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a) \(\frac{5}{6}\)x - \(\frac{3}{8}\)x - 10 = 12
=> \(\left(\frac{5}{6}-\frac{3}{8}\right)\)x = 12 + 10
=> \(\frac{11}{24}\)x = 22
=> x = 22 : \(\frac{11}{24}\)
=> x = 48
Vậy x = 48.
b) (\(\left|x\right|\) - \(\frac{1}{8}\)) . \(\left(-\frac{1}{8}\right)^5\) = \(\left(-\frac{1}{8}\right)^7\)
=> \(\left|x\right|\) - \(\frac{1}{8}\) = \(\left(-\frac{1}{8}\right)^7\) : \(\left(-\frac{1}{8}\right)^5\)
=> \(\left|x\right|\) - \(\frac{1}{8}\) = \(\left(-\frac{1}{8}\right)^{7-5}\)
=> \(\left|x\right|\) - \(\frac{1}{8}\) = \(\frac{1}{64}\)
=> \(\left|x\right|\) = \(\frac{1}{64}\) + \(\frac{1}{8}\)
=> \(\left|x\right|\) = \(\frac{9}{64}\)
=> x = \(\frac{9}{64}\) hoặc x = \(\frac{-9}{64}\)
Vậy x = \(\frac{9}{64}\) hoặc x = \(\frac{-9}{64}\)
hên quá làm đúng hì hì, cảm ơn nhen, hết sợ bị sai ồi
1) a) \(x^2=2x\Leftrightarrow x^2-2x=0\Leftrightarrow x\left(x-2\right)=0\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\) vậy \(x=0;x=2\)
b) \(x^3=x\Leftrightarrow x^3-x=0\Leftrightarrow x\left(x^2-1\right)=0\) \(\Leftrightarrow x\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+1=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-1\\x=1\end{matrix}\right.\) vậy \(x=0;x=-1;x=1\)
\(x^2=2x\Rightarrow x^2-2x=0\Rightarrow x\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-2=0\Rightarrow x=2\end{matrix}\right.\)
\(x^3=x\Rightarrow x^3-x=0\Rightarrow x\left(x^2-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-1=0\Rightarrow x^2=1\Rightarrow x=\pm1\end{matrix}\right.\)
\(A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)\left(\dfrac{1}{25}-1\right)...\left(\dfrac{1}{121}-1\right)\)
\(A=\dfrac{-3}{4}.\dfrac{-8}{9}.\dfrac{-15}{16}.\dfrac{-24}{25}...\dfrac{-120}{121}\)
\(A=\dfrac{3.8.15.24....120}{4.9.16.25...121}\)
\(A=\dfrac{1.3.2.4.3.5.4.6....10.12}{2.2.3.3.4.4.5.5....11.11}\)
\(A=\dfrac{1.2.4....10}{2.3.4.5...11}.\dfrac{3.4.5....12}{2.3.4.5....11}\)
\(A=\dfrac{1}{11}.6=\dfrac{6}{11}\)
3) Áp dụng tính chất:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(B=\dfrac{8^{2017}+1}{8^{2018}+1}< 1\)
\(B< \dfrac{8^{2017}+1+8}{8^{2018}+1+8}\)
\(B< \dfrac{8^{2017}+8}{8^{2018}+8}\)
\(B< \dfrac{8\left(8^{2016}+1\right)}{8\left(8^{2017}+1\right)}\)
\(B< \dfrac{8^{2016}+1}{8^{2017}+1}=A\)
\(B< A\)
a,(=)\(3^{x+1}.\left(3+4\right)=7.3^6\)
(=)\(3^{x+1}=3^6\)
=>x+1=6(=)x=5
b
làm cho 1 cái những cái sau tương tự mà lm nha bạn
\(\frac{x}{5}=-\frac{6}{7}\)
\(=>7x=-6\cdot5\)
\(7x=-30\)
\(x=-\frac{30}{7}\)
\(\frac{x}{2}=-\frac{8}{-x}\)
\(=>\frac{x}{2}=\frac{8}{x}\)
\(=>xx=8\cdot2\)
\(x^2=16\)
\(=>x\in\left\{-4;4\right\}\)
Ta có : x(x - 2) - x(x - 1) - 15 = 0
<=> x2 - 2x - x2 + x - 15 = 0
<=> -x - 15 = 0
=> -x = 15
=> x = -15
a, (2x+1)^2=49
=>2x+1 =7
2x =6
x=3
Vậy x=3.
b, x^2=\(\sqrt{\left(-4\right)^2}\)
=>x^2=4
=>x^2=2^2
=>x = 2
Vậy x=2.
-Xin lỗi bẹn nhoa phần c mình hem pít làm, mình chỉ pít làm 2 phần đó hoy à. Xin lỗi bẹn nhiều nhoa !