bài 1: tìm x, biết
a) |$\frac{1}{3}$x – $\frac{3}{4}$| – |-$\frac{4}{5}$| =-$\frac{1}{5}$
b) (2x-$5)^{4}$ = 81
d) (x+$3)^{5}$ = 32
e) ($\frac{2}{3}$)$^{x+1}$ = $\frac{16}{81}$
g) ($\frac{3}{4}$)$^{2}$ . ($\frac{3}{4}$)$^{3}$ . ($\frac{3}{4}$)$^{5}$ . ($\frac{3}{4}$) $^{0}$ = ($\frac{3}{4}$)$^{x-2}$
bài 1: tìm x, biết a) |$\frac{1}{3}$x – $\frac{3}{4}$| – |-$\frac{4}{5}$| =-$\frac{1}{5}$ b) (2x-$5)^{4}$ = 81 d) (x+$3)^{5}$ = 32 e) ($\frac{2}{3}$
By Athena
Đáp án:
a, `|1/3 x – 3/4| – |-4/5| = -1/5`
`<=> |1/3 x – 3/4| – 4/5 = -1/5`
`<=> |1/3 x – 3/4| = -1/5 + 4/5`
`<=> |1/3 x – 3/4| = 3/5`
<=> \(\left[ \begin{array}{l}1/3 x – 3/4 = 3/5 \\1/3 x – 3/4 = -3/5\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x= 81/20\\x=9/20\end{array} \right.\)
b, `(2x – 5)^4 = 81`
`<=> (2x – 5)^4 = 3^4`
<=> \(\left[ \begin{array}{l}2x – 5 = 3\\2x – 5 = -3\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=4\\x=1\end{array} \right.\)
d, `(x + 3)^5 = 32`
`<=> (x + 3)^5 = 2^5`
<=> \(\left[ \begin{array}{l}x + 3 = 2\\x + 3 = -2\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=-1\\x=-5\end{array} \right.\)
e, `(2/3)^{x+1} = 16/81`
`<=> (2/3)^{x+1} = (2/3)^4`
`<=> x + 1 = 4`
`<=> x = 3`
g, `(3/4)^2 . (3/4)^3 . (3/4)^5 . (3/4)^0 = (3/4)^{x – 2}`
`<=> (3/4)^{10} = (3/4)^{x – 2}`
`<=> x – 2 = 10`
`<=> x = 12`
Giải thích các bước giải:
Đáp án:
$a) {\left[\begin{aligned}x=\dfrac{81}{20}\\x=\dfrac{9}{20}\end{aligned}\right.}\\
b) {\left[\begin{aligned}x=4\\x=1\end{aligned}\right.}\\
d) x=-1\\
e) x=3\\
g)
x=12$
Giải thích các bước giải:
$a) \left | \dfrac{1}{3}x-\dfrac{3}{4} \right |-\left | -\dfrac{4}{5} \right |=-\dfrac{1}{5}\\
\Leftrightarrow \left | \dfrac{1}{3}x-\dfrac{3}{4} \right |-\dfrac{4}{5}=\dfrac{-1}{5}\\
\Leftrightarrow \left | \dfrac{1}{3}x-\dfrac{3}{4} \right |=\dfrac{-1}{5}+\dfrac{4}{5}\\
\Leftrightarrow \left | \dfrac{1}{3}x-\dfrac{3}{4} \right |=\dfrac{3}{5}\\
\Leftrightarrow {\left[\begin{aligned}\dfrac{1}{3}x-\dfrac{3}{4}=\dfrac{3}{5}\\\dfrac{1}{3}x-\dfrac{3}{4}=\dfrac{-3}{5}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}\dfrac{1}{3}x=\dfrac{3}{5}+\dfrac{3}{4}\\\dfrac{1}{3}x=\dfrac{-3}{5}+\dfrac{3}{4}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}\dfrac{1}{3}x=\dfrac{12}{20}+\dfrac{15}{20}\\\dfrac{1}{3}x=\dfrac{-12}{20}+\dfrac{15}{20}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}\dfrac{1}{3}x=\dfrac{27}{20}\\\dfrac{1}{3}x=\dfrac{3}{20}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=3.\dfrac{27}{20}\\x=3.\dfrac{3}{20}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{81}{20}\\x=\dfrac{9}{20}\end{aligned}\right.}\\
b) (2x-5)^4=81\\
\Leftrightarrow (2x-5)^4=3^4\\
\Leftrightarrow 2x-5=\pm 3\\
\Leftrightarrow {\left[\begin{aligned}2x-5=3\\2x-5=-3\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}2x=8\\2x=2\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=4\\x=1\end{aligned}\right.}\\
d) (x+3)^5=32\\
\Leftrightarrow (x+3)^5=2^5\\
\Leftrightarrow x+3=2\\
\Leftrightarrow x=2-3=-1\\
e) \left ( \dfrac{2}{3} \right )^{x+1}=\dfrac{16}{81}\\
\Leftrightarrow \left ( \dfrac{2}{3} \right )^{x+1}=\left (\dfrac{2}{3} \right )^4\\
\Leftrightarrow x+1=4\\
\Leftrightarrow x=4-1=3\\
g)
\left ( \dfrac{3}{4} \right )^2.\left ( \dfrac{3}{4} \right )^3.\left ( \dfrac{3}{4} \right )^5.\left ( \dfrac{3}{4} \right )^0=\left (\dfrac{3}{4} \right )^{x-2}\\
\Leftrightarrow \left (\dfrac{3}{4} \right )^{10}=\left (\dfrac{3}{4} \right )^{x-2}\\
\Leftrightarrow 10=x-2\\
\Leftrightarrow x=10+2=12$