\(B< -1\\ \Leftrightarrow\dfrac{x+5}{-2}< -1\\ \Rightarrow x+5>2\\ \Rightarrow x>-3\)

Để B<-1 thì x+5<2

=>x<-3

1) \(206^2-36=206^2-6^2=\left(206-6\right)\left(206+6\right)\)

\(=200.212=42400\)

2) \(\left(x-4\right)^2.2-\left(12x+x^2\right).2=6\)

\(\Rightarrow x^2-16x+32-24x-2x^2=6\)

\(\Rightarrow40x=26\Rightarrow x=\dfrac{13}{20}\)

\(\Leftrightarrow\left[3x+3-2x-6\right]^3+\left[2x+6-x+5\right]^3+\left[x-5-3x-3\right]^3=0\)

=>(x-3)^3+(x+11)^3+(-2x-8)^3=0

Đặt x-3=a; x+11=b

=>a^3+b^3-(a+b)^3=0

=>3ab(a+b)=0

=>(x-3)(x+11)(2x+8)=0

hay \(x\in\left\{3;-11;-4\right\}\)

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)

a. x:(1/2+2/3)=6/5

=>x:7/6=6/5

=>x=6/5*7/6=>x=7/5

b.(x-1/2)-5(x-2/3)=3/2x

=>x-1/2-5x+10/3=3/2x

=>-4x+17/6=3/2x

=>17/6x=3/2x--4x

=>17/6=x(3/2+4)=>17/6=11/2x=>x=17/33

c.-5(x+1/5)-1/2(x-2/3)=x

=>-5x-1-1/2x+1/3=x=>-11/2x-2/3=x

=>-2/3=x+11/2x=>-2/3=x(1+11/2)=>-2/3=13/2x

=>x=-4/39

a) x : (1/2 + 2/3) = 6/5

=> x : 7/6 = 6/5

=> x = 6/5 x 7/6

      x = 7/5

b) , c) ko bít hihi