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a: \(2^{x^2-2x+1}=1\)
=>\(2^{\left(x-1\right)^2}=2^0\)
=>\(\left(x-1\right)^2=0\)
=>x-1=0
=>x=1
b: \(7^{x^2+7x}=5764801\)
=>\(7^{x^2+7x}=7^8\)
=>\(x^2+7x=8\)
=>\(x^2+7x-8=0\)
=>(x+8)(x-1)=0
=>\(\left[{}\begin{matrix}x+8=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=1\end{matrix}\right.\)
c: \(6^{x^2+12x}=6^{7x}\)
=>\(x^2+12x=7x\)
=>\(x^2+5x=0\)
=>x(x+5)=0
=>\(\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
d: \(\left(\dfrac{1}{3}\right)^{x-1}=3^{2x-5}\)
=>\(3^{-x+1}=3^{2x-5}\)
=>-x+1=2x-5
=>-x-2x=-5-1
=>-3x=-6
=>x=2
e: \(\left(\dfrac{1}{5}\right)^{3x+5}=5^{2x+1}\)
=>\(5^{-3x-5}=5^{2x+1}\)
=>-3x-5=2x+1
=>-5x=6
=>\(x=-\dfrac{6}{5}\)
a: \(2^{x^2-1}=256\)
=>\(2^{x^2-1}=2^8\)
=>\(x^2-1=8\)
=>\(x^2=9\)
=>\(x\in\left\{3;-3\right\}\)
b: \(3^{x^2+3x}=81\)
=>\(3^{x^2+3x}=3^4\)
=>\(x^2+3x=4\)
=>\(x^2+3x-4=0\)
=>(x+4)(x-1)=0
=>\(\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\)
c: \(2^{x^2-5x}=64\)
=>\(2^{x^2-5x}=2^6\)
=>\(x^2-5x=6\)
=>\(x^2-5x-6=0\)
=>(x-6)(x+1)=0
=>\(\left[{}\begin{matrix}x-6=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-1\end{matrix}\right.\)
d: \(\left(\dfrac{1}{3}\right)^x=243\)
=>\(\left(\dfrac{1}{3}\right)^x=3^5=\left(\dfrac{1}{3}\right)^{-5}\)
=>x=-5
e: \(\left(\dfrac{1}{3}\right)^{x+5}=3^{2x+1}\)
=>\(3^{-x-5}=3^{2x+1}\)
=>-x-5=2x+1
=>-3x=6
=>x=-2
a. Làm gọn 1 chút xíu:
\(y=\left(x^{11}+2x^7-3x^5-6x\right)\left(3x^7+6x^2-2\right)\)
\(y'=\left(11x^{10}+14x^6-15x^4-6\right)\left(3x^7+6x^2-2\right)+\left(21x^6+12x\right)\left(x^{11}+2x^7-3x^5-6x\right)\)
b.
\(y'=5\left(x^4-\dfrac{2}{3x}\right)^4\left(4x^3+\dfrac{2}{3x^2}\right)\Rightarrow y'\left(10\right)=5\left(10^4-\dfrac{2}{30}\right)^4\left(4.10^3+\dfrac{2}{300}\right)=?\)
c.
