\(\frac{244}{537}\)+\(\frac{132}{354}\)+\(\frac{4343}{4342}\)+\(\frac{4324}{4545}\) * 0 = ?
1.Tìm y
a) y3 + 3y = 12 x 11
b) 64 x y + 225 = 700 - 75 - 36 x y
2.Tính
a)\(\frac{4343}{7171}+\frac{4343}{7171}+\frac{4343}{7171}+\frac{4343}{7171}\)
b) A = \(\frac{1}{33}\times\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)
bài 2
a] = 3 x \(\frac{4343}{7171}\)= \(\frac{17372}{7171}\)= \(\frac{172}{71}\)
b] = \(\frac{1}{33}\)x \(\frac{44}{7}\)= \(\frac{1}{3}\)x \(\frac{4}{7}\)=\(\frac{4}{21}\)
bài 1
a] y là 9
b] <=> 64y + 36y = 700 - 75 - 225
<=> 100y = 400
<=> y = 4
trên lớp cô sửa rồi nên mình giải luôn:
1) Tìm y
a) y3 + 3y = 12 x 11
y3 + 3y = 132
y x 10 + 3 + 3 x 10 + y = 132
( y x 10 + y ) + ( 3 x 10 + 3 ) = 132
11 x y + 33 = 132
11 x y = 132 - 33
11 x y = 99
y = 99 : 11
y = 9
b) 64 x y + 225 = 700 - 75 - 36 x y
64 x y + 225 = 625 - 36 x y
64 x y + 36 x y = 625 -225
64 x y + 36 x y = 400
( 64 + 36 ) x y = 400
100 x y = 400
y = 400 : 100
y = 4
2) Tính
a) \(\frac{4343}{7171}+\frac{4343}{7171}+\frac{4343}{7171}+\frac{4343}{7171}\)
\(=\frac{4343}{7171}\times4\)
\(=\frac{43}{71}\times4\)
\(=\frac{172}{71}\)
b) A = \(\frac{1}{33}\times\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)
Ta có:
\(\frac{3333}{2020}=\frac{3333:101}{2020:101}=\frac{33}{20}\)
\(\frac{333333}{303030}=\frac{333333:10101}{303030:10101}=\frac{33}{30}\)
\(\frac{33333333}{42424242}=\frac{33333333:1010101}{42424242:1010101}=\frac{33}{42}\)
A = \(\frac{1}{33}\times\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
A = \(\frac{1}{33}\times33\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
A = 1 x \(\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
A = 1 x \(\left(\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}\right)\)
A = 1 x \(\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
A = 1 x \(\left(\frac{1}{3}-\frac{1}{7}\right)\)
A = 1 x \(\left(\frac{7}{21}-\frac{3}{21}\right)\)
A = 1 x \(\frac{4}{21}\)
A = \(\frac{4}{21}\)
Tính giá trị biểu thức sau đây 1 cách hợp lý:
A=\(\left[\left(\frac{2}{537}-\frac{3}{1074}\right).\frac{537}{17}+\frac{33}{34}\right]:\left[\left(\frac{7}{1975}+\frac{11}{3950}\right).\frac{1975}{25}+\frac{5}{2}\right]\)
Không dùng máy tính, hãy so sánh A và B
A=\(\frac{352}{353}_{ }\)\(+\frac{353}{354}+\frac{354}{255}\); B=\(\frac{352+353+354}{353+354+355}\)
Bài này được cái dễ lộn số =.=
Ta có :
\(B=\frac{352+353+354}{353+354+355}=\frac{352}{343+354+355}+\frac{353}{353+354+355}+\frac{354}{353+354+355}\)
Vì :
\(\frac{352}{343}>\frac{352}{353+354+355}\)
\(\frac{353}{354}>\frac{353}{353+354+355}\)
\(\frac{354}{355}>\frac{354}{353+354+355}\)
Nên \(\frac{352}{353}+\frac{353}{354}+\frac{354}{355}>\frac{352+353+354}{353+354+355}\)
Hay \(A>B\)
Vậy \(A>B\)
Chúc bạn học tốt ~
Ta có : \(B=\frac{352+353+354}{353+354+355}\)
\(\Rightarrow B=\frac{352}{353+354+355}+\frac{353}{353+354+355}+\frac{354}{353+354+355}\)
Ta có : \(\frac{352}{353}>\frac{352}{353+354+355}\)
\(\frac{353}{354}>\frac{353}{353+354+355}\)
\(\frac{354}{355}>\frac{354}{353+354+355}\)
Cộng vế theo vế, ta có : \(\frac{352}{353}+\frac{353}{354}+\frac{354}{355}>\frac{352+353+354}{353+354+355}\)
\(\Rightarrow A>B\)
\(\frac{7575}{4545}\)*\(\frac{22222}{33333}\)
36 + [16 - 3 * (x + 2)] = 36
13 * (x + 2) = 36 - 36
13 * (x + 2) = 0
x + 2 = 0 : 13
x + 2 = 0
x = 0 - 2
x = -2
36+[16-3*(x+2)]=36
[16-3*(x-2)] =36-36
16-3*(x-2) =0
13*(x-2) =0
(x-2) =0:13
(x-2) =0
x =0+2
x =2
suy ra x=2
nho kich minh nhe!!!!!!!!!!!!
