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\(\Leftrightarrow12a^2-4b^2=3a^2+3b^2\)
\(\Leftrightarrow9a^2=7b^2\)
\(\Leftrightarrow\dfrac{a^2}{b^2}=\dfrac{7}{9}\)
hay \(\dfrac{a}{b}\in\left\{\dfrac{\sqrt{7}}{3};-\dfrac{\sqrt{7}}{3}\right\}\)
\(\dfrac{3a^2-b^2}{a^2+b^2}=\dfrac{3}{4}\)
\(\Leftrightarrow4.\left(3a^2-b^2\right)=3\left(a^2+b^2\right)\)
\(\Leftrightarrow12a^2-4b^2=3a^2+3b^2\)
\(\Leftrightarrow12a^2-3a^2=3b^2+4b^2\)
\(\Leftrightarrow9a^2=7b^2\)
\(\Leftrightarrow\dfrac{a^2}{b^2}=\dfrac{7}{9}\)
\(\text{hoặc }\dfrac{a}{b}=\pm\dfrac{\sqrt{7}}{3}\)
\(\dfrac{3a^2-b^2}{a^2+b^2}=\dfrac{3}{4}\)
\(\Rightarrow3\left(a^2+b^2\right)=4\left(3a^2-b^2\right)\)
\(\Rightarrow3a^2+3b^2=12a^2-4b^2\)
\(\Rightarrow-9a^2=-7b^2\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{-7}{-9}=\dfrac{7}{9}\Rightarrow\dfrac{a}{b}=\dfrac{\sqrt{7}}{3}\)
Vậy............
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(a=bk;c=dk\)
1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)
\(\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k\left(2b-3d\right)}{2b-3d}=k\)
Do đó: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)
\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)
Do đó: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
3: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
4: \(\dfrac{3a-7b}{b}=\dfrac{3bk-7b}{b}=\dfrac{b\left(3k-7\right)}{b}=3k-7\)
\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)
Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
a) Ta có:
+) a/2=b/3
=>a=2b/3
+) b/5=c/4
=>c=4b/5
Lại có:
a-b+c=49
=> 2b/3 -b + 4b/5 =49
=> 7b/15==49
=> b= 105
Khi đó:
+) a=2b/3=2.105/3=70
+)c=4b/5=4.105/5=84
Vậy a=70; b=105; c=84...
chúc bạn học tốt
\(Q=6a^2b-3a^2=6\cdot\dfrac{1}{9}\cdot\dfrac{11}{4}-3\cdot\dfrac{1}{9}=\dfrac{3}{2}\)
a) Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{b}{a}=\dfrac{d}{c}\)
\(\Leftrightarrow\dfrac{b}{a}-1=\dfrac{d}{c}-1\)
\(\Leftrightarrow\dfrac{b-a}{a}=\dfrac{d-c}{c}\)
\(\Leftrightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\)
\(\Leftrightarrow\dfrac{a}{a-b}=\dfrac{c}{c-d}\)(đpcm)
Lời giải:
Xét thấy \(a=0\Rightarrow \frac{-b^2}{b^2}=\frac{3}{4}\Leftrightarrow -1=\frac{3}{4}\) (vô lý)
\(b=0\Rightarrow \frac{3a^2}{a^2}=\frac{3}{4}\Leftrightarrow 3=\frac{3}{4}\) (vô lý)
Do đó \(a,b\neq 0\)
Khi đó, đặt \(a=tb\)
Ta có \(\frac{3}{4}=\frac{3a^2-b^2}{a^2+b^2}=\frac{3b^2t^2-b^2}{b^2t^2+b^2}=\frac{b^2(3t^2-1)}{b^2(t^2+1)}=\frac{3t^2-1}{t^2+1}\)
\(\Leftrightarrow 3(t^2+1)=4(3t^2-1)\Leftrightarrow t^2=\frac{7}{9}\)
\(\Rightarrow \frac{a}{b}=t=\pm \sqrt{\frac{7}{9}}\)
\(\dfrac{3a^2-b^2}{a^2+b^2}=\dfrac{3}{4}\)
=> \(\dfrac{3a^2-b^2}{a^2+b^2}=\dfrac{3}{4}\)= \(4.\left(3a^2+b^2\right)=3.\left(a^2+b^2\right)\)
=> \(12a^2+4b^2=3a^2+3b^2\)
=> \(12a^2+\left(-3a\right)^2=\left(-4b\right)^2+3b^2\)
=> \(9a^2=-1b^2\)
=> \(\left(\dfrac{a}{b}\right)^2=-\dfrac{1}{9}\)
=> \(\left(\dfrac{a}{b}\right)^2=\left(-\dfrac{1}{9}\right)^{ }\)
=> \(\dfrac{a}{b}=-\dfrac{1}{3}\)
Vậy:..........