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{27}\\ \left(3x-\dfrac{1}{2}\right)^3=\left(\dfrac{-1}{3}\right)^3\\ \Leftrightarrow3x-\dfrac{1}{2}=\dfrac{-1}{3}\\ 3x=\dfrac{-1}{3}+\dfrac{1}{2}\\ 3x=\dfrac{1}{6}\\ x=\dfrac{1}{6}:3\\ x=\dfrac{1}{18}\)

\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=1-\frac{1}{100}\)

\(A=\frac{99}{100}< 2\left(đpcm\right)\)

Áp đụng tính chất dãy tỷ số bằng nhau ta được

\(\frac{a}{b}=\frac{c}{d}\)

\(\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\left(1\right)\)

Ta lại có: 

\(\frac{a}{b}=\frac{c}{d}=\frac{a+b}{c+d}\)

\(\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\frac{a^2+b^2}{c^2+d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)

Ta có:

+) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\left(\frac{a}{b}\right)^2=\left(\frac{c}{d}\right)^2\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}\)(1)

+) \(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}=\frac{\left(a+c\right)^2}{\left(b+d\right)^2}\)(2)

Từ (1)(2)

\(\Rightarrow\frac{a^2+c^2}{b^2+d^2}=\frac{\left(a+c\right)^2}{\left(b+d\right)^2}\left(dpcm\right)\)

a) \(B=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{302\cdot305}\)

\(B=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{302\cdot305}\right)\)

\(B=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{302}-\frac{1}{305}\right)\)

\(B=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{305}\right)=\frac{1}{3}\cdot\frac{303}{610}=\frac{101}{610}\)

b) \(C=\frac{6}{1\cdot4}+\frac{6}{4\cdot7}+....+\frac{6}{202\cdot205}\)

\(C=2\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{202\cdot205}\right)=2\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{202}-\frac{1}{205}\right)\)

\(=2\left(1-\frac{1}{205}\right)=2\cdot\frac{204}{205}=\frac{408}{205}\)

c) \(D=\frac{5^2}{1\cdot6}+\frac{5^2}{6\cdot11}+...+\frac{5^2}{266\cdot271}\)

\(D=5\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+...+\frac{5}{266\cdot271}\right)\)

\(D=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{266}-\frac{1}{271}\right)=5\left(1-\frac{1}{271}\right)=5\cdot\frac{270}{271}=\frac{1350}{271}\)

d) \(E=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{5}{16}\cdot...\cdot\frac{9999}{10000}=\frac{3\cdot8\cdot15\cdot...\cdot9999}{4\cdot9\cdot16\cdot...\cdot10000}=\frac{3}{10000}\)

e) \(F=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{50^2}\right)\)

\(F=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{2500}\right)\)

\(F=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{2499}{2500}=\frac{3\cdot8\cdot15\cdot...\cdot2499}{4\cdot9\cdot16\cdot...\cdot2500}=\frac{3}{2500}\)

a. \(B=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{302.305}\)

\(\Rightarrow3B=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{302.305}\)

\(\Rightarrow3B=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{302}-\frac{1}{305}\)

\(\Rightarrow3B=\frac{1}{2}-\frac{1}{305}\)

\(\Rightarrow3B=\frac{303}{610}\)

\(\Rightarrow B=\frac{101}{610}\)

b. \(C=\frac{6}{1.4}+\frac{6}{4.7}+...+\frac{6}{202.205}\)

\(\Rightarrow\frac{1}{2}C=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{202.205}\)

\(\Rightarrow\frac{1}{2}C=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{202}-\frac{1}{205}\)

\(\Rightarrow\frac{1}{2}C=1-\frac{1}{205}\)

\(\Rightarrow\frac{1}{2}C=\frac{204}{205}\)

\(\Rightarrow C=\frac{408}{205}\)

c. \(D=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{266.271}\)

\(\Rightarrow\frac{1}{5}D=\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{266.271}\)

