Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\left(x-5\right)^{12}=\left(x-5\right)^{10}\)
\(\Rightarrow\left(x-5\right)^{12}-\left(x-5\right)^{10}=0\)
\(\Rightarrow\left(x-5\right)^{10}\left[\left(x-5\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-5\right)^{10}=0\\\left(x-5\right)^2-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\left(x-5\right)^{10}=0^{10}\\\left(x-5\right)^2=0+1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\\left(x-5\right)^2=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0+5\\\left(x-5\right)^2=1^2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x-5=\pm1\end{cases}}\)
\(\Rightarrow x=5;\orbr{\begin{cases}x-5=1\\x-5=-1\end{cases}}\)
\(\Rightarrow x=5;\orbr{\begin{cases}x=1+5\\x=-1+5\end{cases}}\)
\(\Rightarrow x=5;\orbr{\begin{cases}x=4\\x=6\end{cases}}\)
Vậy x = 4 hoặc x = 5 hoặc x = 6
\(a)\left(x-5\right)^{12}=\left(x-5\right)^{10}\)
\(\Leftrightarrow\left(x-5\right)^{12}-\left(x-5\right)^{10}=0\)
\(\Leftrightarrow\left(x-5\right)^{10}\left[\left(x-5\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-5\right)^{10}=0\\\left(x-5\right)^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\\left(x-4\right)\left(x-6\right)=0\end{cases}}\)
[ ra \(\left(x-4\right)\left(x-6\right)\)do \(\left(x-5\right)^2-1=\left(x-5-1\right)\left(x-5+1\right)=\left(x-6\right)\left(x-4\right)\)]
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=4;x=6\end{cases}}\)
_Minh ngụy_
a) \(\left|x\right|-2=5\Leftrightarrow\left|x\right|=7\Leftrightarrow x=\pm7\)
b) \(\left|x-2\right|=5\Leftrightarrow\orbr{\begin{cases}x-2=5\\x-2=-7\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-5\end{cases}}}\)
c) \(2\left(x+7\right)=-16\Leftrightarrow x+7=8\Leftrightarrow x=1\)
d) \(\left(x-5\right)\left(x+7\right)=0\Leftrightarrow\orbr{\begin{cases}x-5=0\\x+7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-7\end{cases}}}\)
a, \(\left(x+1982+172\right)+\left(-1982\right)-162\)
\(=x+1982+172-1982-162=x+10\)
b, \(2x-\left[\left(x+b-c\right)-\left(x+a-2\right)\right]+\left(2b+c-3\right)\)
\(=2x-\left(x+b-c-x-a+2\right)+2b+c-3\)
\(=2x-\left(-a+b-c+2\right)+2b+c-3\)
\(=2x+a-b+c-2+2b+c-3=2x+a+b+2c-5\)
c, \(235+x-\left(65+x\right)+x=235+x-65-x+x=170+x\)
d, \(\left(a+b+1\right)-\left(a-c+1\right)-\left(b+c\right)\)
\(=a+b+1-a+c-1-b-c=0\)
e, \(a-\left(b+c-d\right)+\left(-d\right)-a=a-b-c+d-d-a=-b-c\)
f, \(\left(a+b-c-2019\right)-\left(c-b+a-2020\right)+c\)
\(=a+b-c-2019-c+b-a+2020+c=2b+1\)
\(a,-7624+\left(1543+7624\right)-x=25\)
\(\Leftrightarrow-7624+9167-x=25\)
\(\Leftrightarrow1543-x=25\)
\(\Leftrightarrow x=1518\)
Vậy ................
\(b,\left(27-514\right)-\left(486-73\right)+x=7\)
\(\Leftrightarrow-487-413+x=7\)
\(\Leftrightarrow-900+x=7\)
\(\Leftrightarrow x=907\)
\(c,14-12+x=10-\left|-15\right|+\left|-3\right|\)
\(\Leftrightarrow2+x=10-15+3\)
\(\Leftrightarrow2+x=-2\)
\(\Leftrightarrow x=-4\)
Vậy ....
a) - 7624 + (1543 + 7624) - x = 25
-7624 + 9167 - x = 25
1543 - x = 25
x = 1543 - 25
x = 1518
b) (27 - 514) - (486 - 73) + x = 7
(-487) - 413 + x = 7
-900 + x = 7
x = 7 - (-900)
x = 907
c) 14 - 12 + x = 10 - |-15| + |-3|
14 - 12 + x = 10 - 15 + 3
14 - 12 + x = (-5) + 3
14 - 12 + x = -2
2 + x = -2
x = (-2) - 2
x = -4
A=|x-3|+2020
|x-3|≥0 với mọi x=> dấu ''='' xẩy ra khi x=3
Vậy Min A là 2020
B=(x+2)²-2019
(x-2)²≥0 với mọi x=> dấu ''='' xẩy ra khi x=2
Vậy Min B =-2019
\(a)\)
\(-\frac{7}{12}+\frac{11}{8}-\frac{5}{9}\)
\(=-\frac{42}{72}+\frac{99}{72}-\frac{40}{72}\)
\(=\frac{57}{72}-\frac{40}{72}\)
\(=\frac{17}{72}\)
\(b)\)
\(\frac{1}{7}-\frac{8}{7}:8-3:\frac{3}{4}.\left(-2\right)^3\)
\(=\frac{1}{7}-\frac{8}{7}.\frac{1}{8}-3.\frac{4}{3}.6\)
\(=\frac{1}{7}-\frac{1}{7}-4.6\)
\(=0-24\)
\(=-24\)
\(c)\)
\(1,4.\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right):2\frac{1}{5}\)
\(=\frac{7}{5}.\frac{15}{49}-\left(\frac{12}{15}+\frac{10}{15}\right):\frac{11}{5}\)
\(=\frac{3}{7}-\frac{22}{15}.\frac{5}{11}\)
\(=\frac{3}{7}-\frac{2}{3}\)
\(=\frac{9}{21}-\frac{14}{21}\)
\(=-\frac{5}{21}\)
\(A=\left|x+19\right|+\left|y-5\right|+2020\)
Ta có : \(\left|x+19\right|\ge0\forall x;\left|y-5\right|\ge0\forall y;2020>0\)
Suy ra : \(\left|x+19\right|+\left|y-5\right|+2020\ge2020\)
Dấu ''='' xảy ra : \(\hept{\begin{cases}x+19=0\\y-5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-19\\y=5\end{cases}}}\)
Vậy GTNN A = 2020 <=> x = -19 ; y = 5