Ta có AH=DE ( vì ADHE là hcn)

mà AH²=BH.BC

=> AH⁴=HB².HC²=BD.CE.BC.BA

=> AH³=BD.CE.BC

a) ta có \(\frac{1234}{1235}=1-\frac{1}{1235}\)

\(\frac{4319}{4320}=1-\frac{1}{4320}\)

vì \(\frac{1}{1235}>\frac{1}{4320}\Rightarrow1-\frac{1}{1235}< 1-\frac{1}{4320}\)

\(\Rightarrow\frac{1234}{1235}< \frac{4319}{4320}\)

b) ta có \(\frac{1234}{1244}=1-\frac{10}{1244}\)

\(\frac{4321}{4331}=1-\frac{10}{4331}\)

vì \(\frac{10}{1244}>\frac{10}{4331}\Rightarrow1-\frac{10}{1244}< 1-\frac{10}{4331}\)

\(\Rightarrow\frac{1234}{1244}< \frac{4321}{4331}\)

\(\Rightarrow\frac{-1234}{1244}>\frac{-4321}{4331}\) 

A)

1- 1234/1235= 1/1235

1- 4319/4320= 1/4320

Vậy 1/1235 > 1/4320

1234/1235 > 4329/4320

Đ/s:...

Ko chắc, sai bỏ qa

\(1:12\)

\(2:15\)

giải ra đi bạn

\(A=\frac{2}{1+2}+\frac{2+3}{1+2+3}+...+\frac{2+3+...+20}{1+2+3+...+20}\)

\(A=\frac{2}{3}+\frac{5}{6}+...+\frac{209}{210}\)

\(A=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{210}\right)\)

\(A=\left(1+1+....+1\right)\left(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{210}\right)\)

\(A=19-\left(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{420}\right)\)

\(A=19-\left(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{20.21}\right)\)

\(A=19-2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{20}-\frac{1}{21}\right)\)

\(A=19-2\cdot\left(\frac{1}{2}-\frac{1}{21}\right)\)

\(A=19-2\cdot\frac{19}{42}=19-\frac{19}{21}=\frac{380}{21}\)

Vậy A= \(\frac{380}{21}\)

\(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2005}\right)\left(1-\frac{1}{2006}\right)\)

\(B=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{2004}{2005}\cdot\frac{2005}{2006}\)

\(B=\frac{1\cdot2\cdot...\cdot2004\cdot2005}{2\cdot3\cdot...\cdot2005\cdot2006}\)

\(B=\frac{1}{2006}\)

Vậy \(B=\frac{1}{2006}\)

 dễ ơt mik ko bik thì mới hỏi nên mik ko bik hhihi! ^^ ^^

ai giúp mik đi mà !

a) \(\left(x+5\right)\left(2x-1\right)=\left(2x-3\right)\left(x+1\right)\)

\(\Leftrightarrow x\left(2x-1\right)+5\left(2x-1\right)=x\left(2x+3\right)+\left(2x+3\right)\)

\(\Leftrightarrow2x^2-x+10x-5=2x^2+3x+2x+3\)

\(\Leftrightarrow9x-5=5x+3\)

\(\Leftrightarrow9x-5x=3+5\)

\(\Leftrightarrow4x=8\)

\(\Leftrightarrow x=8\div4\)

\(\Leftrightarrow x=2\)

Vậy nghiệm duy nhất của phương trình là 2.

b) \(\left(x+1\right)\left(x+9\right)=\left(x+3\right)\left(x+5\right)\)

\(\Leftrightarrow x\left(x+1\right)+9\left(x+1\right)=x\left(x+3\right)+5\left(x+3\right)\)

\(\Leftrightarrow x^2+x+9x+9=x^2+3x+5x+15\)

\(\Leftrightarrow10x+9=8x+15\)

\(\Leftrightarrow10x-8x=15-9\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=6\div3\)

\(\Leftrightarrow x=2\)

Vậy nghiệm duy nhất của phương trình là 2

\(5^x+5^{x+2}=650\)

\(\Leftrightarrow5^x+5^x.5^2=650\)

\(\Leftrightarrow5^x.\left(1+5^2\right)=650\)

\(\Leftrightarrow5^x.26=650\)

\(\Leftrightarrow5^x=25\)

\(\Leftrightarrow5^x=5^2\)

\(\Leftrightarrow x=2\)

Vậy x=2

\(5^x+5^{x+2}=650\)

\(\Leftrightarrow5^x.1+5^x.5^2=650\)

\(\Leftrightarrow5^x\left(1+5^2\right)=650\)

\(\Leftrightarrow5^x\left(1+25\right)=650\)

\(\Leftrightarrow5^x.26=650\)

\(\Leftrightarrow5^x=650\div26\)

\(\Leftrightarrow5^x=25\)

\(\Leftrightarrow5^x=5^2\)

\(\Leftrightarrow x=2\)

Vậy x = 2

3,01;6,02;9,03

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