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29 tháng 9 2018
a, \(\left(x-2\right)^{2n+1}=1\)
\(\text{Vì 2n+1 lẻ }\Rightarrow x-2=1\)
\(\Rightarrow x=3\)
b,\(\left(\left(x-2\right)^5\right)^7=1\)
\(\Leftrightarrow\left(x-2\right)^{5.7}=1\)
\(\Leftrightarrow\left(x-2\right)^{35}=1\)
\(\Leftrightarrow x-2=1\)
\(\Leftrightarrow x=3\)
c, \(\left(x-1\right)^{5^2}=1\)
\(\Leftrightarrow\left(x-1\right)^{10}=1\)
\(\text{Do mũ chẵn nên có 2 trường hợp}\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}}\)
T
29 tháng 9 2018
Vì số 1 với bất kì số mũ nào cũng là chính nó nên ta dễ dàng giải bài tập trên
a) \(\left(x-2\right)^{2n+1}=1\)
\(\Leftrightarrow\left(x-2\right)^{2n+1}=1^{2n+1}\)
\(\Leftrightarrow x-2=1\Leftrightarrow x=3\)
b) \(\left(\left(x-2\right)^5\right)^7=1\Leftrightarrow\left(\left(x-2\right)^5\right)^7=1^7\)
\(\Leftrightarrow\left(x-2\right)^5=1\Leftrightarrow\left(x-2\right)^5=1^5\)
\(\Leftrightarrow x-2=1\Leftrightarrow x=3\)
c) \(\left(x-1\right)^{5^2}=1\Leftrightarrow\left(x-1\right)^{25}=1\)
\(\Leftrightarrow\left(x-1\right)^{25}=1^{25}\Leftrightarrow x-1=1\Leftrightarrow x=2\)
P/s: Bài này dễ,không cần phải làm như bạn đâu! Với lại bạn sai câu c) đấy. Ở đây là lũy thừa tầng (có dạng \(x^{m^n}\)) nó khác với dạng (xm)n. Mong bạn phân biệt rõ hai cái này! Dù sao cũng khuyến khích bạn tham gia hỏi đáp nhiều để tiếp các giáo viên và các CTV mình cho bạn ! Cảm ơn bạn đã tham gia hỏi đáp!
16 tháng 8 2019
1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)
=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)
b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c) TT
16 tháng 8 2019
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)
=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)
=> \(\left|50x-140\right|=\left|25x+24\right|\)
=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)
Bài 2 : a. |2x - 5| = x + 1
TH1 : 2x - 5 = x + 1
=> 2x - 5 - x = 1
=> 2x - x - 5 = 1
=> 2x - x = 6
=> x = 6
TH2 : -2x + 5 = x + 1
=> -2x + 5 - x = 1
=> -2x - x + 5 = 1
=> -3x = -4
=> x = 4/3
Ba bài còn lại tương tự
LB
22 tháng 1 2017
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NH
10 tháng 7 2019
B1:
Ta có: a - b = ab => a = ab + b = b(a + 1)
Thay a = b(a + 1) vào a - b = a : b ta có: \(a-b=\frac{b\left(a+1\right)}{b}=a+1\)
=> a - b = a + 1 => a - a - b = 1 => -b = 1 => b = -1
Lại có: ab = a - b
<=> a x (-1) = a - (-1) <=> -a = a + 1 <=> -a - a = 1 <=> -2a = 1 <=> a = -1/2
Vậy...
B2:
a, \(3y\left(y-\frac{2}{5}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3y=0\\y-\frac{2}{5}=0\end{cases}\Rightarrow\orbr{\begin{cases}y=0\\y=\frac{2}{5}\end{cases}}}\)
b, \(7\left(y-1\right)+2y\left(y-1\right)=0\)
\(\Rightarrow\left(y-1\right)\left(7+2y\right)=0\)
\(\Rightarrow\orbr{\begin{cases}y-1=0\\7+2y=0\end{cases}\Rightarrow}\orbr{\begin{cases}y=1\\2y=7\end{cases}\Rightarrow}\orbr{\begin{cases}y=1\\y=\frac{7}{2}\end{cases}}\)
B3: \(K=\frac{-2}{3}+\frac{3}{4}-\frac{-1}{6}+\frac{-2}{5}\)
\(K=\left(-\frac{2}{3}+\frac{1}{6}\right)+\left(\frac{3}{4}-\frac{2}{5}\right)\)
\(K=\left(\frac{-4}{6}+\frac{1}{6}\right)+\left(\frac{15}{20}-\frac{8}{20}\right)\)
\(K=\frac{-1}{2}+\frac{7}{20}=\frac{-10}{20}+\frac{7}{20}=\frac{-3}{20}\)
Ta có: |b-5|+(b-5) luôn là số chẵn
=> 2a+7 chẵn, mà 7 lẻ => 2a lẻ => a=0
=> 2a+7=8=|b-5|+(b-5)
xét b lớn hơn hoặc = 5, ta có:
|b-5|+(b-5)=b-5+b-5=2b-10=8 => 2b=18 => b=9 (TM)
xét b bé hơn 5
|b-5|+(b-5)=-b+5+b-5=0=8(vô lí)
Vậy a=0, b=8