M=-\(^{\left(x+\dfrac{1}{8^{ }}\right)26}\)-\(\left(x-y=\dfrac{3}{8}\right)442\)+5,98
26 va 442 la mu nha cac ban
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Vì:\(-\left(x+\frac{1}{8}\right)^{26}\ge0,-\left(x-y+\frac{3}{8}\right)\ge0\) nên:
\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{1}{8}=0\\x-y+\frac{3}{8}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-1}{8}\\x-y+\frac{3}{8}=0\end{matrix}\right.\)
Thay: x=\(\frac{-1}{8}\) vào x-y+\(\frac{3}{8}\) =0, ta có:
\(\frac{-1}{8}\) -y+\(\frac{3}{8}\) =0
\(\frac{-1}{8}-y=0-\frac{3}{8}\)
\(\Rightarrow\frac{-1}{8}-y=\frac{-3}{8}\)
\(\Rightarrow y=\frac{-1}{8}-\frac{-3}{8}\)
\(y=\frac{-1}{8}+\frac{3}{8}\)
\(y=\frac{2}{8}=\frac{1}{4}\)
Vậy:x=\(\frac{-1}{8}\) ,y=\(\frac{1}{4}\) thì M đạt giá trị lớn nhất bằng 5,98
CHÚC BẠN HỌC TỐT NHÉ!!!
Mấy bài này bạn tự làm đi, chuyển vế tìm x gần giống cấp I mà.
b)\(\dfrac{-3}{5}.x=\dfrac{1}{4}+0,75\)
=>\(\dfrac{-3}{5}.x=1\)
=>\(x=1:\dfrac{-3}{5}\)
=>\(x=\dfrac{-5}{3}\)
Vậy \(x=\dfrac{-5}{3}\)
\(a,\left|x\right|< 2\dfrac{3}{4}\)
\(\Rightarrow-2\dfrac{3}{4}< x< 2\dfrac{3}{4}\)
mà \(x\in Z\)
\(\Rightarrow x\in\left\{-2;-1;0;1;2\right\}\)
\(b,\left|x\right|>1\dfrac{3}{8}\)
\(\Rightarrow-1\dfrac{3}{8}< x< 1\dfrac{3}{8}\)
mà \(x\in Z\)
\(\Rightarrow x\in\left\{;-1;0;1;\right\}\)
\(\)
c: =>7/6<|x-2/3|<26/9
=>7/6<x-2/3<26/9 hoặc -7/6>x-2/3>-26/9
=>11/6<x<32/9 hoặc -1/2>x>-20/9
=>\(x\in\left\{2;3;-1;-2\right\}\)
1.Tính
a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)
b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)
c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)
d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)
e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)
Bài 2
a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)
\(x=\dfrac{13}{49}\)
b.\(\left|x-1,5\right|=2\)
Xảy ra 2 trường hợp
TH1
\(x-1,5=2\)
\(x=3,5\)
TH2
\(x-1,5=-2\)
\(x=-0,5\)
Vậy \(x=3,5\) hoặc \(x=-0,5\) .
Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.
\(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\cdot\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{1-10}{6}-\dfrac{20}{3}\cdot\dfrac{8-5}{20}+\dfrac{2}{3}\cdot\dfrac{12-45}{10}\)
\(=\dfrac{-1}{2}\cdot\dfrac{-9}{6}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{66}{30}=\dfrac{-1}{4}-\dfrac{11}{5}=\dfrac{-5-44}{20}=-\dfrac{49}{20}\)
\(\dfrac{3}{x-5}-\dfrac{x+1}{x\left(x-5\right)}\left(dkxd:x\ne0,x\ne5\right)\\ =\dfrac{3x-x-1}{x\left(x-5\right)}=\dfrac{2x-1}{x^2-5x}\)
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\(\dfrac{8\left(y+2\right)}{3x^2}.\dfrac{15x^5}{4\left(y+2\right)^2}\left(dkxd:x\ne0,y\ne-2\right)\\ =\dfrac{8}{4}.\dfrac{15x^2.x^3}{3x^2}=10x^3\)
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\(\dfrac{8\left(y-1\right)}{3x^2-3}:\dfrac{4\left(y-1\right)^3}{x^2-2x+1}\left(dkxd:x\ne1,x\ne-1\right)\\ =\dfrac{8\left(y-1\right)}{3\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)^2}{4\left(y-1\right)^3}\\ =\dfrac{2\left(x-1\right)}{3\left(x+1\right)\left(y-1\right)^2}\)