Tìm x biết |x+1|+|x+5|=4
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\(ĐK1:2005< 5x-2\Leftrightarrow2007< 5x\Leftrightarrow x>401,4\)
\(\text{Đ}K2:5x-2< 2006\Rightarrow5x< 2008\Rightarrow x< 401,6\)
\(\Rightarrow401,4< x< 401,6\)
\(2005< 5x-2< 2006\)
\(2005+2< 5x< 2006+2\)
\(2007< 5x< 2008\)
\(\frac{2007}{5}< x< \frac{2008}{5}\)
\(\text{a)}X=\frac{a-5}{a}\in Z\Rightarrow\frac{a}{a}-\frac{5}{a}\in Z\Rightarrow1-\frac{5}{a}\in Z\)
\(\Rightarrow\frac{5}{a}\in Z\Rightarrow a\in\text{Ư}\left\{5\right\}\)
\(\Rightarrow a\in\left\{-5;-1;1;5\right\}\)
\(\text{b)}X\in Z^+\Rightarrow\orbr{\begin{cases}a-5>0;a>0\\a-5< 0;a< 0\end{cases}\Rightarrow\orbr{\begin{cases}a>0\\a< 5\end{cases}}\Rightarrow\orbr{\begin{cases}a\in\left\{1;5\right\}\\a\in\varnothing\end{cases}}}\)
\(\Rightarrow a\in\left\{1;5\right\}\text{thì }X\in Z^+\)
\(\text{c)}X\in Z^-\Rightarrow\orbr{\begin{cases}a-5>0;a< 0\\a-5< 0;a>0\end{cases}\Rightarrow\orbr{\begin{cases}a>5;a< 0\\a< 5;a>0\end{cases}\Rightarrow}\orbr{\begin{cases}a\in\varnothing\\a\in\left\{1\right\}\end{cases}}}\)
\(\Rightarrow a=1\text{thì }X\in Z^-\)
\(|x+1|+\left|2x-3\right|=\left|3x-2\right|\)
\(\Leftrightarrow|x+1|+\left|2x-3\right|=\left|2x-3\right|+\left|x+1\right|\)
Vậy \(x\in\left\{\infty;-\infty\right\}\)
Hok Tốt !!!!!!!!!!!!!!
\(A=\frac{2020}{9-x}\left(x\ne9\right)\)
Để A đạt GTLN thì 9-x bé nhất
=> 9-x=1
=> x=8
Vậy \(A_{max}=\frac{2020}{9-8}=2020\)tại x=8
Hok Tốt !!!!!!!!!!!!!!
\(A=\frac{2020}{9-x}\)
A đạt giá trị lớn nhất
\(\Leftrightarrow\frac{2020}{9-x}\) lớn nhất
\(9-x\) nhỏ nhất ( vì 2020 là hằng số )
Vì 9 - x khác 0
\(\Rightarrow9-x=1\)
\(x=9-1\)
\(x=8\)
\(A=\frac{2020}{9-x}=\frac{2020}{9-8}=2020\)
Vật Giá trị lớn nhất cả A là 2020 khi và chỉ khi x = 8
\(+,x=1\Rightarrow y=22-y\text{ nên }y=11\left(\text{thỏa mãn}\right)\)
\(+,x=2\Rightarrow2^y+y^2=23-2y\text{ nên }\left(y+1\right)^2+2^y=24\text{ do đó: }2^y=8;\left(y+1\right)^2=16\text{ hay }y=3\left(tm\right)\)
\(+,x\ge3\text{ thì: }x^y+y^x\le20;x^y+y^x\ge3^y+y^3\text{ nên }y=1\text{ hoặc }y=2\text{ thử lại thấy }y=1;x=11;y=2;x=3\)
1/ \(2^{24}=\left(2^3\right)^8=8^8\)
\(3^{16}=\left(3^2\right)^8=9^8>8^8\Rightarrow3^{16}>2^{24}\)
2/ \(n^{200}=\left(n^2\right)^{100}\)
\(5^{300}=\left(5^3\right)^{100}=125^{100}\)
Để \(\left(n^2\right)^{100}< 125^{100}\)thì \(n^2< 125\Rightarrow n_{max}=11\left(n^2=121\right)\)
3/ a) \(\left(0,125\right)^3.512=\frac{1}{8^3}.8^3=1\)
b) \(\left(0,25\right)^4.1024=\frac{1}{4^4}.4^5=4\)
4/ a) \(\left(2^2\right)^{\left(2^2\right)}=4^{2^2}=16^2=256\)
\(\left(-1\right)^{5^2}+1^{2^5}=\left[\left(-1\right)^2\right]^5+1=1^5+1=2\)
1)
\(2^{24}=\left(2^3\right)^8=8^8\)
\(3^{16}=\left(3^2\right)^8=9^8\)
Vậy \(2^{24}< 3^{16}\)
2)
\(n^{200}=\left(n^2\right)^{100}\)
\(5^{300}=\left(5^3\right)^{100}=125^{100}\)
\(\left(n^2\right)^{100}< 125^{100}\)
\(n^2< 125\)
Xét số lớn nhất nên mk chỉ lấy phần dương
\(n< 11,1\)
Vậy \(n\) lớn nhất = 11
3)
a.
