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a)\(\left(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\right).120+x:\frac{1}{3}=-4\)
\(\Rightarrow\left(\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\right).120+x:\frac{1}{3}=-4\)
\(\Rightarrow\left(\frac{1}{24}-\frac{1}{30}\right).120+x:\frac{1}{3}=-4\)
\(\Rightarrow\frac{1}{120}.120+x:\frac{1}{3}=-4\)
\(\Rightarrow1+x:\frac{1}{3}=-4\)
\(\Rightarrow x:\frac{1}{3}=-4-1=-5\)
\(\Rightarrow x=-5.\frac{1}{3}=\frac{-5}{3}\)
b)\(1\frac{3}{5}+\left(\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{37}}{\frac{5}{7}+\frac{5}{17}+\frac{5}{37}}\right).x=\frac{16}{5}\)
\(\Rightarrow\frac{8}{5}+\left[\frac{2.\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}{5.\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}\right].x=\frac{16}{5}\)
\(\Rightarrow\frac{8}{5}+\frac{2}{5}.x=\frac{16}{5}\)
\(\Rightarrow\frac{2}{5}.x=\frac{16}{5}-\frac{8}{5}=\frac{8}{5}\)
\(\Rightarrow x=\frac{8}{5}:\frac{2}{5}=\frac{8}{5}.\frac{5}{2}=\frac{8}{2}=4\)
\(\Rightarrow x=4\)
Bài 2
a. \(-1\frac{2}{3}-|2x-1|:\frac{3}{5}=-2\)
\(|2x-1|:\frac{3}{5}=\frac{5}{3}-2\)
\(|2x-1|:\frac{3}{5}=-\frac{1}{3}\)
\(|2x-1|=-\frac{1}{5}\)
Vì giá trị tuyệt đối luôn \(\ge0\)với mọi x
mà \(-\frac{1}{5}< 0\)
=> \(x\in\varnothing\)
Bài 2:
a) \(\frac{4}{9}+x=\frac{-5}{3}\)
\(\Leftrightarrow x=\frac{-5}{3}-\frac{4}{9}\)
\(\Leftrightarrow x=\frac{-15}{9}-\frac{4}{9}\)\(=\frac{-19}{9}\)
Vậy: \(x=\frac{-19}{9}\)
b) \(2,4:\left(\frac{1}{2}.x-\frac{3}{4}\right)=\frac{3}{10}\)
\(\Leftrightarrow\frac{24}{10}:\left(\frac{1}{2}x-\frac{3}{4}\right)=\frac{3}{10}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{3}{4}=\frac{24}{10}:\frac{3}{10}=\frac{24}{10}.\frac{10}{3}\)\(=8\)
\(\Leftrightarrow\frac{1}{2}x=8+\frac{3}{4}=\frac{35}{4}\)
\(\Leftrightarrow x=\frac{35}{4}:\frac{1}{2}=\frac{35}{4}.2=\frac{35}{2}\)
c) \(\frac{x+1}{-8}=\frac{-2}{x+1}\)
\(\Rightarrow\left(x+1\right).\left(x+1\right)=\left(-2\right).\left(-8\right)\)
\(\Leftrightarrow\left(x+1\right)^2=16=4^2=\left(-4\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
Vậy: \(x\in\left\{3;-5\right\}\)
Từng bài 1 thôi nha bn!!!
a) Xét hiệu: A = 9.(7x+4y) - 2. (13x+18y)
A = 63x + 36y - 26x - 36y
A = 37x \(\Rightarrow A⋮37\) Vì 7x + 4y chia hết cho 37
9.(7x+4y) chia hết cho 37
Mà A chia hết cho 37
\(2\left(13x+18y\right)⋮37\)
Do 2 và 37 là nguyên tố cùng nhau
13x+18y chia hết cho 37
Vậy nếu 7x+4y chia hết cho 37 thì 13x+18y chia hết cho 37
a)\(\left(x-2,5\right)^2=\frac{4}{9}\\ \left(x-\frac{5}{2}\right)^2=\left(\pm\frac{2}{3}\right)^2\\\Leftrightarrow\left\{{}\begin{matrix}x-\frac{5}{2}=\frac{2}{3}\\x-\frac{5}{2}=\frac{-2}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{19}{6}\\x=\frac{11}{6}\end{matrix}\right. \)
vậy....
b)\(\left(2x+\frac{1}{3}\right)^3=\frac{-8}{27}\\ \left(2x+\frac{1}{3}\right)^3=\left(\frac{-2}{3}\right)^3\\ 2x+\frac{1}{3}=\frac{-2}{3}\\ x=\frac{-1}{2}\)
vậy...
a) \(14:\frac{0,4x+0,6}{x}=7\)
\(\frac{0,4x+0,6}{x}=2\)
0,4x + 0,6 = 2.x
2x - 0,4x = 0,6
1,6x = 0,6
x = 0,375
b) \(\left(160\%+\frac{2}{3}x-x\right).12=660\)
\(\left(160\%+\frac{2}{3}x-x\right)=55\)
\(x\left(\frac{2}{3}-1\right)=53,4\)
\(-\frac{1}{3}x=\frac{267}{5}\)
\(x=\frac{267}{5}.\frac{3}{-1}\)
\(x=-160,2\)
c) \(1:\frac{1.2.3.4.....31}{2.2.2.3.2.4.....2.32}=2^x\)
\(1:\frac{1.2.3.4.....31}{2^{31}.2.3.4.....31.2^5}=2^x\)
\(1:\frac{1}{2^{36}}=2^x\)
\(2^{36}=2^x\)
\(x=36\)
Giải:
\(1+\left(x-\frac{2}{3}\right)^2=\frac{37}{64}\Rightarrow\left(x-\frac{2}{3}\right)^2=-\frac{27}{64}\)
\(\Leftrightarrow x-\frac{2}{3}=-\sqrt{\frac{27}{64}}\Leftrightarrow x=-\sqrt{\frac{27}{64}}+\frac{2}{3}\)
Vậy \(x=\sqrt{\frac{27}{64}}-\frac{2}{3}\)