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\(\Rightarrow\)\(\frac{a}{10}=\frac{b}{15};\frac{b}{15}=\)\(\frac{c}{12}\)
\(\Leftrightarrow\frac{a}{10}=\frac{b}{15}=\frac{c}{12}\)
\(\Rightarrow\frac{a-b+c}{10-15+12}\frac{-49}{-7}=-7\)
\(\Rightarrow a=10.-7=-70\)
\(b=15.-7=-105\)
\(c=12.-7=-84\)
ta có \(\frac{a}{2}=\frac{b}{3}\Rightarrow\frac{a}{10}=\frac{b}{15};\frac{b}{5}=\frac{c}{4}\Rightarrow\frac{b}{15}=\frac{c}{12}\Rightarrow\frac{a}{10}=\frac{b}{15}=\frac{c}{12}=\frac{a-b+c}{10-15+12}=\frac{-49}{7}=-7\)
\(\frac{a}{10}=-7\Rightarrow a=-7.10=-71\)
\(\frac{b}{15}=-7\Rightarrow b=-7.15=-105\)
\(\frac{c}{12}=-7\Rightarrow c=-7.12=-84\)
Ta có : a/b=b/c
suy ra ac= b^2 thay vào ta có
a^2+ ac/ ac+c^2 = a(a+c)/ c(a+c) = a/c
vậy a^2+b^2/ b^2 + c^2 = a/c
\(từ\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Đặt \(\frac{a}{c}=\frac{b}{d}=k\Rightarrow a=ck,b=dk\)
\(\Rightarrow\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}=\frac{ck+c}{dk+d}=\frac{c^2k^2+c^2}{d^2k^2+d^2}=\frac{c^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{c^2}{d^2}=\frac{a^2}{b^2}=\frac{a}{c}\)(đpcm)
HỌCtốt
a)\(\left(\frac{1}{5}\right)^{3n-1}=\frac{1}{25}\)
\(\Leftrightarrow\left(\frac{1}{5}\right)^{3n-1}=\left(\frac{1}{5}\right)^2\)
\(\Leftrightarrow3n-1=2\)
\(\Leftrightarrow3n=3\)
\(\Leftrightarrow n=1\)
b)\(\left(\frac{4}{7}\right)^{n+2}=\frac{7}{4}\)
\(\Leftrightarrow\left(\frac{4}{7}\right)^{n+2}=\left(\frac{4}{7}\right)^{-1}\)
\(\Leftrightarrow n+2=-1\)
\(\Leftrightarrow n=-3\)
c)\(\left(\frac{2}{3}\right)^{-n+1}=\frac{3^3}{2^3}\)
\(\Leftrightarrow\left(\frac{2}{3}\right)^{-n+1}=\left(\frac{3}{2}\right)^3\)
\(\Leftrightarrow\left(\frac{2}{3}\right)^{-n+1}=\left(\frac{2}{3}\right)^{-3}\)
\(\Leftrightarrow-n+1=-3\)
\(\Leftrightarrow n=-4\)
c)\(\left(0,7\right)^{3n+1}=10^3:7^3\)
\(\Leftrightarrow\left(\frac{7}{10}\right)^{3n+1}=\left(\frac{10}{7}\right)^3\)
\(\Leftrightarrow\left(\frac{7}{10}\right)^{3n+1}=\left(\frac{7}{10}\right)^{-3}\)
\(\Leftrightarrow3n+1=-3\)
\(\Leftrightarrow3n=-4\)
\(\Leftrightarrow n=-\frac{4}{3}\)
A=5-3(2x+1)^2
Ta có : (2x+1)^2\(\ge\)0
\(\Rightarrow\)-3(2x-1)^2\(\le\)0
\(\Rightarrow\)5+(-3(2x-1)^2)\(\le\)5
Dấu = xảy ra khi : (2x-1)^2=0
=> 2x-1=0 =>x=\(\frac{1}{2}\)
Vậy : A=5 tại x=\(\frac{1}{2}\)
Ta có : (x-1)^2 \(\ge\)0
=> 2(x-1)^2\(\ge\)0
=>2(x-1)^2+3 \(\ge\)3
=>\(\frac{1}{2\left(x-1\right)^2+3}\)\(\le\)\(\frac{1}{3}\)
Dấu = xảy ra khi : (x-1)^2 =0
=> x = 1
Vậy : B = \(\frac{1}{3}\)khi x = 1
\(\frac{x^2+8}{x^2+2}\)= \(\frac{x^2+2+6}{x^2+2}=1+\frac{6}{x^2+2}\)
Làm như câu B GTNN = 4 khi x =0
k vs nha
Nhầm
\(\frac{c}{4}\)