2^3 . 5 -[141-(5+4)^2 ]
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a. 7/9 - 16/9 = -9/9 = -1
b. 2/-15 + 7/10 = 17/30
c. (4 2/3 - 4 3/4) : -5/12 - 4/5
= (14/3 - 19/4) : (-5/12) - 4/5
= -1/12 : (-5/12) - 4/5
= 1/5 - 4/5
= -3/5
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`3x-16:2^3=31`
`=>3x-16:8=31`
`=>3x-2=31`
`=>3x=31+2`
`=>3x=33`
`=>x=11`
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`2^10:2^8+3[4.7+3.4]`
`=2^2+3[4(3+7)]`
`=4+3[4.10]`
`=4+3.40`
`=4+120`
`=124`
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`4^6:4^3-2^2 . 2^3`
`=4^3-2^5`
`=64-32`
`=32`
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`141+2^5 . 2^4-3^1 . 3^2`
`=141+2^9-3^3`
`=141+512-9`
`=644`
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`x+2^5:2^4=4.4^2`
`=>x+2=4^3`
`=>x=64-2`
`=>x=62`
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160-(2^3.5^2-6.25)= 160-(8.25-150)=160-50=110
4.5^2-32.2^4= 4.25 - 32.16= 100-512=-412
5871:[928-(247-82).5]=5871:[(928-165).5=5871: ( 763.5)=5871:103=57
777:7+1331:11^3 = 111+1331:1331 = 111+1=112
17.85+15.17-120=17(85+15)-120=17.100-120=1700-120=1580
15.141+59.15=15(141+59)=15.200=3000
20-[30-(5-1)^2]=20-[30-4^2]=20-(30-16)=20-11=9
37-11.3.(24-23)+22.10= 37-33.1+220=37-253=-216
160-(2^3.5^2-6.25)= 160-(8.25-150)=160-50=110
4.5^2-32.2^4= 4.25 - 32.16= 100-512=-412
5871:[928-(247-82).5]=5871:[(928-165).5=5871: ( 763.5)=5871:103=57
777:7+1331:11^3 = 111+1331:1331 = 111+1=112
17.85+15.17-120=17(85+15)-120=17.100-120=1700-120=1580 15.141+59.15=15(141+59)=15.200=3000
20-[30-(5-1)^2]=20-[30-4^2]=20-(30-16)=20-11=9
37-11.3.(24-23)+22.10= 37-33.1+220=37-253=-216
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Bài 1:
\(\frac{\frac{5}{131}+\frac{5}{141}-\frac{5}{191}-\frac{5}{4011}}{\frac{7}{131}+\frac{7}{141}+\frac{7}{-191}-\frac{7}{4011}}=\frac{5\left(\frac{1}{131}+\frac{1}{141}-\frac{1}{191}-\frac{1}{4011}\right)}{7\left(\frac{1}{131}+\frac{1}{141}-\frac{1}{191}-\frac{1}{4011}\right)}=\frac{5}{7}\)
Bài 2:
a) \(\frac{x}{7}+\left(\frac{-3}{7}\right)^2=\frac{2}{7}:\frac{4}{3}\)
\(\Rightarrow\frac{x}{7}+\frac{9}{49}=\frac{3}{14}\)
\(\Rightarrow\frac{x}{7}=\frac{3}{98}\)
\(\Rightarrow98x=21\)
\(\Rightarrow x=\frac{3}{14}\)
Vậy \(x=\frac{3}{14}\)
b) \(\left(x-1\right)^{x+6}=\left(x-1\right)^{x+4}\)
\(\Rightarrow\left(x-1\right)^{x+6}-\left(x-1\right)^{x+4}=0\)
\(\Rightarrow\left(x-1\right)^{x+4}.\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\left(x-1\right)^{x+1}=0\) hoặc \(\left(x-1\right)^2-1=0\)
+) \(\left(x-1\right)^{x+1}=0\Rightarrow x-1=0\Rightarrow x=1\)
+) \(\left(x-1\right)^2-1=0\)
\(\Rightarrow\left(x-1\right)^2=1\)
\(\Rightarrow\left(x-1\right)=\pm1\)
+ \(x-1=1\Rightarrow x=2\)
+ \(x-1=-1\Rightarrow x=0\)
Vậy \(x\in\left\{0;2;1\right\}\)
1)
\(\frac{\frac{5}{131}+\frac{5}{141}-\frac{5}{191}-\frac{5}{4011}}{\frac{7}{131}+\frac{7}{141}+\frac{7}{-191}-\frac{7}{4011}}\)
\(=\frac{5\left(\frac{1}{131}+\frac{1}{141}-\frac{1}{191}-\frac{1}{4011}\right)}{7\left(\frac{1}{131}+\frac{1}{141}-\frac{1}{191}-\frac{1}{4011}\right)}\)
\(=\frac{5}{7}\)
2) \(\frac{x}{7}+\left(-\frac{3}{7}\right)^2=\frac{2}{7}:\frac{4}{3}\)
\(=\frac{x}{7}+\frac{9}{49}=\frac{3}{14}\)
\(=\frac{x}{7}=\frac{3}{14}-\frac{9}{49}=\frac{3}{98}\)
\(\Rightarrow98x=21\)
\(\Rightarrow x=\frac{3}{14}\)
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a, 3 . 52 - 27 : 32
= 75 - 3
= 72
b, 24 . 19 - 24 . 16
= 24 . ( 19 - 16 )
= 24 . 3
= 48
c, 15 - 141 + 59 . 15 - 120
= -126 + 885 - 120
= 639
d, 50 - [ 30 - ( 7 - 3 )2 ]
= 50 - [ 30 - 42 ]
= 50 - 14 = 36
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\(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}\\ \dfrac{y}{5}=\dfrac{z}{6}\Rightarrow\dfrac{y}{20}=\dfrac{z}{24}\\ \Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}\)
Áp dụng t/c dtsbn ta có:
\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}=\dfrac{x-2y+3z}{15-2.20+3.24}=\dfrac{141}{47}=3\)
\(\dfrac{x}{15}=3\Rightarrow x=45\\ \dfrac{y}{20}=3\Rightarrow y=60\\ \dfrac{z}{24}=3\Rightarrow z=72\)
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a) 60-3(x-2)=51
3(x-2)=60-51
3(x-2)=9
x-2 = 9:3
x-2 = 3
x = 3+2
x 5
23.5-[141-(5+4)2 ]
=8.5-[141-92 ]
=40-(141-81)
=40-60
=-20