`#040911`

a,

\(\dfrac{1}{2}\cdot\left(x-4\right)-\dfrac{1}{4}\cdot\left(x-\dfrac{4}{3}\right)=2\cdot\left(x-\dfrac{1}{2}\right)\)

\(\Rightarrow\dfrac{1}{2}x-2-\dfrac{1}{4}x+\dfrac{1}{3}=2x-1\\\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{4}x-2x\right)=2-\dfrac{1}{3}-1\\ \Rightarrow-\dfrac{7}{4}x=\dfrac{2}{3}\\ \Rightarrow x=\dfrac{2}{3}\div\left(-\dfrac{7}{4}\right)\\ \Rightarrow x=-\dfrac{8}{21}\)

Vậy, \(x=-\dfrac{8}{21}\)

b,

\(\dfrac{3}{4}-\left(x-\dfrac{1}{2}\right)^2=-\dfrac{11}{2}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{3}{4}-\left(-\dfrac{11}{2}\right)\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{25}{4}\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\left(\pm\dfrac{5}{2}\right)^2\)

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

Vậy, \(x\in\left\{-2;3\right\}\)

c,

\(\dfrac{3}{16}+1\dfrac{1}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}\)

\(\Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}-\dfrac{3}{16}\\ \Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\\ \Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\div\dfrac{17}{16}\\ \Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{17}\)

Bạn xem lại đề có sai kh nhỉ?

c) \(\dfrac{3}{16}+\dfrac{1}{\dfrac{1}{16}}\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}\)

\(\Rightarrow16\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}-\dfrac{3}{16}\)

\(\Rightarrow16\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\)

\(\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}:16\)

\(\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{256}=\left(\dfrac{3}{16}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{2}{3}=\dfrac{3}{16}\\x-\dfrac{2}{3}=-\dfrac{3}{16}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{16}+\dfrac{2}{3}\\x=-\dfrac{3}{16}+\dfrac{2}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{41}{48}\\x=\dfrac{23}{48}\end{matrix}\right.\)

`#040911`

\(4\cdot\left(\dfrac{1}{2}\right)^3+3\cdot\left(\dfrac{1}{2}\right)^2-2\cdot\left(\dfrac{1}{2}\right)^0\\ =4\cdot\dfrac{1}{8}+3\cdot\dfrac{1}{4}-2\cdot1\\ =\dfrac{4}{8}+\dfrac{3}{4}-2\\ =\dfrac{2}{4}+\dfrac{3}{4}-2\\ =\dfrac{5}{4}-2\\ =-\dfrac{3}{4}\)

4.1/8 + 3. 1/4 - 2. 1

= 4/8 + 3/4 - 2

= 1/2 + 3/4 - 2 

= -3 /4

Lời giải:
$\frac{2x-y}{x+y}=\frac{1}{3}$

$\Rightarrow 3(2x-y)=x+y$

$\Leftrightarrow 6x-3y=x+y$

$\Leftrightarrow 5x=4y$

$\Leftrightarrow x=\frac{4}{5}y$. Thay vào biểu thức A:

$A=\frac{\frac{4}{5}y+y}{2.\frac{4}{5}y+y}=\frac{\frac{9}{5}y}{\frac{13}{5}y}=\frac{9}{13}$

Bạn xem lại đề bài, k có yêu cầu gì?

bg chi tiết đây bạn nhé. tick cho mình nhé.

Bài giải

a) Phân số chỉ số phần hai lần cửa hàng bán là:

$\genfrac{}{}{0ex}{}{1}{4}+\genfrac{}{}{0ex}{}{2}{7}=\genfrac{}{}{0ex}{}{15}{28}$(tấm vải)

Lúc đầu tấm vải dài là:

$30:\genfrac{}{}{0ex}{}{15}{28}=56\left(m\right)$

b) Lần thứ nhất cửa hàng bán:

$56:4=14\left(m\right)$

Lần thứ hai cửa hàng bán:

