{[(2x 14):23-3]:2}-1= 0
{[(2x 14):23-3]:2}-1= 0
{[(2x 14):23-3]:2}-1= 0
tìm x
17x – ( -16x – 37) = 2x +
-2x –3. (x – 17) = 34 – 2(-x + 25
17x + 3. ( -16x – 37) = 2x + 43 - 4x
103 -57: [-2. (2x – 1)2 – (-9)0] = -106
3x – 32 > -5x + 1
15 + 4x < 2x – 145
-3. (2x + 5) -16 < -4. (3 – 2x)
-2x + 15 < 3x – 7 < 19 – x
x + (x+1) + (x+2) + (x+3) + .... + 13 + 14 = 14
25 + 24 + 23 +...+ x + (x - 2) + (x – 3) = 25
17x + 3. ( -16x – 37) = 2x + 43 - 4x
<=>17x-48x-111=-2x+43
<=>-29x=154
<=> \(x=-\frac{154}{29}\)
-3. (2x + 5) -16 < -4. (3 – 2x)
\(\Leftrightarrow-6x-31< -12+8x.\)
\(\Leftrightarrow-14x< 19\Rightarrow x< -\frac{19}{14}\)
lên mạng xem ik
hỏi google là đc hết mak
4.(5-x)-3.(1-x)=-32
3.|2x-3|=7-(-14)
22x-1-32=23
Ta có: \(4.\left(5-x\right)-3.\left(1-x\right)=-3^2\)
\(\Leftrightarrow20-4x-3+3x=9\)
\(\Leftrightarrow-4x+3x=9-20+3\)
\(\Leftrightarrow-x=-8\)
\(\Leftrightarrow x=8\)
Vậy: \(x=8\)
Ta có: \(3.\left|2x-3\right|=7-\left(-14\right)\)
\(\Leftrightarrow3.\left|2x-3\right|=7+14=21\)
\(\Leftrightarrow\left|2x-3\right|=\dfrac{21}{3}=7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
Vậy: \(x=-2;5\)
Ta có: \(2^{2x-1}-3^2=23\)
\(\Leftrightarrow2^{2x-1}-9=23\)
\(\Leftrightarrow2^{2x-1}=32\)
\(\Leftrightarrow2^{2x-1}=2^5\)
\(\Leftrightarrow2x-1=5\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
Vậy: \(x=3\)
_Chúc bạn học tốt_
Giải các phương trình sau:
a) 5 x + 3 x − 1 = 5 x + 7 ; b) 2 x − 1 4 + 3 = 1 − 3 x 6 ;
c) x − 3 2 − x x + 4 + 5 = 0 ; d) 3 x − 1 x + 2 3 − 2 x 2 + 1 2 = 11 2 .
a) x = 10 3 b) x = - 31 12
c) x = 7 5 d) x = 4
bài 1: thực hiện phép tính
1) (13.17\(^4\)+4.17\(^4\)):17\(^3\)-(14.\(3^3\)-14.3\(^2\)):9
2) \(2^3\).\(5^2\)-[131-(23-2\(^3\))\(^2\)]
bài 2: tìm x
1) 6\(^2\).x+14.x-3\(^4\)=69
2) 3\(^x\)=81
3) 3. (2x+1)\(^2\)=75
2:
1: =>36x+14x=69+81=150
=>50x=150
=>x=3
2: 3^x=81
=>3^x=3^4
=>x=4
3: 3(2x+1)^2=75
=>(2x+1)^2=25
=>2x+1=5 hoặc 2x+1=-5
=>x=-3 hoặc x=2
1:
1: \(\dfrac{13\cdot17^4+4\cdot17^4}{17^3}-\dfrac{14\cdot3^3-14\cdot3^2}{9}\)
\(=\dfrac{17^4\cdot\left(13+4\right)}{17^3}-\dfrac{14\cdot3^2\left(3-1\right)}{9}\)
\(=17\cdot17-14\cdot2\)
=289-28
=261
2:
\(2^3\cdot5^2-\left[131-\left(23-2^3\right)^2\right]\)
\(=8\cdot25-131+\left(-1\right)^2\)
=69+1
=70
-2x+15 < 3x - 17 < 19 -x