\(0,125^3\cdot512\)
\(=\left(\frac{1}{8}\right)^3\cdot8^3\)
\(=\left(\frac{1}{8}\cdot8\right)^3\)
\(=1^3=1\)
b.
\(0,25^4\cdot1024\)
\(=\left(\frac{1}{4}\right)^4\cdot256\cdot4\)
\(=\left(\frac{1}{4}\right)^4\cdot4^4\cdot4\)
\(=\left(\frac{1}{4}\cdot4\right)^4\cdot4\)
\(=1^4\cdot4\)
\(=1\cdot4\)
\(=4\)
4)
a.
\(\left(2^2\right)^{\left(2^2\right)}\)
\(=4^4\)
\(=256\)
b.
\(\left(-1\right)^{5^2}+1^{2^5}\)
\(=-1^{25}+1^{32}\)
\(=1-1=0\)
Đặt \(\widehat{xOy}=a,\widehat{yOz}=b\Rightarrow a+b=180\)
a) \(\hept{\begin{cases}a+b=180\\3a+4b=640\end{cases}\Leftrightarrow}\hept{\begin{cases}a=80\\b=100\end{cases}}\)
b) \(\hept{\begin{cases}a+b=180\\a-b=20\end{cases}\Leftrightarrow\hept{\begin{cases}a=100\\b=80\end{cases}}}\)
c) \(\hept{\begin{cases}a+b=180\\2a-b=60\end{cases}\Leftrightarrow\hept{\begin{cases}a=80\\b=100\end{cases}}}\)
Bài giải
Ta có : \(\widehat{xOy}+\widehat{yOz}=180^o\)
a, Ta có :
\(3\widehat{xOy}+4\widehat{yOz}=3\left(\widehat{xOy}+\widehat{yOz}\right)+\widehat{yOz}=3\cdot180^o+\widehat{yOz}\)
\(=540^o+\widehat{yOz}=640^o\text{ }\Rightarrow\text{ }\widehat{yOz}=100^o\)
\(\Rightarrow\text{ }\widehat{xOy}=80^o\)
b, \(\widehat{xOy}= ( 180^o+20^o )\text{ : }2=100^o\)
\(\widehat{yOz}=100^o-20^o=80^o\)
c,
\(\widehat{xOy}+\widehat{yOz}=180^o\text{ }\Rightarrow\text{ }\widehat{xOy}=180^o-\widehat{yOz}\)
\(2\left(180^o-\widehat{yOz}\right)-\widehat{yOz}=360^o-3\widehat{yOz}=60^o\)
\(3\widehat{yOz}=300^o\)
\(\widehat{yOz}=100^o\)
\(\widehat{xOy}=80^o\)
\(|x+1|+|x+5|=4\)
\(|-\left(x+1\right)|+|x+5|=4\)
\(|-x-1|+|x+5|=4\)
Ta có :
\(|-x-1|+|x+5|\ge|-x-1+x+5|\)
\(|-x-1|+|x+5|\ge4\)
Dấu = xảy ra
\(\Leftrightarrow\left(-x-1\right)\left(x+5\right)\ge0\)
TH1
\(\hept{\begin{cases}-x-1\ge0\\x+5\ge0\end{cases}}\)
\(\hept{\begin{cases}-x\ge-1\\x\ge-5\end{cases}}\)
\(\hept{\begin{cases}x\le1\\x\ge-5\end{cases}}\) \(\Rightarrow-5\le x\le1\)
TH 2
\(\hept{\begin{cases}-x-1\le0\\x+5\le0\end{cases}}\)
\(\hept{\begin{cases}-x\le-1\\x\le-5\end{cases}}\)
\(\hept{\begin{cases}x\ge1\\x\le-5\end{cases}}\) \(\Rightarrow x=\varnothing\)
vậy \(-5\le x\le1\) là nghiệm