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: \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)

\(\Leftrightarrow2\cdot\widehat{C}+\widehat{C}=180^0\)

\(\Leftrightarrow\widehat{C}=60^0\)

\(\widehat{A}+\widehat{B}=180^0-60^0=120^0\)

\(\widehat{A}=120^0\cdot\dfrac{3}{5}=72^0\)

=>\(\widehat{B}=48^0\)

b: Đặt \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=k\)

=>a=2k; b=3k; c=4k

\(M=\dfrac{\left(2a+3b+4c\right)^2}{a^2+b^2+c^2}=\dfrac{\left(4k+9k+16k\right)^2}{4k^2+9k^2+16k^2}\)

\(=\dfrac{\left(29k\right)^2}{29k^2}=29\)

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 \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=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 \(\left(8;12;16\right);\left(-8;-12;-16\right)\)

cảm ơn bạn nhiều ạ!

Bạn có làm trong này rồi nhé Câu hỏi của Phạm Vũ Ngọc Duy

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\\ \Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}\\ =\dfrac{a^2+3b^2-2c^2}{4+27-32}=-\dfrac{16}{-1}=16\\ \Rightarrow a=\pm8;b=\pm12;c=\pm16\)

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

Câu 1:

Đặt a/b=c/d=k

=>a=bk; c=dk

a: \(\dfrac{a^2}{c^2}=\dfrac{b^2k^2}{d^2k^2}=\dfrac{b^2}{d^2}\)

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

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

b: \(\dfrac{2a-3c}{c}=\dfrac{2bk-3dk}{dk}=\dfrac{2b-3d}{d}\)

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)