x+(x+1) + (x+2) + (x+3) + ... +13 +14 = 14
25 + 24 + 23 + ... + x + (x-2) + (x-3) = 25
Tìm x sao cho x thuộc tập hợp số nguyên:
1) x - 43 = (35 - x) - 48
2) 305 - x + 14 = 48 + (x + 23)
3) - (x - 6 + 85) = (x + 51) - 54
4) - (35 - x - 37 - x) = 33 - x
5) 13 - | x | = | -4 |
6) | x | - 3 + 6 = 16
7) 35 - | 2x - 1 | = 14
8) | 3x - 2 | + 5 = 9 - x
9) x - ( -25 + 7 ) > 12 - ( 15 - 14 )
10) | 17 + ( x - 15 ) | < 4
11) x2 - 5x = 0
12) | x-9 | . (-8) = -16
13) | 4 - 5x = 24 với x < hoặc = 0
14) x . ( x - 2 ) > 0
15) x . ( x - 2 ) < 0
16) (x-1) . (y+1) = 5
17) x . ( y +2 ) = -8
18) xy - 2x - 2y = 0
19) 2x - 5 chia hết cho x - 1
1) x - 43 = (35 - x) - 48
=> x + x = 35 - 48 + 43
=> x + x = 30
=> x = 30 : 2
=> x = 15
2) 305 - x + 14 = 48 + (x + 23)
=> 305 - x + 14 = 48 + x + 23
=> -x - x = 48 + 23 - 14 - 305
=> -x - x = -248
=> -x = -248 : 2
=> -x = -124
=> x = 124
3) - (x - 6 + 85) = (x + 51) - 54
=> -x + 6 - 85 = x + 51 - 54
=> -x - x = 51 - 54 + 85 - 6
=> -x - x = 76
=> -x = 76 : 2
=> -x = 38
=> x = -38
4) - (35 - x - 37 - x) = 33 - x
=> -35 + x + 37 + x = 33 - x
=> x + x + x = 33 + 35 - 37
=> x + x + x = 31
=> x = 31 : 3
=> x \(=\dfrac{31}{3}\)
Vì x \(\in\) Z nên không có giá trị x nào thỏa mãn trong câu này.
5) 13 - | x | = | -4 |
=> 13 - |x| = 4
=> |x| = 13 - 4
=> |x| = 9
=> \(\left[{}\begin{matrix}x=9\\x=-9\end{matrix}\right.\)
6) | x | - 3 + 6 = 16
=> |x| = 16 - 6 + 3
=> |x| = 13
=> \(\left[{}\begin{matrix}x=13\\x=-13\end{matrix}\right.\)
7) 35 - | 2x - 1 | = 14
=> |2x - 1| = 35 - 14
=> |2x - 1| = 21
=> \(\left[{}\begin{matrix}2x-1=21\\2x-1=-21\end{matrix}\right.=>\left[{}\begin{matrix}2x=21+1\\2x=-21+1\end{matrix}\right.=>\left[{}\begin{matrix}2x=22\\2x=-20\end{matrix}\right.=>\left[{}\begin{matrix}x=22:2\\x=-20:2\end{matrix}\right.=>\left[{}\begin{matrix}x=11\\x=-10\end{matrix}\right.\)
8) | 3x - 2 | + 5 = 9 - x
=> |3x - 2| = 9 - 5 - x
=> |3x - 2| = 4 - x
=> \(\left[{}\begin{matrix}3x-2=4-x\\3x-2=x-4\end{matrix}\right.=>\left[{}\begin{matrix}3x+x=4+2\\3x-x=-4+2\end{matrix}\right.=>\left[{}\begin{matrix}4x=6\\2x=-2\end{matrix}\right.=>\left[{}\begin{matrix}x=6:4\\x=-2:2\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{6}{4}\\x=-1\end{matrix}\right.\)
Vì x \(\in\) Z nên x = -1.
