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\(\frac{11.12+33.36+55.60}{22.24+66.72+110.120}=\frac{11.12+33.36+55.60}{2.11.2.12+2.33.2.36+2.55.2.60}=\frac{11.12+33.36+55.60}{4.11.12+4.33.36+4.55.60}=\frac{11.12+33.36+55.60}{4.\left(11.12+33.36+55.60\right)}=\frac{1}{4.1}=\frac{1}{4}\)
A= \(\dfrac{10.11.\left(1+5.5+7.7\right)}{11.12.\left(1+5.5+7.7\right)}=\dfrac{10}{12}=\dfrac{5}{6}\)
\(a,\frac{7}{x}=\frac{x}{28}=>x\cdot x=28\cdot7=>x^2=196=>x^2=14^2\)\(=>x=14\)
\(b,\frac{10+x}{x+17}=\frac{3}{4}=>\left(10+x\right)\cdot4=\left(x+17\right)\cdot3=>40+x4=x3+51\)\(=>x4-x3=51-40=>x=11\)
\(\dfrac{10+x}{17+x}=\dfrac{10\cdot11+50\cdot55+70\cdot77}{11\cdot12+55\cdot66+77\cdot84}\)
\(\Leftrightarrow\dfrac{10+x}{17+x}=\dfrac{10\cdot11+5\cdot10\cdot111+7\cdot10\cdot11}{11\cdot12+5\cdot11\cdot12+7\cdot11\cdot12}\)
\(\Leftrightarrow\dfrac{10+x}{17+x}=\dfrac{10\cdot11\left(1+5+7\right)}{11\cdot12\left(1+5+7\right)}\)
\(\Leftrightarrow\dfrac{10+x}{17+x}=\dfrac{5}{6}\)
\(\Leftrightarrow6\left(10+x\right)=5\left(17+x\right)\)
\(\Leftrightarrow60+6x=85+5x\)
\(\Leftrightarrow6x-5x=85-60\)
\(\Leftrightarrow x=25\) (TM x\(\ne17\))
Vậy ...................................
11 . 12 + 33 . 36 + 55 . 60
= 11 . 12 + 11 . 3 . 12 . 3 + 11 . 5 . 12 . 5
= (11 + 12) . 1 . (11 + 12) . 3 . (11 + 12) . 5
= 23 . 1 . 23 . 3 . 23 . 5
= 23 . (1 + 3 + 5)
= 23 . 9
= 207
\(11\cdot12+33\cdot36+55\cdot60\)
\(=11\cdot12+3\cdot11\cdot3\cdot12+5\cdot11\cdot5\cdot12\)
\(=11\cdot12\left(1+9+25\right)\)
\(=132\cdot35\)
\(=4620\)