Vì \(\left|x+2\right|+\left|x+\dfrac{3}{5}\right|+\left|x+\dfrac{1}{2}\right|>0\) nên \(4x>0\) hay \(x>0\)

\(\Rightarrow x+2+x+\dfrac{3}{5}+x+\dfrac{1}{2}=4x\)

\(3x+2+\dfrac{3}{5}+\dfrac{1}{2}=4x\)

\(3x+\dfrac{31}{10}=4x\)

\(\Rightarrow4x-3x=\dfrac{31}{10}\)

\(\Rightarrow x=\dfrac{31}{10}\)

Lời giải:

Vì $|x+2|+|x+\frac{3}{5}|+|x+\frac{1}{2}|\geq 0$ với mọi $x$
$\Rightarrow 4x\geq 0\Rightarrow x\geq 0$. 

Khi đó:
$x+2>0; x+\frac{3}{5}>0; x+\frac{1}{2}>0$

$\Rightarrow |x+2|+|x+\frac{3}{5}|+|x+\frac{1}{2}|=4x$

$\Rightarrow x+2+x+\frac{3}{5}+x+\frac{1}{2}=4x$

$\Rightarrow 3x+\frac{31}{10}=4x$

$\Rightarrow x=\frac{31}{10}$ (tm)

giúp mk vs xin các bạn 

please

a) 1/5 - (1/2 + 3/4 ) : 5/2

= 1/5 - ( 1/4 + 3/4 ) : 5/2

=1/5 - 1 : 5/2

= 1/5 - 1 . 2/5

= 1/5 - 2/5

= -1/5

b) 1,5 . (1/3 - 2/3)

=3/2 . ( -1/3)

=-1/2

c) 9/10 . 23/11 - 1/11 . 9/10 + 9/10

= 9/10 . ( 23/11 - 1/11 ) + 9/10

= 9/10 . 1 + 9/10

= 9/10 + 9/10

= 18/10 = 9/5

Lời giải:
$G=\frac{2x^2+15}{x^2+3}=\frac{2(x^2+3)+9}{x^2+3}=2+\frac{9}{x^2+3}$

Vì $x^2\geq 0$ với mọi $x\in\mathbb{R}$

$\Rightarrow x^2+3\geq 3$

$\Rightarrow \frac{9}{x^2+3}\leq 3$

$\Rightarrow G=2+\frac{9}{x^2+3}\leq 2+3=5$.

Vậy $G_{\max}=5$. Giá trị này đạt được khi $x=0$

Biểu thức này không có giá trị min bạn nhé.

\(\dfrac{6}{5}\sqrt{1\dfrac{9}{16}}-\left(-\dfrac{3}{4}\right)^2:0,25\)

\(=\dfrac{6}{5}\cdot\sqrt{\dfrac{25}{16}}-\dfrac{9}{16}:0,25\)

\(=\dfrac{6}{5}\cdot\sqrt{\left(\dfrac{5}{4}\right)^2}-\dfrac{9}{16}:\dfrac{1}{4}\)

\(=\dfrac{6}{5}\cdot\dfrac{5}{4}-\dfrac{9\cdot4}{16}\)

\(=\dfrac{6}{4}-\dfrac{9}{4}\)

\(=\dfrac{6-9}{4}\)

\(=-\dfrac{3}{4}\)

Lời giải:
$A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{1998}{1999}=\frac{1.2.3....1998}{2.3.4...1999}=\frac{1}{1999}$

tự lm haha

giải thích nhé

`#3107.101107`

Đặt $A = 1 + 2 + 2^2 + 2^3 + ... + 2^{50}$

$2A = 2 + 2^2 + 2^3 + ... + 2^{51}$

$2A - A = (2 + 2^2 + 2^3 + ... + 2^{51}) - (1 + 2 + 2^2 + ... + 2^{50})$

$A = 2 + 2^2 + 2^3 + ... + 2^{51] - 1 - 2 - 2^2 - ... - 2^{50}$

$A = 2^{51} - 1$

Vậy, `A =` $2^{51} - 1.$

\(1\dfrac{2}{3}x+x=\left(-1,6\right)\\ \Rightarrow\dfrac{5}{3}x+x=\dfrac{18}{5}\\ \Rightarrow\dfrac{8}{3}x=\dfrac{18}{5}\\ \Rightarrow x=\dfrac{20}{27}\)

Sửa:

\(1\dfrac{2}{3}x+x=\left(-1,6\right)\\ \Rightarrow\dfrac{5}{3}x+x=-\dfrac{8}{5}\\ \Rightarrow\dfrac{8}{3}x=-\dfrac{8}{5}\\ \Rightarrow x=-\dfrac{20}{27}\)