9) x - ( -25 + 7 ) > 12 - ( 15 - 14 )
=> x - (-18) > 12 - 1
=> x + 18 > 11
=> x > 11 - 18
=> x > -7
10) | 17 + ( x - 15 ) | < 4
=> \(\left[{}\begin{matrix}17+\left(x-15\right)< 4\\17+\left(x-15\right)< -4\end{matrix}\right.=>\left[{}\begin{matrix}x-15< 4-17\\x-15< -4-17\end{matrix}\right.=>\left[{}\begin{matrix}x-15< -15\\x-15< -21\end{matrix}\right.=>\left[{}\begin{matrix}x< -15+15\\x< -21+15\end{matrix}\right.=>\left[{}\begin{matrix}x< 0\\x< -6\end{matrix}\right.=>x< -6\)
11) x2 - 5x = 0
=> x . (2 - 5) = 0
=> x . (-3) = 0
=> x = 0 : (-3)
=> x = 0
12) | x-9 | . (-8) = -16
=> |x - 9| = (-16) : (-8)
=> |x - 9| = 3
=> \(\left[{}\begin{matrix}x-9=3\\x-9=-3\end{matrix}\right.=>\left[{}\begin{matrix}x=3+9\\x=-3+9\end{matrix}\right.=>\left[{}\begin{matrix}x=12\\x=6\end{matrix}\right.\)
13) | 4 - 5x | = 24 với x < hoặc = 0
=> \(\left[{}\begin{matrix}4-5x=24\\4-5x=-24\end{matrix}\right.=>\left[{}\begin{matrix}5x=4-24\\5x=4-\left(-24\right)\end{matrix}\right.=>\left[{}\begin{matrix}5x=-20\\5x=28\end{matrix}\right.=>\left[{}\begin{matrix}x=-20:5\\x=28:5\end{matrix}\right.=>\left[{}\begin{matrix}x=-4\\x=\dfrac{28}{5}\end{matrix}\right.\)
Vì x \(\le\) 0 nên x = -4
14) x . ( x - 2 ) > 0
=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x-2< 0\end{matrix}\right.\end{matrix}\right.=>\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x< 2\end{matrix}\right.\end{matrix}\right.=>\left[{}\begin{matrix}x>2\\x< 2\end{matrix}\right.\)
15) x . ( x - 2 ) < 0
=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\x-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x>0\\x-2< 0\end{matrix}\right.\end{matrix}\right.=>\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x>0\\x< 2\end{matrix}\right.\end{matrix}\right.=>\left[{}\begin{matrix}2>x< 0\left(loại\right)\\0< x< 2\left(chọn\right)\end{matrix}\right.=>0< x< 2\)
16) (x-1) . (y+1) = 5
=> \(\left[{}\begin{matrix}x-1=5\\y+1=1\end{matrix}\right.=>\left[{}\begin{matrix}x=5+1\\y=1-1\end{matrix}\right.=>\left[{}\begin{matrix}x=6\\y=0\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}x-1=1\\y+1=5\end{matrix}\right.=>\left[{}\begin{matrix}x=1+1\\y=5-1\end{matrix}\right.=>\left[{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}x-1=-1\\y+1=-5\end{matrix}\right.=>\left[{}\begin{matrix}x=-1+1\\y=-5-1\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\y=-6\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}x-1=-5\\y+1=-1\end{matrix}\right.=>\left[{}\begin{matrix}x=-5+1\\y=-1-1\end{matrix}\right.=>\left[{}\begin{matrix}x=-4\\y=-2\end{matrix}\right.\)
17) x . ( y +2 ) = -8
