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Bài 1:
\(a,8.6+288.\left(x+3\right)^2=50\\ \Leftrightarrow48+288\left(x+3\right)^2=50\\ \Leftrightarrow\left(x+3\right)^2=\dfrac{1}{144}\\ \Leftrightarrow x+3\in\left\{-\dfrac{1}{12};\dfrac{1}{12}\right\}\\ \Leftrightarrow x\in\left\{-\dfrac{37}{12};-\dfrac{35}{12}\right\}\\ Vậy.....\)
\(b,\left(x+1\right)+\left(x+2\right)+...+\left(x+100\right)=5750\)
=>Số lượng số hạng của tổng trên là (x+100-x-1):1+1=100(số hạng)
\(\Rightarrow\dfrac{\left(2x+101\right).100}{2}=5750\\ \Rightarrow2x+101=\dfrac{5750.2}{100}\\ \Rightarrow2x+101=115\\ \Rightarrow2x=14\\ \Rightarrow x=7\\ Vậy........\)
288:(x-3)2=2
=> (x-3)2=144
=>x-3=12(vì x thuộc N)
=> x=15
\(8.6+288:\left(x-3\right)^2=50\)
\(\Rightarrow48+288:\left(x-3\right)^2=50\)
\(\Rightarrow288:\left(x-3\right)^2=50-48=2\)
\(\Rightarrow\left(x-3\right)^2=288:2=144\)
Mà \(\left(x-3\right)^2=144=12^2\)
\(\Rightarrow x-3=12\)
\(\Rightarrow x=12+3\)
\(\Rightarrow x=15\)
a) \(47-\left[\left(45\cdot2^4-5^2\cdot12\right):14\right]\)
\(=47-\left[\left(45\cdot16-25\cdot12\right):14\right]\)
\(=47-\left[\left(720-300\right):14\right]\)
\(=47-\left[420:14\right]\)
\(=47-30=17\)
b) \(50-\left[\left(20-2^3\right):2+34\right]\)
\(=50-\left[\left(20-8\right):2+34\right]\)
\(=50-\left[12:2+34\right]\)
\(=50-\left[6+34\right]\)
\(=50-40=10\)
c) \(10^2-\left[60:\left(5^6:5^4-3\cdot5\right)\right]\)
\(=100-\left[60:\left(5^{6-4}-15\right)\right]\)
\(=100-\left[60:\left(5^2-15\right)\right]\)
\(=100-\left[60:\left(25-15\right)\right]\)
\(=100-\left[60:10\right]\)
\(=100-6=94\)
d) \(50-\left[\left(50-2^3\cdot5\right):2+3\right]\)
\(=50-\left[\left(50-8\cdot5\right):2+3\right]\)
\(=50-\left[\left(50-40\right):2+3\right]\)
\(=50-\left[10:2+3\right]\)
\(=50-\left[5+3\right]\)
\(=50-8=42\)
e) \(10-\left[\left(8^2-48\right)\cdot5+\left(2^3\cdot10+8\right)\right]:28\)
\(=10-\left[\left(64-48\right)\cdot5+\left(8\cdot10+8\right)\right]:28\)
\(=10-\left[16\cdot5+\left(80+8\right)\right]:28\)
\(=10-\left[80+88\right]:28\)
\(=10-168:28\)
\(=10-6=4\)
f) \(8697-\left[3^7:3^5+2\left(13-3\right)\right]\)
\(=8697-\left[3^{7-5}+2\cdot10\right]\)
\(=8697-\left[3^2+20\right]\)
\(=8697-\left[9+20\right]\)
\(=8697-29=8668\)
CHÚC BẠN HỌC TỐT!!!!!!!!!!!
\(\dfrac{-2}{3}.x+\dfrac{1}{5}=\dfrac{3}{10}\)
\(\dfrac{-2}{3}.x=\dfrac{3}{10}-\dfrac{1}{5}\)
\(\dfrac{-2}{3}.x=\dfrac{3}{10}-\dfrac{2}{10}\)
\(\dfrac{-2}{3}.x=\dfrac{1}{10}\)
\(x=\dfrac{1}{10}:\dfrac{-2}{3}\)
\(x=\dfrac{1}{10}.\dfrac{-3}{2}\)
\(x=\dfrac{-3}{20}\)
\(8.6+288:\left(x-3\right)^2=50\)
\(48+288:\left(x-3\right)^2=50\)
\(288:\left(x-3\right)^2=50-48\)
\(288:\left(x-3\right)^2=2\)
\(\left(x-3\right)^2=288:2\)
\(\left(x-3\right)^2=144\)
\(\left(x-3\right)^2=12^2=\left(-12\right)^2\)
\(=>\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.=>\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)
Vậy x = 15 hoặc x = -9
\(\Leftrightarrow288:\left(x-3\right)^2=50-\dfrac{8}{6}=\dfrac{292}{6}\)
\(\Leftrightarrow\left(x-3\right)^2=\dfrac{432}{73}\)
hay \(x\in\left\{\dfrac{12\sqrt{219}}{73}+3;\dfrac{-12\sqrt{219}}{73}+3\right\}\)