2, 3 (3-x/8)+2 (5-x)/3=1-x/2-2
giúp mình nha
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a) \(2^x.4=128\Rightarrow2^x=32=2^5\Rightarrow x=5\)
b) \(x^{17}=x\Rightarrow x^{17}-x=0\Rightarrow x\left(x^{16}-1\right)=0\Rightarrow x=0\) hay \(x=1\)
c) \(\left(2x-2\right)^3=8\Rightarrow\left(2x-2\right)^3=2^3\Rightarrow2x-2=2\Rightarrow2x=4\Rightarrow x=2\)
d) \(\left(x-6\right)^3=\left(x-6\right)^2\Rightarrow\left(x-6\right)^3-\left(x-6\right)^2=0\)
\(\Rightarrow\left(x-6\right)^2\left(x-6-1\right)=0\Rightarrow\Rightarrow\left(x-6\right)^2\left(x-7\right)=0\)
\(\Rightarrow x-6=0\) hay \(x-7=0\Rightarrow x=6\) hay \(x=7\)
e) \(\left(7x-11\right)^3=2^5.5^2+200\Rightarrow\left(7x-11\right)^3=32.25+200\)
\(\Rightarrow\left(7x-11\right)^3=1000=10^3\Rightarrow7x-11=10\Rightarrow7x=21\Rightarrow x=3\)
f) \(3+2^{x-1}=24-\left[4^2-\left(2^2-1\right)\right]\Rightarrow2^{x-1}=24-\left[16-3\right]-3\)
\(\Rightarrow2^{x-1}=24-13-3\Rightarrow2^{x-1}=8=2^3\Rightarrow2x-1=3\Rightarrow2x=4\Rightarrow x=2\)
Bài 1:
a) Ta có: \(x\left(x^2-4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;2;-2\right\}\)
b) Ta có: \(\left(2x-3\right)+\left(-3x\right)-\left(x-5\right)=40\)
\(\Leftrightarrow2x-3-3x-x+5=40\)
\(\Leftrightarrow-2x+2=40\)
\(\Leftrightarrow-2x=38\)
hay x=-19
Vậy: x=-19
Bài 2:
a) Ta có: \(-45\cdot12+34\cdot\left(-45\right)-45\cdot54\)
\(=-45\cdot\left(12+34+54\right)\)
\(=-45\cdot100\)
\(=-4500\)
b) Ta có: \(43\cdot\left(57-33\right)+33\cdot\left(43-57\right)\)
\(=43\cdot57-43\cdot33+43\cdot33-33\cdot57\)
\(=43\cdot57-33\cdot57\)
\(=57\cdot\left(43-33\right)\)
\(=57\cdot10=570\)
1/ -8+13-...+x=250
2/ 1-3+5-7+...-x=-600
3/ -2+4-6+8-...+x=100
4/ 2-6=10-14+...-x=-800
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1. a
\(\dfrac{8}{5}-\dfrac{5}{6}\cdot\dfrac{3}{4}\)
\(=\dfrac{8}{5}-\dfrac{5\cdot3}{3\cdot2\cdot4}\)
\(=\dfrac{8}{5}-\dfrac{5}{8}=\dfrac{39}{40}\)
1.b
\(=\dfrac{7}{8}+\dfrac{5}{6}\cdot\dfrac{3}{2}\)
\(=\dfrac{7}{8}+\dfrac{5\cdot3}{3\cdot2\cdot2}\)
\(=\dfrac{7}{8}+\dfrac{5}{4}=\dfrac{17}{8}\)
2.a
\(\dfrac{4}{5}+x=\dfrac{11}{10}\)
\(x=\dfrac{11}{10}-\dfrac{4}{5}=\dfrac{3}{10}\)
2.b
\(x-\dfrac{3}{4}=\dfrac{5}{7}\)
\(x=\dfrac{5}{7}+\dfrac{3}{4}=\dfrac{41}{28}\)
a) \(3^{x-1}+3x+3^{x+1}=1053\)
\(=3^x:3+3^x+3^x.3=1053\)
\(=3^x.\dfrac{1}{3}+1+3=1053\)
\(=3^x.\dfrac{13}{5}=1053\)
\(=3^x=243\)
\(\Rightarrow x=5\)
Vậy \(x=5\)