giúp mình với 2^x+^3*2^x=144 mình cảm ơn
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* Check đề câu a) giúp mình
\(b)\)
\(2^x+2^{x+3}=144\)
\(\rightarrow2^x+2^x.2^3=144\)
\(\rightarrow2^x.\left(1+8\right)=144\)
\(\rightarrow2^x=16\)
\(\rightarrow x=4\)
\(3^x+3^x=810\Leftrightarrow3^x\left(1+1\right)=810\)
\(\Leftrightarrow3^x=405\Leftrightarrow x=log_3405\)
cái đáp án mình đánh bên cymath hoặc đề sai rồi nhé
Em dùng công thức toán học hoặc viết ra giấy, chụp ảnh rồi up lên chứ thế này cô không đúng đề bài để giúp em được.
Lời giải:
$x-\frac{x}{3}\times \frac{3}{2}=2-\frac{1}{2}$
$x-x\times \frac{1}{2}=\frac{3}{2}$
$x\times (1-\frac{1}{2})=\frac{3}{2}$
$x\times \frac{1}{2}=\frac{3}{2}$
$x=\frac{3}{2}: \frac{1}{2}=3$
ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
Ta có: \(\dfrac{x-3}{x+1}=\dfrac{x^2}{x^2-1}\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}\)
Suy ra: \(x^2-4x+3-x^2=0\)
\(\Leftrightarrow-4x=-3\)
hay \(x=\dfrac{3}{4}\)(thỏa ĐK)
Vậy: \(S=\left\{\dfrac{3}{4}\right\}\)
`(x+3)(x^2-5x+8)=(x+3).x^2`
`<=>(x+3)(x^2-5x+8-x^2)=0`
`<=>(x+3)(8-5x)=0`
`<=>` \(\left[ \begin{array}{l}x+3=0\\8-5x=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac85\\x=-3\end{array} \right.\)
Vậy `S={-3,8/5}`
`(x+3)(x^2-5x+8)=(x+3).x^2`
`<=>(x+3)(x^2-5x+8-x^2)=0`
`<=>(x+3)(-5x+8)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\-5x+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{8}{5}\end{matrix}\right.\)
Vậy `S={-3;8/5}`.
\(\dfrac{3}{2}\)(\(x\) - \(\dfrac{5}{3}\)) - \(\dfrac{4}{5}\) = \(x\) + 1
\(\dfrac{3}{2}\) \(x\) - \(\dfrac{15}{6}\) - \(\dfrac{4}{5}\) = \(x\) + 1
\(\dfrac{3}{2}\)\(x\) - \(x\) = 1 + \(\dfrac{15}{6}\) + \(\dfrac{4}{5}\)
\(\dfrac{1}{2}\)\(x\) =\(\dfrac{43}{10}\)
\(x\) = \(\dfrac{43}{10}\) \(\times\) 2
\(x\) = \(\dfrac{43}{5}\)
\(\dfrac{3}{2}\left(x-\dfrac{5}{3}\right)-\dfrac{4}{5}=x+1\\ \Rightarrow\dfrac{3.\left(x-\dfrac{5}{3}\right)}{2}-\dfrac{4}{5}=x+1\\ \Rightarrow\dfrac{3x-5}{2}-\dfrac{4}{5}=x+1\Rightarrow\dfrac{5\left(3x-5\right)}{10}-\dfrac{8}{10}=x+1\\ \Rightarrow\dfrac{15x-33}{10}=x+1\\ \Rightarrow\dfrac{15x-33}{10}-x=x+1\\ \Rightarrow\dfrac{15x-33}{10}=x+1-x\\ \Rightarrow5x-33=10\\ \Rightarrow5x=10+33\\\Rightarrow5x=43\\ \Rightarrow x=\dfrac{43}{5} \)
Mình nhầm, để làm lại:
2x + 2x + 3 = 144
<=> 2x + 2x . 23 = 144
<=> 2x(23 + 1) = 144
<=> 2x . 9 = 144
<=> 2x = 144 : 9
<=> 2x = 16
<=> 16 = 24
<=> x = 4
Vậy x = 4
\(\Leftrightarrow2x\left(x+5\right)-3\left(x-2\right)=7x+1\)
\(\Leftrightarrow2x^2+10x-3x+6-7x-1=0\)
\(\Leftrightarrow2x^2+5=0\)(vô lý)
ĐKXĐ:\(\left\{{}\begin{matrix}x\ne2\\x\ne-5\end{matrix}\right.\)
\(\dfrac{2x}{x-2}-\dfrac{3}{x+5}=\dfrac{7x+1}{x^2+3x-10}\\ \Leftrightarrow\dfrac{2x\left(x+5\right)}{\left(x+5\right)\left(x-2\right)}-\dfrac{3\left(x-2\right)}{\left(x+5\right)\left(x-2\right)}=\dfrac{7x+1}{x^2-2x+5x-10}\\ \Leftrightarrow\dfrac{2x^2+10x}{\left(x+5\right)\left(x-2\right)}-\dfrac{3x-6}{\left(x+5\right)\left(x-2\right)}=\dfrac{7x+1}{x\left(x-2\right)+5\left(x-2\right)}\\ \Leftrightarrow\dfrac{2x^2+10x}{\left(x+5\right)\left(x-2\right)}-\dfrac{3x-6}{\left(x+5\right)\left(x-2\right)}-\dfrac{7x+1}{\left(x+5\right)\left(x-2\right)}=0\)
\(\Leftrightarrow\dfrac{2x^2+10x-3x+6-7x-1}{\left(x+5\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{2x^2+5}{\left(x+5\right)\left(x-2\right)}=0\\ \Rightarrow2x^2+5=0\left(vô.lí\right)\)
Vậy pt vô nghiệm