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30 tháng 5 2016
cách 1:=> (x - 7)^(x+1)= (x-7)^(x+11)
TH1: x-7=0 => x=7 => 0^8=0^18 (TM)
TH2: x-7=1 => x=8 (TM)
TH3: x khác 7 và 8 => x+1=x+11 => vô lý => loại
KL: x = 7 hoặc x=8
30 tháng 5 2016
( x-7)^( x+1) - ( x-7)^(x+11) = 0
( x-7)^( x+1) - ( x-7)^(x+1)*x^10 = 0
( x-7)^( x+1) (1-x^10) = 0
tới đây dễ òi
AN
10 tháng 11 2018
1.a)\(2.x-\dfrac{5}{4}=\dfrac{20}{15}\)
\(\Leftrightarrow2.x=\dfrac{20}{15}+\dfrac{5}{4}=\dfrac{4}{3}+\dfrac{5}{4}=\dfrac{16+15}{12}=\dfrac{31}{12}\)
\(\Leftrightarrow x=\dfrac{31}{12}:2=\dfrac{31}{12}.\dfrac{1}{2}=\dfrac{31}{24}\)
b)\(\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{8}\right)\)
\(\Leftrightarrow\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{2}\right)^3\)
\(\Leftrightarrow x+\dfrac{1}{3}=-\dfrac{1}{2}\)
\(\Leftrightarrow x=-\dfrac{1}{2}-\dfrac{1}{3}=-\dfrac{5}{6}\)
2.Theo đề bài, ta có: \(\dfrac{a}{2}=\dfrac{b}{3}\) và \(a+b=-15\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{a+b}{2+3}=\dfrac{-15}{5}=-3\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=-3\Rightarrow a=-6\\\dfrac{b}{3}=-3\Rightarrow b=-9\end{matrix}\right.\)
3.Ta xét từng trường hợp:
-TH1:\(\left\{{}\begin{matrix}x+1>0\\x-2< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>-1\\x< 2\end{matrix}\right.\)\(\Rightarrow x\in\left\{0;1\right\}\)
-TH2:\(\left\{{}\begin{matrix}x+1< 0\\x-2>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\)\(\Rightarrow x\in\varnothing\)
Vậy \(x\in\left\{0;1\right\}\)
4.\(B=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^9=\left(\dfrac{3}{7}\right)^{21}:\left[\left(\dfrac{3}{7}\right)^2\right]^9=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^{18}=\left(\dfrac{3}{7}\right)^3=\dfrac{27}{343}\)
C
21 tháng 12 2017
Giải:
a) \(5< 5^x < 625\)
\(\Leftrightarrow5< 5^x< 5^4\)
Vì \(5=5=5\)
Nên \(1< x< 4\)
\(\Leftrightarrow x\in\left\{2;3\right\}\)
Vậy ...
b) \(2^{x-1}=16\)
\(\Leftrightarrow2^{x-1}=2^4\)
Vì \(2=2\)
Nên \(x-1=4\)
\(\Leftrightarrow x=4+1=5\)
Vậy ...
c) \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)^{x+2}}{\left(x-1\right)^{x+2}}=\dfrac{\left(x-1\right)^{x+6}}{\left(x-1\right)^{x+2}}\)
\(\Leftrightarrow1=\left(x-1\right)^{x+4}\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=1+1=2\)
Vậy ...
