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1.
a) \(\frac{16}{24}-\frac{1}{3}=\frac{16}{24}-\frac{8}{24}=\)\(\frac{8}{24}=\frac{1}{3}\)
b) \(\frac{4}{5}-\frac{12}{60}=\frac{48}{60}-\frac{12}{60}=\frac{36}{60}=\frac{9}{15}\)
3.
a)\(\frac{17}{6}-\frac{2}{6}=\frac{17-2}{6}=\frac{15}{6}\)
b) \(\frac{16}{15}-\frac{11}{15}=\frac{16-11}{15}=\frac{5}{15}=\frac{1}{3}\)
c) \(\frac{19}{12}-\frac{13}{12}=\frac{19-13}{12}=\frac{6}{12}=\frac{1}{2}\)
đề bài sai rồi
Ta cóA=a3+a2-b3+b2+ab-3ab(a-b+1)
=(a3-b3)+(a2+ab+b2)-24ab(do a-b=7)
=(a-b)(a2+ab+b2)+(a2+ab+b2)-24ab
=(a2+ab+b2)(a-b+1)-24ab
mà a-b=7=>A=8a2+8ab+8b2-24ab
=8a2-16ab+8b2
=8(a-b)2=8 . 72=8 . 49=392
a)\(\frac{42}{15.7}\)= \(\frac{2.3.7}{3.5.7}\)= \(\frac{2}{5}\)
b) \(\frac{35.6}{336}\)= \(\frac{5.7.6}{6.7.8}\)= \(\frac{5}{8}\)
c) \(\frac{4.33}{11.12}\)= \(\frac{4.3.11}{11.3.4}\)= 1
d) \(\frac{9.4+9.11+5.9}{63}\)= \(\frac{9.\left(4+11+5\right)}{9.7}\)= \(\frac{20}{7}\)
#Hk tốt nhé
Bài 1 :
\(a)\) Ta có :
\(3x=4y=6z\)
\(\Leftrightarrow\)\(\frac{3x}{12}=\frac{4y}{12}=\frac{6z}{12}\)
\(\Leftrightarrow\)\(\frac{x}{4}=\frac{y}{3}=\frac{z}{2}\)
\(\Leftrightarrow\)\(\frac{2x}{8}=\frac{y}{3}=\frac{5z}{10}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{2x}{8}=\frac{y}{3}=\frac{5z}{10}=\frac{2x-5z}{8-10}=\frac{-36}{-2}=18\)
Do đó :
\(\frac{x}{4}=18\)\(\Rightarrow\)\(x=18.4=72\)
\(\frac{y}{3}=18\)\(\Rightarrow\)\(y=18.3=54\)
\(\frac{z}{2}=18\)\(\Rightarrow\)\(z=18.2=36\)
Vậy \(x=72\)\(;\)\(y=54\) và \(z=36\)
Chúc bạn học tốt ~
2) Ta có: \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{b+c+c+a+a+b}=\frac{a+b+c}{2.\left(a+b+c\right)}=\frac{1}{2}\)
\(\Rightarrow\frac{a}{b+c}=\frac{1}{2}\Rightarrow2a=b+c\)
\(\frac{b}{c+a}=\frac{1}{2}\Rightarrow2b=c+a\)
\(\frac{c}{a+b}=\frac{1}{2}\Rightarrow2c=a+b\)
Ta có: \(\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{b+a}{b}.\frac{c+b}{c}.\frac{a+c}{a}=\frac{2c.2a.2b}{b.c.a}=8\)
Vậy \(\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=8\)
a) \(\frac{200620062006}{200520052005}=\frac{2006\times100010001}{2005\times100010001}=\frac{2006}{2005}\)
b) \(\frac{352352}{470470}=\frac{352\times1001}{470\times1001}=\frac{352}{470}=\frac{176}{235}\)
a, \(\frac{2006}{2005}\)
b, \(\frac{176}{235}\)
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TL
A \(\frac{1}{4}+\frac{1}{7}=\frac{1\times7}{4\times7}+\frac{1\times4}{7\times4}=\frac{7}{24}+\frac{4}{28}=\frac{11}{28}\)
B\(\frac{1}{3}+\frac{1}{6}=\frac{1\times2}{3\times2}=\frac{2}{6}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}\)
C\(\frac{1}{4}+\frac{1}{5}=\frac{1\times5}{4\times5}+\frac{1\times4}{5\times4}=\frac{5}{20}+\frac{4}{20}=\frac{9}{20}\)
\(\frac{11}{28},\frac{1}{2},\frac{9}{20}\)MSC 140
\(\frac{11}{28}=\frac{11\times5}{28\times5}=\frac{55}{140}\)
\(\frac{1}{2}=\frac{1\times70}{2\times70}=\frac{70}{140}\)
\(\frac{9}{20}=\frac{9\times7}{20\times7}=\frac{63}{140}\)
\(\frac{55}{140}< \frac{63}{140}< \frac{70}{140}\)hoặc \(\frac{1}{4}+\frac{1}{7}< \frac{1}{4}+\frac{1}{5}< \frac{1}{3}+\frac{1}{6}\)
Vậy cho B
HT
3636/4545+x=4848/1515
x=4848/1515-3636/4545
x=14544/4545-3636/4545
x=10908/4545
3535/5050-x=8/25
x=3535/5050-1616/5050
x=1919/5050
\(\frac{a}{ac+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\)
\(=\frac{a}{ac+a+abc}+\frac{b}{bc+b+1}+\frac{bc}{abc+bc+b}\)
\(=\frac{1}{bc+b+1}+\frac{b}{bc+b+1}+\frac{bc}{bc+b+1}\)
\(=\frac{bc+b+1}{bc+b+1}\)
\(=1\)
Ta có:
\(N=\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{abc+ab+a}+\frac{c}{ac+c+abc}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{1+ab+a}+\frac{c}{c\left(a+1+ab\right)}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{1+ab+a}+\frac{1}{a+1+ab}\)
\(=\frac{a+ab+1}{ab+a+1}=1\)
Vậy N = 1