1: =>\(5^{x-2}-9=2^4-\left(6^2-6^2\right)\)

=>\(5^{x-2}=16+9=25\)

=>x-2=2

=>x=4

2: \(\Leftrightarrow3^x+16=19^6:19^5-3=19-3=16\)

=>3^x=0

=>x=0

3: \(\Leftrightarrow2^x+2^x\cdot16=272\)

=>2^x*17=272

=>2^x=16

=>x=4

4: \(\Leftrightarrow2^{x-1}+3=24-\left(4^2-2^2+1\right)=24-\left(16-4+1\right)\)

=>\(2^{x-1}+3=24-16+4-1=8+4-1=12-1=11\)

=>2^x-1=8

=>x-1=3

=>x=4

a: \(280-\left(x-140\right):35=270\)

=>(x-140):35=280-270=10

=>x-140=350

=>x=350+140

=>x=490

b: \(\left(190-2x\right):35-32=16\)

=>\(\left(190-2x\right):35=32+16=48\)

=>\(190-2x=35\cdot48=1680\)

=>2x=190-1680=-1490

=>x=-745

c: \(720:\left\lbrack41-\left(2x-5\right)\right\rbrack=2^3\cdot5\)

=>\(720:\left\lbrack41-\left(2x-5\right)\right\rbrack=8\cdot5=40\)

=>41-(2x-5)=720:40=18

=>2x-5=41-18=23

=>2x=28

=>x=14

d: \(\left(x:23+45\right)\cdot37-22=2^4\cdot105\)

=>\(\left(\frac{x}{23}+45\right)\cdot37=16\cdot105+22=1702\)

=>\(\frac{x}{23}+45=46\)

=>\(\frac{x}{23}=1\)

=>x=23

e: \(\left(3x-4\right)\left(x-1\right)^3=0\)

=>\(\left[\begin{array}{l}3x-4=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac43\\ x=1\end{array}\right.\)

f: \(2^{2x-1}:4=8^3\)

=>\(2^{2x-1-2}=2^9\)

=>2x-3=9

=>2x=12

=>x=6

g: \(x^{17}=x\)

=>\(x^{17}-x=0\)

=>\(x\left(x^{16}-1\right)=0\)

=>\(\left[\begin{array}{l}x=0\\ x^{16}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x^{16}=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=1\\ x=-1\end{array}\right.\)

h: \(\left(x-5\right)^4=\left(x-5\right)^6\)

=>\(\left(x-5\right)^6-\left(x-5\right)^4=0\)

=>\(\left(x-5\right)^4\cdot\left\lbrack\left(x-5\right)^2-1\right\rbrack=0\)

=>\(\left(x-5\right)^4\cdot\left(x-4\right)\left(x-6\right)=0\)

=>x∈{4;5;6}

i: \(\left(x+2\right)^5=2^{10}\)

=>\(\left(x+2\right)^5=\left(2^2\right)^5=4^5\)

=>x+2=4

=>x=2

k: 1+2+3+...+x=78

=>\(\frac{x\left(x+1\right)}{2}=78\)

=>x(x+1)=156

=>\(x^2+x-156=0\)

=>(x+13)(x-12)=0

=>x=-13(loại) hoặc x=12(nhận)

l: \(\left(3x-2^4\right)\cdot7^3=2\cdot7^4\)

=>\(3x-16=2\cdot\frac{7^4}{7^3}=2\cdot7=14\)

=>3x=16+14=30

=>\(x=\frac{30}{3}=10\)

n: \(5^{x}:5^2=125\)

=>\(5^{x-2}=5^3\)

=>x-2=3

=>x=5

m: \(\left(x+1\right)^2=\left(x+1\right)^0\)

=>\(\left(x+1\right)^2=1\)

=>\(\left[\begin{array}{l}x+1=1\\ x+1=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-2\end{array}\right.\)

o: Số số hạng của dãy số 2;4;..;52 là:

(52-2):2+1=50:2+1=25+1=26(số)

Tổng của dãy số 2;4;...;52 là:

\(\left(52+2\right)\cdot\frac{26}{2}=54\cdot13=702\)

(2+x)+(4+x)+...+(52+x)=780

=>26x+702=780

=>26x=78

=>x=3

p: \(70=2\cdot5\cdot7;80=2^4\cdot5\)

=>ƯCLN(70;80)=\(2\cdot5=10\)

