Tìm x biết :
a)3x+1.15=135
b)x+2.x+22.x+23.x+...+22016.x=22017-1
c)x.(x-1)+(x-1)2=0
d)4.32\(\le\)2x-5\(\le\)1024(x ϵ N)
Tìm số tự nhiên x biết :
a, 2.(x – 5)+7 = 77
b, x - 1 3 - 3 5 : 3 4 + 2 . 2 3 = 14
c, 1 + 2 + 2 2 + 2 3 + . . . + 2 2016 = 2 x - 1 - 1
d, 5 2 x - 3 - 2 . 5 2 = 5 2 . 3
a, 2.(x – 5)+7 = 77
<=> 2.(x – 5) = 70 <=> x – 5 = 35 <=> x = 40
b, x - 1 3 - 3 5 : 3 4 + 2 . 2 3 = 14
<=> x - 1 3 - 3 + 2 4 = 14
<=> x - 1 3 = 14 + 3 - 16 = 1
<=> x – 1 = 1 <=> x = 2
c, 1 + 2 + 2 2 + 2 3 + . . . + 2 2016 = 2 x - 1 - 1
Đặt: A = 1 + 2 + 2 2 + 2 3 + . . . + 2 2016 => 2A = 2 + 2 2 + 2 3 + . . . + 2 2017
=> 2A – A = ( 2 + 2 2 + 2 3 + . . . + 2 2017 ) – ( 1 + 2 + 2 2 + 2 3 + . . . + 2 2016 )
=> A = 2 2017 - 1
Ta có: 1 + 2 + 2 2 + 2 3 + . . . + 2 2016 = 2 x - 1 - 1 => 2 2017 - 1 = 2 x - 1 - 1 => x = 2018
d, 5 2 x - 3 - 2 . 5 2 = 5 2 . 3
<=> 5 2 x - 3 = 5 2 . 3 + 5 2 . 2
<=> 5 2 x - 3 = 5 2 . ( 3 + 2 )
<=> 5 2 x - 3 = 5 3
<=> 2x – 3 = 3 => x = 3
Tìm x ϵ N biết :
a) 280 – ( x – 140 ) : 35 = 270 ;
b)( 190 – 2x ) : 35 – 32 = 16;
c) 720 : [ 41 – ( 2x – 5 ) ] = 23 .5
d) ( x : 23 + 45 ) . 37 – 22 = 24 . 105 ;
e) ( 3x – 4 ) . ( x – 1 ) 3 = 0 ;
f) 22x-1 : 4 = 83
g) x17 = x ;
h) ( x – 5 ) 4 = ( x – 5 ) 6 ;
i) ( x + 2 ) 5 = 210 ;
k ) 1 + 2 + 3 + … + x = 78
l) ( 3 .x – 24) . 73 = 2 .74 ;
n) 5x : 52 = 125 ;
m) ( x + 1) 2 = ( x + 1) 0 ;
o) ( 2 + x ) + ( 4 + x ) + ( 6 + x ) + ……+ ( 52 + x ) = 780 ;
p) 70⋮ x , 80 ⋮ x và x > 8
q) x ⋮ 12 , x ⋮ 25 , x ⋮ 30 và 0 < x < 500
`a,x(x-1)-(x+2)^2=1`
`<=>x^2-x-x^2-4x-4=1`
`<=>-5x=5`
`<=>x=-1`
`b,(x+5)(x-3)-(x-2)^2=-1`
`<=>x^2+2x-15-x^2+4x-4+1=0`
`<=>6x-18=0`
`<=>x-3=0`
`<=>x=3`
`c,x(2x-4)-(x-2)(2x+3)=0`
`<=>2x(x-2)-(x-2)(2x+3)=0`
`<=>(x-2)(2x-2x-3)=0`
`<=>-3(x-2)=0`
`<=>x-2=0`
`<=>x=2`
`d,x(3x+2)+(x+1)^2-(2x-5)(2x+5)=-12`
`<=>3x^2+2x+x^2+2x+1-4x^2+25=-12`
`<=>4x+26=-12`
`<=>4x=-38`
`<=>x=-19/2`
Bài 2 tìm x ϵ z
a)-4/3/5 . 2 /4 /23 ≤ x ≤ -2/2/5 : 1/6/15
b)-4/1/3 . ( 1/2 - 1/6) ≤ x ≤ -2/3 . ( 1/3 . 1/2 .3/4)
a: \(\Leftrightarrow-\dfrac{23}{5}\cdot\dfrac{50}{23}< =x< =-\dfrac{12}{5}:\dfrac{7}{5}=\dfrac{-12}{7}\)
=>-10<=x<=-12/7
