Tìm x,biết
( x + 3 ) + ( x + 5 ) + ( x + 7 ) - ( x + 9 ) = 52
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a, X = 273
b, X = 156
c, X = 4
d, X = 505
e, X = \(\frac{37}{245}\)
c) \(\left|x\right|=3,5\Rightarrow\left[{}\begin{matrix}x=3,5\\x=-3,5\end{matrix}\right.\)
d) \(\left|x\right|=-2,7\Rightarrow x\in\varnothing\)
l) \(\left|x+\dfrac{3}{4}\right|-5=-2\Rightarrow\left|x+\dfrac{3}{4}\right|=3\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=3\\x+\dfrac{3}{4}=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3-\dfrac{3}{4}\\x=-3-\dfrac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=\dfrac{15}{4}\end{matrix}\right.\)
Đính chính câu l \(x=-\dfrac{15}{4}\) không phải \(x=\dfrac{15}{4}\)
Cần bổ sung điều kiện \(x;y\inℤ\)
a) \(\left(x-7\right)\left(xy+1\right)=9=1\cdot9=3\cdot3=\left(-1\right)\cdot\left(-9\right)=\left(-3\right)\cdot\left(-3\right)\)
Xét bảng :
x-7 | 1 | 9 | -1 | -9 | 3 | -3 |
xy+1 | 9 | 1 | -9 | -1 | 3 | -3 |
x | 8 | 16 | 6 | -2 | 10 | 4 |
y | 1 | 0 | -1,(6) | 1 | 0,2 | -1 |
Vì x,y thuộc Z nên ta có (x;y)={(8;1),(16;0),(-2;1),(4;-1)
b) \(\left(x+5\right)\left(3x-12\right)>0\)
TH1 : \(\hept{\begin{cases}x+5>0\\3x-12>0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-5\\x>4\end{cases}\Leftrightarrow x>4}}\)
TH2 : \(\hept{\begin{cases}x+5< 0\\3x-12< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< -5\\x< 4\end{cases}\Leftrightarrow x< -5}}\)
Vậy....
\(1,\\ \left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\\ \Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)
\(2,\\ a,\left|2x-3\right|>5\Leftrightarrow\left[{}\begin{matrix}2x-3< -5\\2x-3>5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -1\\x>4\end{matrix}\right.\\ b,\left|3x-1\right|\le7\Leftrightarrow\left[{}\begin{matrix}3x-1\le7\\1-3x\le7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le\dfrac{8}{3}\\x\ge-2\end{matrix}\right.\\ c,\cdot x< -\dfrac{3}{2}\\ \Leftrightarrow5-3x+\left(-2x-3\right)=7\Leftrightarrow2-5x=7\Leftrightarrow x=-1\left(ktm\right)\\ \cdot-\dfrac{3}{2}\le x\le\dfrac{5}{3}\\ \Leftrightarrow\left(5-3x\right)+\left(2x+3\right)=7\Leftrightarrow8-x=7\Leftrightarrow x=1\left(tm\right)\\ \cdot x>\dfrac{5}{3}\\ \Leftrightarrow\left(3x-5\right)+\left(2x+3\right)=7\Leftrightarrow5x-2=7\Leftrightarrow x=\dfrac{9}{5}\left(tm\right)\\ \Leftrightarrow S=\left\{1;\dfrac{9}{5}\right\}\)
a) \(19\times\left(52+38\right)-9\times\left(52+38\right)=\)
\(=\) \(19\times90-9\times90\)
\(=90\times\left(19-9\right)\)
\(=90\times10\)
\(=900\)
b) \(\dfrac{2}{5}\div\dfrac{3}{7}\times\dfrac{3}{7}\div\dfrac{2}{5}+2019=\)
\(=\left(\dfrac{2}{5}\div\dfrac{2}{5}\right)\times\left(\dfrac{3}{7}\div\dfrac{3}{7}\right)+2019\)
\(=1\times1+2019\)
\(=2019\)
Bài 5 :
S = 1 + 3 - 5 - 7 + 9 + 11 - ... - 397 - 399
S = 1 + (3 - 5 - 7 + 9) + (11 - 13 - 15 + 17) + ... + (387 - 389 - 391 + 393) + (395 - 397 - 399)
S = 1 + 0 + 0 + ... + 0 + (- 401)
S = 1 - 401
S = - 400
Bài 5
A= 1+3-5-7+9+11-13-15+...-397-399
A= ( 1+3-5-7)+( 9+11-13-15)+...+( 393+395-397-399)
A= -8 -8 -...-8
A = -8.50 ( từ 1 đến 399 có 200 số, chia làm 4 cặp)
A= -400
a)x-63:9=18
x-63 =18.9
x-63 =162
x =162+63
x =225
b)3(29-x)+52=103
3(29-x) =103-52
3(29-x) =51
29-x =51:3
29-x =17
x =12
c)|x-1|=4
=>x-1=4 hoặc x-1= -4
=>x=5 hoặc x= -3
d)(-7)-x= -21
x=(-7)-(-21)
x=14
a) x-63;9=18
x-7=18
x=18+7=25
b)3(29-x)=103-52
87-3x=51
-3x=51-87=-36
x=12
c)
* x-1 =4
x=5
* x-1=-4
x=-3
d)(-7)-x=-21
-x=-21+7=14
x=14
a x - 4/5 = 7/25
x = 7/25 + 4/5
x = 27/25
b x :3/5 = 7/9
x = 7/9 x 3/5
x = 7/15
c x x 15 = 2850
x = 2850 : 15
x = 190
d x : 52 = 113
x = 113 x 52
x = 5876
a: x=7/25+4/5=7/25+20/25=27/25
b: x:3/5=7/9
nên x=7/9x3/5=21/45=7/16
c: 15x=2850
nên x=190
x = 23
x + 3 + x + 5 + x + 7 - x - 9 = 52
2x + 6 = 52
2x = 46
x = 23