\(b,6x^3-8=40\)
\(a,5^{x+1}:5=5^4\)
K
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20 tháng 2 2021
Bài 1:
a) Ta có: \(x\left(x^2-4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;2;-2\right\}\)
b) Ta có: \(\left(2x-3\right)+\left(-3x\right)-\left(x-5\right)=40\)
\(\Leftrightarrow2x-3-3x-x+5=40\)
\(\Leftrightarrow-2x+2=40\)
\(\Leftrightarrow-2x=38\)
hay x=-19
Vậy: x=-19
Bài 2:
a) Ta có: \(-45\cdot12+34\cdot\left(-45\right)-45\cdot54\)
\(=-45\cdot\left(12+34+54\right)\)
\(=-45\cdot100\)
\(=-4500\)
b) Ta có: \(43\cdot\left(57-33\right)+33\cdot\left(43-57\right)\)
\(=43\cdot57-43\cdot33+43\cdot33-33\cdot57\)
\(=43\cdot57-33\cdot57\)
\(=57\cdot\left(43-33\right)\)
\(=57\cdot10=570\)
15 tháng 3 2021
Bài 1:
c) ĐKXĐ: \(x\notin\left\{\dfrac{1}{4};-\dfrac{1}{4}\right\}\)
Ta có: \(\dfrac{3}{1-4x}=\dfrac{2}{4x+1}-\dfrac{8+6x}{16x^2-1}\)
\(\Leftrightarrow\dfrac{-3\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}=\dfrac{2\left(4x-1\right)}{\left(4x+1\right)\left(4x-1\right)}-\dfrac{6x+8}{\left(4x-1\right)\left(4x+1\right)}\)
Suy ra: \(-12x-3=8x-2-6x-8\)
\(\Leftrightarrow-12x-3-2x+10=0\)
\(\Leftrightarrow-14x+7=0\)
\(\Leftrightarrow-14x=-7\)
\(\Leftrightarrow x=\dfrac{1}{2}\)(nhận)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
3 tháng 6 2021
Giải:
a) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(x=\dfrac{-13}{12}\)
b) \(2.\left(x-\dfrac{1}{3}\right)=\left(\dfrac{1}{3}\right)^2+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{1}{9}+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{2}{3}\)
\(x-\dfrac{1}{3}=\dfrac{2}{3}:2\)
\(x-\dfrac{1}{3}=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}+\dfrac{1}{3}\)
\(x=\dfrac{2}{3}\)
c) \(\left|2x-\dfrac{3}{4}\right|-\dfrac{3}{8}=\dfrac{1}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{8}+\dfrac{3}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{1}{2}\\2x-\dfrac{3}{4}=\dfrac{-1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{1}{8}\end{matrix}\right.\)
d) \(\dfrac{2}{3}x+\dfrac{1}{6}x=3\dfrac{5}{8}\)
\(x.\left(\dfrac{2}{3}+\dfrac{1}{6}\right)=\dfrac{29}{8}\)
\(x.\dfrac{5}{6}=\dfrac{29}{8}\)
\(x=\dfrac{29}{8}:\dfrac{5}{6}\)
\(x=\dfrac{87}{20}\)
AH
Akai Haruma
Giáo viên
4 tháng 8 2021
Bài 2:
a.
$(6x+1)^2+(6x-1)^2-2(6x+1)(6x-1)$
$=[(6x+1)-(6x-1)]^2=2^2=4$
b.
$3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$
$=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$
$=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)$
$=(2^8-1)(2^8+1)(2^{16}+1)$
$=(2^{16}-1)(2^{16}+1)=2^{32}-1$
c.
$2C=(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^{16}+1)$
$=(5^4-1)(5^4+1)(5^8+1)(5^{16}+1)$
$=(5^8-1)(5^8+1)(5^{16}+1)$
$=(5^{16}-1)(5^{16}+1)=5^{32}-1$
$\Rightarrow C=\frac{5^{32}-1}{2}$
\(6x^3-8=40\\ 6x^3=40+8=48\\x^3=\dfrac{48}{6}=8=2^3\\ Vậy:x=2\\ ---\\ 5^{x+1}:5=5^4\\5^{x+1}:5^1=5^4\\ 5^{x+1-1}=5^4\\ 5^x=5^4\\ Vậy:x=4\)
\(a,\) \(6x^3-8=40\)
\(6x^3=40+8\)
\(6x^3=48\)
\(x^3=8\)
\(x^3=8\)
\(x^3=2^3\)
\(x=2\)
Vậy \(x=2\)
\(b,\) \(5^{x+1}:5=5^4\)
\(5^{x+1-1}=5^4\)
\(5^x=5^4\)
\(x=4\)
Vậy \(x=4\)