$56\times \genfrac{}{}{0ex}{}{2}{7}=16\left(m\right)$

Lần thứ ba cửa hàng bán:

$56-14-16=26\left(m\right)$

Bài giải

a) Phân số chỉ số phần hai lần cửa hàng bán là:

$\genfrac{}{}{0ex}{}{1}{4}+\genfrac{}{}{0ex}{}{2}{7}=\genfrac{}{}{0ex}{}{15}{28}$(tấm vải)

Lúc đầu tấm vải dài là:

$30:\genfrac{}{}{0ex}{}{15}{28}=56\left(m\right)$

b) Lần thứ nhất cửa hàng bán:

$56:4=14\left(m\right)$

Lần thứ hai cửa hàng bán:

$56\times \genfrac{}{}{0ex}{}{2}{7}=16\left(m\right)$

Lần thứ ba cửa hàng bán:

$56-14-16=26\left(m\right)$

Đ/S : 26m

\(B=-\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^6+3\)

vì \(B=-\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^6\le0,\forall x\inℝ\)

\(\Rightarrow B=-\left(\dfrac{4}{9}x-\dfrac{2}{15}\right)^6+3\le3\)

Dấu "=" xảy ra khi và chỉ khi

\(\dfrac{4}{9}x-\dfrac{2}{15}=0\Rightarrow\dfrac{4}{9}x=\dfrac{2}{15}\Rightarrow x=\dfrac{9}{15}\)

Vậy \(GTLN\left(B\right)=3\left(tạix=\dfrac{9}{15}\right)\)

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

vì \(\left(2x+\dfrac{1}{3}\right)^4\ge0,\forall x\inℝ\)

\(\Rightarrow A=\left(2x+\dfrac{1}{3}\right)^4-1\ge-1\)

Dấu "=" xảy ra khi và chỉ khi

\(2x+\dfrac{1}{3}=0\Rightarrow2x=-\dfrac{1}{3}\Rightarrow x=-\dfrac{1}{6}\)

\(\Rightarrow GTNN\left(A\right)=-1\left(tạix=-\dfrac{1}{6}\right)\)

ab=3 cm

\(AB=5\left(cm\right)< CD=k^2\)

\(MN=11\left(cm\right)>CD=k^2\)

\(\Rightarrow CD=k^2=9\left(cm\right)\)

k bình là gì vậy

\(A=\dfrac{3x-1}{x+2}\inℕ\left(x\inℕ;x\ne-2\right)\)

\(\Rightarrow3x-1⋮x+2\)

\(\Rightarrow3x-1-3\left(x+2\right)⋮x+2\)

\(\Rightarrow3x-1-3x-6⋮x+2\)

\(\Rightarrow-7⋮x+2\)

\(\Rightarrow x+2\in U\left(7\right)=\left\{1;7\right\}\)

\(\Rightarrow x\in\left\{-1;5\right\}\)

\(\Rightarrow x\in\left\{5\right\}\left(x\inℕ\right)\)

          \(\dfrac{x-6}{1998}\) + \(\dfrac{x-4}{2000}\) = \(\dfrac{x-2000}{4}\) + \(\dfrac{x-1998}{6}\)

\(\dfrac{x-6}{1998}\) - 1 + \(\dfrac{x-4}{2000}\) - 1 = \(\dfrac{x-2000}{4}\) - 1 + \(\dfrac{x-1998}{6}\) - 1

\(\dfrac{x-6-1998}{1998}\) + \(\dfrac{x-4-2000}{2000}\)  = \(\dfrac{x-2000-4}{4}\) + \(\dfrac{x-1998-6}{6}\)

\(\dfrac{x-2004}{1998}\)  + \(\dfrac{x-2004}{2000}\) = \(\dfrac{x-2004}{4}\) + \(\dfrac{x-2004}{6}\)

(\(x-2004\)).[\(\dfrac{1}{1998}\) + \(\dfrac{1}{2000}\) - \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\)] = 0

\(x\) - 2004 = 0

\(x\)             = 2004