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\(y+30\%y=-1,3\\ 130\%y=-1,3\\ \Rightarrow y=\dfrac{-1,3}{130\%}=-1\)
\(x:\dfrac{4}{28}=\dfrac{13}{-19}+\dfrac{8}{25}\\ 7x=-\dfrac{173}{475}\\ x=-\dfrac{\dfrac{173}{475}}{7}=-\dfrac{173}{3325}\)
Bài 1:
a: \(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2.15+3.75=\dfrac{8}{5}\)
=>x+4/15=8/5 hoặc x+4/15=-8/5
=>x=4/3 hoặc x=-28/15
b: \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{3}x=-\dfrac{1}{6}\\\dfrac{5}{3}x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{6}:\dfrac{5}{3}=\dfrac{-3}{30}=\dfrac{-1}{10}\\x=\dfrac{1}{10}\end{matrix}\right.\)
c: \(\Leftrightarrow\left|x-1\right|-1=1\)
=>|x-1|=2
=>x-1=2 hoặc x-1=-2
=>x=3 hoặc x=-1
Bài 2:
b: \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Leftrightarrow x=y=-\dfrac{9}{25}\)
Bài 3:
a: \(A=\left|x+\dfrac{15}{19}\right|-1>=-1\)
Dấu '=' xảy ra khi x=-15/19
b: \(\left|x-\dfrac{4}{7}\right|+\dfrac{1}{2}>=\dfrac{1}{2}\)
Dấu '=' xảy ra khi x=4/7
\(\frac{27}{4}=\frac{-x}{3}=>x=-\frac{81}{4}\notinℤ\)
\(^{y^2=\frac{4}{9}=\left(\frac{2}{3}\right)^2=>y=\pm\frac{2}{3}\notinℤ}\)
\(\frac{27}{4}=\frac{\left(z+3\right)}{-4}=\left(z+3\right)=-27=\left(-3\right)^3=>z+3=-3=>z=-6\)
\(+)|t|-2=-54=>|t|=-52\)(vô lí)
\(+)|t|-2=54=>|t|=56=>t=\pm56\)
a) Để phân số \(\dfrac{3}{n-2}\) là số nguyên thì n - 2 \(⋮\) 3
\(\Rightarrow\) n - 2 \(\in\) Ư(3)
\(\Rightarrow\) n - 2 \(\in\){3; -3; 1;-1}
n \(\in\){5; -1; 3; 2}
c) \(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+......+\dfrac{1}{28.29}\)
\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{29}-\dfrac{1}{30}\)
\(=\dfrac{1}{3}-\dfrac{1}{30}\)
\(=\dfrac{10}{30}-\dfrac{1}{30}\)
\(=\dfrac{9}{30}\)
=\(\dfrac{3}{10}\)
-x/3=24/4=6
=>x=-18
3/y2=6
=>y2=1/2
hay \(y=\pm\dfrac{\sqrt{2}}{2}\)
\(\dfrac{\left(z+3\right)^3}{-4}=6\)
=>(z+3)3=-24
\(\Leftrightarrow z+3=-\sqrt[3]{24}\)
hay \(z=-\sqrt[3]{24}-3\)
||t|-2|/8=6
=>||t|-2|=48
=>|t|-2=48
=>t=50 hoặc t=-50
a: =>-3/2+x-7=5-1/3x+4/15
=>4/3x=413/30
hay x=413/40
b: \(\Leftrightarrow5-\dfrac{3}{2}x=-\dfrac{22}{3}\cdot\dfrac{-11}{8}=\dfrac{121}{12}\)
=>3/2x=-61/12
hay x=-61/18
c: (3x+2)2+|3x+2y|=0
=>3x+2=0 và 3x=-2y
=>x=-2/3 và -2y=-2
=>(x,y)=(-2/3;1)
Ta có: \(A=\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{6}-1\right)\left(\dfrac{1}{10}-1\right)\left(\dfrac{1}{15}-1\right)\left(\dfrac{1}{21}-1\right)\left(\dfrac{1}{28}-1\right)\left(\dfrac{1}{36}-1\right)\)
\(=\dfrac{-2}{3}.\dfrac{-5}{6}.\dfrac{-9}{10}.\dfrac{-14}{15}.\dfrac{-20}{21}.\dfrac{-27}{28}.\dfrac{-35}{36}\)
\(=\dfrac{-2.\left(-5\right).3.\left(-3\right).2.\left(-7\right).\left(-4\right).5.\left(-3\right).9.5.\left(-7\right)}{3.2.3.2.5.3.5.3.7.4.7.4.9}\)
\(=\dfrac{-5}{3.4}=\dfrac{-5}{12}\)
Vậy \(A=\dfrac{-5}{12}.\)
\(C=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}\)
\(2C=2\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}\right)\)
\(2C=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+....+\dfrac{1}{2^{2015}}\)
\(2C-C=\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2015}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2016}}\right)\)
\(C=2-\dfrac{1}{2^{2016}}\)
\(\dfrac{0,8\left(\dfrac{4}{5}.1,25\right)}{0,64-\dfrac{1}{25}}+\dfrac{\left(10,8-\dfrac{2}{25}\right):\dfrac{4}{7}}{\left(6\dfrac{5}{9}-3\dfrac{1}{4}.2\dfrac{2}{17}\right)}\)
\(=\dfrac{0,8}{0,6}+\dfrac{\dfrac{268}{25}.\dfrac{7}{4}}{\left(\dfrac{59}{9}-\dfrac{13}{4}.\dfrac{36}{17}\right)}\)
\(=\dfrac{4}{3}+\dfrac{\dfrac{469}{25}}{\dfrac{-50}{153}}\)
\(=\dfrac{4}{3}-57,4056\)
\(=-56,07\)
Phần nào không hiểu bạn có thể nhắn hỏi mình nhe
Ta có : mẫu số 1 : 4 . 1
mẫu số hai : 4.7
... mẫu số thứ 96 = 100.103 = 10300
=> Số số hạng y là 100
Ta có :
\((y+..+y) + (\frac{3}{1.4} + \frac{3}{4.7} + ...+ \frac{3}{100.103})\)
\(= ( y+...+y) + [1. (\frac{1}{1.4} + \frac{1}{4.7} + ..+ \frac{1}{100.103})]\)
\(= (y+...y) + [1.(\frac{1}{1} - \frac{1}{4} + \frac{1}{4} - \frac{1}{7} + ...+ \frac{1}{100} - \frac{1}{103}) ]\)
\(= (y+...+y) + (1 - \frac{1}{103})\)
\(= (y+...+y) + \frac{102}{103}\)
\(=> (y+...+y) = \frac{308}{103} - \frac{102}{103} = \frac{206}{103}\)
\(=> y = \frac{206}{103} : 100 = \frac{206}{10300} = \frac{103}{5150}\) ( Chia 100 vì có 100 số hạng y)
Vậy \(y = \frac{103}{5150}\)