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a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)
\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)
=>x=10
b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)
hay \(x\in\left\{0;1;2\right\}\)
c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)
\(\Leftrightarrow6-x=0\)
hay x=6
a: =>5x=3x-6
=>2x=-6
hay x=-3
b: \(\Leftrightarrow\left(x-3\right)^2=4\cdot5^2=100\)
=>x-3=10 hoặc x-3=-10
=>x=13 hoặc x=-7
c: \(\left|x^3+1\right|+2\ge2\forall x\)
Dấu '=' xảy ra khi x=-1
Bài 1:
a) Ta có \(\left|x\right|\ge0\) (với mọi \(x\))
Mà \(\left|x\right|\le3\)
\(\Rightarrow0\le\left|x\right|\le3\)
\(\Rightarrow\left|x\right|\in\left\{0;1;2;3\right\}\)
\(\Rightarrow x\in\left\{0;1;2;3;-1;-2;-3\right\}\)
b) Ta có: \(\left|x-1\right|\ge0\) (với mọi \(x\))
Mà \(\left|x-1\right|\le4\)
\(\Rightarrow0\le\left|x-1\right|\le4\)
\(\Rightarrow\left|x-1\right|\in\left\{0;1;2;3;4\right\}\)
\(\Rightarrow x-1\in\left\{0;1;-1;2;-2;3;-3;4;-4\right\}\)
\(\Rightarrow x\in\left\{1;2;0;3;-1;4;-2;5;-3\right\}\)
Bài 2:
\(A=4+2^2+2^3+2^4+...+2^{20}\)
\(\Rightarrow A=2+2+2^2+2^3+2^4+...+2^{20}\)
Đặt \(B=2+2^2+2^3+...+2^{20}\)
\(\Rightarrow2B=2^2+2^3+2^4+...+2^{21}\)
\(\Rightarrow2B-B=\left(2^2+2^3+2^4+...+2^{21}\right)-\left(2+2^2+2^3+...+2^{20}\right)\)
\(\Rightarrow B=2^{21}-2\)
\(\Rightarrow A=2+2^{21}-2\)
\(\Rightarrow A=2^{21}\)
Bg
c) 9 < 3x : 3 < 81
=> 32 < 3x - 1 < 34
=> x - 1 = {2; 3; 4}
=> x = {3; 4; 5}
d) 5x . 5x + 1 . 5 x + 2 < 218 . 518 : 218
=> 5x + x + 1 + x + 2 < 218 : 218 . 518
=> 53x + 3 < 1.518
=> 53.(x + 1) < 518
=> 3.(x + 1) < 18
=> x + 1 < 18 : 3
=> x + 1 < 6
=> x < 6 - 1
=> x < 5
c. \(9\le3^x:3\le81\)
\(\Rightarrow3^2\le3^{x-1}\le3^4\)
\(\Rightarrow3^{x-1}\in\left\{3^2;3^3;3^4\right\}\)
\(\Rightarrow x-1\in\left\{2;3;4\right\}\)
\(\Rightarrow x\in\left\{3;4;5\right\}\)
d. Thêm đk : x thuộc N
\(5^x.5^{x+1}.5^{x+2}\le2^{18}.5^{18}:2^{18}\)
\(\Rightarrow5^{x+x+1+x+2}\le5^{18}\)
\(\Rightarrow x+x+x+1+2\le18\)
\(\Rightarrow3x+3\le18\)
\(\Rightarrow3\left(x+1\right)\le18\)
\(\Rightarrow x+1\le6\)
\(\Rightarrow x\le5\)
\(\Rightarrow x\in\left\{1;2;3;4;5\right\}\)
\(\dfrac{1}{-2}< \dfrac{x}{2}\le0\)
\(\Rightarrow\dfrac{-1}{2}< \dfrac{x}{2}\le\dfrac{0}{2}\)
\(\Rightarrow-1< x\le0\)
Vì \(x\in Z\Rightarrow x=0\)
a) (-5) . (3-4) = ( -5) x ( -1 ) = 5
b) (-2)2 . (-1)3 = 4 x -1 = -4
c) Các số nguyên thỏa mãn là : -6 ; -5 ; -4 ; -3 ; -2 ; -1 ; 0 ; 1;2;3;4
Ta có -6x-5x-4x-3x-2x-1x0x1x2x3x4 = 0 ( vì trong tích có thừa số 0 nên tích bằng 0)
a)35
b)=-4
c)=0