hay \(x\in\left\{-10;-9;-8;-7;-6;-5;-4;-3;-2\right\}\)
b: \(\Leftrightarrow-\dfrac{13}{3}\cdot\dfrac{1}{3}< =x< =-\dfrac{2}{3}\cdot\dfrac{1}{8}\)
=>-13/9<=x<=-1/12
hay \(x=-1\)
Tìm x ϵ Z biết:
a) |x| = 1005
b) |x| + 15 = 22; x>0
c) |x| + 12 = 25; x<0
d) |3x-15| = 0
e) |-x + 2| = 4
f) |-x + |31-(-89)|| = 14
g) |x+1| + |x-1| = 2
h) |x| < 5
i) 5< |x| \(\le\) 7
Giải chi tiết giúp mình nhé
a) |x|=2005
=> x=2005 hoặc -2005
vì giá trị tuyệt đối của 1 số nguyên dương là chính số đó nên x=2005
vì giá trị tuyệt đối của 1 số nguyên âm là số đối của số đó
=> -2005 có số đối là 2005 nên x cũng có thể bằng -2005
a) x = 1005 hoặc x = -1005
b) x + 15 = 22
x = 22 - 15
x = 7
Vì x là |x| nen cung co the = -7
Nhung vi theo de bai thi x>0 nen x = 7
c) x + 12 = 25
x = 25 - 12
x = 13
Vi x la |x| nen cung co the = -13
Vi theo de thi x<0 nen x = -13
a) \(\left|x\right|=1005\)
\(\Rightarrow\) x = 1005 hoặc x = -1005
b) \(\left|x\right|\) + 15 = 22 ; x > 0
x > 0 \(\Rightarrow\)\(\left|x\right|=x\)
\(\Rightarrow\left|x\right|+15=22\)
\(\Rightarrow x+15=22\)
\(\Rightarrow x=22-15\)
\(\Rightarrow x=7\)
c) \(\left|x\right|+12=25;x< 0\)
\(x< 0\Rightarrow\left|x\right|=-x\)
\(\Rightarrow\left|x\right|+12=25\)
\(\Rightarrow-x+12=25\)
\(\Rightarrow-x=25-12\)
\(\Rightarrow-x=13\)
\(\Rightarrow x=-13\)
d) \(\left|3x-15\right|=0\)
\(\Rightarrow3x-15=0\)
\(\Rightarrow3x=0+15\)
\(\Rightarrow3x=15\)
\(\Rightarrow x=15:3\)
\(\Rightarrow x=5\)
e) \(\left|-x+2\right|=4\)
\(\Rightarrow-x+2=4\) hoặc \(-x+2=-4\)
TH1: \(-x+2=4\)
\(-x=4-2\)
\(-x=2\)
\(\Rightarrow x=-2\)
TH2: \(-x+2=-4\)
\(-x=-4-2\)
\(-x=-6\)
\(\Rightarrow x=6\)
f) \(\left|-x+\left|31-\left(-89\right)\right|\right|=14\)
\(\Rightarrow\left|-x+\left|120\right|\right|=14\)
\(\Rightarrow\left|-x+120\right|=14\)
\(\Rightarrow-x+120=14\) hoặc \(-x+120=-14\)
TH1: \(-x+120=14\)
\(\Rightarrow-x=14-120\)
\(\Rightarrow-x=-106\)
\(\Rightarrow x=106\)
TH2: \(-x+120=-14\)
\(\Rightarrow-x=-14-120\)
\(\Rightarrow-x=-134\)
\(\Rightarrow x=134\)
bài 1
a)(x-1)(x+2)-(x-3)(x+1)=5x-3
b)(2x-1)(x+3)-(x-2)(x+2)=3x+1
c)x^2(x-1)-x(x-1)(x+1)=0
d)4x(x-5)-(2x-3)(2x+3)=9
Lời giải:
a.