29 tháng 8 2017
a) \(\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\) vậy \(x=1\)
b) \(\left(x-2\right)^2-1=0\Leftrightarrow\left(x-2\right)^2=1\) \(\Leftrightarrow\left\{{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\) vậy \(x=3;x=1\)
c) \(\left(2x-1\right)^3=-8\Leftrightarrow2x-1=\sqrt[3]{-8}\Leftrightarrow2x-1=-2\)
\(\Leftrightarrow2x=-1\Leftrightarrow x=\dfrac{-1}{2}\) vậy \(x=\dfrac{-1}{2}\)
d) \(\left(x+2\right)^2+1=0\Leftrightarrow\left(x+2\right)^2=-1\) (vô lí)
vậy phương trình vô nghiệm
29 tháng 8 2017
a) (x-1)2 = 0
<=> x-1 = 0
<=> x = 1
b) (x-2)2 - 1 = 0
<=> (x-2)2 = 1
<=> \(\left\{{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
c) (2x-1)3 = -8
<=> (2x-1)3 = -23
<=> 2x - 1 = -2
<=> 2x = -1
<=> x = \(-\dfrac{1}{2}\)
d) (x+2)2 + 1 = 0
<=> (x+2)2 = -1
<=> x+2 = -1
<=> x = -3
PT
3 tháng 8 2017
a) \(x+\dfrac{3}{10}=\dfrac{-2}{5}\)
\(x=\dfrac{-2}{5}-\dfrac{3}{10}\)
\(x=\dfrac{-7}{10}\)
b) \(x+\dfrac{5}{6}=\dfrac{2}{5}-\left(-\dfrac{2}{3}\right)\)
\(x+\dfrac{5}{6}=\dfrac{2}{5}+\dfrac{2}{3}\)
\(x+\dfrac{5}{6}=\dfrac{16}{15}\)
\(x=\dfrac{16}{15}-\dfrac{5}{6}\)
\(x=\dfrac{7}{30}\)
c) \(1\dfrac{2}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x=-\dfrac{4}{5}-\dfrac{3}{7}\)
\(\dfrac{7}{5}x=\dfrac{-43}{35}\)
\(\Rightarrow x=\dfrac{-43}{49}\)
d) \(\left[x+\dfrac{3}{4}\right]-\dfrac{1}{3}=0\)
\(\left[x+\dfrac{3}{4}\right]=0+\dfrac{1}{3}\)
\(\left[x+\dfrac{3}{4}\right]=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}-\dfrac{3}{4}\)
\(x=\dfrac{-5}{12}\)
e) \(\left[x+\dfrac{4}{5}\right]-\left(-3,75\right)=-\left(-2,15\right)\)
\(\left[x+\dfrac{4}{5}\right]+3,75=2,15\)
\(x+\dfrac{4}{5}=2,15-3,75\)
\(x+\dfrac{4}{5}=-\dfrac{8}{5}\)
\(x=\dfrac{-8}{5}-\dfrac{4}{5}\)
\(x=\dfrac{-12}{5}\)
f) \(\left(x-2\right)^2=1\)
\(\Rightarrow x=1\)
Sức chịu đựng có giới hạn -.-
NH
3 tháng 8 2017
- Mình tiếp tục cho Nguyễn Phương Trâm nhé.
g, \(\left(2x-1\right)^3=-27\)
\(\Rightarrow\left(2x-1\right)^3=\left(-3\right)^3\)
\(\Rightarrow2x-1=-3\)
\(\Rightarrow2x=-2\)
=> \(x=-1\)
- Vậy x = -1
h,\(\dfrac{x-1}{-15}=-\dfrac{60}{x-1}\)
\(\Rightarrow\left(x-1\right)^2=-60.\left(-15\right)\)
\(\Rightarrow\left(x-1\right)^2=900 \)
\(\Rightarrow\left(x-1\right)^2=30^2\Rightarrow x-1=30\)
=> x = 31
i,\(x:\left(\dfrac{-1}{2}\right)^3=\dfrac{-1}{2}\)
=> \(x:\left(-\dfrac{1}{8}\right)=-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{16}\)
- Vậy x=\(\dfrac{1}{16}\)
j, \(\left(\dfrac{3}{4}\right)^5.x=\left(\dfrac{3}{4}\right)^7\)
\(\Rightarrow \left(\dfrac{3}{4}\right).x=\left(\dfrac{3}{4}\right)^2\)
\(\Rightarrow x=\left(\dfrac{3}{4}\right)^2:\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{4}\)
- Vạy x = \(\dfrac{3}{4}\)
k, \(8^x:2^x=4\Rightarrow\left(8:2\right)^x=4\)
=>\(4^x=4\)
=> x = 1
- Vậy x = 1