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a ) \(\left|x-2\right|=2\)
\(\Rightarrow\orbr{\begin{cases}x-2=2\\x-2=-2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2+2\\x=-2+2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=4\\x=0\end{cases}}\)
Vậy \(x=4\) hoặc \(x=0\)
b ) \(\left|x-\frac{4}{5}\right|=\frac{3}{4}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{4}{5}=\frac{3}{4}\\x-\frac{4}{5}=-\frac{3}{4}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{4}+\frac{4}{5}\\x=-\frac{3}{4}+\frac{4}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{15}{20}+\frac{16}{20}\\-\frac{15}{20}+\frac{16}{20}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{31}{20}\\x=\frac{1}{20}\end{cases}}\)
Vậy \(x=\frac{31}{20}\) hoặc \(x=\frac{1}{20}\)
c ) \(6-\left|\frac{1}{2}-x\right|=\frac{2}{5}\)
\(\left|\frac{1}{2}-x\right|=6-\frac{2}{5}\)
\(\left|\frac{1}{2}-x\right|=\frac{28}{5}\)
\(\Rightarrow\orbr{\begin{cases}\frac{1}{2}-x=\frac{28}{5}\\\frac{1}{2}-x=-\frac{28}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}-\frac{28}{5}\\x=\frac{1}{2}-\left(-\frac{28}{5}\right)\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{51}{10}\\x=\frac{61}{10}\end{cases}}\)
Vậy \(x=-\frac{51}{10}\) hoặc \(x=\frac{61}{10}\)
d ) \(\frac{1}{5}-\left|\frac{1}{5}.x\right|=\frac{1}{5}\)
\(\left|\frac{1}{5}.x\right|=\frac{1}{5}-\frac{1}{5}\)
\(\left|\frac{1}{5}.x\right|=0\)
\(\Rightarrow\frac{1}{5}.x=0\)
\(x=0:\frac{1}{5}\)
\(x=0\)
Vậy \(x=0\)
e ) \(-2,5+\left|3x+5\right|=-1,5\)
\(\left|3x+5\right|=-1,5-\left(-2,5\right)\)
\(\left|3x+5\right|=1\)
\(\Rightarrow\orbr{\begin{cases}3x+5=1\\3x+5=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=1-5\\3x=-1-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=-4\\3x=-6\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-4:3\\x=-6:3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{4}{3}\\x=-2\end{cases}}\)
Vậy \(x=-\frac{4}{3}\) hoặc \(x=-2\)
g ) \(\left(\frac{2}{3}x-1\right)+\left(\frac{3}{4}x+\frac{1}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\left(\frac{2}{3}x-1\right)=0\\\left(\frac{3}{4}x+\frac{1}{2}\right)=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}x-1=0\\\frac{3}{4}x+\frac{1}{2}=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}x=0+1\\\frac{3}{4}x=0-\frac{1}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}x=1\\\frac{3}{4}x=-\frac{1}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1:\frac{2}{3}\\x=-\frac{1}{2}:\frac{3}{4}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{2}{3}\end{cases}}\)
Vậy \(x=\frac{3}{2};x=-\frac{2}{3}\)
Bạn bạn hơi dài nhưng đã xong
Chúc bạn học tốt !!!
Trả lời:
a, | x - 2 | = 2
=> x - 2 = 2 hoặc x - 2 = -2
=> x = 4 x = 0
Vậy x = 4; x = 0
b, \(\left|x-\frac{4}{5}\right|=\frac{3}{4}\)
\(\Rightarrow x-\frac{4}{5}=\frac{3}{4}\)hoặc \(x-\frac{4}{5}=\frac{-3}{4}\)
\(\Rightarrow x=\frac{31}{20}\)hoặc \(x=\frac{1}{20}\)
Vậy \(x=\frac{31}{20};x=\frac{1}{20}\)
c, \(6-\left|\frac{1}{2}-x\right|=\frac{2}{5}\)
\(\Rightarrow\left|\frac{1}{2}-x\right|=\frac{28}{5}\)
\(\Rightarrow\frac{1}{2}-x=\frac{28}{5}\)hoặc \(\frac{1}{2}-x=\frac{-28}{5}\)
\(\Rightarrow x=\frac{-51}{10}\)hoặc \(x=\frac{61}{10}\)
Vậy \(x=\frac{-51}{10};x=\frac{61}{10}\)
d, \(\frac{1}{5}-\left|\frac{1}{5}\cdot x\right|=\frac{1}{5}\)
\(\Rightarrow\left|\frac{1}{5}\cdot x\right|=0\)
\(\Rightarrow\frac{1}{5}\cdot x=0\)
\(\Rightarrow x=0\)
Vậy x = 0
e, \(-2,5+\left|3x+5\right|=-1,5\)
\(\Rightarrow\left|3x+5\right|=1\)
\(\Rightarrow3x+5=1\)hoặc \(3x+5=-1\)
\(\Rightarrow3x=-4\)hoặc \(3x=-6\)
\(\Rightarrow x=\frac{-4}{3}\)hoặc \(x=-2\)
Vậy \(x=\frac{-4}{3};x=-2\)
g, \(\left(\frac{2}{3}x-1\right)+\left(\frac{3}{4}x+\frac{1}{2}\right)=0\)
\(\Rightarrow\frac{2}{3}x-1+\frac{3}{4}x+\frac{1}{2}=0\)
\(\Rightarrow\frac{17}{12}x-\frac{1}{2}=0\)
\(\Rightarrow\frac{17}{12}x=\frac{1}{2}\)
\(\Rightarrow x=\frac{6}{17}\)
Vậy \(x=\frac{6}{17}\)
a) \(\frac{3}{4}+\frac{1}{4}.x=\frac{1}{2}+\frac{1}{2}x\)
