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a) x = 2.
b) x = 7.
c) x= 12.
d) x= 45.
e) x = 18.
f) x = 10.
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1
S₂ = 21 + 23 + 25 + ... + 1001
Số số hạng của S₂:
(1001 - 21) : 2 + 1 = 491
⇒ S₂ = (1001 + 21) . 491 : 2 = 250901
--------
S₄ = 15 + 25 + 35 + ... + 115
Số số hạng của S₄:
(115 - 15) : 10 + 1 = 11
⇒ S₄ = (115 + 15) . 11 : 2 = 715
Bài 2
a) 2x - 138 = 2³.3²
2x - 138 = 8.9
2x - 138 = 72
2x = 72 + 138
2x = 210
x = 210 : 2
x = 105
b) 5.(x + 35) = 515
x + 35 = 515 : 5
x + 35 = 103
x = 103 - 35
x = 78
c) 814 - (x - 305) = 712
x - 305 = 814 - 712
x - 305 = 102
x = 102 + 305
x = 407
d) 20 - [7.(x - 3) + 4] = 2
7(x - 3) + 4 = 20 - 2
7(x - 3) + 4 = 18
7(x - 3) = 18 - 4
7(x - 3) = 14
x - 3 = 14 : 7
x - 3 = 2
x = 2 + 3
x = 5
e) 9ˣ⁻¹ = 9
x - 1 = 1
x = 1 + 1
x = 2
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có : 2 x : 2 2 = 2 5 nên x = 7.
b) Ta có: 3 x : 3 2 = 3 5 nên x = 7.
c) Ta có : 4 4 : 4 x = 4 2 nên x = 2.
d) Ta có : 5 x : 5 2 = 5 2 nên x = 4,
e) Ta có: 5 x + 1 : 5 = 5 4 nên x = 4.
f) Ta có : 4 2 x - 1 : 4 = 4 2 nên x = 2
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(4x^3+12=120\)
=>\(4x^3=108\)
=>\(x^3=27=3^3\)
=>x=3
b: \(\left(x-4\right)^2=64\)
=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)
c: (x+1)^3-2=5^2
=>\(\left(x+1\right)^3=25+2=27\)
=>x+1=3
=>x=2
d: 136-(x+5)^2=100
=>(x+5)^2=36
=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)
e: \(4^x=16\)
=>\(4^x=4^2\)
=>x=2
f: \(7^x\cdot3-147=0\)
=>\(3\cdot7^x=147\)
=>\(7^x=49\)
=>x=2
g: \(2^{x+3}-15=17\)
=>\(2^{x+3}=32\)
=>x+3=5
=>x=2
h: \(5^{2x-4}\cdot4=10^2\)
=>\(5^{2x-4}=\dfrac{100}{4}=25\)
=>2x-4=2
=>2x=6
=>x=3
i: (32-4x)(7-x)=0
=>(4x-32)(x-7)=0
=>4(x-8)*(x-7)=0
=>(x-8)(x-7)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)
k: (8-x)(10-2x)=0
=>(x-8)(x-5)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)
m: \(3^x+3^{x+1}=108\)
=>\(3^x+3^x\cdot3=108\)
=>\(4\cdot3^x=108\)
=>\(3^x=27\)
=>x=3
n: \(5^{x+2}+5^{x+1}=750\)
=>\(5^x\cdot25+5^x\cdot5=750\)
=>\(5^x\cdot30=750\)
=>\(5^x=25\)
=>x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
a, \(x-14=3x+18\)
\(\Rightarrow x-3x=18+14\)
\(\Rightarrow-2x=32\Rightarrow x=\frac{32}{-2}=-16\)
b, \(\left(x+7\right).\left(x-9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+7=0\\x-9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-7\\x=9\end{cases}}}\)
c, \(\left|2x-5\right|-7=22\)
\(\Rightarrow\left|2x-5\right|=22+7\)
\(\Rightarrow\left|2x-5\right|=29\)
\(\Rightarrow\orbr{\begin{cases}2x+5=29\\2x-5=29\end{cases}}\Rightarrow\orbr{\begin{cases}2x=24\\2x=34\end{cases}\Rightarrow}\orbr{\begin{cases}x=12\\x=17\end{cases}}\)
d, \(\left(\left|2x\right|-5\right)-7=22\)
\(\Rightarrow\left(\left|2x\right|-5\right)=29\)
\(\Rightarrow\left|2x\right|=29+5\Rightarrow\left|2x\right|=34\Rightarrow x=\pm17\)
