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1/h=1/2(1/a+1/b)=1/2a+1/2b=(a+b)/2ab
=>(a+b/)2ab-1/h=0
quy dong len ta co
(a+b)h/2abh-2ab/2abh=0=> (ah+bh-2ab)/2abh=0 =>ah+bh-2ab=0
=>ah+bh-ab-ab=0
=>a(h-b)-b(a-h)=0
=>a(h-b)=b(a-h)
=>a/b=(a-h)(h-b)
\(\frac{a}{b}=\frac{c}{d}\)
=> \(\frac{a}{c}=\frac{b}{d}=>\frac{a^2}{c^2}=\frac{c^2}{d^2}=\frac{a.b}{b.c}=\frac{a}{c}\)
=> \(\frac{a^2+b^2}{b^2+c^2}=\frac{a}{c}\)
=> dpcm
a/
\(x-y=\frac{a}{b}-\frac{c}{d}=\frac{ad-cb}{bd}=\frac{1}{bd}.\) (1)
\(y-z=\frac{c}{d}-\frac{e}{h}=\frac{ch-de}{dh}=\frac{1}{dh}\)(2)
+ Nếu d>0 => (1)>0 và (2)>0 => x>y; y>x => x>y>z
+ Nếu d<0 => (1)<0 và (2)<0 => x<y; y<z => x<y<z
b/
\(m-y=\frac{a+e}{b+h}-\frac{c}{d}=\frac{ad+de-cb-ch}{d\left(b+h\right)}=\frac{\left(ad-cb\right)-\left(ch-de\right)}{d\left(b+h\right)}=\frac{1-1}{d\left(b+h\right)}=0\)
=> m=y
+
cảm ơn bn nha Nguyễn Ngoc Anh Minh mk k cho bn r đó kb vs mk nha
a) \(\left(x+\frac{1}{3}\right)^3=\frac{-8}{27}\)
\(\left(x+\frac{1}{3}\right)^3=\left(\frac{-2}{3}\right)^3\)
\(x+\frac{1}{3}=\frac{-2}{3}\)
\(x=-1\)
b) \(\left(\frac{1}{3}x+\frac{4}{3}\right)^2=\frac{25}{9}\)
\(\left(\frac{1}{3}x+\frac{4}{3}\right)^2=\left(\frac{5}{3}\right)^2\)
\(\frac{1}{3}x+\frac{4}{3}=\frac{5}{3}\)
\(\frac{1}{3}x=\frac{1}{3}\)
\(x=1\)
c) \(2^x+2^{x+1}=24\)
\(2^x+2^x.2=24\)
\(2^x.\left(1+2\right)=24\)
\(2^x.3=24\)
\(2^x=8\)
\(2^x=2^3\)
\(x=3\)
a, (x+1/3)^3 = -8/27
=>(x+1/3)^3 = (-2/3)^3
=>x+1/3 = -2/3
=>x = -1
b, (1/3x+4/3)^2 = 25/9
=>(1/3x+4/3)^2 = (5/3)^2
=>(1/3x+4/3) = 5/3
=>1/3x = 1/3
=> x = 1
c, 2^x + 2^x+1 = 24
=>2^x + 2^x . 2 = 24
=>2^x.(1+2) = 24
=>2^x . 3 = 24
=>2^x =8
=>2^x = 2^3
=> x = 3
Ta có a+2/b+2 = a/b+2 + 2/b+2 = a(b+2)/b+2 + 2(b+2)/b+2 = a+2
Do a+2 > a/b => a+2/b+2 >a/b
mik làm đại ko bik đúng hay sai đâu nha
Xét tích : \(a\left(b+2\right)=ab+2a\)
\(b\left(a+2\right)=ab+2b\)
Nếu \(a>b\)thì \(ab+2a>ab+2b\)
hay \(a\left(b+2\right)>b\left(a+2\right)\)
\(\Rightarrow\frac{a}{b}>\frac{a+2}{b+2}\)
Nếu \(a< b\)thì \(ab+2a< ab+2b\)
hay \(a\left(b+2\right)< b\left(a+2\right)\)
\(\Rightarrow\frac{a}{b}< \frac{a+2}{b+2}\)
Nếu \(a=b\)thì \(ab+2a=ab+2b\)
hay \(a\left(b+2\right)=b\left(a+2\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a+2}{b+2}\)