\(\Rightarrow3.\frac{1}{4}+\frac{1}{4}.x=\frac{1}{2}.\left(x+1\right)\)
\(\Rightarrow\frac{1}{4}.\left(x+3\right)=\frac{1}{2}.\left(x+1\right)\)
\(\Rightarrow\frac{x+1}{x+3}=\frac{1}{4}:\frac{1}{2}=\frac{1}{2}\)\(\Rightarrow\left(x+1\right).2=x+3\Rightarrow2x+2=x+3\)
\(\Rightarrow2x-x=3-2\Rightarrow x=1\)
vay x=1
a ) \(\left(\frac{2}{5}-x\right):1\frac{1}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-4-\frac{1}{2}\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-\frac{9}{2}\)
\(\frac{2}{5}-x=-\frac{9}{2}.\frac{4}{3}\)
\(\frac{2}{5}-x=-3\)
\(x=\frac{2}{5}-\left(-3\right)\)
\(x=\frac{2}{5}+3\)
\(x=\frac{3}{5}-\frac{15}{5}\)
\(x=-\frac{12}{5}\)
Vay \(x=-\frac{12}{5}\)
b ) \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(1+\frac{2}{5}+\frac{2}{3}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15}{15}+\frac{6}{15}+\frac{10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15+6+10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\frac{31}{15}=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{5}{4}.\frac{31}{15}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{1}{4}.\frac{31}{3}\)
\(-3+\frac{3}{x}-\frac{1}{3}=-\frac{31}{12}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{1}{2}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{6}{12}\)
\(-3+\frac{3}{x}=\frac{-25}{12}\)
\(\frac{3}{x}=\frac{-25}{12}+3\)
\(\frac{3}{x}=\frac{-25}{12}+\frac{36}{12}\)
\(\frac{3}{x}=\frac{5}{6}\)
\(\frac{18}{6x}=\frac{5x}{6x}\)
Đèn dây , bạn tự làm tiếp nhé , de rồi chứ
Bạn ghi ra nhiều vậy người khác nhìn rối mắt không trả lời được đâu ghi từng bài ra thôi
Mình chỉ làm được vài bài thôi, kiến thức có hạn :>
Bài 1:
Câu a và c đúng
Bài 2:
a) |x| = 2,5
=>x = 2,5 hoặc
x = -2,5
b) |x| = 0,56
=>x = 0,56
x = - 0,56
c) |x| = 0
=. x = 0
d)t/tự
e) |x - 1| = 5
=>x - 1 = 5
x - 1 = -5
f) |x - 1,5| = 2
=>x - 1,5 = 2
x - 1,5 = -2
=>x = 2 + 1,5
x = -2 + 1,5
=>x = 3,5
x = - 0,5
các câu sau cx t/tự thôi
Bài 3: Ko hỉu :)
Bài 4: Kiến thức có hạn :)
+) \(5\frac{2}{3}x+1\frac{2}{3}=4\frac{1}{2}\Leftrightarrow\frac{17}{3}x+\frac{5}{3}=\frac{9}{2}\Leftrightarrow\frac{17}{3}x=\frac{17}{6}\Leftrightarrow x=\frac{1}{2}\)
+) \(\frac{x}{27}=\frac{-2}{9}\Leftrightarrow x=\frac{-2}{9}.27=-6\)
+) \(\left|x+1,5\right|=2\Leftrightarrow\orbr{\begin{cases}x+1,5=2\\x+1,5=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0,5\\x=-3,5\end{cases}}}\)
+) \(A=\left|x-1004\right|-\left|x+1003\right|\)
Ta có BĐT \(\left|x\right|-\left|y\right|\le\left|x-y\right|,\)dấu "=" xảy ra khi và chỉ khi x,y cùng dấu hay \(xy\ge0\)
Áp dụng: \(A=\left|x-1004\right|-\left|x+1003\right|\le\left|x-1004-x-1003\right|=\left|-2007\right|=2007\)
Vậy \(maxA=2007\Leftrightarrow\left(x-1004\right)\left(x+1003\right)\ge0\Leftrightarrow\orbr{\begin{cases}x\ge1004\\x\le-1003\end{cases}}\)
\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)
\(=\frac{7}{2}-2\)
\(=\frac{7}{2}-\frac{4}{2}\)
\(=\frac{3}{2}\)
\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)
\(=\frac{3}{7}.\left(2-9\right)\)
\(=\frac{3}{7}.\left(-7\right)\)
\(=-3\)
\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( tính nhẩm, ko chắc đúng )
1
a) \(3\frac{1}{2}-\frac{1}{2}\cdot\left(-4,25-\frac{3}{4}\right)^2\) : \(\frac{5}{4}\)
= \(3\cdot25:\frac{5}{4}\)
= \(3\cdot\left(25:\frac{5}{4}\right)\)
=\(3\cdot20\)
=60
b)=\(\frac{3}{7}\cdot\left(1\frac{1}{2}+0,5-9\right)\)
=\(\frac{3}{7}\cdot\left(-7\right)\)
=\(-3\)
c) =
\(a,\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)
\(\Leftrightarrow\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{4}:x=\frac{8}{20}-\frac{15}{20}\)
\(\Leftrightarrow\frac{1}{4}:x=\frac{-7}{20}\)
\(\Leftrightarrow x=\frac{1}{4}.\frac{20}{-7}\)
\(\Leftrightarrow x=\frac{20}{-28}\)
\(b,|x-2,5|=1,5\)
\(\Leftrightarrow\orbr{\begin{cases}x-2,5=1,5\\x-2,5=-1,5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1,5+2,5\\x=-1,5+2,5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)