chung minh : neu \(\dfrac{a}{b}=\dfrac{c}{d}\) thi :
a,\(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)
b,\(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)
chung minh : neu \(\dfrac{a}{b}=\dfrac{c}{d}\) thi :
a,\(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)
b,\(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)
đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
a) thay \(a=bk;c=dk\) ta có
\(\dfrac{5a+3b}{5a-3b}=\dfrac{5bk+3b}{5bk-3b}=\dfrac{b\left(5k+3\right)}{b\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\)(1)
\(\dfrac{5c+3d}{5c-3d}=\dfrac{5dk+3d}{5dk-3d}=\dfrac{d\left(5k+3\right)}{d\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\)(2)
từ (1);(2)\(\Rightarrow\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)
b) thay \(a=bk;c=dk\) ta có
\(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7(bk)^2+3bkb}{11(bk)^2-8b^2}=\dfrac{7b^2k^2+3b^2k}{11b^2k^2-8b^2}\)
\(=\dfrac{b^2\left(7k^2+3k\right)}{b^2\left(11k^2-8\right)}=\dfrac{7k^2+3k}{11k^2-8}\)(3)
\(\dfrac{7c^2+3cd}{11c^2-8d^2}=\dfrac{7\left(dk\right)^2+3dkd}{11\left(dk\right)^2-8d^2}=\dfrac{7d^2k^2+3d^2k}{11d^2k^2-8d^2}\)
\(=\dfrac{d^2\left(7k^2+3k\right)}{d^2\left(11k^2-8\right)}=\dfrac{7k^2+3k}{11k^2-8}\)(4)
từ (3);(4)\(\Rightarrow\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)
Tìm x,y:
\(\dfrac{x-9}{3}\)=\(\dfrac{x+y}{13}\)=\(\dfrac{xy}{200}\)
đề có sai ko bạn
bạn coi kĩ lại đi
cái chỗ \(x-9\) ấy hay là x-y vậy? bạn coi kĩ lại giúp mk
CMR:Nếu \(\dfrac{a}{b}\)=\(\dfrac{b}{c}\)=\(\dfrac{c}{a}\)thì a=b=c
Ta có:\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{b+c+a}=1\)
\(\Rightarrow\dfrac{a}{b}=1\Leftrightarrow a=b\)
\(\Rightarrow\dfrac{b}{c}=1\Leftrightarrow b=c\)
\(\Rightarrow\dfrac{c}{a}=1\Leftrightarrow a=c\)
vậy a=b=c => đpcm
https://hoc24.vn/hoi-dap/question/646533.html
Tính giá trị của biểu thức:
\(A=\dfrac{2014x+2013y}{2014x-2013y}.\)Biết \(\dfrac{x}{y}=\dfrac{2}{3}\)
Vì \(\dfrac{x}{y}=\dfrac{2}{3}->x=2,y=3\)
A =\(\dfrac{2014.2+2014.3}{2014.2-2014.3}=\dfrac{4028+6042}{4028-6042}=\dfrac{10070}{-2014}=-5\)
Tính:\(A=3\dfrac{1}{417}.\dfrac{1}{762}-\dfrac{1}{139}+\dfrac{761}{762}-\dfrac{4}{417.762}+\dfrac{5}{139}\)
LƯU Ý:\(3\dfrac{1}{417}\)là hỗn số
\(A=3\dfrac{1}{417}.\dfrac{1}{762}-\dfrac{1}{139}+\dfrac{761}{762}-\dfrac{4}{417.762}+\dfrac{5}{139}\)
\(=\left(\dfrac{1252}{417.762}-\dfrac{4}{417.762}\right)+\left(-\dfrac{1}{139}+\dfrac{5}{139}\right)+\dfrac{761}{762}\)
\(=\dfrac{1248}{417.762}+\dfrac{4}{139}+\dfrac{761}{762}=\dfrac{1248}{417.672}+\dfrac{12.762}{417.762}+\dfrac{761.417}{417.762}\)
\(=\dfrac{327729}{317754}\)
tim x, y, z biet :
a, \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\) va 2x + 3y - z = 186
b, \(\dfrac{x}{3}=\dfrac{y}{4}\) va \(\dfrac{y}{5}=\dfrac{z}{7}\) va 2x + 3y - z = 327
c, 2x = 3y = 5z va x + y - z = 95
d, \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\) va xyz = 810
a)Vì \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)nên \(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{x}{28}\).
