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Bài làm
m) (x + 2).(3 - x) = 0;
=> x + 2 = 0 hoặc 3 - x = 0
=> x = -2 hoặc x = 3
Vậy x = -2 hoặc x = 3
d) 511.712 + 511.711
= 511 . ( 712 + 711 )
= 511 . [ 711 . ( 7 + 1 ) ]
= 511 . 711 . 8
= ( 5 . 7 )11 . 8
= 3511 . 8
512.712 + 9.511.711
= 511 ( 5 . 712 + 9 . 1 . 711 )
= 511 [ 711 ( 5 . 7 + 9 . 1 . 1 ) ]
= 511 ( 711 . 44 )
= 511 . 711 . 44
= 3511 . 44
m. \(\left(x+2\right)\left(3-x\right)=0\Leftrightarrow\orbr{\begin{cases}x+2=0\\3-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)
d. \(\frac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}=\frac{5^{11}.\left(7^{12}+7^{11}\right)}{5^{11}.\left(5.7^{12}+9.7^{11}\right)}=\frac{7^{12}+7^{11}}{5.7^{12}+9.7^{11}}=\frac{1}{5.9}=\frac{1}{45}\)
q. \(\left(x-3\right)+\left(x-2\right)+\left(x-1\right)+...+10+11=11\)
\(\Rightarrow\left(x-3\right)+\left(x-2\right)+\left(x-1\right)+...+10=0\)
\(\Rightarrow\left[\left(x-3\right)+\left(x-2\right)+\left(x-1\right)\right]+(1+2+3+...+10)=0\)
\(\Rightarrow\left(x-3\right)+\left(x-2\right)+\left(x-1\right)+55=0\)
\(\Rightarrow x-3+x-2+x-1=-55\)
\(\Rightarrow3x-6=-55\)
\(\Rightarrow3x=-49\)
\(\Rightarrow x=-\frac{49}{3}\)
a) 102 : 108 . 117
k biết làm :
164 : 43
= 48 : 43
= 45
39 : 38 . 32 . 15
= 3 . 32
= 33
51 . 516 : 514 . 50
= 517 : 514 . 1
= 53
1143 : 1142 . 112
= 11 . 112
= 113
125 . 124 : 127 . 12
= 129 : 127 . 12
= 122 . 12
= 123
sau 58 ,7,50 ,2,-60 là phép tính khác .
mấy b giúp mk nha ?
a) Đặt A = \(\frac{5^{12}+1}{5^{13}+1}\Rightarrow5A=\frac{5^{13}+5}{5^{13}+1}=1+\frac{4}{5^{13}+1}\)
Đặt \(B=\frac{5^{11}+1}{5^{12}+1}\Rightarrow5B=\frac{5^{12}+5}{5^{12}+1}=1+\frac{4}{5^{12}+1}\)
Vì \(\frac{4}{5^{13}+1}< \frac{4}{5^{12}+1}\Rightarrow1+\frac{4}{5^{13}+1}< 1+\frac{4}{5^{12}+1}\Rightarrow5A< 5B\Rightarrow A< B\)
Áp dụng công thức : \(\frac{a}{b}< 1\Leftrightarrow\frac{a}{b}< \frac{a+m}{b+m}\left(a;b;m\in N\right)\)
Ta có : \(A=\frac{5^{12}+1}{5^{13}+1}< 1\)
\(\Leftrightarrow A=\frac{5^{12}+1}{5^{13}+1}< \frac{5^{12}+1+4}{5^{13}+1+4}=\frac{5^{12}+5}{5^{13}+5}=\frac{5\left(5^{11}+1\right)}{5\left(5^{12}+1\right)}=B\)
\(\Leftrightarrow A< B\)
a)2/3 + 5+3/7:2+4/49.(-7)+1/2.(-2)\(^4\)
= 2/3+5+3/14+-4/7+8
=(2/3+3/14+-4/7)+(5+8)
=13/42+13
=559/42
b)2+3/4.(-0,4)-1-3/5.2,75+(-1/2):4/11
=2+(-0,3) - 1 - 1,65 + (-11/8)
= -2,325
c)( 2 + 1/3 +3 +1/2 ) : ( -4 - 1/6 +3 +1/7 ) +7 +1/2
=35/6 : -43/42 + 7 + 1/2
= -245/43 +7 +1/2
= 155/86
d) 2/3+ -5/12
= 8/12 + -5/12
= 3/12
e) -2/3 . -9/12
= \(\frac{-2.-9}{3.12}\)
= 18/36
=1/2
f) -2/9 - 7/12
=\(\frac{-24}{108}\)- \(\frac{14}{108}\)
= \(\frac{-38}{108}\)
=\(\frac{-19}{54}\)
g) 2/3 : -4/9
= \(\frac{2}{3}\). \(\frac{-9}{4}\)
= -3/2
h) 4 . ( -1/3)\(^3\)- 2 . ( -1/2)\(^2\)+3 .( 1/2 ) +1
=- 4 . 1/27 - 2 . 1/4 + 3 .1/2 +1
= -4/27 - 1/2 + 3/2 + 1
= 50 / 27
k ) ( 4 - 5/12 ) : 2 +5/24
= 43/12 : 2 +5/24
= 43/24 + 5/24
= 48/24
= 2
\(a,\left(\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}\right):\left(\dfrac{5}{12}+1-\dfrac{7}{11}\right)\)
\(=\dfrac{115}{132}:\dfrac{103}{132}=\dfrac{115}{132}.\dfrac{132}{103}=\dfrac{115}{103}\)
\(b,\left(-\dfrac{1}{2}\right)^2-\left(-2\right)^2-5^0\)
\(=\dfrac{1}{4}-4-1=-\dfrac{19}{4}\)
\(c,12\dfrac{1}{3}-\dfrac{5}{7}:\left(24-23\dfrac{5}{7}\right)\)
\(=\dfrac{37}{3}-\dfrac{5}{7}.\dfrac{2}{7}=\dfrac{37}{3}-\dfrac{10}{49}=\dfrac{1783}{147}\)