=> \(\left[{}\begin{matrix}x=1\\y+2=-8\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\y=-8-2\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\y=-10\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}x=-1\\y+2=8\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\y=8-2\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\y=4\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}x=-8\\y+2=1\end{matrix}\right.=>\left[{}\begin{matrix}x=-8\\y=1-2\end{matrix}\right.=>\left[{}\begin{matrix}x=-8\\y=-1\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}x=8\\y+2=-1\end{matrix}\right.=>\left[{}\begin{matrix}x=8\\y=-1-2\end{matrix}\right.=>\left[{}\begin{matrix}x=8\\y=-3\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}x=2\\y+2=-4\end{matrix}\right.=>\left[{}\begin{matrix}x=2\\y=-4-2\end{matrix}\right.=>\left[{}\begin{matrix}x=2\\y=-6\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}x=-2\\y+2=4\end{matrix}\right.=>\left[{}\begin{matrix}x=-2\\y=4-2\end{matrix}\right.=>\left[{}\begin{matrix}x=-2\\y=2\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}x=4\\y+2=-4\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\y=-4-2\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\y=-6\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}x=-4\\y+2=2\end{matrix}\right.=>\left[{}\begin{matrix}x=-4\\y=2-2\end{matrix}\right.=>\left[{}\begin{matrix}x=-4\\y=0\end{matrix}\right.\)
18) xy - 2x - 2y = 0
=> x . (y - 2) - 2y = 0
=> x . (y - 2) - 2y - 4 = -4
=> x . (y - 2) - 2 . (y - 2) = -4
=> (y - 2) . (x - 2) = -4
=> \(\left[{}\begin{matrix}y-2=1\\x-2=-4\end{matrix}\right.=>\left[{}\begin{matrix}y=1+2\\x=-4+2\end{matrix}\right.=>\left[{}\begin{matrix}y=3\\x=-2\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}y-2=-1\\x-2=4\end{matrix}\right.=>\left[{}\begin{matrix}y=-1+2\\x=4+2\end{matrix}\right.=>\left[{}\begin{matrix}y=1\\x=6\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}y-2=2\\x-2=-2\end{matrix}\right.=>\left[{}\begin{matrix}y=2+2\\x=-2+2\end{matrix}\right.=>\left[{}\begin{matrix}y=4\\x=0\end{matrix}\right.\)
hoặc
=> \(\left[{}\begin{matrix}y-2=-2\\x-2=2\end{matrix}\right.=>\left[{}\begin{matrix}y=-2+2\\x=2+2\end{matrix}\right.=>\left[{}\begin{matrix}y=0\\x=4\end{matrix}\right.\)
19) 2x - 5 \(⋮\) x - 1
=> (2x - 2) - (5 - 2) \(⋮\) x - 1
=> 2(x - 1) - 3 \(⋮\) x - 1
Vì 2(x - 1) \(⋮\) x - 1 nên 3 \(⋮\) x - 1
=> x - 1 \(\in\) Ư(3) = {-3; -1; 1; 3}
=> x \(\in\) {-2; 0; 2; 4}
P/s: Mình không bảo đảm là đúng hết nên câu nào sai thì bạn thông cảm nha~
Tìm x
1) (2x-1)(x+3)(2-x)=0
2)x^3 + x^2 + x + 1 = 0
3) 2x(x-3)+5(x-3) =0
4)x(2x-7)-(4x-14)=0
5) 2x^3 + 3x^2 + 2x + 3 = 0
1) (2x-1)(x+3)(2-x)=0
=>2x-1 =0 hoặc x+3=0 hoặc 2-x=0
=>x=1/2 hoặc x=-3 hoặc x=2
2)x^3 + x^2 + x + 1 = 0
=>.x^2(x+1)+(x+1)=0
=>(x^2+1)(x+1)=0
=>x^2+1=0 hoặc x+1=0
=> x =-1
3) 2x(x-3)+5(x-3) =0
=>(2x+5)(x-3)=0
=>2x+5=0 hoặc x-3=0
=>x=-5/2 hoặc x=3
4)x(2x-7)-(4x-14)=0
=> (x-2)(2x-7)=0
=> x-2 =0 hoặc 2x-7=0
=>x=2 hoặc x=7/2
5)2x^3+3x^2+2x+3=0
=>x^2(2x+3)+2x+3=0
=>(x^2+1)(2x+3)=0
=>x^2+1=0 hoặc 2x+3=0
=> x =-3/2