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\(=81.\dfrac{12.\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}{4.\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}:\dfrac{5.\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}{6.\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}.\dfrac{158}{711}\)
\(=81.\dfrac{12}{4}:\dfrac{5}{6}.\dfrac{2}{9}\)
\(=243:\dfrac{5}{6}.\dfrac{2}{9}\)
\(=\dfrac{1458}{5}.\dfrac{2}{9}\)
\(=\dfrac{324}{5}\)
\(-\left(x-1\right)^2\) = / \(\dfrac{1}{4}-\dfrac{1}{2}-\dfrac{3}{4}\)/
\(-\left(x-1\right)^2\)= / \(\dfrac{-4}{4}\)/ = /-1/ = 1
\(-\left(x-1\right)^2\)= \(1^2\) (vì \(1^2\) = 1 )
\(-\left(x-1\right)\)= 1 hoặc -(x - 1) = -1
x - 1 = -1 hoặc x - 1 = 1
x = -1 + 1 hoặc x = 1 + 1
x = 0 hoặc x = 2
Bài 1:
Ta có: \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{4}\right)^2\ge0\\\left|y+\dfrac{1}{4}\right|\ge0\end{matrix}\right.\Rightarrow\left(x-\dfrac{1}{4}\right)^2+\left|y+\dfrac{1}{4}\right|\ge0\)
\(\Rightarrow A=\left(x-\dfrac{1}{4}\right)^2+\left|y+\dfrac{1}{4}\right|+\dfrac{13}{14}\ge\dfrac{13}{14}\)
Dấu " = " khi \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{4}\right)^2=0\\\left|y+\dfrac{1}{4}\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{4}=0\\y+\dfrac{1}{4}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=\dfrac{-1}{4}\end{matrix}\right.\)
Vậy \(MIN_A=\dfrac{13}{14}\) khi \(x=\dfrac{1}{4};y=-\dfrac{1}{4}\)
Trời ơi cái đề bài !!!
Thoy thì làm từng câu vậy
a) \(I=10101.\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{111111}\right)\)
\(I=10101.\left(\dfrac{10}{222222}+\dfrac{5}{222222}-\dfrac{8}{222222}\right)\)
\(I=10101.\left(\dfrac{15}{222222}-\dfrac{8}{222222}\right)\)
\(I=10101.\dfrac{7}{222222}\)
\(I=\dfrac{7}{22}\)
e) \(\dfrac{-3}{5}.\dfrac{2}{7}+\dfrac{-3}{5}.\dfrac{5}{7}+2\dfrac{3}{5}\)
= \(\dfrac{-3}{5}.\left(\dfrac{2}{7}+\dfrac{5}{7}\right)+\dfrac{13}{5}\)
= \(\dfrac{-3}{5}.1+\dfrac{13}{5}\)
= \(\dfrac{-3}{5}+\dfrac{13}{5}\)
= 2
ừ Vy Nguyễn, mik làm nè:
e, \(\dfrac{-2}{3}-\dfrac{1}{3}\left(2x-5\right)=\dfrac{3}{2}.\)
\(\dfrac{1}{3}\left(2x-5\right)=\dfrac{-2}{3}-\dfrac{3}{2}.\)
\(\dfrac{1}{3}\left(2x-5\right)=\dfrac{-4}{6}+\dfrac{-9}{6}.\)
\(\dfrac{1}{3}\left(2x-5\right)=\dfrac{-13}{6}.\)
\(2x-5=\dfrac{-13}{6}:\dfrac{1}{3}.\)
\(2x-5=\dfrac{-13}{6}.3.\)
\(2x-5=\dfrac{-13}{2}.\)
\(2x=\dfrac{-13}{2}+5.\)
\(2x=\dfrac{-13}{2}+\dfrac{10}{2}.\)
\(2x=\dfrac{-3}{2}.\)
\(x=\dfrac{-3}{2}:2.\)
\(x=\dfrac{-3}{2.2}=\dfrac{-3}{4}.\)
g, \(\dfrac{2}{5}x+\dfrac{1}{2}=\dfrac{-3}{4}.\)
\(\dfrac{2}{5}x=\dfrac{-3}{4}-\dfrac{1}{2}.\)
\(\dfrac{2}{5}x=\dfrac{-3}{4}+\dfrac{-2}{4}.\)
\(\dfrac{2}{5}x=\dfrac{-5}{4}.\)
\(x=\dfrac{-5}{4}:\dfrac{2}{5}.\)
\(x=\dfrac{-5}{4}.\dfrac{5}{2}.\)
\(x=\dfrac{-25}{8}.\)
h, \(\left(2x-2\dfrac{4}{5}\right):3\dfrac{1}{8}=1\dfrac{3}{5}.\)
\(\left(2x-2\dfrac{4}{5}\right)=\dfrac{8}{5}.\dfrac{25}{8}.\)
\(\left(2x-2\dfrac{4}{5}\right)=5.\)
\(2x=5+2\dfrac{4}{5}.\)
\(2x=7\dfrac{4}{5}.\)
\(x=7\dfrac{4}{5}:2.\)
\(x=\dfrac{39}{10}.\)
(còn tiếp ở phần sau!!!)
Tiếp:
i, \(3,2x-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):3\dfrac{2}{3}=\dfrac{7}{20}.\)
\(\dfrac{16}{5}x-\left(\dfrac{4}{5}+\dfrac{2}{3}\right)=\dfrac{7}{20}.\dfrac{11}{3}.\)
\(\dfrac{16}{5}x-\left(\dfrac{4}{5}+\dfrac{2}{3}\right)=\dfrac{77}{60}.\)
\(\dfrac{16}{5}x-\left(\dfrac{12}{15}+\dfrac{10}{15}\right)=\dfrac{77}{60}.\)
\(\dfrac{16}{5}x-\dfrac{22}{15}=\dfrac{77}{60}.\)
\(\dfrac{16}{5}x=\dfrac{77}{60}+\dfrac{22}{15}.\)
\(\dfrac{16}{5}x=\dfrac{77}{60}+\dfrac{88}{60}.\)
\(\dfrac{16}{5}x=\dfrac{165}{60}=\dfrac{11}{4}.\)
\(x=\dfrac{11}{4}:\dfrac{16}{5}.\)
\(x=\dfrac{11}{4}.\dfrac{5}{16}=\dfrac{55}{64}.\)
k, \(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{28}.\)
\(\left(\dfrac{3x}{7}+1\right)=\dfrac{-1}{28}.\left(-4\right).\)
\(\left(\dfrac{3x}{7}+1\right)=\dfrac{1}{7}.\)
\(\dfrac{3x}{7}=\dfrac{1}{7}-1.\)
\(\dfrac{3x}{7}=\dfrac{1}{7}-\dfrac{7}{7}.\)
\(\dfrac{3x}{7}=\dfrac{-6}{7}.\)
\(\Rightarrow3x=-6.\)
\(\Rightarrow x=-6:3=-2.\)
~ Chúc bn học tốt!!! ~
Bài mik đúng thì nhớ tik mik nha!!!