70⋮x; 80⋮x

=>x∈ƯC(70;80)

=>x∈Ư(10)

mà x>8

nên x=10

q: \(12=2^2\cdot3;25=5^2;30=2\cdot3\cdot5\)

=>BCNN(12;25;30)=\(2^2\cdot3\cdot5^2=300\)

x⋮12; x⋮25; x⋮30

=>x∈BC(12;25;30)

=>x∈B(300)

mà 0<x<500

nên x=300

a, 2009 - |x - 2009| = x 

=> |x - 2009| = 2009 - x 

=> x = 2009

b/100x+(1+2+3+...+100)=205550

100x+5050=205550

100x=205550-5050

100x=200500

x=200500/100

x=2005

d/(3x-2⁴).7⁵=2.7⁶.1/2009⁰

(3x-2⁴).7⁵=2.7⁶.1

(3x-2⁴)=2.7⁶:7⁵

(3x-2⁴)=2.7

3x-16=14

3x=14+16

3x=30

x=30/10=3

3 + 2x - 1= 24 - [42 - (22 - 1)

3 + 2x - 1= 24 - [42 - 21]

3 + 2x - 1= 24 - 21

3 + 2x - 1= 3

3 + 2x = 3 + 1

3 + 2x = 4

2x = 4 - 3

2x =1

x = 1:2

x = 0,5

Vậy x = 0,5

3 + 2x - 1= 24 - [42 - (22 - 1)

3 + 2x - 1= 24 - [42 - 21]

3 + 2x - 1= 24 - 21

3 + 2x - 1= 3

3 + 2x = 3 + 1

3 + 2x = 4

2x = 4 - 3

2x =1

x = 1:2

x = 0,5

suy ra x = 0,5

Lời giải:
a. $15-(-2x)=22+3x$

$15+2x=22+3x$

$15-22=3x-2x$

$-7=x$

b.

$5(17-3x)+24=4$

$5(17-3x)=4-24=-20$

$17-3x=-20:5=-4$

$3x=17-(-4)=21$

$x=21:3=7$

c.

$42:(x^2+5)=3$

$x^2+5=42:3=14$

$x^2=14-5=9=3^2=(-3)^2$

$\Rightarrow x=3$ hoặc $x=-3$

d.

$73-3x^2=5^6:(-5)^4=(-5)^6:(-5)^4=(-5)^2=25$

$3x^2=73-25=48$
$x^2=48:3=16=4^2=(-4)^2$

$\Rightarrow x=4$ hoặc $x=-4$

a: Ta có: \(\left(3x-2\right)\left(2x-1\right)-\left(6x^2-3x\right)=0\)

\(\Leftrightarrow2x-1=0\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x^3-\left(x+1\right)\left(x^2-x+1\right)=x\)

\(\Leftrightarrow x^3-x^3-1=x\)

hay x=-1

c: Ta có: \(56x^4+7x=0\)

\(\Leftrightarrow7x\left(8x^3+1\right)=0\)

\(\Leftrightarrow x\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)

d: Ta có: \(x^2-5x-24=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)

a, 2.x + 7 = 15

   2x         = 8

   x           = 4

b, 25 – 3.(6 – x) = 22

            3.(6-x)    = 3

               6-x      = 1

                  x      = 5

c, [(2x – 11) : 3 + 1].5 = 20

      (2x-11) : 3+1         = 4

       (2x-11):3              = 3

        2x-11                  = 1

        2x                       = 12

        x                         = 6

  e, 2 . 3x = 10 . 312 + 8 . 274

      6x      = 3120 + 2192

      6x      = 5312

        x      = 5312/6

g, x – 12 = (–8) + (–17)

    x  - 12 = -25

    x         = -13

Lần sau tách nhỏ nội dung câu hỏi ra nha em, chứ trả lời thế này biếng lắm '^^ Chị làm chỉ mang tính tham khảo kết quả thôi, còn cụ thể thì em tách từng bước một ra he :>

a, \(2\cdot x+7=15\)

\(\Leftrightarrow2\cdot x=8\)

\(\Leftrightarrow x=4\)

Vậy x = 4.

b, \(25-3\cdot\left(6-x\right)=22\)

\(\Leftrightarrow3\cdot\left(6-x\right)=3\)

\(\Leftrightarrow6-x=1\)

\(\Leftrightarrow x=5\)

