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a) \(\left(-2\right)^3+2^2+\left(-1\right)^{20}+\left(-2\right)^0\)
\(=-8+4+1+1=-2\)
b) \(\left(3^2\right)^2-\left(-5^2\right)^2+\left[\left(-2\right)^3\right]^2\)
\(=9^2-\left(-25\right)^2+\left(-8\right)^2\)
\(=81-625+64=-480\)
c) Bạn sửa lại đề!
Bài 2:
1. \(8^n:2^n=4\)
⇔ \(2^{3n}:2^n=2^2\)
⇔ \(2^{2n}=2^2\)
⇔ \(2n=2\)
⇔ \(n=2:2\)
⇒ \(n=1\)
Vậy \(n=1.\)
Chúc bạn học tốt!
Bài 1:
1. \(x:-\left(-\frac{1}{2}\right)=-\frac{1}{2}\)
⇒ \(x:\frac{1}{2}=-\frac{1}{2}\)
⇒ \(x=\left(-\frac{1}{2}\right).\frac{1}{2}\)
⇒ \(x=-\frac{1}{4}\)
Vậy \(x=-\frac{1}{4}.\)
3. \(\frac{16}{2^n}=2\)
⇒ \(2^n=16:2\)
⇒ \(2^n=8\)
⇒ \(2^n=2^3\)
⇒ \(n=3\)
Vậy \(n=3.\)
4. \(\frac{-3^n}{81}=-27\)
⇒ \(\left(-3\right)^n=\left(-27\right).81\)
⇒ \(\left(-3\right)^n=-2187\)
⇒ \(\left(-3\right)^n=\left(-3\right)^7\)
⇒ \(n=7\)
Vậy \(n=7.\)
Chúc bạn học tốt!
\(\left(2x+3\right)^2+\left(3x-2\right)^4=0\)
vì \(\left(2x+3\right)^2\ge0;\left(3x-2\right)^4\ge0\)
nên\(\Rightarrow\hept{\begin{cases}\left(2x+3\right)^2=0\\\left(3x-2\right)^4=0\end{cases}\Rightarrow\hept{\begin{cases}2x+3=0\\3x-2=0\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}x=-\frac{3}{2}\\x=\frac{2}{3}\end{cases}}\)
1/vì (1,782x-2-1,78x):1,78x=0
nên 1,78x2-2-1,78x=0
=>1,782x-2=1,78x
=>2x-2=x
2x=x+2
=>x=2
2/vì cơ số bằng nhau nên ta có
x-2=1;-1;0
ta có: x-2=1 => x=3
x-2=-1 => x=1
x-2=0 => x=2
3/ta có
(x+2)3=33 =>x+2=3 =>x=1
mik mệt rồi bạn cứ gải tiếp đi
\(\left(\frac{1}{2}\right)^5\times x=\left(\frac{1}{2}\right)^7\)
\(x=\left(\frac{1}{2}\right)^7\div\left(\frac{1}{2}\right)^5\)
\(x=\left(\frac{1}{2}\right)^{7-5}=\left(\frac{1}{2}\right)^2=\frac{1}{4}\) .
