Bài 1 : Cho \(\dfrac{U+2}{U-2}\) = \(\dfrac{V+3}{V-3}\) và \(U^2\) + \(V^2\) = 52 .
Tính U ; V .
Bài 2 : Cho \(\dfrac{x}{y}=\dfrac{z}{t}\) . Cmr \(\dfrac{x.y}{z.t}=\dfrac{\left(x+y\right)^2}{\left(z+t\right)^2}\) .
Bài 3 : Cho \(\dfrac{a}{a'}=\dfrac{b}{b'}=\dfrac{c}{c'}=\text{4}\) . Tính M \(\dfrac{a-3b+2c}{a'-3b'+2c'}\) .
Bài 4 : Cho \(\left(a_2\right)^2=a_1.a_3;\left(a_3\right)^2=a_2.a_4\) .
Cmr \(\dfrac{\left(a_1\right)^2+\left(a_2\right)^2+\left(a_3\right)^2}{\left(a_2\right)^2+\left(a_3\right)^2+\left(a_4\right)^2}=\dfrac{a_1}{a_3}\) .
Bài 5 : Cho \(\dfrac{a}{c}=\dfrac{c}{b}\) . Cmr :
a) \(\dfrac{a^2+c^2}{b^2+c^2}=\dfrac{a}{b}\)
b) \(\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{b-c}{a}\)
5a
Ta có \(\dfrac{a}{b}=\dfrac{a^2}{b^2}\) ; \(\dfrac{c}{d}=\dfrac{c^2}{d^2}\)
\(\dfrac{a}{b}=\dfrac{c}{d}\)=> \(\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}\)=>\(\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}\)=\(\dfrac{a^2+c^2}{b^2+d^2}\)(T/c cuả dãy tỉ số bằng nhau)
=> ĐPCM
Bài 1:
Ta có: \(\dfrac{U+2}{U-2}=\dfrac{V+3}{V-3}\)
\(\Rightarrow\left(U+2\right)\left(V-3\right)=\left(U-2\right)\left(V+3\right)\)
\(\Rightarrow\left(U+2\right)V-\left(U+2\right)3=\left(U-2\right)V+\left(U-2\right)3\)
\(\Rightarrow UV+2V-3U-6=UV-2V+3U-6\)
\(\Rightarrow2V-3U=3U-2V\)
\(\Rightarrow2U+2U=3V+3V\)
\(\Rightarrow2\times2U=2\times3V\)
\(\Rightarrow2U=3V\)
\(\Rightarrow\dfrac{U}{3}=\dfrac{V}{2}\)
(Như mình đã nói ở Câu hỏi của ARMY, cách này tuy dài nhg chặt hơn)
(Tiếp tục nè)
\(\Rightarrow\dfrac{U^2}{9}=\dfrac{V^2}{4}=\dfrac{U^2+V^2}{9+4}=\dfrac{52}{13}=4\)
\(\Rightarrow U^2=4\times9=36\Rightarrow U=\pm6\)
\(V^2=4\times4=16\Rightarrow V=\pm4\)
Vì \(\dfrac{U}{3}=\dfrac{V}{2}\Rightarrow U,V\) cùng dấu
\(\Rightarrow\left(U,V\right)=\left(6;4\right)\) và \(\left(U,V\right)=\left(-6;-4\right)\)
Vậy...
Bài 2:
Vì:\(\dfrac{x}{y}=\dfrac{z}{t}\Rightarrow\dfrac{x}{z}=\dfrac{y}{t}=\dfrac{x+y}{z+t}\Rightarrow\left(\dfrac{x}{z}\right)^2=\left(\dfrac{y}{t}\right)^2=\left(\dfrac{x+y}{z+t}\right)^2\left(1\right)\)
Ta còn có: \(\dfrac{x}{z}=\dfrac{y}{t}\Rightarrow\left(\dfrac{x}{z}\right)^2=\left(\dfrac{y}{t}\right)^2=\dfrac{xy}{zt}\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\Rightarrow\).....
Chúc bn học tốt
P/S: Những chỗ mình ghi chấm chấm có nghĩa là bn tự kết nốt