Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Áp dụng tính chất dãy tỉ số bằng nhau , ta có
\(\dfrac{a}{10}=\dfrac{b}{6}=\dfrac{c}{21}=\dfrac{5a+b-2c}{50+6-42}=\dfrac{28}{14}=2\)
=)\(\dfrac{a}{10}=2\) =)a = 20
=)\(\dfrac{b}{6}=2\) =)b = 12
=)\(\dfrac{c}{21}=2\) =) c = 42
Vậy ta tìm được a ; b ; c lần lượt là 20 , 12 , 42
\(A=4,8.\left(3,1-1,5\right)+1,5.\left(4,8-3,1\right)\)
\(A=4,8.3,1-4,8.1,5+1,5.4,8-1,5.3,1\)
\(A=3,1.\left(4,8-1,5\right)-4,8\left(1,5+1,5\right)\)
\(A=3,1.3,3-4,8.3\)
\(A=10,23-14,4=-4,17\)
\(B=\dfrac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}=\dfrac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(3.2^2\right)^{10}}=\dfrac{2^{19}.3^9+3.5.2^{18}.3^8}{2.3^9.2^{10}+3^{10}.\left(2^2\right)^{10}}=\dfrac{2^{19}.3^9+3^9.2^{18}.5}{2^{11}.3^9+3^{10}.2^{20}}=\dfrac{2^{18}.3^9\left(2+5\right)}{2^{11}.3^9\left(1+3.2^9\right)}=\dfrac{2^7.7}{1+3.2^9}\)
\(\Leftrightarrow D=1-\dfrac{1}{3}-\dfrac{1}{6}-\dfrac{1}{10}-\dfrac{1}{15}-\dfrac{1}{21}-\dfrac{1}{28}\)
\(\Rightarrow\dfrac{1}{2}D=\dfrac{1}{2}-\dfrac{1}{2.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}-\dfrac{1}{5.6}-\dfrac{1}{6.7}-\dfrac{1}{7.8}\)
\(\Rightarrow D\dfrac{1}{2}=\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{7}+\dfrac{1}{8}\)
\(\Rightarrow D=\dfrac{1}{8}.2=\dfrac{1}{4}\)
Vậy D=1/4
\(\dfrac{x}{10}=\dfrac{y}{6}=\dfrac{z}{21}\) ⇒ \(\dfrac{5x}{50}=\dfrac{y}{6}=\dfrac{2z}{42}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{5x}{50}\) = \(\dfrac{y}{6}=\dfrac{2z}{42}\) = \(\dfrac{5x+y-2z}{50+6-42}\) = \(\dfrac{28}{14}\)= 2
x = 2.50:5 = 10; y = 2.6=12; z =2.42:2= 42
Kết luận (x; y; z) =( 20; 12; 42)
Bài 1:
$\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt; c=dt$. Khi đó:
\(\frac{2a^2-3ab+5b^2}{2a^2+3ab}=\frac{2(bt)^2-3.bt.b+5b^2}{2(bt)^2+3bt.b}=\frac{b^2(2t^2-3t+5)}{b^2(2t^2+3t)}\)
$=\frac{2t^2-3t+5}{2t^2+3t}(1)$
\(\frac{2c^2-3cd+5d^2}{2c^2+3cd}=\frac{2(dt)^2-3.dt.d+5d^2}{2(dt)^2+3dt.d}=\frac{d^2(2t^2-3t+5)}{d^2(2t^2+3t)}=\frac{2t^2-3t+5}{2t^2+3t}(2)\)
Từ $(1);(2)$ suy ra đpcm.
Bài 2:
Từ $\frac{a}{c}=\frac{c}{b}\Rightarrow c^2=ab$. Khi đó:
$\frac{b^2-c^2}{a^2+c^2}=\frac{b^2-ab}{a^2+ab}=\frac{b(b-a)}{a(a+b)}$ (đpcm)
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{2a+b}{2a-b}=\dfrac{2bk+b}{2bk-b}=\dfrac{2k+1}{2k-1}\)
\(\dfrac{2c+d}{2c-d}=\dfrac{2dk+d}{2dk-d}=\dfrac{2k+1}{2k-1}\)
=>\(\dfrac{2a+b}{2a-b}=\dfrac{2c+d}{2c-d}\)
b: \(\dfrac{2a+b}{a-2b}=\dfrac{2bk+b}{bk-2b}=\dfrac{2k+1}{k-2}\)
\(\dfrac{2c+d}{c-2d}=\dfrac{2dk+d}{dk-2d}=\dfrac{2k+1}{k-2}\)
=>\(\dfrac{2a+b}{a-2b}=\dfrac{2c+d}{c-2d}\)
\(A=1-\frac{1}{10}-\frac{1}{15}-\frac{1}{3}-\frac{1}{28}-\frac{1}{6}-\frac{1}{21}\)
\(=1-\frac{1}{3}-\frac{1}{6}-\frac{1}{10}-\frac{1}{15}-\frac{1}{21}-\frac{1}{28}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{2.3}-\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-\frac{1}{6.7}-\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{7}+\frac{1}{8}\)\(=\frac{1}{8}\)
\(\Rightarrow A=\frac{1}{8}.2=\frac{1}{4}\)
Vậy tổng của biểu thức cần tính là \(\frac{1}{4}\)