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a) \(3\dfrac{1}{4}=\dfrac{2}{3}:\left(\dfrac{-x}{2}\right)\Leftrightarrow\dfrac{13}{4}=\dfrac{2}{3}.\dfrac{-2}{x}\Leftrightarrow\dfrac{-2}{x}=\dfrac{39}{8}\Leftrightarrow x=-\dfrac{16}{39}\)
b) \(1-2\left(x+\dfrac{1}{3}\right)=\left|-\dfrac{2}{3}+\dfrac{1}{5}\right|\Leftrightarrow1-2x-\dfrac{2}{3}=\dfrac{7}{15}\Leftrightarrow2x=-\dfrac{2}{15}\Leftrightarrow x=-\dfrac{1}{15}\)
c) \(\left(2x-1\right)\left(\dfrac{2}{5}-\dfrac{1}{3}x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\\dfrac{2}{5}-\dfrac{1}{3}x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\\dfrac{1}{3}x=\dfrac{2}{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{6}{5}\end{matrix}\right.\)
d) \(-4\dfrac{3}{5}.2\dfrac{4}{23}\le x\le-2\dfrac{3}{5}:1\dfrac{6}{15}\Leftrightarrow-10\le x\le-\dfrac{13}{7}\Leftrightarrow x\in\left\{-10;-9;-8;-7;-6;-5;-4;-3;-2;-1\right\}\)(do \(x\in Z\))
Bài 2:
c: Ta có: \(\left(2x-1\right)\left(\dfrac{2}{5}-\dfrac{1}{3}x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\\dfrac{2}{5}-\dfrac{1}{3}x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\\dfrac{1}{3}x=\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{6}{5}\end{matrix}\right.\)
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\(-23.63+23.21-58.23\\ =23.\left(-63\right)+23.21-58.23\\ =23\left(-63+21-58\right)\\ =23.\left(-100\right)=-2300\)
-23 . 63 + 23 . 21 - 58 . 23
= 23 . (-63 + 21 - 58) = 23 . (-100) = -2300
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3 + 1,3 = a,b = 4,3
Vậy số a,b là 4,3
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Đặt \(A=1+3^2+3^4+...+3^{100}\)
\(9A=3^2+3^4+3^6+...+3^{102}\)
\(9A-A=\left(3^2+3^4+3^6+...+3^{102}\right)-\left(1+3^2+3^4+...+3^{100}\right)\)
\(8A=3^{102}-1\)
\(A=\frac{3^{102}-1}{8}\)
Vậy \(A=\frac{3^{102}-1}{8}\)
Chúc bạn học tốt ~
Đặt A = 1 + 3^2 + 3^4 + 3^6 + ....+ 3^100
3^2A = 3^2 + 3^4 + 3^6 + ..+3^102
8A=3^2A - A = 3102 - 1
A = 3102 - 1/8
=. A = 3102 - 1 /8
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<=> 10x-9,9=0,1.x+9,9
<=> 100x-99=x+99
<=> 99x=99+99
<=> 99x=198 => x=198:99 => x=2
Đáp số: x=2