\(y'=\dfrac{7}{\left(x+1\right)^2}\Rightarrow y'\left(4\right)=\dfrac{7}{25}\)
a:
ĐKXĐ: \(x\notin\left\{\dfrac{3}{2};1\right\}\)
\(y=\dfrac{\left(x-2\right)^2}{\left(2x-3\right)\left(x-1\right)}=\dfrac{x^2-4x+4}{2x^2-2x-3x+3}\)
=>\(y=\dfrac{x^2-4x+4}{2x^2-5x+3}\)
=>\(y'=\dfrac{\left(x^2-4x+4\right)'\left(2x^2-5x+3\right)-\left(x^2-4x+4\right)\left(2x^2-5x+3\right)'}{\left(2x^2-5x+3\right)^2}\)
=>\(y'=\dfrac{\left(2x-4\right)\left(2x^2-5x+3\right)-\left(2x-5\right)\left(x^2-4x+4\right)}{\left(2x^2-5x+3\right)^2}\)
=>\(y'=\dfrac{4x^3-10x^2+6x-8x^2+20x-12-2x^3+8x^2-8x+5x^2-20x+20}{\left(2x^2-5x+3\right)^2}\)
=>\(y'=\dfrac{2x^3-5x^2-2x+8}{\left(2x^2-5x+3\right)^2}\)
b:
ĐKXĐ: x<>-3
\(y=\left(x+3\right)+\dfrac{4}{x+3}\)
=>\(y'=\left(x+3+\dfrac{4}{x+3}\right)'=1+\left(\dfrac{4}{x+3}\right)'\)
\(=1+\dfrac{4'\left(x+3\right)-4\left(x+3\right)'}{\left(x+3\right)^2}\)
=>\(y'=1+\dfrac{-4}{\left(x+3\right)^2}=\dfrac{\left(x+3\right)^2-4}{\left(x+3\right)^2}\)
y'=0
=>\(\left(x+3\right)^2-4=0\)
=>\(\left(x+3+2\right)\left(x+3-2\right)=0\)
=>(x+5)(x+1)=0
=>x=-5 hoặc x=-1
c:
ĐKXĐ: x<>-2
\(y=\dfrac{\left(5x-1\right)\left(x+1\right)}{x+2}\)
=>\(y=\dfrac{5x^2+5x-x-1}{x+2}=\dfrac{5x^2+4x-1}{x+2}\)
=>\(y'=\dfrac{\left(5x^2+4x-1\right)'\left(x+2\right)-\left(5x^2+4x-1\right)\left(x+2\right)'}{\left(x+2\right)^2}\)
=>\(y'=\dfrac{\left(5x+4\right)\left(x+2\right)-\left(5x^2+4x-1\right)}{\left(x+2\right)^2}\)
=>\(y'=\dfrac{5x^2+10x+4x+8-5x^2-4x+1}{\left(x+2\right)^2}\)
=>\(y'=\dfrac{10x+9}{\left(x+2\right)^2}\)
\(y'\left(-1\right)=\dfrac{10\cdot\left(-1\right)+9}{\left(-1+2\right)^2}=\dfrac{-1}{1}=-1\)
d:
ĐKXĐ: x<>2
\(y=x-2+\dfrac{9}{x-2}\)
=>\(y'=\left(x-2+\dfrac{9}{x-2}\right)'=1+\left(\dfrac{9}{x-2}\right)'\)
\(=1+\dfrac{9'\left(x-2\right)-9\left(x-2\right)'}{\left(x-2\right)^2}\)
=>\(y'=1+\dfrac{-9}{\left(x-2\right)^2}=\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}\)
y'=0
=>\(\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}=0\)
=>\(\left(x-2\right)^2-9=0\)
=>(x-2-3)(x-2+3)=0
=>(x-5)(x+1)=0
=>x=5 hoặc x=-1
a: \(6^x=5\)
=>\(x=log_65\)
b: \(7^{3-x}=5\)
=>\(3-x=log_75\)
=>\(x=3-log_75\)
c: \(\left(\dfrac{3}{5}\right)^{x-2}=\dfrac{27}{125}\)
=>\(\left(\dfrac{3}{5}\right)^{x-2}=\left(\dfrac{3}{5}\right)^3\)
=>x-2=3
=>x=5
d: \(\left(\dfrac{4}{5}\right)^x=\dfrac{5}{4}\)
=>\(\left(\dfrac{4}{5}\right)^x=\left(\dfrac{4}{5}\right)^{-1}\)
=>x=-1
a.
\(6^x=5\Rightarrow x=log_65\)
b.
\(7^{3-x}=5\Rightarrow3-x=log_75\)
\(\Rightarrow x=3-log_75\)
c.
\(\left(\dfrac{3}{5}\right)^{x-2}=\dfrac{27}{125}\Rightarrow x-2=log_{\dfrac{3}{5}}\left(\dfrac{27}{125}\right)\)
\(\Rightarrow x-2=3\Rightarrow x=5\)
d.
