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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
Câu 2
(a+3)(b-4)-(a-3)(b+4)=0
=>ab-4a+3b-12-ab-4a+3b+12=0
=>-8a=-6b
=>a/b=3/4
=>a/3=b/4
A=\(\dfrac{5}{4}\).(5-\(\dfrac{4}{3}\)).(\(-\dfrac{1}{11}\))
= \(\dfrac{5}{4}\).\(\dfrac{11}{3}\).(\(-\dfrac{1}{11}\))
=\(\dfrac{5}{4}\).[\(\dfrac{11}{3}.\left(-\dfrac{1}{11}\right)\text{]}\)
=\(\dfrac{5}{4}.\dfrac{1}{3}\)
=\(\dfrac{5}{12}\) (1)
B=\(\dfrac{3}{4}:\left(-12\right).\left(-\dfrac{2}{3}\right)\) =\(\dfrac{3}{4}:\text{[}\left(-12\right).\left(-\dfrac{2}{3}\right)\text{]}\)
=\(\dfrac{3}{4}:8\) =\(\dfrac{3}{4}.\dfrac{1}{8}\)=\(\dfrac{3}{32}\)(2)
C=\(\dfrac{5}{4}:\left(-15\right).\left(-\dfrac{2}{5}\right)\) =\(\dfrac{5}{4}:\text{[}\left(-15\right).\left(-\dfrac{2}{5}\right)\text{]}\)
=\(\dfrac{5}{4}:6=\dfrac{5}{4}.\dfrac{1}{6}=\dfrac{5}{24}\left(3\right)\)
D=(-3).\(\left(\dfrac{2}{3}-\dfrac{5}{4}\right):\left(-7\right)\) =(-3).\(\left(-\dfrac{7}{12}\right)\):(-7)=\(\dfrac{7}{4}:\left(-7\right)\)=\(\dfrac{7}{4}\).\(\left(\dfrac{-1}{7}\right)\)=\(\dfrac{-1}{4}\) (4)
Từ (1),(2),(3)và(4)=>Ta có thể sắp xếp các kết quả trên theo thứ tự tăng dần là:
(Bạn tự làm nhé! mình bận đi học rồi
)
\(A=\dfrac{5}{4}\cdot\dfrac{15-4}{3}\cdot\dfrac{-1}{11}=\dfrac{5}{4}\cdot\dfrac{11}{3}\cdot\dfrac{-1}{11}=\dfrac{-5}{12}\)=-50/120
\(B=\dfrac{3}{4}\cdot\dfrac{-1}{12}\cdot\dfrac{-2}{3}=\dfrac{3\cdot2}{4\cdot12\cdot3}=\dfrac{2}{4\cdot12}=\dfrac{1}{24}\)=5/120
\(C=\dfrac{5}{4}\cdot\dfrac{-1}{15}\cdot\dfrac{-2}{5}=\dfrac{2}{4\cdot15}=\dfrac{1}{30}\)=4/120
\(D=3\cdot\dfrac{8-15}{12}\cdot\dfrac{-1}{7}=\dfrac{1}{4}\)=30/120
Vì -50<4<5<30
nên A<C<B<D
giả sử a/2002 = b/2003 = c/2004 = k
=> a = 2002k ; b=2003k và c=2004k
=> 4(a-b)(b-c) = 4(2002k - 2003k)(2003k - 2004k)
=> 2(a-b)(b-c) = 4k^2 (1)
Ta có (c-a)^2 = (2004k - 2002k)^2 = 4k^2 (2)
từ (1) và (2) ta có 2(a-b)(b-c) = (c-a)^2
Bài 1:
Áp dụng t.c của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\\ =\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a^3}{b^3}=\dfrac{a.b.c}{b.c.d}=\dfrac{a}{d}\left(dpcm\right)\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\) (*)
a) Từ (*)suy ra:
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\dfrac{b^2.k^2+b^2}{d^2.k^2+d^2}=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}\)\(=\dfrac{b^2}{d^2}\) (1)
\(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\dfrac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\dfrac{b^2.\left(k+1\right)^2}{d^2.\left(k+1\right)^2}=\dfrac{b^2}{d^2}\)(2)
Từ (1) và (2) suy ra: \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\) (đpcm)
b) Tương tự câu a nhé bạn!
\(A=\dfrac{5}{4}\left(5-\dfrac{4}{3}\right)\left(-\dfrac{1}{11}\right)\)
\(A=\dfrac{5}{4}.\dfrac{11}{3}.\left(-\dfrac{1}{11}\right)\)
\(A=-\dfrac{5}{12}\)
\(B=\dfrac{3}{4}:\left(-12\right).\left(-\dfrac{2}{3}\right)\)
\(B=\dfrac{3}{4}.\left(-\dfrac{1}{12}\right).\left(-\dfrac{2}{3}\right)\)
\(B=\dfrac{1}{24}\)
\(C=\dfrac{5}{4}:\left(-15\right).\left(-\dfrac{2}{5}\right)\)
\(C=\dfrac{5}{4}.\left(-\dfrac{1}{15}\right).\left(-\dfrac{2}{5}\right)\)
\(C=\dfrac{1}{30}\)
\(D=\left(-3\right)\left(\dfrac{2}{3}-\dfrac{5}{4}\right):\left(-7\right)\)
\(D=\left(-3\right)\left(-\dfrac{7}{12}\right)\left(-\dfrac{1}{7}\right)\)
\(D=-\dfrac{1}{4}\)
Sắp xếp theo thứ tự tăng dần:
\(A,D,C,B\)
4.a
\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\\ \Leftrightarrow\left(3x-y\right).4=3\left(x+y\right)\\ \Rightarrow12x-4y=3x+3y\\ \Rightarrow12x-3x=4y+3y\\ \Rightarrow9x=7y\\ \Rightarrow\dfrac{x}{y}=\dfrac{7}{9}\)
Đặt \(\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{5}=k\Rightarrow a=3k;b=4k;c=5k\)
\(\Rightarrow4\left(a-b\right)\left(b-c\right)=4\left(3k-4k\right)\left(4k-5k\right)\)
\(=4.\left[\left(3-4\right).k\right].\left[\left(4-5\right).k\right]\)
\(=4.\left[-k\right].\left[-k\right]=4k^2\left(1\right)\)
\(\Rightarrow\left(a-c\right)^2=\left(3k-5k\right)^2=\left[\left(3-5\right).k\right]^2=\left[-2k\right]^2=4k^2\left(2\right)\)
Từ \(\left(1\right),\left(2\right)\Rightarrow4\left(a-b\right)\left(b-c\right)=\left(a-c\right)^2\)
Vậy \(4\left(a-b\right)\left(b-c\right)=\left(a-c\right)^2\left(dpcm\right)\)