e, \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\)
Vì \(\left|x+3\right|\ge0;\left|x+9\right|\ge0;\left|x+5\right|\ge0;4x\ge0\)
Nên \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\ge0\)
\(\Rightarrow\left|x+3\right|>0\Rightarrow\left|x+3\right|=x+3\)
\(\left|x+9\right|>0\Rightarrow\left|x+9\right|=x+9\)
\(\left|x+5\right|>0\Rightarrow\left|x+5\right|=x+5\)
Ta có :
\(x+3+x+9+x+5=4x\)
\(\Rightarrow3x+\left(3+9+5\right)=4x\)
\(\Rightarrow4x-3x=17\)
\(\Rightarrow x=17\)
2. a , b sai đề bn
c, \(\left(5x+1\right).\left(y-1\right)=4\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)
\(\text{ }Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Ta có bảng sau :
5x+1 | 1 | -1 | 2 | -2 | 4 | -4 |
y-1 | -4 | 4 | -2 | 2 | -1 | 1 |
x | 0 | -2/5 | 1/5 | -3/5 | 3/5 | -1 |
y | -3 | 5 | -1 | 3 | 0 | 2 |
d, \(5xy-5x+y=5\)
\(\Rightarrow\left(5xy-5x\right)+y=5\)
\(\Rightarrow5x.\left(y-1\right)+y=5\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)=4\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)
\(Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Ta có bảng sau :
5x+1 | 1 | -1 | 2 | -2 | 4 | -4 |
y-1 | -4 | 4 | -2 | 2 | -1 | 1 |
x | 0 | -2 | 1/5 | -3/5 | 3/5 | -1 |
y | -3 | 5 | -1 | 3 | 0 | 2 |
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1\)) \(5-\left(10-x\right)=7\)
\(10-x=5-7\)
\(10-x=-2\)
\(x=10-\left(-2\right)\)
\(x=12\)
\(2\)) \(-32-\left(x-5\right)=0\)
\(x-5=-32-0\)
\(x-5=-32\)
\(x=-32+5\)
\(x=-27\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
b) \(\dfrac{39}{7}:x=13\)
\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)
c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)
\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)
\(\dfrac{14}{5}x=34+50=84\)
\(x=\dfrac{84}{\dfrac{14}{5}}=30\)
d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)
\(\dfrac{1}{6}x=\dfrac{5}{12}\)
\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)
g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)
\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)
\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)
\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)
h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)
\(x=1\)
Mỏi tay woa bn làm nốt nha!!
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>2x^3=58-4=54
=>x^3=27
=>x=3
b; =>(5-x)^5=2^5
=>5-x=2
=>x=3
c: =>(5x-6)^3=4^3
=>5x-6=4
=>5x=10
=>x=2
d: (3x)^3=(2x+1)^3
=>3x=2x+1
=>x=1
1=>2x3=54
=>x3=27 =>x=3
2=>(5-x)5=25
=>5-x=2
=>x=3
3=>(5x-6)3=43
=>5x-6=4
=>5x=10=>x=2
4=>3x=2x+1
=>x=1
a) (5-x)5=32
=>(5-x)5=25
=>5-x=2
=> x = 3
b) x5=x7
=> x7-x5=0
=> x5(x2-1)=0
\(\Rightarrow\left[{}\begin{matrix}x^5=0\\x^2-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)
c) (2x+1):7=22+32
=> (2x+1):7=4+9
=> (2x+1):7=13
=> 2x+1=91
=> 2x=90
=> x = 45
d) 5x+52=650
=> 52.(5x-2+1)=650
=> 5x-2+1=26
=> 5x-2=25
=> 5x-2=52
=> x-2=2
=>x=4