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a)\(\left(\frac{25}{49}+\frac{17}{39}+\frac{22}{39}\times\frac{25}{49}\right)\times\frac{41}{25}\)
\(=\left(\frac{25}{49}+\left(\frac{17}{39}+\frac{22}{39}\right)\times\frac{25}{49}\right)\times\frac{41}{25}\)
\(=\left(\frac{25}{49}+\frac{39}{39}\times\frac{25}{49}\right)\times\frac{41}{25}\)
\(=\left(\frac{25}{49}+1\times\frac{25}{49}\right)\times\frac{41}{25}\)
\(=\left(\frac{25}{49}+\frac{25}{49}\right)\times\frac{41}{25}\)
\(=\frac{50}{49}\times\frac{41}{25}\)
\(=\frac{2050}{1225}\)
a) Có : 23/27>22/27
Mà 22/27>22/29
=> 23/27>22/29
b) Có : 12/25=24/50
24/50<24/49;24/49<25/49
=> 12/25<25/49
a,23/27>23/29
22/29<23/29
=>22/29<23/29<23/27
=>22/29<23/27
kl:....(kết luận)
a) \(\left(15-x\right)+\left(x-12\right)=7-\left(-8+x\right)\)
\(15-x+x-12=7+8-x\)
\(x=12\)
b) \(x-\left\{55-\left[49+\left(-28-x\right)\right]\right\}=13-\left\{47+\left[25-\left(32-x\right)\right]\right\}\)
\(x-55+49-28-x=13-47-25+32-x\)
\(x=7\)
\(a,\left(15-x\right)+\left(x-12\right)=7-\left(-8+x\right)\)
\(15-x+x-12=7+8-x\)
\(15-x=3\)
\(x=12\)
\(b,x-\left\{55-\left[49+\left(-28-x\right)\right]\right\}=13-\left\{47+\left[25-\left(32-x\right)\right]\right\}\)
\(x-\left[55-\left(49-28-x\right)\right]=13-\left[47+\left(25-32+x\right)\right]\)
\(x-\left(55-49+28+x\right)=13-\left(47+25-32+x\right)\)
\(x-\left(34+x\right)=13-\left(40+x\right)\)
\(x-34-x=13-40-x\)
\(-27-x=-34\)
\(x=-27--34\)
\(x=7\)
\(c,\left|x-10\right|-\left|-12\right|=4\)
\(\left|x-10\right|-12=4\)
\(\left|x-10\right|=16\)
\(\Rightarrow x-10=16\)hoặc \(x-10=-16\)
\(x=26\)hoặc \(x=-6\)
5/9 x 12/25 = 4/15
12/11 : 36/ 55 = 5/3
9/55 x 22 = 18/5
144/7 : 36 = 4/7
a, 556892 : 2 =278446
b, 50 + 20 + 11 + 22 + 33 =136
c, 22 + 44 + 55 + 66 + 88 = 275
d, 57 + 43 + 25 + 68 + 75 + 32 =300
\(49.x+55.x-x.4=178,6+121,4\)
\(\Rightarrow x\left(49+55-4\right)=300\)
\(\Rightarrow x100=300\)
\(\Rightarrow x=300:100=3\)
vậy x = 3
a) Ta có \(\dfrac{23}{27}>\dfrac{23}{29};\dfrac{23}{29}>\dfrac{22}{29}\)
Vậy \(\dfrac{23}{27}>\dfrac{22}{29}\)
b) Ta có \(\dfrac{15}{25}=1-\dfrac{2}{5}\)
\(\dfrac{25}{49}=1-\dfrac{24}{49}\)
Vì \(\dfrac{2}{5}=\dfrac{24}{60}< \dfrac{24}{49}\)
Vậy \(\dfrac{15}{25}>\dfrac{25}{49}\)
\(\frac{22}{49}\cdot\frac{25}{2}\cdot\frac{7}{55}=\frac{2\cdot5\cdot11\cdot7}{7\cdot7\cdot2\cdot11\cdot5}=\frac{1}{7}\)