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{9}{21}\right)^2\)
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{3}{7}\right)^4\)
\(x=\left(\frac{3}{7}\right)^4\div\left(\frac{3}{7}\right)^2\)
\(x=\left(\frac{3}{7}\right)^{4-2}=\left(\frac{3}{7}\right)^2=\frac{9}{49}\)
\(2^x=2\Rightarrow x=1\)
\(3^x=3^4\Rightarrow x=4\)
\(7^x=7^7\Rightarrow x=7\)
\(\left(-3\right)^x=\left(-3\right)^5\Rightarrow x=5\)
\(\left(-5\right)^x=\left(-5\right)^4\Rightarrow x=4\)
\(2^x=4\Leftrightarrow2^x=2^2\Rightarrow x=2\)
\(2^x=8\Leftrightarrow2^x=2^3\Rightarrow x=3\)
\(2^x=16\Leftrightarrow2^x=2^4\Rightarrow x=4\)
\(3^{x+1}=3^2\Leftrightarrow x+1=2\Leftrightarrow x=2-1\Rightarrow x=1\)
\(5^{x-1}=5\Leftrightarrow x-1=1\Leftrightarrow x=1+1\Rightarrow x=2\)
\(6^{x+4}=6^{10}\Leftrightarrow x+4=10\Leftrightarrow x=10-4\Rightarrow x=6\)
\(5^{2x-7}=5^{11}\Leftrightarrow2x-7=11\Leftrightarrow2x=11+7\Leftrightarrow2x=18\Leftrightarrow x=18\div2\Rightarrow x=9\)
\(\left(-2\right)^{4x+2}=64\)
\(2^{-4x+2}=2^6\Leftrightarrow-4x+2=6\Leftrightarrow-4x=6-2\Leftrightarrow-4x=4\Leftrightarrow x=4\div\left(-4\right)\Rightarrow x=-1\)
\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^5\Rightarrow x=5\)
\(\left(\frac{5}{6}\right)^{2x}=\left(\frac{5}{6}\right)^5\Rightarrow2x=5\Rightarrow x=\frac{5}{2}\)
\(\left(\frac{3}{4}\right)^{2x-1}=\left(\frac{3}{4}\right)^{5x-4}\Rightarrow2x-1=5x-4\)
\(2x-5x=-4+1\)
\(-3x=-3\Rightarrow x=1\)
\(\left(\frac{-1}{10}\right)^x=\frac{1}{100}\)
\(\left(\frac{1}{10}\right)^{-x}=\left(\frac{1}{10}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{2}\right)^x=\frac{9}{4}\)
\(\left(\frac{3}{2}\right)^{-x}=\left(\frac{3}{2}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{5}\right)^{2x}=\frac{9}{25}\)
\(\left(\frac{3}{5}\right)^{-2x}=\left(\frac{3}{5}\right)^2\Rightarrow-2x=2\Rightarrow x=-1\)
\(\left(\frac{-2}{3}\right)^x=\frac{-8}{27}\)
\(\left(\frac{-2}{3}\right)^x=\left(\frac{-2}{3}\right)^3\Rightarrow x=3\).
hehe.
đánh tới què tay, hoa mắt lun r nekkk!!
a) Ta có : 2017 - |x - 2017| = x
=> |x - 2017| = 2017 - x (1)
Điều kiện xác định : \(2017-x\ge0\Rightarrow2017\ge x\Rightarrow x\le2017\)
Khi đó (1) <=> \(\orbr{\begin{cases}x-2017=2017-x\\x-2017=-\left(2017-x\right)\end{cases}\Rightarrow\orbr{\begin{cases}2x=2017+2017\\x-2017=-2017+x\end{cases}\Rightarrow}\orbr{\begin{cases}2x=4034\\0x=0\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}x=2017\\x\text{ thỏa mãn }\Leftrightarrow x\le2017\end{cases}}\Rightarrow x\le2017\)
b) Ta có : \(\hept{\begin{cases}\left(2x-1\right)^{2016}\ge0\forall x\\\left(y-\frac{2}{5}\right)^{2016}\ge\\\left|x+y+z\right|\ge0\forall x;y;z\end{cases}0\forall y}\Rightarrow\left(2x-1\right)^{2016}+\left(y-\frac{2}{5}\right)^{2016}+\left|x+y+z\right|\ge0\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}2x-1=0\\y-\frac{2}{5}=0\\x+y+z=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{2}{5}\\\frac{1}{2}+\frac{2}{5}+z=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{2}{5}\\z=-\frac{9}{10}\end{cases}}}\)
\(a,35:\left(\frac{-5}{3}\right)+2\frac{1}{2}:\left(-\frac{5}{3}\right)=35.\left(-\frac{3}{5}\right)+\frac{5}{2}.\left(-\frac{3}{5}\right)\)
\(=-21+-\frac{3}{2}\)
\(=\frac{-42-3}{2}=-\frac{45}{2}\)
\(b,\frac{10^3+2.5^3+5^3}{55}=\frac{\left(2.5\right)^3+2.5^3+5^3}{5.11}\)
\(=\frac{5^3\left(2^3+2+1\right)}{5.11}\)
\(=\frac{5^2\left(8+2+1\right)}{11}\)
\(=\frac{5^2.11}{11}=5^2=25\)
\(C,\frac{27^2.2^5}{6^6.32^3}=\frac{\left(3^3\right)^2.2^5}{\left(2.3\right)^6.2^5}\)
\(=\frac{3^6}{2^6.3^6}=\frac{1}{2^6}=\frac{1}{64}\)