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a, \(\left(2600+6400\right)-3x=1200\)
\(\Rightarrow9000-3x=1200\)
\(\Rightarrow3x=7800\)
\(\Rightarrow x=2600\)
b, \(\left[\left(6x-72\right):2-84\right].28=5628\)
\(\Rightarrow\left(6x-72\right):2-84=201\)
\(\Rightarrow\left(6x-72\right):2=285\)
\(\Rightarrow6x-72=570\)
\(\Rightarrow6x=642\)
\(\Rightarrow x=107\)
a) (2600 + 6400) - 3x = 1200
9000 - 3x = 1200
3x = 9000 - 1200
3x = 7800
x = 2600
Vậy x = 2600
b) [(6x - 72) : 2 - 48].28 = 5628
(6x - 72) : 2 - 48 = 5628 : 28
(6x - 72) : 2 - 48 = 201
(6x - 72) : 2 = 201 + 48
(6x - 72) : 2 = 249
6x - 72 = 249.2
6x - 72 = 498
6x = 498 + 72
6x = 570
x = 570 : 6
x = 95
Vậy x = 95
a.
\(10⋮\left(x-1\right)\)
\(\Rightarrow x-1=Ư\left(10\right)\)
\(\Rightarrow x-1=\left\{-10;-5;-2;-1;1;2;5;10\right\}\)
\(\Rightarrow x=\left\{-9;-4;-1;0;2;3;6;11\right\}\)
b.
\(\left(x+5\right)⋮\left(x-2\right)\Rightarrow\left(x-2\right)+7⋮x-2\)
\(\Rightarrow7⋮x-2\)
\(\Rightarrow x-2=Ư\left(7\right)=\left\{-7;-1;1;7\right\}\)
\(\Rightarrow x=\left\{-5;1;3;9\right\}\)
c.
\(\left(3x+8\right)⋮\left(x-1\right)\)
\(\Rightarrow\left(3x-3+11\right)⋮\left(x-1\right)\)
\(\Rightarrow3\left(x-1\right)+11⋮x-1\)
\(\Rightarrow11⋮\left(x-1\right)\)
\(\Rightarrow x-1=Ư\left(11\right)=\left\{-11;-1;1;11\right\}\)
\(\Rightarrow x=\left\{-10;0;2;12\right\}\)
a)\(\left(\frac{1}{2}-\frac{1}{3}\right).6^x+6^{x+2}=6^{15}+6^{18}\)
\(\frac{1}{6}.6^x+6^{x+2}=6^{15}\left(1+6^3\right)\)
\(\frac{1}{6}.6^x\left(1+6^3\right)=6^{15}.217\)
\(6^{x-1}.217=6^{15}.217\)
\(6^{x-1}=6^{15}\)
\(x-1=15\)
\(x=16\)
b) \(\left(\frac{1}{2}-\frac{1}{6}\right).3^{x+4}-4.3^x=3^{16}-4.3^{13}\)
\(\frac{1}{3}.3^x.4\left(3^4-1\right)=3^{13}.4\left(3^3-1\right)\)
\(3^x.4.\left(3^3-1\right)=3^{13}.4.\left(3^3-1\right)\)
\(3^x=3^{13}\)
\(x=13\)
\(\left(\frac{1}{2}-\frac{1}{6}\right).\left(3^x.3^4\right)-4.3^x=3^{16}-4.3^{13}\)
=> \(\frac{1}{3}.3^x.3^4-4.3^x=3^{16}-4.3^{13}\)
=> \(3^x.3^4-4.3^x=\left(3^{16}-4.3^{13}\right):\frac{1}{3}\)
=> \(3^x.3^4-4.3^x=-386339074,3\)
=> \(3^x.\left(3^4-4\right)=-386339074,3\)
=> \(3^x.77=-386339074,3\)
=> \(3^x=-386339074,3:77\)
=> \(3^x=-5017390,575\)
=> x = ... chắc tự ngồi tính đc
\(\frac{|x-2|}{12}\)\(+\)\(\frac{|x-2|}{20}+\)\(\frac{|x-2|}{30}+\)\(\frac{|x-2|}{42}\)\(=\frac{70^5}{2^3.21^6}\)
\(\Rightarrow|x-2|.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=\frac{2^5.5^5.7^5}{2^3.7^6.3^6}\)
\(\Rightarrow|x-2|.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)=\frac{2^2.5^5}{7.3^6}\)
\(\Rightarrow|x-2|.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)=\frac{4.5^5}{21.3^5}\)
\(\Rightarrow|x-2|\left(\frac{1}{3}-\frac{1}{7}\right)=\frac{4.5^5}{21.3^5}\)\(\Rightarrow|x-2|=\frac{5^5}{3^5}\)
ĐẾN ĐÂY DỄ RÙI TỰ GIẢI TIẾP
1. \(x⋮12,x⋮10\Rightarrow x\in BC(12,10)\)và -200 < x < 200
Theo đề bài , ta có :
\(12=2^2\cdot3\)
\(10=2\cdot5\)
\(\Rightarrow BCNN(10,12)=2^2\cdot3\cdot5=60\)
\(\Rightarrow BC(10,12)=B(60)=\left\{0;60;-60;120;-120;180;-180;240;...\right\}\)
Mà \(x\in BC(10,12)\)và -200 < x < 200 => \(x\in\left\{0;60;-60;120;-120;180;-180\right\}\)
Học tốt
\(-84-\left(-18-6\right)=x+\left(-20\right)\)
\(\Rightarrow-84+18+6=x-20\)
\(\Rightarrow-60=x-20\)
\(\Rightarrow x=-60+20\)
\(\Rightarrow x=-40\)
\(-84-\left(-18-6\right)=x+\left(-20\right)\)
\(\Rightarrow\)\(-84+18+6=x+\left(-20\right)\)
\(\Rightarrow\)\(66+6=x+\left(-20\right)\)
\(\Rightarrow\)\(72=x+\left(-20\right)\)
\(\Rightarrow\)\(x=72-\left(-20\right)\)
\(\Rightarrow\)\(x=92\)