\(\left(\dfrac{4}{5}\right)^x=\dfrac{5}{4}\Rightarrow\left(\dfrac{4}{5}\right)^x=\left(\dfrac{4}{5}\right)^{-1}\)
\(\Rightarrow x=-1\)
c.
\(\Leftrightarrow x^2+3-\left(3x+1\right)\sqrt{x^2+3}+2x^2+2x=0\)
Đặt \(\sqrt{x^2+3}=t>0\)
\(\Rightarrow t^2-\left(3x+1\right)t+2x^2+2x=0\)
\(\Delta=\left(3x+1\right)^2-4\left(2x^2+2x\right)=\left(x-1\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{3x+1-x+1}{2}=x+1\\t=\dfrac{3x+1+x-1}{2}=2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+3}=x+1\left(x\ge-1\right)\\\sqrt{x^2+3}=2x\left(x\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+3=x^2+2x+1\left(x\ge-1\right)\\x^2+3=4x^2\left(x\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow x=1\)
a.
Đề bài ko chính xác, pt này ko giải được
b.
ĐKXĐ: \(x\ge-\dfrac{7}{2}\)
\(2x+7-\left(2x+7\right)\sqrt{2x+7}+x^2+7x=0\)
Đặt \(\sqrt{2x+7}=t\ge0\)
\(\Rightarrow t^2-\left(2x+7\right)t+x^2+7x=0\)
\(\Delta=\left(2x+7\right)^2-4\left(x^2+7x\right)=49\)
\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{2x+7-7}{2}=x\\t=\dfrac{2x+7+7}{2}=x+7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x+7}=x\left(x\ge0\right)\\\sqrt{2x+7}=x+7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-7=0\left(x\ge0\right)\\x^2+12x+42=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow x=1+2\sqrt{2}\)
a: ĐKXĐ: \(4x-3>0\)
=>x>3/4
\(log_5\left(4x-3\right)=2\)
=>\(log_5\left(4x-3\right)=log_525\)
=>4x-3=25
=>4x=28
=>x=7(nhận)
b: ĐKXĐ: \(x\ne0\)
\(log_2x^2=2\)
=>\(log_2x^2=log_24\)
=>\(x^2=4\)
=>\(\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-2\left(nhận\right)\end{matrix}\right.\)
c: ĐKXĐ: \(x\notin\left\{-\dfrac{1}{2};\dfrac{3}{2}\right\}\)
\(\log_52x+1=\log_5-2x+3\)
=>2x+1=-2x+3
=>4x=2
=>\(x=\dfrac{1}{2}\left(nhận\right)\)
d: ĐKXD: \(x\notin\left\{3\right\}\)
\(ln\left(x^2-6x+7\right)=ln\left(x-3\right)\)
=>\(x^2-6x+7=x-3\)
=>\(x^2-7x+10=0\)
=>(x-2)(x-5)=0
=>\(\left[{}\begin{matrix}x=2\left(nhận\right)\\x=5\left(nhận\right)\end{matrix}\right.\)
e: ĐKXĐ: \(x\notin\left\{\dfrac{1}{5};2\right\}\)
\(log\left(5x-1\right)=log\left(4-2x\right)\)
=>5x-1=4-2x
=>7x=5
=>\(x=\dfrac{5}{7}\left(nhận\right)\)
a.
\(2^x=2^{3x-1}\Leftrightarrow x=3x-1\)
\(\Rightarrow x=\dfrac{1}{2}\)
b.
\(7^{x-5}=49\Leftrightarrow x-5=log_749=2\)
\(\Rightarrow x=7\)
c.
\(3^{5x-3}=1\Rightarrow5x-3=log_31=0\)
\(\Rightarrow x=\dfrac{3}{5}\)
d.
\(\left(\dfrac{1}{7}\right)^{5x}=7^{x+6}\Leftrightarrow7^{-5x}=7^{x+6}\)
\(\Leftrightarrow-5x=x+6\)
\(\Rightarrow x=-1\)