B=(1/11+1/12+...+1/20)+(1/21+...+1/30)+(1/31+1/32+...+1/40)+...+(1/61+1/62+...+1/70)

=>B<1/10*10+1/20*10+...+1/60*10

=>B<1+1/2+...+1/6

=>B<1+1/2+1/3+1/6+1/4+1/5
=>B<5/2

A

A

A

Là \(\left(\dfrac{1}{2}\right)^2\) hay \(\dfrac{1}{2^2}\) vậy bạn

Những cái sau tương tự

Mk xin phép ko vt lại đề nx

\(\Rightarrow A=\left[\left(3x-2\right)\left(x+1\right)-\left(2x+5\right)\left(x^2-1\right)\right]\div x+1\)

\(\Rightarrow A=3x-2-\left(2x-5\right)\left(x-1\right)\)

\(\Rightarrow x=\dfrac{1}{2}\)

\(\Rightarrow A=\dfrac{3}{2}-2-\left(1-5\right)\left(\dfrac{1}{2}-1\right)=-\dfrac{5}{2}\)

\(3^{x+1}=\left(\dfrac{1}{81}\right)^2=\left(3^{-4}\right)^2=3^{-8}\)

\(\Rightarrow x+1=-8\Rightarrow x=-9\)

\(\dfrac{2}{3}+\dfrac{1}{3}:x=\dfrac{1}{2}\)
\(\dfrac{1}{3}:x=\dfrac{1}{2}-\dfrac{2}{3}\)
\(\dfrac{1}{3}:x=-\dfrac{1}{6}\)
\(x=\dfrac{1}{3}:\left(-\dfrac{1}{6}\right)\)
\(x=-2\)
Vậy ...
#AvoidMe

`2/3 +1/3 : x=1/2`

`=> 1/3 : x= 1/2 -2/3`

`=> 1/3 : x= 3/6 -4/6`

`=> 1/3 : x=-1/6`

`=> x=1/3 :(-1/6)`

`=> x=1/3 xx (-6)`

`=> x= -2`

Vậy `x=-2`

(x+2).5=210

a) \(\left|4x-1\right|-\left|3x-\dfrac{1}{2}\right|=0\\ \Leftrightarrow\left|4x-1\right|=\left|3x-\dfrac{1}{2}\right|\\ \Leftrightarrow\left[{}\begin{matrix}4x-1=3x-\dfrac{1}{2}\\4x-1=\dfrac{1}{2}-3x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}4x-3x=1-\dfrac{1}{2}\\4x+3x=\dfrac{1}{2}+1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\7x=\dfrac{3}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{14}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{1}{2};\dfrac{3}{14}\right\}\) là nghiệm của pt.

b) \(\left|x-1\right|-2x=\dfrac{1}{2}\\ \Leftrightarrow\left|x-1\right|=2x+\dfrac{1}{2}\left(ĐK:x\ge\dfrac{-1}{4}\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2x+\dfrac{1}{2}\\x-1=-2x-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-2x=1+\dfrac{1}{2}\\x+2x=1-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-x=\dfrac{3}{2}\\3x=\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\left(ktmđk\right)\\x=\dfrac{1}{6}\left(tmđk\right)\end{matrix}\right.\)

Vậy \(x=\dfrac{1}{6}\) là nghiệm của pt.

Lời giải:

a.

$|4x-1|-|3x-\frac{1}{2}|=0$

$\Leftrightarrow |4x-1|=|3x-\frac{1}{2}$

\(\Leftrightarrow \left[\begin{matrix} 4x-1=3x-\frac{1}{2}\\ 4x-1=\frac{1}{2}-3x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{1}{2}\\ x=\frac{3}{14}\end{matrix}\right.\)

b. Nếu $x\geq 1$ thì:

$|x-1|-2x=\frac{1}{2}$

$\Leftrightarrow x-1-2x=\frac{1}{2}$
$\Leftrightarrow -x-1=\frac{1}{2}$

$\Leftrightarrow x=\frac{-3}{2}$ (vô lý vì $x\geq 1$)

Nếu $x< 1$ thì:

$1-x-2x=\frac{1}{2}$

$\Leftrightarrow x=\frac{1}{6}$ (tm)

\(|-2x+1,5|=\dfrac{1}{4}\Rightarrow-2x+1,5=\pm\dfrac{1}{4}\)

\(-2x+1,5=\dfrac{1}{4}\Rightarrow-2x=1,5-0,25\Rightarrow-2x=1,25\Rightarrow x=1,25:\left(-2\right)\Rightarrow x=...\)

\(-2x+1,5=-\dfrac{1}{4}\Rightarrow-2x=-0,25-1,5\Rightarrow-2x=1,75\Rightarrow x=1,75:\left(-2\right)\Rightarrow x=...\)

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

\(\Rightarrow1.\dfrac{1}{4}+3x=\pm\dfrac{5}{4}\)

\(1.\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow3x=\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=1\Rightarrow x=3\)

\(1.\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow3x=-\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=-\dfrac{3}{2}x=...\)

a: ta có: \(\left|-2x+\dfrac{3}{2}\right|=\dfrac{1}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x+\dfrac{3}{2}=\dfrac{1}{4}\\-2x+\dfrac{3}{2}=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-2x=-\dfrac{5}{4}\\-2x=-\dfrac{7}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{7}{8}\end{matrix}\right.\)

b: Ta có: \(\dfrac{3}{2}-\left|\dfrac{5}{4}+3x\right|=\dfrac{1}{4}\)

\(\Leftrightarrow\left|3x+\dfrac{5}{4}\right|=\dfrac{5}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{5}{4}=\dfrac{5}{4}\\3x+\dfrac{5}{4}=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=0\\3x=-\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{6}\end{matrix}\right.\)

\(n^4-1\\ =\left(n^2-1\right)\left(n^2+1\right)\\ =\left(n-1\right)\left(n+1\right)\left(n^2+1\right)\)

\(Vì.n\notin B_{\left(2\right)}\)

\(\Rightarrow\left(n-1\right)⋮2\\ \left(n+1\right)⋮2\\ \left(n^2+1\right)⋮2\\ \Rightarrow\left(n-1\right)\left(n+1\right)\left(n^2+1\right)⋮2.2.2=2^3=8\)