lx-1,5l+l2,5-xl=0

=>lx-1,5l=-l2,5-xl

mà lx-1,5l>(=)0=>-l2,5-xl>(=)0

=>l2,5-xl=0=>x=2,5

=>lx-1,5l+l2,5-xl=1(trái giả thiết)

Vậy không có x thỏa mãn lx-1,5l+l2,5-xl=0

|x-1,5| + | 2,5 - x| = 0

=> |x  - 1,5| > hoặc = 0 và | 2.5 - x| > hoặc = 0, vs mọi x 

Nên |x - 1,5 | =0 và | 2,5 - x| = 0

=> x-1,5 = 0 và 2,5 - x =0

=>x = 1,5 và x = 2,5

Vậy x vô nghiệm

\(a,0,1^{2-x}>0,1^{4+2x}\\ \Leftrightarrow2-x>2x+4\\ \Leftrightarrow3x< -2\\ \Leftrightarrow x< -\dfrac{2}{3}\)

\(b,2\cdot5^{2x+1}\le3\\ \Leftrightarrow5^{2x+1}\le\dfrac{3}{2}\\ \Leftrightarrow2x+1\le log_5\left(\dfrac{3}{2}\right)\\ \Leftrightarrow2x\le log_5\left(\dfrac{3}{2}\right)-1\\ \Leftrightarrow x\le\dfrac{1}{2}log_5\left(\dfrac{3}{2}\right)-\dfrac{1}{2}\\ \Leftrightarrow x\le log_5\left(\dfrac{\sqrt{30}}{10}\right)\)

c, ĐK: \(x>-7\)

\(log_3\left(x+7\right)\ge-1\\ \Leftrightarrow x+7\ge\dfrac{1}{3}\\ \Leftrightarrow x\ge-\dfrac{20}{3}\)

Kết hợp với ĐKXĐ, ta có:\(x\ge-\dfrac{20}{3}\)

d, ĐK: \(x>\dfrac{1}{2}\)

\(log_{0,5}\left(x+7\right)\ge log_{0,5}\left(2x-1\right)\\ \Leftrightarrow x+7\le2x-1\\ \Leftrightarrow x\ge8\)

Kết hợp với ĐKXĐ, ta được: \(x\ge8\)

Bỏ | | đi và giải bình thường (vì x>0)

dễ chỉ có 2TH thôi

Nguyen Sy Hoc x dương rồi còn 2 TH gì

a, (19x+2.5²) : 14 = (13-8)² - 4²

(19x + 2.25) : 14 = 5² - 4²

(19x + 50) : 14 = 25 - 16

(19x + 50) : 14 = 9

19 x + 50 = 9.14

19x + 50 = 126 

19x = 126 - 50

19x = 76

x = 76 : 19

x = 4

vậy____

b) x + (x + 1) + (x + 2) + (x + 3)+.....+(x+30) = 1240

(x+x+x+...+x) + (1+2+3+...+30) = 1240

31x + 465 = 1240

31x = 1240 - 465

31x = 775

x = 775 : 31

x = 25

vậy____

c) |x + 7| = 20 + 5.(-3)

|x + 7| = 20 + (-15)

|x + 7| = 5

\(\Rightarrow\orbr{\begin{cases}x+7=5\\x+7=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=5-7\\x=-5-7\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-2\\x=-12\end{cases}}\)

vậy_____

`@` `\text {Ans}`

`\downarrow`

`3^3 * x^2 - 2^4 * x^2 = 8^2 * 5 - 4^2 * 3^2`

`=> x^2 . (3^3 - 2^4) = 2^6 . 5 - 2^4 . 3^2`

`=> x^2 . 11 = 2^4 . (2^2 . 5 - 3^2)`

`=> x^2 . 11 = 2^4 . 11`

`=> x^2 . 11 - 2^4 . 11 = 0`

`=> 11 . (x^2 - 16) = 0`

`=> x^2 - 16 = 0`

`=> x^2 = 16`

`=> x^2 = (+-4)^2`

`=> x = `\(\pm4\)

Vậy, `x \in`\(\left\{4;-4\right\}\)

_____

\(\left[\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{3}\right)^3\right]x+3^2\cdot2^2=4^2\cdot3\)

`=>`\(\left(\dfrac{1}{4}-\dfrac{1}{27}\right)x+\left(3\cdot2\right)^2=48\)

`=>`\(\dfrac{23}{108}\cdot x+6^2=48\)

`=>`\(\dfrac{23}{108}x=48-6^2\)

`=>`\(\dfrac{23}{108}x=48-36\)

`=>`\(\dfrac{23}{108}x=12\)

`=>`\(x=\dfrac{1296}{23}\)

Vậy, `x = `\(\dfrac{1296}{23}\)

\(3^3.x^2-2^4.x^2=8^2.5-4^3.3^2\)

\(\Leftrightarrow x^2\left(27-16\right)=2^6.5-2^6.9\)

\(\Leftrightarrow11x^2=2^6.\left(5-9\right)=-4.2^6=-2^8\)

\(\Leftrightarrow x^2=-\dfrac{2^6}{11}< 0\)

\(\Rightarrow x\in\varnothing\)

\(\left[\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{3}\right)^3\right]x+3^2.2^2=4^2.3\)

\(\Leftrightarrow\left(\dfrac{1}{4}-\dfrac{1}{27}\right)x+36=48\)

\(\Leftrightarrow\dfrac{23}{108}x=12\Leftrightarrow x=\dfrac{12.108}{23}=\dfrac{1296}{23}\)

\(A=\left(\dfrac{6}{1.4}\right)\left(\dfrac{12}{2.5}\right)\left(\dfrac{20}{3.6}\right)\left(\dfrac{x^2+3x+2}{x\left(x+3\right)}\right)\)

\(A=\dfrac{2.3}{1.4}.\dfrac{3.4}{2.5}.\dfrac{4.5}{3.6}...\dfrac{\left(x+1\right)\left(x+2\right)}{x\left(x+3\right)}\)

\(A=\dfrac{2.3.4...\left(x+1\right)}{1.2.3...x}.\dfrac{3.4.5...\left(x+2\right)}{4.5.6...\left(x+3\right)}=\left(x+1\right)\dfrac{3}{x+3}=\dfrac{3\left(x+1\right)}{x+3}\)

\(\left|x+\dfrac{1}{1.5}\right|+\left|x+\dfrac{1}{5.9}\right|+\left|x+\dfrac{1}{9.14}\right|+...+\left|x+\dfrac{1}{397.401}\right|\ge0\)

\(\Rightarrow101x\ge0\)

\(\Rightarrow x\ge0\)

\(\Rightarrow x+\dfrac{1}{1.5}+x+\dfrac{1}{5.9}+...+x+\dfrac{1}{397.401}=101x\)

\(\Rightarrow101x+\left(\dfrac{1}{1.5}+\dfrac{1}{5.9}+...+\dfrac{1}{397.401}\right)=x\)

\(\Rightarrow\dfrac{1}{4}\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{397.401}\right)=x\)

\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+....+\dfrac{1}{397}-\dfrac{1}{401}\right)\)

\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{401}\right)\)

\(\Rightarrow x=\dfrac{1}{4}.\dfrac{400}{401}\)

\(\Rightarrow x=\dfrac{100}{401}\)