a. $(x-1)(x+2)-(x-3)(x+1)=5x-3$
$\Leftrightarrow (x^2+x-2)-(x^2-2x-3)=5x-3$
$\Leftrightarrow 3x+1=5x-3$
$\Leftrightarrow 4=2x$
$\Leftrightarrow x=2$
b.
$(2x-1)(x+3)-(x-2)(x+3)=3x+1$
$\Leftrightarrow (2x^2+5x-3)-(x^2-4)=3x+1$
$\Leftrightarrow x^2+5x+1=3x+1$
$\Leftrightarrow x^2+2x=0$
$\Leftrightarrow x(x+2)=0$
$\Leftrightarrow x=0$ hoặc $x=-2$
c.
$x^2(x-1)-x(x-1)(x+1)=0$
$\Leftrightarrow x^2(x-1)-(x^2+x)(x-1)=0$
$\Leftrightarrow (x-1)[x^2-(x^2+x)]=0$
$\Leftrightarrow (x-1)(-x)=0$
$\Leftrightarrow x-1=0$ hoặc $-x=0$
$\Leftrightarrow x=1$ hoặc $x=0$
d.
$4x(x-5)-(2x-3)(2x+3)=9$
$\Leftrightarrow 4x^2-20x-(4x^2-9)=9$
$\Leftrightarrow -20x=0$
$\Leftrightarrow x=0$
a: Ta có: \(\left(x-1\right)\left(x+2\right)-\left(x-3\right)\left(x+1\right)=5x-3\)
\(\Leftrightarrow x^2+2x-x-2-x^2-x+3x+3-5x+3=0\)
\(\Leftrightarrow-2x+4=0\)
\(\Leftrightarrow2x=4\)
hay x=2
b: Ta có: \(\left(2x-1\right)\left(x+3\right)-\left(x-2\right)\left(x+2\right)=3x+1\)
\(\Leftrightarrow2x^2+6x-x-3-x^2+4-3x-1=0\)
\(\Leftrightarrow x^2+2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
c: Ta có: \(x^2\left(x-1\right)-x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x-x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
d: Ta có: \(4x\left(x-5\right)-\left(2x-3\right)\left(2x+3\right)=9\)
\(\Leftrightarrow4x^2-20x-4x^2+9=9\)
hay x=0
Tìm x ϵ Z biết:
\(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
`-1/5<=x/8<=1/4`
`=>8* -1/5<=x<=1/4*8`
`=>-8/5<=x<=2`
Mà `x in ZZ`
`=>x in {-1,0,1,2}`
−1/5≤x8≤1/4-15≤x8≤14
⇒8⋅−1/5≤x≤14⋅8⇒8⋅-15≤x≤14⋅8
⇒−85≤x≤2⇒-85≤x≤2
Mà x∈Zx∈ℤ
⇒x∈{−1,0,1,2}
\(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\)
\(\Rightarrow5x\in\left\{-5;0;5;10\right\}\)
\(\Rightarrow x\in\left\{-1;0;1;2\right\}\)
Giai các bpt
a, 2x+2>4
b, 3x+2>-5
c,10-2x>2
d,1-2x<3
e,10x+3-5≤14x+12
f/ (3x-1)< 2x+4
g 4x-8 ≥3(2x-1)-2x+1
h/ x^x -x(x+2)> 3x-1
i/ x+8 >3x-1
j/ 3x- (2x+5) ≤(2x-3)
k/ (x-3) (x+3)<x(x+2)+3
l/ 2(3x-1) -2x<2x+1
m, (3-2x/5)> (2-x/3)
n, (x-2/6)-(x-1/3)≤x/2
o, (x+1/3)>(2x-1/6) -2
p, 1+ (2x+1)/3) >(2x-1/6) -2
q, (x+5/6)-(2x+1/3)≤ (x+3/2)
r, (5x+4/6) -(2x-1/12)≥4
a \(2x+2>4\\ \Leftrightarrow2\left(x+1\right)>4\\ \Leftrightarrow x+1>2\\ \Leftrightarrow x>1\)
b \(3x+2>-5\\ \Leftrightarrow3x>-7\\ \Leftrightarrow x>\dfrac{-7}{3}\)
c \(10-2x>2\\ \Leftrightarrow2\left(5-x\right)>2\\ \Leftrightarrow5-x>1\\ \Leftrightarrow-x>-4\\ \Leftrightarrow x< 4\)
d \(1-2x< 3\\ \Leftrightarrow-2x< 2\\ \Leftrightarrow2x>2\\ \Leftrightarrow x>1\)
a)2x+2>4
<=> 2x>4-2
<=>2x>2
<=>x>1
Vậy...
