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1, Tìm x :
a, \(x^7.x^5=3^{12}\)
\(\Rightarrow x^{12}=3^{12}\)
\(\Rightarrow x=3\)
Vậy x = 3
b, \(\left(x+1\right)^4=5^8\div25^4\)
\(\left(x+1\right)^4=5^8\div\left(5^2\right)^4\)
\(\left(x+1\right)^4=5^8\div5^8\)
\(\left(x+1\right)^4=1\)
\(\Rightarrow x\in\left\{0;1\right\}\)
Vậy \(x\in\left\{0;1\right\}\)
c, \(x^6=x\)
\(\Rightarrow x^6-x=0\)
\(\Rightarrow x.x^5-x.1=0\)
\(\Rightarrow x\left(x^5-1\right)=0\)
x = 0 hoặc x5 - 1 = 0
x = 0 hoặc x5= 1
x = 0 hoặc x5 = 1
\(\Rightarrow x\in\left\{0;1\right\}\)
Vậy \(x\in\left\{0;1\right\}\)
2, Tính :
\(\left(4^{20}+4^{15}\right)\div\left(4^{10}+4^5\right)\)
\(=4^{15}.\left(4^5+1\right)\div4^5.\left(4^5+1\right)\)
\(=4^{15}\div4^5\)
\(=4^{10}\)
Vậy giá trị biểu thức trên bằng 410
\(A=2^0+2^1+2^2+...+2^{2016}\)
\(2A=2+2^2+2^3+...+2^{2017}\)
\(2A-A=\left(2+2^2+2^3+...+2^{2017}\right)-\left(2^0+2^1+2^2+...+2^{2016}\right)\)
\(\Rightarrow A=2^{2017}-1\)
Vậy : \(A=2^{2017}-1\)
1) \(\Leftrightarrow x+11-15+x+20=0\)
\(\Leftrightarrow2x+16=0\)
\(\Leftrightarrow x=-8\)
2) \(\Leftrightarrow2x-16+x-13=16\)
\(\Leftrightarrow3x-45=0\)
\(\Leftrightarrow x=15\)
Những câu dưới bạn làm tương tự như vậy nhé
1)(x+11)–(15–x) =–20
x+11 - 15 + x = -20
x + ( 11 -15 ) = -20
x + ( -4 ) = -20
x = -20 - ( -4 )
x = -16
Ta có :1.2+2.4+3.6+4.8+5.10/3.4+6.8+9.12+12.16+15.20=[1.2(1+4+9+...+25)]/[3.4(1+4+9+16)]
=(1.2)/(3.4)=2/12=1/6
CHÚC BẠN HỌC TỐT!!
cho tớ nhé!!!!
Bài 1:
19920 < 20020 = (23 . 52)20 = 260 . 540
200315 > 200015 = (24 . 53)15 = 260 . 545
Do: 260 . 540 < 260 . 545 => 19920 < 260 . 540 < 260 . 545 < 200315 => 19920 < 200315
Bài 2:
a/ (3 . x - 9) . 312 = 315 => 3 . x - 9 = 315 : 312
=> 3 . x - 9 = 27 => 3 . x = 27 + 9
=> 3 . x = 36 => x = 12
b/ (7 . x + 6) . 55 = 58 => 7 . x + 6 = 58 : 55
=> 7 . x + 6 = 125 => 7 . x = 125 - 6
=> 7 . x = 119 => x = 17
c/ (x - 5)4 = (x - 5)6
<=> x - 5 = 1 hoặc x - 5 = -1 hoặc x - 5 = 0
=> x = 6 hoặc x = 4 hoặc x = 5
2n+1/n+1 có gt nguyên
<=>2n+2-1/n+1
2(n+1)-1/n+1
2-(1/n+1)
để 2n+1/n+1 có gt nguyên
<=>1/n+1 có gt nguyên
=>n+1 thuộc {+_1}
lm tiếp nhé
a; - \(\dfrac{10}{13}\) + \(\dfrac{5}{17}\) - \(\dfrac{3}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{11}{20}\)
= - (\(\dfrac{10}{13}\) + \(\dfrac{3}{13}\)) + (\(\dfrac{5}{17}\) + \(\dfrac{12}{17}\)) - \(\dfrac{11}{20}\)
= - 1 + 1 - \(\dfrac{11}{20}\)
= 0 - \(\dfrac{11}{20}\)
= - \(\dfrac{11}{20}\)
b; \(\dfrac{3}{4}\) + \(\dfrac{-5}{6}\) - \(\dfrac{11}{-12}\)
= \(\dfrac{9}{12}\) - \(\dfrac{10}{12}\) + \(\dfrac{11}{12}\)
= \(\dfrac{10}{12}\)
= \(\dfrac{5}{6}\)
c; [13.\(\dfrac{4}{9}\) + 2.\(\dfrac{1}{9}\)] - 3.\(\dfrac{4}{9}\)
= [\(\dfrac{52}{9}\) + \(\dfrac{2}{9}\)] - \(\dfrac{4}{3}\)
= \(\dfrac{54}{9}\) - \(\dfrac{4}{3}\)
= \(\dfrac{14}{3}\)