Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(27^{2-x}< =9\)
=>\(\left(3^3\right)^{2-x}< =3^2\)
=>\(3^{6-3x}< =3^2\)
=>6-3x<=2
=>-3x<=-4
=>\(x>=\dfrac{4}{3}\)
b: \(7^{3-x}< 49\)
=>\(7^{3-x}< 7^2\)
=>3-x<2
=>-x<2-3=-1
=>x>1
c: \(27^{3-x}>9\)
=>\(\left(3^3\right)^{3-x}>3^2\)
=>\(3^{9-3x}>3^2\)
=>9-3x>2
=>-3x>-7
=>\(x< \dfrac{7}{3}\)
d: \(2^{3-x}< 2^3\)
=>3-x<3
=>-x<0
=>x>0
e: \(27^{3-x^2}< 27^{x+1}\)
=>\(3-x^2< x+1\)
=>\(-x^2-x+2< 0\)
=>\(x^2+x-2>0\)
=>(x+2)(x-1)>0
=>\(\left[{}\begin{matrix}x>1\\x< -2\end{matrix}\right.\)
b.
\(\left(x^2+1\right)^2=5-x\sqrt{2x^2+4x}\)
\(\Leftrightarrow x^4+2x^2-4+x\sqrt{2x^2+4x}=0\)
Đặt \(x\sqrt{2x^2+4x}=t\Rightarrow t^2=x^2\left(2x^2+4x\right)=2\left(x^4+2x^2\right)\)
Pt trở thành:
\(\dfrac{t^2}{2}-4+t=0\)
\(\Leftrightarrow t^2+2t-8=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\sqrt{2x^2+4x}=2\left(x>0\right)\\x\sqrt{2x^2+4x}=-4\left(x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^4+2x^2-2=0\left(x>0\right)\\x^4+2x^2-8=0\left(x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\sqrt{3}-1}\\x=-\sqrt{2}\end{matrix}\right.\)
a.
ĐKXĐ: \(x\ne0\)
\(\Leftrightarrow\dfrac{9}{x^2}+2+\dfrac{2x}{\sqrt{2x^2+9}}=3\)
\(\Leftrightarrow\dfrac{2x^2+9}{x^2}+\dfrac{2x}{\sqrt{2x^2+9}}=3\)
Đặt \(\dfrac{x}{\sqrt{2x^2+9}}=t\Rightarrow\dfrac{2x^2+9}{x^2}=\dfrac{1}{t^2}\)
Pt trở thành:
\(\dfrac{1}{t^2}+2t=3\)
\(\Rightarrow2t^3-3t^2+1=0\)
\(\Leftrightarrow\left(t-1\right)^2\left(2t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{\sqrt{2x^2+9}}=1\left(x>0\right)\\\dfrac{x}{\sqrt{2x^2+9}}=-\dfrac{1}{2}\left(x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=2x^2+9\left(vô-nghiệm\right)\\4x^2=2x^2+9\left(x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow x=-\dfrac{3\sqrt{2}}{2}\)
Kiểm tra lại vế trái đề bài câu b
\(a,\left(\dfrac{1}{4}\right)^{x-2}=\sqrt{8}\\ \Leftrightarrow\left(\dfrac{1}{2}\right)^{2x-4}=\left(\dfrac{1}{2}\right)^{-\dfrac{3}{2}}\\ \Leftrightarrow2x-4=-\dfrac{3}{2}\\ \Leftrightarrow2x=\dfrac{5}{2}\\ \Leftrightarrow x=\dfrac{5}{4}\)
\(b,9^{2x-1}=81\cdot27^x\\ \Leftrightarrow3^{4x-2}=3^{4+3x}\\ \Leftrightarrow4x-2=4+3x\\ \Leftrightarrow x=6\)
c, ĐK: \(x-2>0\Rightarrow x>2\)
\(2log_5\left(x-2\right)=log_59\\
\Leftrightarrow log_5\left(x-2\right)^2=log_59\\
\Leftrightarrow\left(x-2\right)^2=3^2\\
\Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\\
\Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
Vậy phương trình có nghiệm là x = 5.
d, ĐK: \(x-1>0\Leftrightarrow x>1\)
\(log_2\left(3x+1\right)=2-log_2\left(x-1\right)\\ \Leftrightarrow log_2\left(3x+1\right)\left(x-1\right)=2\\ \Leftrightarrow3x^2-2x-1=4\\ \Leftrightarrow3x^2-2x-5=0\\ \Leftrightarrow\left(3x-5\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
Vậy phương trình có nghiệm \(x=\dfrac{5}{3}\)
a. Đề bài sai, phương trình không giải được
b.
