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Điều kiện 0=</x/<=11; 0=</y/<7
Ta có: \(IyI=\frac{33-3.IxI}{5}=\frac{3\left(11-IxI\right)}{5}\)
=> Để y nguyên thì (11-/x/) phải chia hết cho 5 và 0</y/<7
=> /x/=(1, 6, 11) => x=(-1,1,-6,6,-11,11)
Và /y/=(6,3,0) => y=(-6,6,-3,3,0)
Các cặp (x,y) nguyên là: (-1,-6); (-1,6); (-6,-3); (-6,3); (-11,0) và (11,0)
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1, <=>33+x=45
<=> x = 12
Vậy x=12
2, <=>x+73 =102
<=> x =29
3,x=2
4, -23+x=25<=>x=2
6, x-8=20<=>x=28
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4x-2x+x-27:9=33
2x+x-27:9=33
3x-27:9=33
3x-27=33×9=297
3x=297+27=324
x=324÷3=108
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Bài 1 :
a) x={2,4}
b) x-1={-3,-2,-1,0,1,2,3,4}
=> x={-2,-1,0,1,2,3,4,5}
c) x+2={-7,-6,-5,-4}
=> x={-9,-8,-7,-6}
Bài 2 :
(x-3)(x+2)=0
=> x-3=0 => x=3
=> x+2=0 => x=-2
Vậy x=-2 hoặc x=3
BÀI 1
A) 3<X<5
=>X=4
B) -4<X+2<5
=>X-1\(\in\left(-3;-2;-1;0;1;2;3;4\right)\)
=> X-1=-3 => X-1=-2 =>X-1=-1 =>X-1=0 => X-1=1
X=-2 X=-1 X= 0 X=1 X=2
=>X-1=2 => X-1=3 =>X-1=4
X=3 X=4 X=5
C) -8<X+2<-3
=> X+2\(\in\left(-7;-6;-5;-4\right)\)
=> X+2=-7 =>X+2=-6 =>X+2=-5 =>X+2=-4
X=-9 X=-8 X=-7 X=-6
BÀI 2
\(\left(X-3\right).\left(X+2\right)=0\)
\(\Rightarrow X-3=X+2=O\)
\(TH1:X-3=0\)
X=3
TH2: X+2=0
X=-2
VẬY X=3 HOẶC X=-2
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\(\Leftrightarrow-\dfrac{2}{9}+\dfrac{2}{3}-\dfrac{4}{9}>=x>=\dfrac{3}{7}-\dfrac{7}{5}+\dfrac{11}{7}+\dfrac{2}{5}\)
=>0>=x>=1
=>\(x\in\varnothing\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
a: \(x\in\left\{-5;-4;-3;-2;-1\right\}\)
b: \(x\in\left\{-3;-2;-1;0;1;2;3;4;5;6\right\}\)
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a, x - 43 = ( 35 - x ) - 48
\(x-43=35-x-48\)
\(x+x=35-48+43\)
\(2x=30\)
\(x=15\)
b, 305 - x + 14 = 48 + ( x - 23 )
\(305-x+14=48+x-23\)
\(-x-x=48-23-14-305\)
\(-2x=-294\)
\(x=147\)
c, ( x - 6 + 85 ) = ( x + 51 ) - 54
\(x-6+85=x+51-54\)
\(x-x=51-54-85+6\)
\(0=-82\)
\(x\in\varnothing\)
d, -( 35 - x ) - ( 37 - x ) = 33 - x
\(-35+x-37+x=33-x\)
\(x+x+x=33+37+35\)
\(3x=105\)
\(x=35\)
e, - 12. ( x - 5 ) + 7 . ( 3 - x ) = 5
\(-12x+60+21-7x=5\)
\(-12x-7x=5-21-60\)
\(-19x=-76\)
\(x=4\)
f, 30. ( x+ 2 ) - 6 . ( x - 5 ) - 24 = 100
\(30x+60-6x+30-24=100\)
\(30x-6x=100+24-30-60\)
\(24x=34\)
\(x=\frac{17}{12}\)
chúc bạn học tốt
Vì \(\hept{\begin{cases}\left|x\right|< 3\\3\le\left|x\right|\le5\end{cases}}\)
\(\Rightarrow\left|x\right|\in\left\{0;1;2\right\}\)
Vậy \(x\in\left\{0;-1;1;2;-2\right\}\)
\(\left|x\right|< 3\)
\(\Leftrightarrow-3< x< 3\)
Vậy \(-3< x< 3\)
\(3\le\left|x\right|\le5\)
\(\Leftrightarrow\hept{\begin{cases}\left|x\right|\le5\\\left|x\right|\ge3\end{cases}}\Leftrightarrow\hept{\begin{cases}-5< x< 5\\x>3ho\text{ặ}cx< -3\end{cases}}\Leftrightarrow-5< x< 5\)
Vậy \(-5< x< 5\)