b)3x+2>-5
<=>3x>-5-2
<=>3x>-7
<=>x>\(\dfrac{-7}{3}\)
Vậy...
c)10-2x>2
<=>-2x>-10+2
<=>-2x>-8
<=>x<4
Vậy...
d)1-2x<3
<=>-2x<3-1
<=>-2x<2
<=>x>-1
Vậy...
e)10x+3-5\(\le\)14x+12
<=>10x-2\(\le\)14x+12
<=>10x-14x\(\le\)2+12
<=>-4x\(\le\)14
<=>x\(\ge\)\(\dfrac{-7}{2}\)
Vậy...
f)(3x-1)<2x+4
<=> 3x-2x<1+4
<=>x<5
Vậy...
Tính tổng các số nguyên x biết:
1/ -10 < x < 11
2/ -8 ≤ x ≤ 7
3/ -23 < x ≤ 23
4/ │x│≤ 2
5/ │-x│< 13
1/ ta co
vi x \(\in Z\Rightarrow x\in\left\{-9;-8;..;9;10\right\}\)
Tong cac so x thoa man la
-9+(-8)+(-7)+....+9+10
=(-9+9)+(-8+8)+...+(-1+1)+0
=0+0+0+..+0+0
=0
vay tong cac so ma x thoa man la 0
2/ ta co
vi x \(\in Z\Rightarrow x\in\left\{-8;-7;..;5;6;7\right\}\)
Tong cac so ma x thoa man la
-8+(-7)+(-6)+...+6+7
=-8+0+(-7+7)+(-6+6)+(-5+5)+...+(-1+1)
=-8+0+0+0+...+0
=-8
vay tong cac gia tri ma x thoa man la -8
3/ ta co
vi x \(\in Z\Rightarrow x\in\left\{-22;-21;...;22;23\right\}\)
Tong cac gia tri ma x thoa man la
(-22)+(-21)+....+22+23
=23+0+(-21+21)+(-22+22)+...+(-1+1)
=23+0+0+0+...+0
=23
vay tong cac gia tri ma x thoa man la 23
4/ ta co :
vi |x|\(\le2\Rightarrow\left|x\right|\in\left\{1;2\right\}hay.x\in\left\{2;1;-1;-2\right\}\)
Tong cac gia tri ma x thoa man la :
2+1+(-1)+(-2)
=3+(-3)
=0
vay tong cac gia tri ma x thoa man la 0
5/ ta co
│-x│< 13 nen |x| \(\in\left\{12;11;10;..;2;1;0;-1;-2;...\right\}\)
hay x \(\in\left\{12;11;10;9;...;1;0;-12;-13;...;-1\right\}\)
Tong cac so ma x thoa man la
12+13+14+15+....+1+0+(-1)+(-2)+....+(-12)
=(-12+12)+(-13+13)+...+(-1+1)+0
=0+0+0+0+...+0+0
=0
Vay tong cac gia tri ma x thoa man la 0