Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt a/3=b/5=k
=>a=3.k
=>a2=9.k2
=>b=5.k
=>b2=25.k2
Ta có: C= 5a2+3b2/10a2-3b2
=> c= 5.9.k2+3.25.k2/10.9.k2-3.25.k2
=> C= k2.(5.9+3.25) / k2.(9.10-3.25)
=> C= 120/15
=> C=8
Nếu đúng tick giúp mik nha
Đặt a/3=b/5=k
=>a=3k; b=5k
\(B=\dfrac{5\cdot9k^2+3\cdot25k^2}{10\cdot9k^2-3\cdot25k^2}=8\)
\(a,a=-\dfrac{3}{2}\)
\(\Rightarrow3\left[2\left(-\dfrac{3}{2}\right)-1\right]+5\left(3+\dfrac{3}{2}\right)=3.\left(-3-1\right)+5.\dfrac{9}{2}=-12+\dfrac{45}{2}=\dfrac{21}{2}\)
\(b,x=2,1\)
\(\Rightarrow25.2,1-4\left(3.2,1-1\right)+7\left(5-2.2,1\right)=52,5-4.5,3+7.0,8=36,9\)
\(c,b=\dfrac{1}{2}\)
\(\Rightarrow12\left(2-3.\dfrac{1}{2}\right)+35.\dfrac{1}{2}-9\left(\dfrac{1}{2}+1\right)=12.\dfrac{1}{2}+\dfrac{35}{2}-9.\dfrac{3}{2}=6+\dfrac{35}{2}-\dfrac{27}{2}=10\)
\(d,a=-0,2\)
\(\Rightarrow4.\left(-0,2\right)^2-2\left(10.\left(-0,2\right)-1\right)+4.\left(-0,2\right)\left(2-\left(-0,2\right)^2\right)\)
\(=4.0,04-2.\left(-3\right)-0,8.1,96\)
\(=0,16+6-1,568\)
\(=4,592\)
a: A=6a-3+15-5a=a+12
Khi a=-3/2 thì A=-3/2+12=10,5
b: B=25x-12x+4+35-8x=5x+39
Khi x=2,1 thì B=10,5+39=49,5
c: C=24-6b+35b-9b-9=20b+15
Khi b=0,5 thì C=10+15=25
d: D=4a^2-20a+2+8a-4a^3=-4a^3+4a^2-12a+2
Khi a=-0,2 thì
D=-4*(-1/5)^3+4*(-1/5)^2-12*(-1/5)+2=4,592
\(1,\\ a,3^{2^3}=3^8>3^6=\left(3^2\right)^3\\ b,\left(-8\right)^9=\left(-2\right)^{27}< \left(-2\right)^{25}=\left(-32\right)^5\\ c,2^{21}=8^7< 9^7=3^{14}\\ 2,\)
\(a,\) Áp dụng tcdtsbn:
\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)
\(b,\) Sửa: \(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow a=bk;c=dk\)
\(\Leftrightarrow\dfrac{ab}{cd}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2};\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\dfrac{b^2}{d^2}\\ \LeftrightarrowĐpcm\)
Bài 1: Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
\(\dfrac{a}{a+c}=\dfrac{ck}{ck+c}=\dfrac{ck}{c\left(k+1\right)}=\dfrac{k}{k+1}\)
\(\dfrac{b}{b+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)
Do đó: \(\dfrac{a}{a+c}=\dfrac{b}{b+d}\)
Bài 1: Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
\(\dfrac{a}{a+c}=\dfrac{ck}{ck+c}=\dfrac{ck}{c\left(k+1\right)}=\dfrac{k}{k+1}\)
\(\dfrac{b}{b+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)
Do đó: \(\dfrac{a}{a+c}=\dfrac{b}{b+d}\)
