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(2/3×x-1/3)=2/3+1/3
(2/3×x-1/3)=3/3
2/3×x=3/3+1/3
2/3×x=4/3
x=4/3:3/2
x=4/3×2/3
x=8/9
\(11-\left(3x-1\right)=\frac{9}{2}-\left(5-3,5x\right)\)
\(=>11-3x+1=\frac{9}{2}-5+3,5x\)
\(=>-3x+12=3,5x-\frac{1}{2}\)
\(=>-3x-3,5x=-\frac{1}{2}-12\)
\(=>-6,5x=-12,5\)
\(=>x=\frac{-12,5}{-6,5}=\frac{25}{13}\)
Ủng hộ nha
\(11-\left(3x-1\right)=\frac{9}{2}-\left(5-3,5x\right)\)
\(11-3x+1=\frac{9}{2}-5+3,5x\)
\(12-3x=-\left(0,5\right)+3,5x\)
\(12,5-3x=3,5x\)
\(12,5=6,5x\)
\(x=12,5:6,5=\frac{25}{13}\)
b)
\(4\frac{5}{9}:2\frac{5}{18}-7< x< \left(3\frac{1}{5}:3,2+4,5.1\frac{31}{45}\right):\left(21.\frac{1}{2}\right)\)
\(\Rightarrow\frac{41}{9}:\frac{41}{18}-7< x< \left(\frac{16}{5}:\frac{16}{5}+\frac{9}{2}.\frac{76}{45}\right):\frac{21}{2}\)
\(\Rightarrow2-7< x< \left(1+\frac{38}{5}\right):\frac{21}{2}\)
\(\Rightarrow-5< x< \frac{43}{5}:\frac{21}{2}\)
\(\Rightarrow-5< x< \frac{86}{105}\)
Vì \(x\in Z\left(gt\right)\)
\(\Rightarrow x\in\left\{-4;-3;-2;-1;0\right\}.\)
Vậy \(x\in\left\{-4;-3;-2;-1;0\right\}.\)
\(\frac{\left(2^3\cdot5\cdot7\right)\cdot\left(5^2\cdot7^3\right)}{\left(2\cdot5\cdot7^2\right)^2}\)
\(=\frac{2^3\left(5\cdot5^2\right)\left(7\cdot7^3\right)}{2^2\cdot5^2\cdot7^4}\)
\(=\frac{2^2\cdot2\cdot5\cdot5^2\cdot7^4}{2^2\cdot5^2\cdot7^4}\)
Triệt tiêu ta còn \(2\cdot5=10\)
\(\frac{\left(2^3.5.7\right).\left(5^2.7^2\right)}{\left(2.5.7^2\right)^2}\)
\(=\frac{2^3.\left(5.7\right).\left(5^2.7^3\right)}{2^2.5^2.7^4}\)
\(=\frac{2^2.2.5.5^2.7.7^3}{2^2.5^2.7^4}\)
\(=\frac{2^2.2.5.5^2.7^4}{2^2.5^2.7^4}\)
\(=2.5\)
\(=10\)
Bài 2
\(a,\)\(\left(x^2+7\right)\left(x^2-49\right)< 0\)
Vì \(x^2+7>0\)\(\Rightarrow x^2-49< 0\)
\(\Rightarrow\left(x-7\right)\left(x+7\right)< 0\)
\(...\)
Bài 2:
a) \(\left(x^2+7\right).\left(x^2-49\right)< 0\)
\(\Leftrightarrow\hept{\begin{cases}x^2+7< 0\\x^2-49>0\end{cases}}\)hoặc \(\hept{\begin{cases}x^2+7>0\\x^2-49< 0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x^2< -7\\x^2>49\end{cases}\left(loai\right)}\)hoặc \(\hept{\begin{cases}x^2>-7\\x^2< 49\end{cases}}\)
\(\Leftrightarrow-7< x^2< 49\)
Mà \(x^2\ge0\)và \(x^2\)là 1 SCP
\(\Rightarrow x^2\in\left\{1;4;9;16;25;36\right\}\)
\(\Rightarrow x\in\left\{1;2;3;4;5;6\right\}\)
Vậy \(x\in\left\{1;2;3;4;5;6\right\}\)