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làm tiếp cái trước(ấn nhầm)
\(x=\frac{5}{42}-\frac{15}{28}\)
\(x=\frac{5.4}{6.4.7}-\frac{15.6}{4.7.6}\)
\(x=\frac{20}{168}-\frac{90}{168}\)
\(x=\frac{-70}{168}\)
\(x=\frac{-5}{12}\)
2.
1.
\(\frac{11}{13}-\left(\frac{5}{42}-x\right)=-\left(\frac{15}{28}-\frac{11}{13}\right)\)
\(\frac{11}{13}-\frac{5}{42}+x=-\frac{15}{28}+\frac{11}{13}\)
\(\frac{11}{13}-\frac{11}{13}-\frac{5}{42}+\frac{15}{28}=-x\)
\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)
A) (x—1)2= | 1/4–1/2–3/4 |
(x—1)2= | 1/4–2/4–3/4 |
(x—1)2=|—1|
(x—1)2=1
==> (x—1)=1 hoặc (x—1)=-1
x=1+1 hoặc x—1=-1+1
x=2 hoặc x=0
b)(xx—8).(x2–15)<0
==> xx—8 <0 và x2> 0
Hay xx—8 >0 và x2<0
Mình chỉ biết tới đó thôi
\(a,\left(x-1\right)^2=\left|\frac{1}{4}-\frac{1}{2}-\frac{3}{4}\right|\)
\(\Rightarrow\left(x-1\right)^2=\left|-1\right|\)
\(\Rightarrow\left(x-1\right)^2=1\)
\(\Rightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}\)
vậy__
b, k bt
\(\frac{1}{20}\left(x-\frac{8}{15}\right)=-\frac{1}{30}\) \(\left(28+\frac{1}{5}\right).\left(\frac{3}{5}.x+\frac{4}{7}\right)=0\)
\(x-\frac{8}{15}=-\frac{1}{30}:\frac{1}{20}\) \(\frac{141}{5}.\left(\frac{3}{5}.x+\frac{4}{7}\right)=0\)
\(x-\frac{8}{15}=-\frac{2}{3}\) \(\frac{3}{5}.x+\frac{4}{7}=0\)
\(x=-\frac{2}{3}+\frac{8}{15}\) \(\frac{3}{5}.x=-\frac{4}{7}\)
\(x=-\frac{2}{15}\) \(x=-\frac{20}{21}\)
\(e,\frac{22}{15}-x=-\frac{8}{27}\)
=> \(x=\frac{22}{15}-\left[-\frac{8}{27}\right]\)
=> \(x=\frac{22}{15}+\frac{8}{27}\)
=> \(x=\frac{198}{135}+\frac{40}{135}=\frac{198+40}{135}=\frac{238}{135}\)
\(g,\left[\frac{2x}{5}-1\right]:\left[-5\right]=\frac{1}{4}\)
=> \(\left[\frac{2x}{5}-\frac{1}{1}\right]=\frac{1}{4}\cdot\left[-5\right]\)
=> \(\left[\frac{2x}{5}-\frac{5}{5}\right]=-\frac{5}{4}\)
=> \(\frac{2x-5}{5}=-\frac{5}{4}\)
=> \(2x-5=-\frac{5}{4}\cdot5=-\frac{25}{4}\)
=> \(2x=-\frac{5}{4}\)
=> \(x=-\frac{5}{8}\)
\(h,-2\frac{1}{4}x+9\frac{1}{4}=20\)
=> \(-\frac{9}{4}x+\frac{37}{4}=20\)
=> \(-\frac{9}{4}x=20-\frac{37}{4}=\frac{43}{4}\)
=> \(x=\frac{43}{4}:\left[-\frac{9}{4}\right]=\frac{43}{4}\cdot\left[-\frac{4}{9}\right]=\frac{43}{1}\cdot\left[-\frac{1}{9}\right]=-\frac{43}{9}\)
\(i,-4\frac{3}{5}\cdot2\frac{4}{23}\le x\le-2\frac{3}{5}:1\frac{6}{15}\)
=> \(-\frac{23}{5}\cdot\frac{50}{23}\le x\le-\frac{13}{5}:\frac{21}{15}\)
=> \(-\frac{1}{1}\cdot\frac{10}{1}\le x\le-\frac{13}{5}\cdot\frac{15}{21}\)
=> \(-10\le x\le-\frac{13}{1}\cdot\frac{3}{21}\)
=> \(-10\le x\le-\frac{13}{1}\cdot\frac{1}{7}\)
=> \(-10\le x\le-\frac{13}{7}\)
Đến đây tìm x
|x-4| > 0với mọi x
=>|x-4|+1>1 với mọi x
=>\(A=\frac{-15}{\left|x-4\right|+1}\le\frac{-15}{1}=-15\)
=>Amax=-15
GTLN chứ?
( giá trị tuyệt đối