ĐKXĐ: \(x\ge-\dfrac{2}{3}\)
\(\left(2x+10\right)\left(\dfrac{1-\left(3+2x\right)}{1+\sqrt{3+2x}}\right)^2=4\left(x+1\right)^2\)
\(\Leftrightarrow\dfrac{\left(2x+10\right)4.\left(x+1\right)^2}{\left(1+\sqrt{3+2x}\right)^2}=4\left(x+1\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}4\left(x+1\right)^2=0\Rightarrow x=-1\\2x+10=\left(1+\sqrt{3+2x}\right)^2\left(1\right)\end{matrix}\right.\)
Xét (1)
\(\Leftrightarrow2x+10=2x+4+2\sqrt{2x+3}\)
\(\Leftrightarrow\sqrt{2x+3}=3\)
\(\Leftrightarrow x=3\)
a) \({3^{{x^2} - 4x + 5}} = 9 \Leftrightarrow {x^2} - 4x + 5 = 2 \Leftrightarrow {x^2} - 4x + 3 = 0 \Leftrightarrow \left( {x - 3} \right)\left( {x - 1} \right) = 0\)
\( \Leftrightarrow \left[ \begin{array}{l}x = 3\\x = 1\end{array} \right.\)
Vậy phương trình có nghiệm là \(x \in \left\{ {1;3} \right\}\)
b) \(0,{5^{2x - 4}} = 4 \Leftrightarrow 2x - 4 = {\log _{0,5}}4 \Leftrightarrow 2x = 2 \Leftrightarrow x = 1\)
Vậy phương trình có nghiệm là x = 1
c) \({\log _3}(2x - 1) = 3\) ĐK: \(2x - 1 > 0 \Leftrightarrow x > \frac{1}{2}\)
\( \Leftrightarrow 2x - 1 = 27 \Leftrightarrow x = 14\) (TMĐK)
Vậy phương trình có nghiệm là x = 14
d) \(\log x + \log (x - 3) = 1\) ĐK: \(x - 3 > 0 \Leftrightarrow x > 3\)
\(\begin{array}{l} \Leftrightarrow \log \left( {x.\left( {x - 3} \right)} \right) = 1\\ \Leftrightarrow {x^2} - 3x = 10\\ \Leftrightarrow {x^2} - 3x - 10 = 0\\ \Leftrightarrow \left( {x + 2} \right)\left( {x - 5} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x = - 2 (loại) \,\,\,\\x = 5 (TMĐK) \,\,\,\,\,\,\,\end{array} \right.\end{array}\)
Vậy phương trình có nghiệm x = 5
\(a,\left(0,3\right)^{x-3}=1\\ \Leftrightarrow x-3=0\\ \Leftrightarrow x=3\\ b,5^{3x-2}=25\\ \Leftrightarrow3x-2=2\\ \Leftrightarrow3x=4\\ \Leftrightarrow x=\dfrac{4}{3}\\ c,9^{x-2}=243^{x+1}\\ \Leftrightarrow3^{2x-4}=3^{5x+5}\\ \Leftrightarrow2x-4=5x+5\\ \Leftrightarrow3x=-9\\ \Leftrightarrow x=-3\)
d, Điều kiện: \(x>-1;x\ne0\)
\(log_{\dfrac{1}{x}}\left(x+1\right)=-3\\ \Leftrightarrow x+1=x^3\\ x\simeq1,325\left(tm\right)\)
e, Điều kiện: \(x>\dfrac{5}{3}\)
\(log_5\left(3x-5\right)=log_5\left(2x+1\right)\\ \Leftrightarrow3x-5=2x+1\\ \Leftrightarrow x=6\left(tm\right)\)
f, Điều kiện: \(x>\dfrac{1}{2}\)
\(log_{\dfrac{1}{7}}\left(x+9\right)=log_{\dfrac{1}{7}}\left(2x-1\right)\\ \Leftrightarrow x+9=2x-1\\ \Leftrightarrow x=10\left(tm\right)\)
\(a,\left(\dfrac{1}{3}\right)^{2x+1}\le9\\ \Leftrightarrow2x+1\ge-2\\ \Leftrightarrow2x\ge-3\\ \Leftrightarrow x\ge-\dfrac{3}{2}\)
\(b,4^x>2^{x-2}\\ \Leftrightarrow2^{2x}>2^{x-2}\\ \Leftrightarrow2x>x-2\\ \Leftrightarrow x>-2\)
a: \(2^{2x-2}>=8\)
=>\(2^{2x-2}>=2^3\)
=>2x-2>=3
=>2x>=5
=>\(x>=\dfrac{5}{2}\)
b: \(4^{2x+2}< =16\)
=>\(4^{2x+2}< =4^2\)
=>2x+2<=2
=>2x<=0
=>x<=0
c: \(5^{x-9}>5^2\)
=>x-9>2
=>x>11
d: \(9^{x+2}< 9\)
=>\(9^{x+2}< 9^1\)
=>x+2<1
=>x<-1
e: \(9^{x-1}>9^{x^2-x-9}\)
=>\(x-1>x^2-x-9\)
=>\(x^2-x-9-x+1< 0\)
=>\(x^2-2x-8< 0\)
=>(x-4)(x+2)<0
=>-2<x<4