Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1) \(5^x+5^{x+2}=650\)
\(\Rightarrow5^x.1+5^x.5^2=650\)
\(\Rightarrow5^x.\left(1+5^2\right)=650\)
\(\Rightarrow5^x.26=650\)
\(\Rightarrow5^x=650:26\)
\(\Rightarrow5^x=25\)
\(\Rightarrow5^x=5^2\)
\(\Rightarrow x=2\)
Vậy \(x=2.\)
Mình chỉ làm câu 1) thôi nhé.
Chúc bạn học tốt!
a. Thay x = -1 vào biểu thức ta được:
\(\left(-1\right)^{10}+\left(-1\right)^9+\left(-1\right)^8+...+\left(-1\right)\)
\(=1-1+1-1+...+1-1\)
\(=0\)
b. Thay x = -1 vào biểu thức ta được:
\(\left(-1\right)^{100}+\left(-1\right)^{99}+\left(-1\right)^{98}+...-1\)
\(=1-1+1-1+...+1-1\)
\(=0\)
a: \(=\left(-1\right)^{10}+\left(-1\right)^9+\left(-1\right)^8+...+\left(-1\right)^2+\left(-1\right)\)
\(=\left(1-1\right)+\left(1-1\right)+...+\left(1-1\right)\)
=0
b: \(=\left(-1\right)^{100}+\left(-1\right)^{99}+...+\left(-1\right)^2+\left(-1\right)\)
\(=\left(1-1\right)+...+\left(1-1\right)\)
=0
c: \(=1^{100}-1^{99}+1^{98}-1^{97}+...+1^2-1\)
=0
f: \(=3\cdot\sqrt{9-5}+7=3\cdot2+7=13\)
a: \(\Leftrightarrow4^x\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)=4^8\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)\)
=>4^x=4^8
=>x=8
b: \(\Leftrightarrow2^x\cdot\dfrac{1}{2}+2^x\cdot2=2^{10}\left(2^2+1\right)\)
=>2^x=2^11
=>x=11
c: =>1/6*6^x+6^x*36=6^15(1+6^3)
=>6^x=6*6^15
=>x=16
d: \(\Leftrightarrow8^x\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)=8^9\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)\)
=>x=9
c: \(=\dfrac{7}{23}\cdot\left(\dfrac{-4}{3}-\dfrac{5}{2}\right)=\dfrac{7}{23}\cdot\dfrac{-8-15}{6}\)
\(=\dfrac{7}{23}\cdot\dfrac{-23}{6}=-\dfrac{7}{6}\)
d: \(=\dfrac{5}{7}\left(23+\dfrac{1}{4}-13-\dfrac{1}{4}\right)=\dfrac{5}{7}\cdot10=\dfrac{50}{7}\)
e: \(=\dfrac{2^5\cdot3^3\cdot5^3}{2^3\cdot3^3\cdot2^2\cdot5^2}=5\)
i: \(=\dfrac{1}{3^{10}}\cdot3^{50}-\dfrac{2^{10}}{3^{10}}:\dfrac{4^5}{3^{10}}\)
\(=3^{40}-1\)
\(\left(\frac{1}{2}\right)^5\times x=\left(\frac{1}{2}\right)^7\)
\(x=\left(\frac{1}{2}\right)^7\div\left(\frac{1}{2}\right)^5\)
\(x=\left(\frac{1}{2}\right)^{7-5}=\left(\frac{1}{2}\right)^2=\frac{1}{4}\) .
