9^x-1= 9
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\(\dfrac{6}{7}\cdot\dfrac{8}{9}+\dfrac{7}{9}\cdot\dfrac{2}{7}-\dfrac{18}{21}\cdot\dfrac{1}{9}+\dfrac{7}{9}\cdot\dfrac{3}{7}\)
\(=\dfrac{6}{7}\cdot\dfrac{8}{9}-\dfrac{6}{7}\cdot\dfrac{1}{9}+\dfrac{7}{9}\left(\dfrac{2}{7}+\dfrac{3}{7}\right)\)
\(=\dfrac{6}{7}\left(\dfrac{8}{9}-\dfrac{1}{9}\right)+\dfrac{7}{9}\cdot\dfrac{5}{7}\)
\(=\dfrac{7}{9}\left(\dfrac{6}{7}+\dfrac{5}{7}\right)=\dfrac{7}{9}\cdot\dfrac{11}{7}=\dfrac{11}{9}\)
`1/36 (x-1/2)^2=1/9`
`(x-1/2)^2=4 = 2^2=(-2)^2`
\(\left[{}\begin{matrix}x-\dfrac{1}{2}=2\\x-\dfrac{1}{2}=-2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(\dfrac{1}{36}.\left(x-\dfrac{1}{2}\right)^2=\dfrac{1}{9}\)
\(\left(x-\dfrac{1}{2}\right)^2=\dfrac{1}{9}:\dfrac{1}{36}\)
\(\left(x-\dfrac{1}{2}\right)^2=4\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=2^2\) hoặc \(\left(x-\dfrac{1}{2}\right)^2=\left(-2\right)^2\)
\(x-\dfrac{1}{2}=2\) hoặc \(x-\dfrac{1}{2}=-2\)
\(x=\dfrac{5}{2}\) hoặc \(x=\dfrac{-3}{2}\)
Chúc bạn học tốt!
c: ĐKXĐ: x<>-1
Để C là số nguyên thì \(2x-7⋮x+1\)
=>\(2x+2-9⋮x+1\)
=>\(x+1\in\left\{1;-1;3;-3;9;-9\right\}\)
=>\(x\in\left\{0;-2;2;-4;8;-10\right\}\)
d: ĐKXĐ: x<>-3
Để D là số nguyên thì \(5x+9⋮x+3\)
=>\(5x+15-6⋮x+3\)
=>\(x+3\inƯ\left(-6\right)\)
=>\(x+3\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
=>\(x\in\left\{-2;-4;-1;-5;0;-6;3;-9\right\}\)
\(\left(x+\dfrac{1}{3}\right)\times\dfrac{9}{14}\times\dfrac{7}{3}-\dfrac{1}{3}=1:\dfrac{9}{5}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)\times\dfrac{3}{2}-\dfrac{1}{3}=\dfrac{5}{9}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)\times\dfrac{3}{2}=\dfrac{5}{9}+\dfrac{1}{3}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)\times\dfrac{3}{2}=\dfrac{8}{9}\\ \Rightarrow x+\dfrac{1}{3}=\dfrac{8}{9}:\dfrac{3}{2}\\ \Rightarrow x+\dfrac{1}{3}=\dfrac{16}{27}\\ \Rightarrow x=\dfrac{16}{27}-\dfrac{1}{3}\\ \Rightarrow x=\dfrac{7}{27}\)
y x 2/3 + 1/3 = 8/9 : 2
Y x 2/3 +1/3 = 4/9
Y x 1/3 = 4/9 - 1/3
y x 1/3 = 1/9
Y = 1/9 : 1/3
y = 1/3
y x 2/3 + 1/3 = 8/9 : 2
y x 2/3 + 1/3 = 4/9
y = (4/9 - 1/3) : 2/3
y = 1/6
( dùng thứ lại biểu thức cộng trừ trước nhân chia sau)
150 : \(x\) = 9 - (-1)
150 : \(x\) = 9 + 1
150 : \(x\) = 10
\(x\) = 150 : 10
\(x\) = 15
a) \(\left|4x-1\right|-\left|3x-\dfrac{1}{2}\right|=0\\ \Leftrightarrow\left|4x-1\right|=\left|3x-\dfrac{1}{2}\right|\\ \Leftrightarrow\left[{}\begin{matrix}4x-1=3x-\dfrac{1}{2}\\4x-1=\dfrac{1}{2}-3x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}4x-3x=1-\dfrac{1}{2}\\4x+3x=\dfrac{1}{2}+1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\7x=\dfrac{3}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{14}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{1}{2};\dfrac{3}{14}\right\}\) là nghiệm của pt.
b) \(\left|x-1\right|-2x=\dfrac{1}{2}\\ \Leftrightarrow\left|x-1\right|=2x+\dfrac{1}{2}\left(ĐK:x\ge\dfrac{-1}{4}\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2x+\dfrac{1}{2}\\x-1=-2x-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-2x=1+\dfrac{1}{2}\\x+2x=1-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-x=\dfrac{3}{2}\\3x=\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\left(ktmđk\right)\\x=\dfrac{1}{6}\left(tmđk\right)\end{matrix}\right.\)
Vậy \(x=\dfrac{1}{6}\) là nghiệm của pt.
Lời giải:
a.
$|4x-1|-|3x-\frac{1}{2}|=0$
$\Leftrightarrow |4x-1|=|3x-\frac{1}{2}$
\(\Leftrightarrow \left[\begin{matrix} 4x-1=3x-\frac{1}{2}\\ 4x-1=\frac{1}{2}-3x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{1}{2}\\ x=\frac{3}{14}\end{matrix}\right.\)
b. Nếu $x\geq 1$ thì:
$|x-1|-2x=\frac{1}{2}$
$\Leftrightarrow x-1-2x=\frac{1}{2}$
$\Leftrightarrow -x-1=\frac{1}{2}$
$\Leftrightarrow x=\frac{-3}{2}$ (vô lý vì $x\geq 1$)
Nếu $x< 1$ thì:
$1-x-2x=\frac{1}{2}$
$\Leftrightarrow x=\frac{1}{6}$ (tm)
9:(x+2)2+9=9,25
⇒9:(x+2)2+9=9,25
⇒9:(x+2)2+9=9,25
⇒9:(x+2)2=0,25
⇒(x+2)2=36
⇒hoặc x+2=√36⇒x+2=6⇒x=4
x+2=-√36⇒x+2=-6⇒x=-8
vậy x={4;-8}
Ta có: \(\dfrac{9}{\left(x+2\right)^2}+9=9.25\)
\(\Leftrightarrow\dfrac{9}{\left(x+2\right)^2}=0.25\)
\(\Leftrightarrow\left(x+2\right)^2=36\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=6\\x+2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)
\(\left(x+5\right)^8=9\left(x+5\right)^6\Leftrightarrow\left(x+5\right)^8-9\left(x+5\right)^6=0\)
\(\Leftrightarrow\left(x+5\right)^6\left[\left(x+5\right)^2-9\right]=0\Leftrightarrow\left(x+5\right)^6\left(x+2\right)\left(x+8\right)=0\Leftrightarrow x=-5;x=-2;x=-8\)
Ta có: \(9^{x-1}=9\)
=>\(9^{x-1}=9^1\)
=>x-1=1
=>x=1+1=2
\(9^{x-1}=9\)
\(x-1=1\)
\(x=1+1\)
\(x=2\)