1: =>\(5^{2x-3}=5^2\cdot3+5^2\cdot2=5^2\cdot5=5^3\)

=>2x-3=3

=>2x=6

=>x=3

2: \(41-2^{x+1}=9\)

=>\(2^{x+1}=32\)

=>x+1=5

=>x=4

3: =>\(4^{x+2}=65-1=64\)

=>x+2=3

=>x=1

\(5^{2x-3}-2.5^2=5^2.3\\ 5^{2x-3}=5^2.3+5^2.2\\ 5^{2x-3}=5^2.\left(3+2\right)\\ 5^{2x-3}=5^2.5\\ 5^{2x-3}=5^3\\ \Rightarrow2x-3=3\\ 2x=3+3\\ 2x=6\\ x=\dfrac{6}{2}\\ Vậy:x=3\)

\(41-2^{x+1}=9\\ 2^{x+1}=41-9\\ 2^{x+1}=32=2^5\\ \Rightarrow x+1=5\\ \Leftrightarrow x=5-1=4\\ ---\\ 65-4^{x+2}=2014^0\\ 65-4^{x+2}=1\\ 4^{x+2}=65-1\\ 4^{x+2}=64=4^3\\ \Rightarrow x+2=3\\ Vậy:x=1\)

a/50

a/50 

đáp án:d/49

a) Ta có: \(\left(x-1\right)^2=1\)

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

Vậy: \(x\in\left\{0;2\right\}\)

b) Ta có: x(3x+9)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)

Vậy: \(x\in\left\{0;-3\right\}\)

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

d) \(\left|x\right|=-2,7\Rightarrow x\in\varnothing\) 

l) \(\left|x+\dfrac{3}{4}\right|-5=-2\Rightarrow\left|x+\dfrac{3}{4}\right|=3\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=3\\x+\dfrac{3}{4}=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3-\dfrac{3}{4}\\x=-3-\dfrac{3}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=\dfrac{15}{4}\end{matrix}\right.\)

Đính chính câu l \(x=-\dfrac{15}{4}\) không phải \(x=\dfrac{15}{4}\)

c) x = 3,5 hoặc -3,5

d) x = -2,7

e) x = -1,27

f) x = 0

g) x = -2

h) x = 0

l) x = 9/4

   x = -15/4

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

Ta có: x . (x + 1) : 2 = 78

=>  x . (x + 1) = 156

=> x  . ( x + 1) = 12. 13

=> x = 12

1) 1 + 2 + 3 + ...+ x = 78

Số lượng số hạng là: 

(x - 1) : 1 + 1 = x (số hạng)

=> (1 + x) . x : 2 = 78     

     (1 + x) . x     =  78 . 2

      x . (1 + x)    =   156

      x . (1 + x)    = 12. 13

=>  x =               = 12

2) (2 + x) + (4 + x) + (6 + x) +...+ ( 52 + x ) = 780

     ( x + x  + x +...+ x ) +  ( 2 + 4 + 6 +...+ 52 ) = 780

     26 . x                       + ( 2 + 4 ) . 26 : 2        = 780

     26 . x                       +          78                    = 780

     26 . x                                                            = 780 -7 8

     26 . x                                                            = 702

          x                                                              =  702 : 26

          x                                                              =       27

              -------------chúc bạn học tốt------------

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

Ta có: x . (x + 1) : 2 = 78

=>  x . (x + 1) = 156

=> x  . ( x + 1) = 12. 13

=> x = 12

\(-\frac{3}{4}\left(-x+52\right)-\frac{1}{3}=-2\left(\frac{1}{3}-x\right)+\frac{1}{4}\)

\(\frac{3}{4}x-39-\frac{1}{3}=2x-\frac{2}{3}+\frac{1}{4}\)

 \(\frac{3}{4}x-\frac{118}{3}=2x-\frac{5}{12}\)

\(\frac{3}{4}x-\frac{118}{3}-2x+\frac{5}{12}=0\)

\(-\frac{467}{12}-\frac{5}{4}x=0\)

\(x=-\frac{467}{15}\)

Bài 1 :

Gọi \(A=5+5^2+5^3+...+5^{98}+5^{99}\\ 5A=5^2+5^3+5^4+...+5^{99}+5^{100}\\ 5A-A=\left(5^2+5^3+5^4+...+5^{99}+5^{100}\right)-\left(5+5^2+5^3+...+5^{98}+5^{99}\right)\\ 4A=5^{100}-5\\ A=\dfrac{5^{100}-5}{4}\)

Bài 2:

\(\left(12x-4\right)\cdot8^{2022}=4\cdot8^{2023}\\ 12x-4=4\cdot8^{2023}:8^{2022}\\ 12x-4=4\cdot8\\ 12x-4=32\\ 12x=36\\ x=3\)

2:

a: =-(x^2-12x-20)

=-(x^2-12x+36-56)

=-(x-6)^2+56<=56

Dấu = xảy ra khi x=6

b: =-(x^2+6x-7)

=-(x^2+6x+9-16)

=-(x+3)^2+16<=16

Dấu = xảy ra khi x=-3

c: =-(x^2-x-1)

=-(x^2-x+1/4-5/4)

=-(x-1/2)^2+5/4<=5/4

Dấu = xảy ra khi x=1/2

1) 

a) \(A=x^2+4x+17\)

\(A=x^2+4x+4+13\)

\(A=\left(x+2\right)^2+13\) 

Mà: \(\left(x+2\right)^2\ge0\) nên \(A=\left(x+2\right)^2+13\ge13\)

Dấu "=" xảy ra: \(\left(x+2\right)^2+13=13\Leftrightarrow x=-2\)

Vậy: \(A_{min}=13\) khi \(x=-2\)

b) \(B=x^2-8x+100\)

\(B=x^2-8x+16+84\)

\(B=\left(x-4\right)^2+84\)

Mà: \(\left(x-4\right)^2\ge0\) nên: \(A=\left(x-4\right)^2+84\ge84\)

Dấu "=" xảy ra: \(\left(x-4\right)^2+84=84\Leftrightarrow x=4\)

Vậy: \(B_{min}=84\) khi \(x=4\)

c) \(C=x^2+x+5\)

\(C=x^2+x+\dfrac{1}{4}+\dfrac{19}{4}\)

\(C=\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}\)

Mà: \(\left(x+\dfrac{1}{2}\right)^2\ge0\) nên \(A=\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)

Dấu "=" xảy ra: \(\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}=\dfrac{19}{4}\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy: \(A_{min}=\dfrac{19}{4}\) khi \(x=-\dfrac{1}{2}\)

1:

a: A=x^2+4x+4+13

=(x+2)^2+13>=13

Dấu = xảy ra khi x=-2

b; =x^2-8x+16+84

=(x-4)^2+84>=84

Dấu = xảy ra khi x=4

c: =x^2+x+1/4+19/4

=(x+1/2)^2+19/4>=19/4

Dấu = xảy ra khi x=-1/2