Vậy x = 5.

c, \(\left[\left(2x-11\right):3+1\right]\cdot5=20\)

\(\Leftrightarrow\left(2x-11\right):3+1=4\)

\(\Leftrightarrow\left(2x-11\right):3=3\)

\(\Leftrightarrow2x-11=9\)

\(\Leftrightarrow2x=20\)

\(\Leftrightarrow x=10\)

Vậy x = 10.

d, \(\left(25-2x\right)\cdot3:5-32=42\)

\(\Leftrightarrow\)\(\frac{3\cdot\left(25-2x\right)}{5}=74\)

\(\Leftrightarrow3\cdot\left(25-2x\right)=370\)

\(\Leftrightarrow25-2x=\frac{370}{3}\)

\(\Leftrightarrow2x=-\frac{295}{3}\)

\(\Leftrightarrow x\approx49\)

Vậy \(x\approx49\) .

e, \(2\cdot3x=10\cdot312+8\cdot274\)

\(\Leftrightarrow6x=5312\)

\(\Leftrightarrow x=5312:6\approx885\)

Vậy \(x\approx885\) .

g, \(x-12=\left(-8\right)+\left(-17\right)\)

\(\Leftrightarrow x-12=-25\)

\(\Leftrightarrow x=-25+12=-13\)

Vậy x = -13.

h, \(7-2x=18-3x\)

\(\Leftrightarrow-2x+3x=18-7\)

\(\Leftrightarrow x=11\)

Vậy \(x=11\) .

i, \(3\cdot\left(x+5\right)-x-11=24\)

\(\Leftrightarrow3x+15-x-11=24\)

\(\Leftrightarrow2x=24+11-15\)

\(\Leftrightarrow2x=20\)

\(\Leftrightarrow x=10\)

Vậy \(x=10\) .

\(21,\frac{2}{x-1}\le\frac{5}{2x-1}\left(x\ne1;x\ne\frac{1}{2}\right)\)

\(\Leftrightarrow\frac{2}{x-1}-\frac{5}{2x-1}\le0\)

\(\Leftrightarrow\frac{4x-2-5x+5}{\left(x-1\right)\left(2x-1\right)}\text{≤}0\)

\(\Leftrightarrow\frac{-x+3}{\left(x-1\right)\left(2x-1\right)}\text{≤}0\)

x -x+3 x-1 2x-1 VT -∞ +∞ 1/2 1 3 0 0 0 | | || | | || | | 0 - + + + + + - - - + + + + + + - -

Vậy \(\frac{-x+3}{\left(x-1\right)\left(2x-1\right)}\le0\Leftrightarrow x\in\left(\frac{1}{2};1\right)\cup[3;+\text{∞})\)

23,24 tương tự 21

\(25,2x^2-5x+2< 0\) (1)

Ta có: \(\left\{{}\begin{matrix}2x^2-5x+2=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{2}\end{matrix}\right.\\a=2>0\end{matrix}\right.\) \(\Leftrightarrow\frac{1}{2}< x< 2\)

\(26,-5x^2+4x+12< 0\)

\(\left\{{}\begin{matrix}-5x^2+4x+12=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\frac{6}{5}\end{matrix}\right.\\a=-5< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>2\\x< -\frac{6}{5}\end{matrix}\right.\)

\(27,16x^2+40x+25>0\)

\(\left\{{}\begin{matrix}16x^2+40x+25=0\Leftrightarrow x=-\frac{5}{4}\\a=16>0\end{matrix}\right.\)

\(\Leftrightarrow x\ne-\frac{5}{4}\)

\(28,-2x^2+3x-7\ge0\)

\(\left\{{}\begin{matrix}-2x^2+3x-7=0\left(vo.nghiem\right)\\a=-2< 0\end{matrix}\right.\)

\(\Rightarrow-2x^2+3x-7< 0\) ∀x

=> bpt vô nghiệm

\(29,3x^2-4x+4\ge0\)

\(\left\{{}\begin{matrix}3x^2-4x+4=0\left(vo.nghiem\right)\\a=3>0\end{matrix}\right.\)

=> \(3x^2-4x+4>0\) => bpt vô số nghiệm

\(30,x^2-x-6\le0\)

\(\left\{{}\begin{matrix}x^2-x-6=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\\a=1>0\end{matrix}\right.\)

\(\Rightarrow-2\le x\le3\)