a)Ta thấy x²+4>0 lđ với mọi x

(x-1)(x²+4)<0 <=> x-1<0 <=> x<1

b) Ta có: \(|x-5|\ge0lđ\Leftrightarrow|x-5|+5\ge5\)

Mà \(|x-5|+5=x\)

=> x\(\ge5\)

bài này mình chắc chắn là

trả lời ở dưới

nữa

dưới nữa

tự đi mà giải hỏi cái dề

Ta có 5 < x³ - 15 \(\le\)16

=> 5 + 15 < x³ - 15 + 15 \(\le\)16 + 15

=> 20 < x³ \(\le\)31 (1)

Vì x nguyên

=> x³ nguyên 

Kết hợp với (1) => x³ = 27 

=> x = 3

Vậy x = 3

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

Bài 1 :

a, Ta có : \(\left(-123\right)+\left|-13\right|+\left(-7\right)\)

= \(\left(-123\right)+13+\left(-7\right)=\left(-117\right)\)

b, Ta có : \(\left|-10\right|+\left|45\right|+\left(-\left|-455\right|\right)+\left|-750\right|\)

= \(10+45-455+750=350\)

c, Ta có : \(-\left|-33\right|+\left(-15\right)+20-\left|45-40\right|-57\)

= \(\left(-33\right)+\left(-15\right)+20-5-57=-90\)

\(12^8:2^8< =6^x< =\left(6^3\right)^3\)

<=> \(1679616< =6^x< =6^9\)

<=> \(6^8< =6^x< =6^9\)

<=> \(8< =x< =9\)

bài 1:

<=> \(x\left(x^2-\frac{9}{16}\right)=0\)

TH1:x=0

TH2: \(x^2-\frac{9}{16}=0\)

=> \(x^2=\frac{9}{16}\)

TH2a: \(\Rightarrow x=\frac34\)

\(TH2b:x=-\frac34\)

bài 2:

1) <=> \(2N=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\cdots+\frac{2}{98\cdot99\cdot100}\)

Áp dụng công thức: \(\frac{2}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+1\right)}\) ta có:

\(2N=\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)+\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)+\cdots+\left(\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\)

\(2N=\frac{1}{1\cdot2}-\frac{1}{99\cdot100}\)

\(2N=\frac{4949}{9900}\)

\(\Rightarrow N=\frac{4949}{19800}\)

2) <=> \(5N=5^2+5^3+5^4+\cdots+5^{101}\)

=> \(5N-N=\left(5^2+5^3+5^4+\cdots+5^{101}\right)-\left(5+5^2+5^3+\cdots+5^{99}+5^{100}\right)\)

\(4N=5^{101}-5\)

=> \(N=\frac{\left(5^{101}-5\right)}{4}\)