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

300

300

= 300 

1) \(\left(x+1\right)^2=x^2+2x+1\)

2) \(\left(2x+1\right)^2=4x^2+4x+1\)

3) \(\left(2x+y\right)^2=4x^2+4xy+y^2\)

4) \(\left(2x+3\right)^2=4x^2+12x+9\)

5) \(\left(3x+2y\right)^2=9x^2+12xy+4y^2\)

6) \(\left(2x^2+1\right)^2=4x^4+4x^2+1\)

7) \(\left(x^3+1\right)^2=x^6+2x^3+1\)

8) \(\left(x^2+y^3\right)^2=x^4+2x^2y^3+y^6\)

9) \(\left(x^2+2y^2\right)^2=x^4+4x^2y^2+4y^4\)

10) \(\left(\dfrac{1}{2}x+\dfrac{1}{3}y\right)^2=\dfrac{1}{4}x^2+\dfrac{1}{3}xy+\dfrac{1}{9}y^2\)

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: \(x^2-5x-24=0\)

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

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

Đáp án cần chọn là: D

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

2:

a: \(2x-138=2^3\cdot3^2\)

=>\(2x-138=8\cdot9=72\)

=>2x=138+72=210

=>x=105

b: \(5\cdot\left(x+35\right)=515\)

=>x+35=103

=>x=103-35=68

c: \(814-\left(x-305\right)=712\)

=>x-305=814-712=102

=>x=102+305=407

d: \(20-\left[7\left(x-3\right)+4\right]=2\)

=>7(x-3)+4=18

=>7(x-3)=14

=>x-3=2

=>x=5

e: \(9^{x-1}=9\)

=>x-1=1

=>x=2

f: \(5^{x-2}-3^2=2^4-\left(2^8\cdot2^2-2^{10}\cdot2^2\right)\)

=>\(5^{x-2}-9=16-1024+4096\)

=>\(5^{x-2}=3097\)

=>\(x-2=log_53097\)

=>\(x=2+log_53097\)

Bài 1

a/ \(x\left(x^2+1\right)+2\left(x^2+1\right)=0\)

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

b/

\(\Leftrightarrow x^3-6x^2+9x+5x^2-30x+45=0\)

\(\Leftrightarrow x\left(x-3\right)^2+5\left(x-3\right)^2=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-3\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)

1.

c/ \(\Leftrightarrow x^3+2x^2+2x+x^2+2x+2=0\)

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

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

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

d/

\(\Leftrightarrow x^4+x^3-2x^2-x^3-x^2+2x+4x^2+4x-8=0\)

\(\Leftrightarrow x^2\left(x^2+x-2\right)-x\left(x^2+x-2\right)+4\left(x^2+x-2\right)=0\)

\(\Leftrightarrow\left(x^2-x+4\right)\left(x^2+x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+4=0\left(vn\right)\\x^2+x-2=0\end{matrix}\right.\)

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

Bài 1:

e/ \(\Leftrightarrow x^4+2x^2-8x+5=0\)

\(\Leftrightarrow x^4-2x^3+x^2+2x^3-4x^2+2x+5x^2-10x+5=0\)

\(\Leftrightarrow x^2\left(x-1\right)^2+2x\left(x-1\right)^2+5\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x^2+2x+5\right)\left(x-1\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x+5=0\left(vn\right)\\x=1\end{matrix}\right.\)

Bài 2:

a/ Đặt \(x^2-5x=t\)

\(t^2+10t+24=0\Rightarrow\left[{}\begin{matrix}t=-4\\t=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x=-4\\x^2-5x=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x+4=0\\x^2-5x+6=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=4\\x=2\\x=3\end{matrix}\right.\)

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 \(x=10\) .

Thanks mn ^v^