\(\Rightarrow\frac{1}{5}D=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{266}-\frac{1}{271}\)

\(\Rightarrow\frac{1}{5}D=1-\frac{1}{271}\)

\(\Rightarrow\frac{1}{5}D=\frac{270}{271}\)

\(\Rightarrow D=\frac{1350}{271}\)

Bài 1:

b) Ta có:

\(16^5=2^{20}\)

\(\Rightarrow B=16^5+2^{15}=2^{20}+2^{15}\)

\(\Rightarrow B=2^{15}.2^5+2^{15}\)

\(\Rightarrow B=2^{15}\left(2^5+1\right)\)

\(\Rightarrow B=2^{15}.33\)

\(\Rightarrow B⋮33\) (Đpcm)

c) \(C=5+5^2+5^3+5^4+...+5^{100}\)

\(\Rightarrow C=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{99}+5^{100}\right)\)

\(\Rightarrow C=1\left(5+5^2\right)+5^2\left(5+5^2\right)+...+5^{98}\left(5+5^2\right)\)

\(\Rightarrow\left(1+5^2+...+5^{98}\right)\left(5+5^2\right)\)

\(\Rightarrow C=Q.30\)

\(\Rightarrow C⋮30\) (Đpcm)

Bài 1 : a, \(A=1+3+3^2+...+3^{118}+3^{119}\)

\(A=\left(1+3+3^2+3^3\right)+...+\left(3^{116}+3^{117}+3^{118}+3^{119}\right)\)

\(A=\left(1+3+3^2+3^3\right)+...+3^{116}\left(1+3+3^2+3^3\right)\)

\(A=1.30+...+3^{116}.30=\left(1+...+3^{116}\right).30⋮3\)

Vậy \(A⋮3\)

b, \(B=16^5+2^{15}=\left(2.8\right)^5+2^{15}\)

\(=2^5.8^5+2^{15}=2^5.\left(2^3\right)^5+2^{15}\)

\(=2^5.2^{15}+2^{15}.1=2^{15}\left(32+1\right)=2^{15}.33⋮33\)

Vậy \(B⋮33\)

c, Tương tự câu a nhưng nhóm 2 số

Bài 2 : a, \(n+2⋮n-1\) ; Mà : \(n-1⋮n-1\)

\(\Rightarrow\left(n+2\right)-\left(n-1\right)⋮n-1\)

\(\Rightarrow n+2-n+1⋮n-1\Rightarrow3⋮n-1\)

\(\Rightarrow n-1\in\left\{1;3\right\}\Rightarrow n\in\left\{2;4\right\}\)

Vậy \(n\in\left\{2;4\right\}\) thỏa mãn đề bài

b, \(2n+7⋮n+1\)

Mà : \(n+1⋮n+1\Rightarrow2\left(n+1\right)⋮n+1\Rightarrow2n+2⋮n+1\)

\(\Rightarrow\left(2n+7\right)-\left(2n+2\right)⋮n+1\)

\(\Rightarrow2n+7-2n-2⋮n+1\Rightarrow5⋮n+1\)

\(\Rightarrow n+1\in\left\{1;5\right\}\Rightarrow n\in\left\{0;4\right\}\)

Vậy \(n\in\left\{0;4\right\}\) thỏa mãn đề bài

c, tương tự phần b

d, Vì : \(4n+3⋮2n+6\)

Mà : \(2n+6⋮2n+6\Rightarrow2\left(2n+6\right)⋮2n+6\Rightarrow4n+12⋮2n+6\)

\(\Rightarrow\left(4n+12\right)-\left(4n+3\right)⋮2n+6\)

\(\Rightarrow4n+12-4n-3⋮2n+6\Rightarrow9⋮2n+6\)

\(\Rightarrow2n+6\in\left\{1;2;9\right\}\Rightarrow2n=3\Rightarrow n\in\varnothing\)

Vậy \(n\in\varnothing\)