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{9}{21}\right)^2\)
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{3}{7}\right)^4\)
\(x=\left(\frac{3}{7}\right)^4\div\left(\frac{3}{7}\right)^2\)
\(x=\left(\frac{3}{7}\right)^{4-2}=\left(\frac{3}{7}\right)^2=\frac{9}{49}\)
\(2^x=2\Rightarrow x=1\)
\(3^x=3^4\Rightarrow x=4\)
\(7^x=7^7\Rightarrow x=7\)
\(\left(-3\right)^x=\left(-3\right)^5\Rightarrow x=5\)
\(\left(-5\right)^x=\left(-5\right)^4\Rightarrow x=4\)
\(2^x=4\Leftrightarrow2^x=2^2\Rightarrow x=2\)
\(2^x=8\Leftrightarrow2^x=2^3\Rightarrow x=3\)
\(2^x=16\Leftrightarrow2^x=2^4\Rightarrow x=4\)
\(3^{x+1}=3^2\Leftrightarrow x+1=2\Leftrightarrow x=2-1\Rightarrow x=1\)
\(5^{x-1}=5\Leftrightarrow x-1=1\Leftrightarrow x=1+1\Rightarrow x=2\)
\(6^{x+4}=6^{10}\Leftrightarrow x+4=10\Leftrightarrow x=10-4\Rightarrow x=6\)
\(5^{2x-7}=5^{11}\Leftrightarrow2x-7=11\Leftrightarrow2x=11+7\Leftrightarrow2x=18\Leftrightarrow x=18\div2\Rightarrow x=9\)
\(\left(-2\right)^{4x+2}=64\)
\(2^{-4x+2}=2^6\Leftrightarrow-4x+2=6\Leftrightarrow-4x=6-2\Leftrightarrow-4x=4\Leftrightarrow x=4\div\left(-4\right)\Rightarrow x=-1\)
\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^5\Rightarrow x=5\)
\(\left(\frac{5}{6}\right)^{2x}=\left(\frac{5}{6}\right)^5\Rightarrow2x=5\Rightarrow x=\frac{5}{2}\)
\(\left(\frac{3}{4}\right)^{2x-1}=\left(\frac{3}{4}\right)^{5x-4}\Rightarrow2x-1=5x-4\)
\(2x-5x=-4+1\)
\(-3x=-3\Rightarrow x=1\)
\(\left(\frac{-1}{10}\right)^x=\frac{1}{100}\)
\(\left(\frac{1}{10}\right)^{-x}=\left(\frac{1}{10}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{2}\right)^x=\frac{9}{4}\)
\(\left(\frac{3}{2}\right)^{-x}=\left(\frac{3}{2}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{5}\right)^{2x}=\frac{9}{25}\)
\(\left(\frac{3}{5}\right)^{-2x}=\left(\frac{3}{5}\right)^2\Rightarrow-2x=2\Rightarrow x=-1\)
\(\left(\frac{-2}{3}\right)^x=\frac{-8}{27}\)
\(\left(\frac{-2}{3}\right)^x=\left(\frac{-2}{3}\right)^3\Rightarrow x=3\).
hehe.
đánh tới què tay, hoa mắt lun r nekkk!!
c: \(=\dfrac{7}{23}\cdot\dfrac{-24-45}{18}=\dfrac{7}{23}\cdot\dfrac{-69}{18}=\dfrac{7}{18}\cdot\left(-3\right)=-\dfrac{7}{6}\)
d: \(=\dfrac{7}{5}\left(23+\dfrac{1}{4}-13-\dfrac{1}{4}\right)=\dfrac{7}{5}\cdot10=14\)
e: \(=\dfrac{2^5\cdot3^3\cdot5^3}{2^3\cdot3^3\cdot2^2\cdot5^2}=5\)
i: \(=\dfrac{1}{3^{10}}\cdot3^{50}-\dfrac{2^{10}}{3^{10}}:\dfrac{4^5}{9^5}=3^{40}-1\)
\(x^2=1\Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
\(x^2=3\Rightarrow\left[{}\begin{matrix}x=-\sqrt{3}\\x=\sqrt{3}\end{matrix}\right.\)
\(x^2=5\Rightarrow\left[{}\begin{matrix}x=-\sqrt{5}\\x=\sqrt{5}\end{matrix}\right.\Rightarrow x=-\sqrt{5}\left(vì.x< 0\right)\)
\(x^2=7\Rightarrow\left[{}\begin{matrix}x=-\sqrt{7}\\x=\sqrt{7}\end{matrix}\right.\Rightarrow x=-\sqrt{7}\left(vì.x< 0\right)\)
\(x^2=9\Rightarrow\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)
\(\left(x-2\right)^2=2\Rightarrow\left[{}\begin{matrix}x-2=-\sqrt{2}\\x-2=\sqrt{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2-\sqrt{2}\\x=2+\sqrt{2}\end{matrix}\right.\)
\(\left(x-4\right)^2=4\Rightarrow\left[{}\begin{matrix}x-2=-2\\x-2=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(\left(x-6\right)^2=6\Rightarrow\left[{}\begin{matrix}x-6=-\sqrt{6}\\x-6=\sqrt{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6-\sqrt{6}\\x=6+\sqrt{6}\end{matrix}\right.\)
\(\left(x-8\right)^2=8\Rightarrow\left[{}\begin{matrix}x-8=-2\sqrt{2}\\x-8=2\sqrt{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8-2\sqrt{2}\\x=2+2\sqrt{2}\end{matrix}\right.\)
\(\left(x-10\right)^2=10\Rightarrow\left[{}\begin{matrix}x-10=-\sqrt{10}\\x-10=\sqrt{10}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=10-\sqrt{10}\\x=10+\sqrt{10}\end{matrix}\right.\)
\(\left(x-\sqrt{3}\right)^2=3\Rightarrow\left[{}\begin{matrix}x-\sqrt{3}=-\sqrt{3}\\x-\sqrt{3}=\sqrt{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=2\sqrt{3}\end{matrix}\right.\)
\(\left(x-\sqrt{5}\right)^2=5\Rightarrow\left[{}\begin{matrix}x-\sqrt{5}=-\sqrt{5}\\x-\sqrt{5}=\sqrt{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=2\sqrt{5}\end{matrix}\right.\)