\(\left(2x-1\right)^{10}=\left(2x-1\right)^{11}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=1\\2x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)

Vậy ...........

\(\Rightarrow\left(2x-1\right)^{11}-\left(2x-1\right)^{10}=0\)

\(\Rightarrow\left(2x-1\right)^{10}.\left[\left(2x-1\right)-1\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(2x-1\right)^{10}=0\\\left(2x-1\right)-1=0\end{cases}\Rightarrow\orbr{\begin{cases}2x-1=0\\2x-1=1\end{cases}}}\)

                                                     \(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}\)

Vậy \(x\in\left\{\frac{1}{2};1\right\}\)

1) \(\left|x\right|< 10\)

\(\Leftrightarrow-10< x< 10\)

2) \(\left|x\right|>11\)

\(\Leftrightarrow\left[{}\begin{matrix}x< -11\\x>11\end{matrix}\right.\)

3) \(\left|x\right|\ge2x\left(\forall x\ge0\right)\)

\(\)\(\Leftrightarrow\left[{}\begin{matrix}x\le-2x\\x\ge2x\end{matrix}\right.\) 

\(\Leftrightarrow\left[{}\begin{matrix}3x\le0\\x\le0\end{matrix}\right.\)

\(\Leftrightarrow x=0\) \(\left(thỏa.đk:x\ge0\right)\)

4) \(\left|x\right|\le-3x\left(\forall x\le0\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\left(-3x\right)\\x\le-3x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x\le0\\4x\le0\end{matrix}\right.\)

\(\Leftrightarrow x\le0\) \(\left(thỏa.đk\right)\)

a)(x+2).(x+3)-(x-2).(x+5)=10

  ( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10

 x^2 +3x+2x+6-x^2 -5x+2x+10-10=0

 2x+6=0

2x=-6

x=-3

Xét cấp số cộng 1, 6, 11, ..., 96.

Ta có: 96 = 1 + 5(n − 1) ⇒ n = 20

Suy ra

Và 2x.20 + 970 = 1010

Từ đó x = 1

\(\Leftrightarrow\)4.(x²+2x+1)+4x+4x²+1-8(x²+10x-x-10)=11

\(\Leftrightarrow\)4x²+8x+4+4x+4x²+1-8x²-80x+8x+80=11

\(\Leftrightarrow\)-60x=-74

\(\Leftrightarrow\)x=\(\frac{37}{30}\)

a/ x = 4

a) 2^x.(1 + 2³) = 144

2^x . 9 = 144

2^x = 16

=> x = 4

b) (2x - 1)¹⁰ = (2x - 1)¹⁰⁰

(2x - 1)¹⁰⁰ - (2x - 1)¹⁰  = 0

 (2x - 1)¹⁰.[ (2x - 1)⁹⁰ - 1] = 0

=>  (2x - 1)¹⁰ = 0 hoặc  (2x - 1)⁹⁰ - 1 = 0

=> 2x = 1   hoặc    (2x - 1)⁹⁰ = 1

=> x = \(\frac{1}{2}\)  hoặc   \(2x-1=\orbr{\begin{cases}1\\-1\end{cases}}\)

=>                             \(2x=\orbr{\begin{cases}2\\0\end{cases}}\)

=> x = {\(\frac{1}{2};1;0\)}

Bài 3:

a: \(S=1+5^2+5^4+\cdots+5^{200}\)

=>25S=\(5^2+5^4+5^6+\cdots+5^{202}\)

=>25S-S=\(5^2+5^4+\cdots+5^{202}-1-5^2-\cdots-5^{200}\)

=>24S=\(5^{202}-1\)

=>\(S=\frac{5^{202}-1}{24}\)

b: \(4^{30}=\left(2^2\right)^{30}=2^{60}=2^{30}\cdot2^{30}=8^{10}\cdot4^{15}\)

\(3\cdot24^{10}=3\cdot3^{10}\cdot8^{10}=8^{10}\cdot3^{11}\)

mà \(4^{15}>3^{11}\)

nên \(4^{30}>3\cdot24^{10}\)

=>\(2^{30}+3^{30}+4^{30}>3\cdot24^{10}\)

Bài 2:

a: |2x-3|>5

=>\(\left[\begin{array}{l}2x-3>5\\ 2x-3<-5\end{array}\right.\Rightarrow\left[\begin{array}{l}2x>8\\ 2x<-2\end{array}\right.\Rightarrow\left[\begin{array}{l}x>4\\ x<-1\end{array}\right.\)

c: |3x-1|<=7

=>-7<=3x-1<=7

=>-6<=3x<=8

=>\(-2\le x\le\frac83\)

d: \(\left|3x-5\right|+\left|2x+3\right|=7\) (1)

TH1: \(x<-\frac32\)

=>2x+3<0; 3x-5<0

(1) sẽ trở thành: -2x-3-3x+5=7

=>-5x+2=7

=>-5x=5

=>x=-1(loại)

TH2: -3/2<=x<5/3

=>2x+3>=0; 3x-5<0

(1) sẽ trở thành: 2x+3-3x+5=7

=>-x+8=7

=>-x=-1

=>x=-1(nhận)

TH3: x>=5/3

=>2x+3>0; 3x-5>=0

(1) sẽ trở thành: 2x+3+3x-5=7

=>5x-2=7

=>5x=9

=>x=9/5(nhận)