36+[16-3*(x+2)]=36
16-3*(x+2)=36-36
16-3*(x+2)=0
3*(x+2)=16-0=16
x+2=16/3
x=16/3-2
x=10/3
\(0,25:\left(10,3-9,8\right)-\frac{3}{4}\\ b,\left(3\frac{4}{5}-2.x\right).1\frac{1}{3}\\ c,\frac{x}{7}=\frac{6}{-21}\\ d,\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{ }{132}\)cái cuối là 1/132
c) x=-2 nha
d) =\(\frac{1}{5.6}\)+\(\frac{1}{6.7}\)+......+\(\frac{1}{11.12}\)
=\(\frac{1}{5}\)-\(\frac{1}{6}\)+\(\frac{1}{6}\)-\(\frac{1}{7}\)+.....+\(\frac{1}{11}\)-\(\frac{1}{12}\)
=\(\frac{1}{5}\)-\(\frac{1}{12}\)= \(\frac{7}{60}\)
\(=\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{11.12}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{5}-\frac{1}{12}\)
\(=\frac{7}{60}\)
GPT :
a, \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}..\right).503x=1+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4023}{2011}+\frac{4024}{2012}\)
b, \(\left(\frac{0,6+\frac{3}{7}-\frac{2}{11}}{1+\frac{5}{7}-\frac{5}{11}}+\frac{\frac{2}{3}-1,5+\frac{2}{9}}{\frac{5}{3}-3,75+\frac{5}{9}}\right)+93x=\left(\frac{3737}{4545}-\frac{954954}{975975}\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)-7.\left(x-3\right)\)
\(VP=1+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4023}{2011}+\frac{4024}{2012}\)
\(=1-1+\left(\frac{2014}{2}-1\right)+\left(\frac{2015}{3}-1\right)+...+\left(\frac{4023}{2011}-1\right)+\left(\frac{40024}{2012}-1\right)+2012\)
\(=\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}+\frac{2012}{1}\)
\(=2012.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)
\(\Rightarrow2012=503.x\Rightarrow x=\frac{2012}{503}=4\)
GPT :
a, \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}..\right).503x=1+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4023}{2011}+\frac{4024}{2012}\)
b, \(\left(\frac{0,6+\frac{3}{7}-\frac{2}{11}}{1+\frac{5}{7}-\frac{5}{11}}+\frac{\frac{2}{3}-1,5+\frac{2}{9}}{\frac{5}{3}-3,75+\frac{5}{9}}\right)+93x=\left(\frac{3737}{4545}-\frac{954954}{975975}\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)-7.\left(x-3\right)\)
hay sap xep cac phan so theo thu tu tu be den lon len
a\(\frac{4343}{4444}\)\(\frac{1515}{1616}\)\(\frac{71}{72}\)\(\frac{93}{94}\)
b\(\frac{1717}{1515}\), \(\frac{127127}{125125}\),\(\frac{93}{91}\),\(\frac{111}{109}\)
\(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)