3+10+17+...+584
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Đặt \(BC=x\left(cm\right)\) (ĐK: \(x>9\))
\(\Rightarrow AC=BC-9=x-9\left(cm\right)\)
Theo định lý Py-ta-go ta có:
\(BC^2=AC^2+AB^2\Rightarrow AB^2=BC^2-AC^2\)
\(\Rightarrow AB=\sqrt{x^2-\left(x-9\right)^2}=\sqrt{x^2-\left(x^2-18x+81\right)}\)
\(\Rightarrow AB=\sqrt{18x-81}\)
Theo đề bài: \(C_{ABC}=70\left(cm\right)\)
\(\Rightarrow AB+AC+BC=70\)
\(\Rightarrow\sqrt{18x-81}+\left(x-9\right)+x=70\)
\(\Rightarrow\sqrt{18x-81}=79-2x\left(x\le\dfrac{79}{2}\right)\)
\(\Rightarrow18x-81=\left(79-2x\right)^2\)
\(\Rightarrow18x-81=6241-316x+4x^2\)
\(\Rightarrow4x^2-334x+6322=0\)
\(\Delta=\left(-334\right)^2-4\cdot4\cdot6322=10404>0\)
\(x_1=\dfrac{334+\sqrt{10404}}{2\cdot4}=\dfrac{109}{2}>\dfrac{79}{2}\left(ktm\right)\)
\(x_2=\dfrac{334-\sqrt{10404}}{2\cdot4}=29\left(tm\right)\)
\(\Rightarrow BC=29\left(cm\right)\)
\(AC=29-9=20\left(cm\right)\)
\(AB=\sqrt{18\cdot29-81}=21\left(cm\right)\)
Vậy: ...
Lời giải:
$A=1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{78}$
$A:2=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{156}$
$A:2=\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+....+\frac{1}{12\times 13}$
$A:2=\frac{2-1}{1\times 2}+\frac{3-2}{2\times 3}+\frac{4-3}{3\times 4}+....+\frac{13-12}{12\times 13}$
$A:2=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{12}-\frac{1}{13}$
$A:2=1-\frac{1}{13}=\frac{12}{13}$
$A=\frac{12}{13}\times 2=\frac{24}{13}$
Đặt \(A=\) \(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{78}\)
\(\dfrac{A}{2}=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{156}\)
\(\dfrac{A}{2}=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+...+\dfrac{1}{12\times13}\)
\(\dfrac{A}{2}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{12}-\dfrac{1}{13}\)
\(\dfrac{A}{2}=1-\dfrac{1}{13}\)
\(\dfrac{A}{2}=\dfrac{13}{13}-\dfrac{1}{13}\)
\(\dfrac{A}{2}=\dfrac{12}{13}\)
\(A=\dfrac{12}{13}\times2\)
\(A=\dfrac{24}{13}\)
Lời giải:
a. Xét tam giác $ABC$ và $HBA$ có:
$\widehat{B}$ chung
$\widehat{BAC}=\widehat{BHA}=90^0$
$\Rightarrow \triangle ABC\sim \triangle HBA$ (g.g)
b.
$BC=\sqrt{AB^2+AC^2}=\sqrt{15^2+20^2}=25$ (cm) - định lý Pitago
$AH=2S_{ABC}:BC=AB.AC:BC=15.20:25=12$ (cm)
$BH=\sqrt{AB^2-AH^2}=\sqrt{15^2-12^2}=9$ (cm) - định lý Pitago
c.
Theo tính chất đường phân giác:
$\frac{DA}{DC}=\frac{AB}{BC}=\frac{15}{25}=\frac{3}{5}$
$DA+DC=AC=20$
$\Rightarrow DA=20:(3+5).3=7,5$ (cm)
$DC=AC-DA=20-7,5=12,5$ (cm)
Diện tích chiếc khăn quàng đó là:
\(\dfrac{5,6\times20}{2}=56\left(cm^2\right)\)
Đáp số: 56 cm2.
a, \(\dfrac{42}{54}=\dfrac{7}{x}\)
Ta có: \(x.42=7.54\)
\(=>x.42=378\)
\(=>x=378:42\)
\(=>x=9\)
Vậy x = 9
b, \(\dfrac{-2}{3}=\dfrac{y}{15}\)
Ta có: \(y.3=\left(-2\right).15\)
\(=>y.3=-30\)
\(=>y=\left(-30\right):3\)
\(=>y=-10\)
Vậy y = -10
c, \(\dfrac{6}{10}=\dfrac{3}{x}=\dfrac{y}{-20}\)
* Ta có: \(x.6=3.10\)
\(=>x.6=30\)
\(=>x=30:6\)
\(=>x=5\)
Vì x = 5 \(\Rightarrow\dfrac{3}{5}=\dfrac{y}{-20}\)
Ta có: \(y.5=3.\left(-20\right)\)
\(=>y.5=-60\)
\(=>y=\left(-60\right):5\)
\(=>y=-12\)
Vậy x = 5 ; y = -12
d, \(\dfrac{-x}{-6}=\dfrac{-5}{6}\Rightarrow\dfrac{x}{6}=\dfrac{-5}{6}\Rightarrow x=-5\) ( Cùng mẫu số )
Vậy x = -5
\(#NqHahh\)
\(a.\) \(\dfrac{42}{54}=\dfrac{7}{x}\)
\(\Rightarrow x\cdot42=7\cdot54\)
