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a ) x +5 = -10
x = -10 -5
x = - 15
b) x - ( - 10 ) = 5
x = 5+(-10)
x = -5
c) \(\left|x\right|\) -5 = 3
\(\left|x\right|=8\)
x ϵ { -8 ; 8 }
d) 15 - ( - x ) = 20
Không có số tự nhiên x nào mà 15 ( - x ) = 20
e ) \(\left|x-4\right|=3-\left(-7\right)\\ \left|x-4\right|=10\\ \left|x\right|=14\\ x\in\left\{\pm14\right\}\)
f ) \(\left|x+5\right|=10-\left(-20\right)\\ \left|x+5\right|=30\\ \left|x\right|=25\\ x\in\left\{\pm25\right\}\)
mình ghi nhầm nên các bạn cứ hết hai phân số là một câu nhé ví dụ như \(\dfrac{-5}{8}\):\(\dfrac{15}{4}\)
Trước hết ta hãy so sánh :
\(\dfrac{10^{100}+1}{10^{101}+1}\)với \(\dfrac{10^{100}+1}{10^{102}+1}\)
Ta có: Cả hai phân số trên cùng tử.
\(\Rightarrow\dfrac{10^{100}+1}{10^{101}+1}>\dfrac{10^{100}+1}{10^{102}+1}\)
Tiếp đó so sánh : \(\dfrac{10^{101}+1}{10^{102}+1}\)với \(1\)
Ta được: \(\dfrac{10^{101}+1}{10^{102}+1}< 1\)
Ta lại so sánh được:\(\dfrac{10^{100}+1}{10^{102}+1}< 1\) (*)
Từ (*) suy ra \(\dfrac{10^{100}+1}{10^{101}+1}< \dfrac{10^{101}+1}{10^{102}+2}< \dfrac{10^{101}+1}{10^{102}+1}< 1\Rightarrow\dfrac{10^{100}+1}{10^{101}+1}< \dfrac{10^{101}+1}{10^{102}+1}\)
Ngoài ra còn một cách như sau:
\(\dfrac{10^{101}+1}{10^{102}+1}=\dfrac{10^{\left(100+1\right)}+1}{10^{\left(101+1\right)}+1}=\dfrac{10}{10}.\dfrac{10^{100}+1}{10^{101}+1}>\dfrac{10^{100}+1}{10^{101}+1}\) hay B > A hay A < B
Bài 1:
d)
\(\dfrac{x+5}{95}+\dfrac{x+10}{90}+\dfrac{x+15}{85}+\dfrac{x+20}{80}=-4\)
\(\Leftrightarrow\dfrac{x+5}{95}+1+\dfrac{x+10}{90}+1+\dfrac{x+15}{85}+1+\dfrac{x+20}{80}+1=-4+1+1+1+1\)
\(\Leftrightarrow\dfrac{x+100}{95}+\dfrac{x+100}{90}+\dfrac{x+100}{85}+\dfrac{x+100}{80}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{95}+\dfrac{1}{90}+\dfrac{1}{85}+\dfrac{1}{80}\right)=0\)
\(\Leftrightarrow x+100=0\) ( vì: \(\dfrac{1}{95}+\dfrac{1}{90}+\dfrac{1}{85}+\dfrac{1}{80}\ne0\))
\(\Leftrightarrow x=-100\)
a , \(\left(\dfrac{-2}{3}+1\dfrac{1}{4}-\dfrac{1}{6}\right):\dfrac{-24}{10}\)
=\(\left(\dfrac{-2}{3}+\dfrac{5}{4}-\dfrac{1}{6}\right):\dfrac{-12}{5}\)
=\(\left(\dfrac{-8}{12}+\dfrac{15}{12}-\dfrac{2}{12}\right)\cdot\dfrac{-5}{12}\)
=\(\dfrac{5}{12}\cdot\dfrac{-5}{12}=\dfrac{-25}{144}\)
b , \(\dfrac{13}{15}\cdot0,25\cdot3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right)1\dfrac{23}{24}\)
=\(\dfrac{13}{15}\cdot\dfrac{1}{4}\cdot3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right)\cdot\dfrac{57}{24}\)
=\(\dfrac{13}{20}-\dfrac{47}{60}\cdot\dfrac{57}{24}\)
=\(\dfrac{13}{20}-\dfrac{893}{480}=\dfrac{312}{480}-\dfrac{893}{480}=\dfrac{-581}{480}\)
c , \(\left(\dfrac{12}{32}+\dfrac{5}{-20}-\dfrac{10}{24}\right):\dfrac{2}{3}\)
=\(\left(\dfrac{180}{480}-\dfrac{120}{480}-\dfrac{200}{480}\right)\cdot\dfrac{3}{2}\)
= \(\dfrac{-7}{24}\cdot\dfrac{3}{2}=\dfrac{-7}{16}\)
d , \(4\dfrac{1}{2}:\left(2,5-3\dfrac{3}{4}\right)+\left(-\dfrac{1}{2}\right)\)
=\(\dfrac{9}{2}:\left(\dfrac{5}{2}-\dfrac{15}{4}\right)-\dfrac{1}{2}\)
