Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ko đúng đấy chứ
mình nhầm :
2) Vì /2x-3y/2015 lớn h+n hoặc bằng 0
và (x+y+x)2014 lớn hơn hoặc bằng 0 (với mọi x , y )
Mà /2x-3y/2015+ (x+y+z)2014 = 0
=) x+y+z = 0 (1)
=)2x- 3y = 0
=) x+y+x =0
=) 2(x+y+x)=0
=) 2x + 2y + 2x = 0
=) 3y+2y+3y = 0
=) 7y=0 =)y=0
thay y =0 vào (1)
=) ta có : x+y+x=0
=)x+0+x = 0
=) 2x=0 =) x=0
Vậy (x,y) = (0,0)
1. Tìm n, biết:
a) \(\dfrac{-32}{\left(-2\right)^n}=4\)
\(\Rightarrow\dfrac{\left(-2\right)^5}{\left(-2\right)^n}=\left(-2\right)^2\)
\(\Rightarrow\left(-2\right)^n.\left(-2\right)^2=\left(-2\right)^5\)
(-2)n + 2 = (-2)5
n + 2 = 5
n = 5 - 2
n = 3.
b) \(\dfrac{8}{2^n}=2\)
\(\Rightarrow\dfrac{2^3}{2^n}=2\)
\(\Rightarrow\) 2n . 2 = 23
n + 1 = 3
n = 3 - 1
n = 2.
c) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{2n-1}=\left(\dfrac{1}{2}\right)^3\)
2n - 1 = 3
2n = 3 + 1
2n = 4
n = 4 : 2
n = 2.
2. Tính:
a) \(\left(\dfrac{1}{2}\right)^3.\left(\dfrac{1}{4}\right)^2\)
\(=\left(\dfrac{1}{2}\right)^3.\left[\left(\dfrac{1}{2}\right)^2\right]^2\)
\(=\left(\dfrac{1}{2}\right)^3.\left(\dfrac{1}{2}\right)^4\)
\(=\left(\dfrac{1}{2}\right)^7\)
\(=\dfrac{1}{128}\)
b) 273 : 93
= (33)3 : (32)3
= 39 : 36
= 33
= 27
c) 1252 : 253
= (53)2 : (52)3
= 56 : 56
= 1
d) \(\dfrac{27^2.8^5}{6^6.32^3}\)
\(=\dfrac{\left(3^3\right)^2.\left(2^3\right)^5}{6^6.\left(2^5\right)^3}\)
\(=\dfrac{3^6.2^{15}}{6^6.2^{15}}\)
\(=\dfrac{3^6}{6^6}\)
\(=\dfrac{1}{64}.\)
B2 :
b) 27\(^3\): 9\(^3\)= (27:9)\(^3\)= 3\(^3\)
c) 125\(^2\): 25\(^3\)= 15625 : 15625 = 1
1. a) Ta có: M = |x + 15/19| \(\ge\)0 \(\forall\)x
Dấu "=" xảy ra <=> x + 15/19 = 0 <=> x = -15/19
Vậy MinM = 0 <=> x = -15/19
b) Ta có: N = |x - 4/7| - 1/2 \(\ge\)-1/2 \(\forall\)x
Dấu "=" xảy ra <=> x - 4/7 = 0 <=> x = 4/7
Vậy MinN = -1/2 <=> x = 4/7
2a) Ta có: P = -|5/3 - x| \(\le\)0 \(\forall\)x
Dấu "=" xảy ra <=> 5/3 - x = 0 <=> x = 5/3
Vậy MaxP = 0 <=> x = 5/3
b) Ta có: Q = 9 - |x - 1/10| \(\le\)9 \(\forall\)x
Dấu "=" xảy ra <=> x - 1/10 = 0 <=> x = 1/10
Vậy MaxQ = 9 <=> x = 1/10
5a.
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+....+\dfrac{1}{19.21}\\ =\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{19}-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}.\dfrac{20}{21}=\dfrac{10}{21}\)
b.