\(\Rightarrow x\cdot42=378\)
\(\Rightarrow x=378:42\)
\(\Rightarrow x=9\)
Vậy \(\dfrac{42}{54}=\dfrac{7}{9}.\)
\(b.\) \(\dfrac{-2}{3}=\dfrac{y}{15}\)
\(\Rightarrow y\cdot3=\left(-2\right)\cdot15\)
\(\Rightarrow y\cdot3=\left(-30\right)\)
\(\Rightarrow y=\left(-30\right):3\)
\(\Rightarrow y=\left(-10\right)\)
Vậy \(\dfrac{-2}{3}=\dfrac{-10}{15}\)
\(c.\) \(\dfrac{6}{10}=\dfrac{3}{x}=\dfrac{y}{-20}\)
\(\Rightarrow x\cdot6=3\cdot10\)
\(\Rightarrow x\cdot6=30\)
\(\Rightarrow x=30:6\)
\(\Rightarrow x=5\)
Vậy: \(\dfrac{6}{10}=\dfrac{3}{5}=\dfrac{y}{-20}\)
Mặt khác: \(\dfrac{3}{5}=\dfrac{y}{-20}\)
\(\Rightarrow y\cdot5=3\cdot\left(-20\right)\)
\(\Rightarrow y\cdot5=\left(-60\right)\)
\(\Rightarrow y=\left(-60\right):5\)
\(\Rightarrow y=\left(-12\right)\)
Vậy \(\dfrac{6}{10}=\dfrac{3}{5}=\dfrac{-12}{-20}\)
\(d.\) \(\dfrac{-x}{-6}=\dfrac{-5}{6}\)
\(\Rightarrow\dfrac{x}{6}=\dfrac{-5}{6}\)
Do cùng mẫu số nên ta xét tử, ta thấy:
\(x=\left(-5\right)\)
Vậy \(\dfrac{-5}{6}=\dfrac{-5}{6}\)
Ta có: \(8b-9a=31\)
\(\Rightarrow8b=31+9a\)
\(\Rightarrow b=\dfrac{31+9a}{8}\)
\(\Rightarrow b=\dfrac{32+8a+a-1}{8}\)
\(\Rightarrow b=\dfrac{8\cdot\left(4+a\right)+a-1}{8}\)
\(\Rightarrow b=4+a+\dfrac{a-1}{8}\)
Để \(b\in N\) thì:
\(\dfrac{a-1}{8}\in N\)
\(\Rightarrow a-1⋮8\)
\(\Rightarrow a-1=8k\left(k\in N\right)\)
\(\Rightarrow a=8k+1\)
Khi đó: \(b=4+8k+1+\dfrac{8k+1-1}{8}\)
\(\Rightarrow b=5+8k+\dfrac{8k}{8}\)
\(\Rightarrow b=5+8k+k\)
\(\Rightarrow b=5+9k\)
Mặt khác: \(\dfrac{11}{17}< \dfrac{a}{b}< \dfrac{23}{29}\)
\(\Rightarrow\dfrac{11}{17}< \dfrac{8k+1}{5+9k}< \dfrac{23}{29}\)
Xét: \(\dfrac{11}{17}< \dfrac{8k+1}{5+9k}\)
\(\Rightarrow11\left(5+9k\right)< 17\left(8k+1\right)\)
\(\Rightarrow55+99k< 136k+17\)
\(\Rightarrow136k-99k>55-17\)
\(\Rightarrow37k>38\)
\(\Rightarrow k>\dfrac{38}{37}\left(1\right)\)
Xét: \(\dfrac{8k+1}{5+9k}< \dfrac{23}{29}\)
\(\Rightarrow29\left(8k+1\right)< 23\left(5+9k\right)\)
\(\Rightarrow232k+29< 115+207k\)
\(\Rightarrow232k-207k< 115-29\)
\(\Rightarrow25k< 86\)
\(\Rightarrow k< \dfrac{86}{25}\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow\dfrac{38}{27}< k< \dfrac{86}{25}\)
Mà \(k\in N\)
\(\Rightarrow k\in\left\{2;3\right\}\)
\(+,\) \(k=2\).
\(\Rightarrow\left\{{}\begin{matrix}a=8\cdot2+1=17\\b=5+9\cdot2=23\end{matrix}\right.\)
\(+,\) \(k=3\)
\(\Rightarrow\left\{{}\begin{matrix}a=8\cdot3+1=25\\b=5+9\cdot3=32\end{matrix}\right.\)
\(\Rightarrow\) Vậy \(\left(a;b\right)\in\left\{\left(17;23\right),\left(25;32\right)\right\}\)
\(\dfrac{x^2}{y}=3\) và \(\dfrac{x}{y}=21\) \(\left(ĐKXĐ:x,y>0\right)\)
\(\Rightarrow\dfrac{x^2}{y}:\dfrac{x}{y}=\dfrac{3}{21}=\dfrac{1}{7}\)
\(\Rightarrow\dfrac{x^2}{y}\cdot\dfrac{y}{x}=\dfrac{1}{7}\)
\(\Rightarrow x=\dfrac{1}{7}.\)
Khi đó: \(y=\dfrac{x}{21}=\dfrac{1}{7}:21=\dfrac{1}{7}\cdot\dfrac{1}{21}=\dfrac{1}{147}\)
\(\Rightarrow y=\dfrac{1}{147}\)
Vậy \(\left(x;y\right)=\left(\dfrac{1}{7};\dfrac{1}{147}\right)\)
Số số hạng của tổng:
(584 - 3) : 7 + 1 = 84 (số)
3 + 10 + 17 + ... + 584 = (584 + 3) . 84 : 2 = 24654
A = 3 + 10 + 17 +...+ 584
Dãy số trên là dãy số cách đều với khoảng cách là: 10 - 3 = 7
Số số hạng của dãy số trên là: (584 - 3) : 7 + 1 = 84
Tổng của dãy số trên là:
A = (584 + 3) x 84 : 2 = 24654
Vậy 3 + `10 + 17 +...+ 584 = 24654