=\(\dfrac{9}{2}:\dfrac{-5}{4}-\dfrac{1}{2}=\dfrac{9}{2}\cdot\dfrac{-4}{5}-\dfrac{1}{2}=\dfrac{-18}{5}-\dfrac{1}{2}=\dfrac{-41}{10}\)
e , \(\dfrac{-5}{2}:\left(\dfrac{3}{4}-\dfrac{1}{2}\right)=\dfrac{-5}{2}\left(\dfrac{3}{4}-\dfrac{2}{4}\right)\)
=\(\dfrac{-5}{2}:\dfrac{1}{4}=\dfrac{-5}{2}\cdot4=-10\)
\(a,70-5\times\left(x-3\right)=45\)
\(5\times\left(x-3\right)=70-45\)
\(5\times\left(x-3\right)=25\)
\(x-3=25:5\)
x - 3 = 5
x = 5 + 3 = 8
vậy x = 8
\(b,12+\left(5+x\right)=20\)
12 + 5 + x = 20
17 + x = 20
x = 20 - 17 = 3
vậy x = 3
c, 130 - (100 + x) = 25
130 - 100 - x = 25
30 - x = 25
x = 30 - 25 = 5
vậy x = 5
\(g,10+2\times x=4^5:4^3\)
\(10+2\times x=4^2\)
\(10+2\times x=16\)
\(2\times x=16-10\)
\(2\times x=6\)
x = 6 : 2 = 3
vậy x = 3
h, \(14\times x+54=82\)
\(14\times x=82-54\)
\(14\times x=28\)
\(x=28:14=2\)
vậy x = 2
\(k,15\times x-133=17\)
\(15\times x=17+133\)
\(15\times x=150\)
\(x=150:15=10\)
vậy x = 10
2) Tinh nhanh:
a) \(\dfrac{5}{23}\) . \(\dfrac{17}{26}\) + \(\dfrac{5}{23}\) . \(\dfrac{10}{26}\) - \(\dfrac{5}{23}\)
= \(\dfrac{5}{23}\) . \(\left(\dfrac{17}{26}+\dfrac{10}{26}-1\right)\)
= \(\dfrac{5}{23}\) . \(\left(\dfrac{27}{26}-1\right)\) = \(\dfrac{5}{23}\) . \(\dfrac{1}{26}\)
= \(\dfrac{5}{598}\)
b) \(\dfrac{1}{7}.\dfrac{5}{9}+\dfrac{5}{9}.\dfrac{2}{7}+\dfrac{5}{9}.\dfrac{1}{7}+\dfrac{5}{9}.\dfrac{3}{7}\)
= \(\dfrac{5}{9}.\left(\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)
= \(\dfrac{5}{9}\) . 1= \(\dfrac{5}{9}\)
B1a)\(11\frac34-\left(6\frac56-4\frac12\right)+1\frac23\)
=\(11\frac34-6\frac56+4\frac12+1\frac23\)
=\(\left(11-6+4+1\right)+\left(\frac34-\frac56+\frac12+\frac23\right)\)
=\(10+\left(\frac{9}{12}-\frac{10}{12}+\frac{6}{12}+\frac{8}{12}\right)\)
=\(10+\left(-\frac{1}{12}+\frac{6}{12}+\frac{8}{12}\right)\)
=10+\(\frac{13}{12}\)
=\(\frac{120}{12}+\frac{13}{12}\)
=\(\frac{133}{12}\)
b)\(2\frac{17}{20}-1\frac{11}{5}+6\frac{9}{20}:3\)
= \(\frac{57}{20}-\frac{16}{5}+\frac{129}{20}\times\frac13\)
=\(\frac{57}{20}-\frac{16}{5}+\frac{129}{60}\)
=\(\frac{171}{60}-\frac{192}{60}+\frac{129}{60}\)
=\(\frac{108}{60}\)
=\(\frac95\)
\(A=\frac{8^{10}+4^{10}}{8^4+4^{11}}\\ =\frac{8^4.8^6+4^{10}}{8^4+4^{10}.4}\\ =\frac{8^6}{4}\\ =\frac{262144}{4}\\ =65536\)
\(\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.\left(-1\right)^{2013}=\dfrac{8}{27}.\dfrac{9}{16}.\left(-1\right)=-\dfrac{1}{6}\)
\(\left(\dfrac{1}{5}\right)^{15}.\left(\dfrac{1}{4}\right)^{20}=\dfrac{1}{5^{12}}.\dfrac{1}{4^{20}}=5^{-12}.4^{-20}=125^{-4}.1024^{-4}=\left(125.1024\right)^{-4}=128000^{-4}\)
\(\dfrac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\dfrac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}=\dfrac{2^{12}.3^{10}+2^{12}.2^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}=\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}\left(2.3+1\right)}=\dfrac{2.6}{3.7}=\dfrac{4}{7}\)
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