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\\ =\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)< \dfrac{1}{2}.1=\dfrac{1}{2}\)
a,\(\dfrac{3}{5}+\dfrac{2}{7}=\dfrac{31}{35}\)
b,\(\dfrac{-4}{3}:\dfrac{2}{15}=\dfrac{-60}{6}=-10\)
c,\(\dfrac{3}{7}.\dfrac{2}{9}+\dfrac{7}{9}.\dfrac{3}{7}=\dfrac{3}{7}.\left(\dfrac{2}{9}+\dfrac{7}{9}\right)=\dfrac{3}{7}\)
d,\(\left(\dfrac{-2}{3}\right)^3.9^2+\left(\dfrac{-3}{4}\right)^2.32\)
\(=\dfrac{\left(-2\right)^3}{3^3}.3^4+\dfrac{\left(-3\right)^2}{4^2}.2^5\)
\(=\left(-8\right).3+\dfrac{3^2}{4^2}.2^5\)
\(=\left(-24\right)+2.9\)
\(=\left(-24\right)+18\)
\(=-6\)
Bài 1:
|\(x\)| = 1 ⇒ \(x\) \(\in\) {-\(\dfrac{1}{3}\); \(\dfrac{1}{3}\)}
A(-1) = 2(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)) + 5
A(-1) = \(\dfrac{2}{9}\) + 1 + 5
A (-1) = \(\dfrac{56}{9}\)
A(1) = 2.(\(\dfrac{1}{3}\) )2- \(\dfrac{1}{3}\).3 + 5
A(1) = \(\dfrac{2}{9}\) - 1 + 5
A(1) = \(\dfrac{38}{9}\)
|y| = 1 ⇒ y \(\in\) {-1; 1}
⇒ (\(x;y\)) = (-\(\dfrac{1}{3}\); -1); (-\(\dfrac{1}{3}\); 1); (\(\dfrac{1}{3};-1\)); (\(\dfrac{1}{3};1\))
B(-\(\dfrac{1}{3}\);-1) = 2.(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)).(-1) + (-1)2
B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) - 1 + 1
B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\)
B(-\(\dfrac{1}{3}\); 1) = 2.(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)).1 + 12
B(-\(\dfrac{1}{3};1\)) = \(\dfrac{2}{9}\) + 1 + 1
B(-\(\dfrac{1}{3}\); 1) = \(\dfrac{20}{9}\)
B(\(\dfrac{1}{3};-1\)) = 2.(\(\dfrac{1}{3}\))2 - 3.(\(\dfrac{1}{3}\)).(-1) + (-1)2
B(\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) + 1 + 1
B(\(\dfrac{1}{3}\); -1) = \(\dfrac{20}{9}\)
B(\(\dfrac{1}{3}\); 1) = 2.(\(\dfrac{1}{3}\))2 - 3.(\(\dfrac{1}{3}\)).1 + (1)2
B(\(\dfrac{1}{3}\); 1) = \(\dfrac{2}{9}\) - 1 + 1
B(\(\dfrac{1}{3}\);1) = \(\dfrac{2}{9}\)
b: \(\left(\dfrac{2}{5}-\dfrac{7}{10}x\right):\dfrac{5}{3}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{2}{5}-\dfrac{7}{10}x=\dfrac{-3}{4}\cdot\dfrac{5}{3}=\dfrac{-5}{4}\)
\(\Leftrightarrow x\cdot\dfrac{7}{10}=\dfrac{2}{5}+\dfrac{5}{4}=\dfrac{8+25}{20}=\dfrac{33}{20}\)
\(\Leftrightarrow x=\dfrac{33}{20}:\dfrac{7}{10}=\dfrac{33}{20}\cdot\dfrac{10}{7}=\dfrac{33}{14}\)
c: \(\dfrac{7}{16}:\left(\dfrac{1}{4}x+\dfrac{9}{2}\right)-\dfrac{11}{6}=0\)
\(\Leftrightarrow\dfrac{7}{16}:\left(\dfrac{1}{4}x+\dfrac{9}{2}\right)=\dfrac{11}{6}\)
\(\Leftrightarrow x\cdot\dfrac{1}{4}+\dfrac{9}{2}=\dfrac{11}{6}:\dfrac{7}{16}=\dfrac{88}{21}\)
\(\Leftrightarrow x\cdot\dfrac{1}{4}=\dfrac{88}{21}-\dfrac{9}{2}=-\dfrac{13}{42}\)
hay \(x=-\dfrac{26}{21}\)
a: \(\left(x+2\right)^2+3>=3\)
=>M<=3/3=1
Dấu '=' xảy ra khi x=-2
b: Để N là số nguyên thì \(3n-12+21⋮n-4\)
\(\Leftrightarrow n-4\in\left\{1;3;7;21;-1;-3;-7;-21\right\}\)
hay \(n\in\left\{5;7;11;25;3;1;-3;-17\right\}\)