Giải:

Đặt \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=k\Rightarrow\left\{{}\begin{matrix}a=2k\\b=3k\\c=4k\end{matrix}\right.\)

Ta có: \(a^2+3b^2-2c^2=-16\)

\(\Rightarrow4k^2+27k^2-32k^2=-16\)

\(\Rightarrow-k^2=-16\)

\(\Rightarrow k^2=16\)

\(\Rightarrow k=\pm4\)

+) \(k=4\Rightarrow a=8,b=12,c=16\)

+) \(k=-4\Rightarrow a=-8;b=-12;c=-16\)

Vậy bộ số \(\left(a;b;c\right)\) là \(\left(8;12;16\right);\left(-8;-12;-16\right)\)

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\dfrac{a^2}{4}=\dfrac{3b^2}{27}=\dfrac{2c^2}{32}=\dfrac{a^2+3b^2-2c^2}{4+27-32}=\dfrac{-16}{-1}=16\)

\(\Rightarrow a^2=64,b^2=144,c^2=256\) hay:

\(\left(a;b;c\right)=\left(8;12;16\right)=\left(-8;-12;-16\right)\)

ĐS: \(\left(a;b;c\right)=\left(8;12;16\right)=\left(-8;-12;-16\right)\)

a, Vì \(\dfrac{a}{c}=\dfrac{c}{b}\Rightarrow ab=c^2\)

Ta có :

\(\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{\left(b+a\right)\left(b-a\right)}{a^2+ab}=\dfrac{\left(b+a\right)\left(b-a\right)}{a\left(a+b\right)}=\dfrac{b-a}{a}\)

Vậy \(\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{b-a}{a}\)

a)đặt \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=k\(\Rightarrow\)a=bk, c=dk
\(\dfrac{2a+3b}{2a-3b}=\dfrac{2bk+3b}{2bk-3b}=\dfrac{b\left(2k+3\right)}{b\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\) (1)
\(\dfrac{2c+3d}{2c-3d}=\dfrac{2dk+3d}{2dk-3d}=\dfrac{d\left(2k+3\right)}{d\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\) (2)
từ (1),(2)\(\Rightarrow\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\)

b)ta có:
\(\dfrac{ab}{cd}=\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2-b^2}{c^2-d^2}\)
câu c bn tự giải nhé dễ mak ahihihichúc bn hc tốt

Giải:

Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=k\Rightarrow\left\{\begin{matrix}a=2k\\b=3k\\c=4k\end{matrix}\right.\)

Ta có: \(a^2+3b^2-2c^2=\left(-16\right)\)

\(\Rightarrow4k^2+27k^2-32k^2=-16\)

\(\Rightarrow\left(-1\right)k^2=-16\)

\(\Rightarrow k^2=16\)

\(\Rightarrow k=\pm4\)

+) \(k=4\Rightarrow a=8;b=12;c=16\)

+) \(k=-4\Rightarrow a=-8;b=-12;c=-16\)

Vậy bộ số \(\left(x;y;z\right)\) là \(\left(8;12;16\right);\left(-8;-12;-16\right)\)

a) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có:

\(\dfrac{2a+3b}{2a-3b}=\dfrac{2bk+3b}{2bk-3b}=\dfrac{b\left(2k+3\right)}{b\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\) (1)

\(\dfrac{2c+3d}{2c-3d}=\dfrac{2dk+3d}{2dk-3d}=\dfrac{d\left(2k+3\right)}{d\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\) (2)

Từ (1) và (2) suy ra \(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\)

b) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=q\Rightarrow\left\{{}\begin{matrix}a=bq\\c=dq\end{matrix}\right.\)

Ta có:

\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bq+b}{dq+d}\right)^2=\left[\dfrac{b\left(q+1\right)}{d\left(q+1\right)}\right]^2=\dfrac{b}{d}\) (1)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bq\right)^2+b^2}{\left(dq\right)^2+d^2}=\dfrac{b^2.q^2+b^2}{d^2.q^2+d^2}=\dfrac{b^2\left(q^2+1\right)}{d^2\left(q^2+1\right)}=\dfrac{b}{d}\) (2)

Từ (1) và (2) suy ra \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)

\(\dfrac{a}{b}=\dfrac{c}{d}\) => \(\dfrac{a}{c}=\dfrac{b}{d}\)

áp dụng tính chất dãy tỉ số = nhau ta có

\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a+3b}{2c+3d}=\dfrac{2a-3b}{2c-3d}\)

= \(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\) (đpcm)

a: Đặt a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{2a-3b}{2a+3b}=\dfrac{2bk-3b}{2bk+3b}=\dfrac{2k-3}{2k+3}\)

\(\dfrac{2c-3d}{2c+3d}=\dfrac{2dk-3d}{2dk+3d}=\dfrac{2k-3}{2k+3}\)

Do đó: \(\dfrac{2a-3b}{2a+3b}=\dfrac{2c-3d}{2c+3d}\)

b: \(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=k\)

\(\dfrac{a-c}{b-d}=\dfrac{bk-dk}{b-d}=k\)

Do đó: \(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)

*a/b=c/d=k=>a=bk;c=dk

Thay a=bk vào 2a+3b/2a-3b=2bk+3b/2bk-3b=2k+3/2k-3

Tương tự thay c=dk vào 2c+3d/2c-3d=2dk+3d/2dk-3d=2k+3/2k-3

=>2a+3b/2a-3b=2c+3d/2c-3d

*a/b=c/d=>a/c=b/d=k

=>k^2=a^2/c^2=c^2/d^2=a^2-b^2/c^2-d^2 (1)

k^2=a/c.b/d=ab/cd (2)

Từ (1) và (2)=>ab/cd=a^2-b^2/c^2-d^2

*a/b=c/d=>a/c=b/d=k=a+b/c+d

=>k^2=(a+b/c+d)^2

k^2=a^2/c^2=b^2/d^2=a^2+b^2/c^2+d^2

=>(a+b/c+d)^2=a^2+b^2/c^2+d^2

Gọi \(\dfrac{a}{b}=\dfrac{c}{d}=k\).\(\Rightarrow a=bk,c=dk\)

a)Ta có:\(\dfrac{2a+3b}{2a-3b}=\dfrac{2bk+3b}{2bk-3b}=\dfrac{b\left(2k+3\right)}{b\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\)(1)

\(\dfrac{2c+3d}{2c-3d}=\dfrac{2dk+3d}{2dk-3d}=\dfrac{d\left(2k+3\right)}{d\left(2k-3\right)}\dfrac{2k+3}{2k-3}\)(2)

Từ (1),(2)ta có:\(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\)

b)Ta có:\(\dfrac{ab}{cd}=\dfrac{bk\times b}{dk\times d}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}\)(1)

\(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b^2}{d^2}\)(2)

Từ (1),(2) ta có:\(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)

c)Ta có:\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\left(\dfrac{b}{d}\right)^2=\dfrac{b^2}{d^2}\)(1)

\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\dfrac{b^2}{d^2}\)(2)

Từ (1), (2) ta có \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)

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

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)