Áp dụng t/c dãy tỉ số = nhau, ta có :
\(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{186}{62}=3\)
⇒2x = 3.30 = 90 ⇒ x = 45
3y = 3.60 = 180 ⇒ y = 60
z = 3.28 = 84
Ý b) có gì đó sai sai ?
c)Ta có :
\(2x=3y=5z\Rightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\)
Áp dụng t/c dãy tỉ số = nhau, ta có :
\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{x+y-z}{15+10-6}=\dfrac{95}{19}=5\)
⇒x = 5.15 = 75
y = 5.10 = 50
z = 5.6 = 30
d)Ta có :
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\left(k\in Z\right)\)
⇒ x = 2k ; y = 3k ; z = 5k
⇒ xyz = 2k.3k.5k = 30k3 = 810
⇒ k = 3 Vậy x = 3.2 = 6; y = 3.3 = 9; z = 3.5 = 15Cho 3 đường thẳng xy;zt;ab cắt nhau tại đỉnh O.Biết xoz=70o và boy=50o.Tính các góc còn lại.
Tìm x,y biết:\(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\)
Áp dụng tính chất dãy tỉ số bằng nhau,ta có:
\(\dfrac{1+2y}{18}=\dfrac{1+6y}{6x}\Rightarrow\dfrac{1+2y+1+6y}{18+6x}\Rightarrow\dfrac{8y+2}{18+6x}\Leftrightarrow\dfrac{2\left(1+4y\right)}{2\left(9+3x\right)}=\dfrac{1+4y}{9+3x}\)
\(\Rightarrow\dfrac{1+4y}{9+3x}=\dfrac{1+4y}{24}\Leftrightarrow9+3x=24\Rightarrow x=\dfrac{24-9}{3}=5\)
Thay x=5 vào biểu thức: \(\dfrac{1+2y}{18}=\dfrac{1+6y}{6x}\),ta đc:
\(\dfrac{1+2y}{18}=\dfrac{1+6y}{6.5}\Leftrightarrow\dfrac{1+2y}{18}=\dfrac{1+6y}{30}\Leftrightarrow\dfrac{5\left(1+2y\right)}{90}=\dfrac{3\left(1+6y\right)}{90}\)
\(\Leftrightarrow5\left(1+2y\right)=3\left(1+6y\right)\Leftrightarrow5+10y=3+18y\)
\(\Leftrightarrow10y-18y=3-5\Leftrightarrow-8y=-2\Leftrightarrow y=\dfrac{1}{4}=0,25\)
vậy x=5 và y=0,25
Tìm x,biết:
\(\dfrac{315-x}{101}+\dfrac{313-x}{103}+\dfrac{311-x}{105}+\dfrac{309-x}{107}=0\)
Lời giải:
PT \(\Leftrightarrow \frac{315-x}{101}+1+\frac{313-x}{103}+1+\frac{311-x}{105}+1+\frac{309-x}{107}+1=4\)
\(\Leftrightarrow \frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=4\)
\(\Leftrightarrow (416-x)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=4\)
\(\Rightarrow 416-x=\frac{4}{\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}}\)
\(\Rightarrow x=416-\frac{4}{\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}}\)
Qua điểm O kẻ 4 đường thẳng phân biệt.
a/Hỏi có bao nhiêu góc tạo thành?Có bao nhiêu góc bẹt?
b/Có bao nhiêu góc đối đỉnh?