\(P=\dfrac{\left(x-1\right)^2+3}{\left(x-1\right)^2+5}=\dfrac{\left(x-1\right)^2+5-2}{\left(x-1\right)^2+5}=1-\dfrac{2}{\left(x-1\right)^2+5}\)

\(\left(x-1\right)^2+5>=5\forall x\)

=>\(\dfrac{2}{\left(x-1\right)^2+5}< =\dfrac{2}{5}\forall x\)

=>\(-\dfrac{2}{\left(x-1\right)^2+5}>=-\dfrac{2}{5}\forall x\)

=>\(P=\dfrac{-2}{\left(x-1\right)^2+5}+1>=-\dfrac{2}{5}+1=\dfrac{3}{5}\forall x\)

Dấu '=' xảy ra khi x-1=0

=>x=1

\(Q=\dfrac{\left(2y+3\right)^2-3}{\left(2y+3\right)^2+4}=\dfrac{\left(2y+3\right)^2+4-7}{\left(2y+3\right)^2+4}=1-\dfrac{7}{\left(2y+3\right)^2+4}\)

\(\left(2y+3\right)^2+4>=4\forall y\)

=>\(\dfrac{7}{\left(2y+3\right)^2+4}< =\dfrac{7}{4}\forall y\)

=>\(-\dfrac{7}{\left(2y+3\right)^2+4}>=-\dfrac{7}{4}\forall y\)

=>\(Q=-\dfrac{7}{\left(2y+3\right)^2+4}+1>=-\dfrac{3}{4}\forall y\)

Dấu '=' xảy ra khi 2y+3=0

=>2y=-3

=>y=-3/2

\(F=\dfrac{\left(x-1\right)^2+5}{\left(x-1\right)^2+2}=\dfrac{\left(x-1\right)^2+2+3}{\left(x-1\right)^2+2}=1+\dfrac{3}{\left(x-1\right)^2+2}\)

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

=>\(\dfrac{3}{\left(x-1\right)^2+2}< =\dfrac{3}{2}\forall x\)

=>\(F=\dfrac{3}{\left(x-1\right)^2+2}+1< =\dfrac{5}{2}\forall x\)

Dấu '=' xảy ra khi x-1=0

=>x=1

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{x\left(x+1\right)}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\\ =1+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+...+\left(\dfrac{1}{x}-\dfrac{1}{x}\right)-\dfrac{1}{x+1}\\ =1-\dfrac{1}{x+1}\\ =\dfrac{x+1}{x+1}-\dfrac{1}{x+1}\\ =\dfrac{x}{x+1}\)

\(5^{2x-1}=125\\ =>5^{2x-1}=5^3\\ =>2x-1=3\\ =>2x=3+1\\ =>2x=4\\ =>x=\dfrac{4}{2}\\ =>x=2\)

Vậy: ... 

5^{2x - 1} = 125

5^{2x - 1} = 5³

2x - 1 = 3

2x      = 3 +1

2x      = 4

  x      = 4 :2

  x      = 2

a, 19 + 18 + 17 + 16 + 14 + 21 + 22 + 23 + 24 + 26

=  ( 19 + 21 ) + ( 18 + 22 ) + ( 17 + 23 ) + ( 16 + 24 ) + ( 14 + 26 )

=   40 + 40 + 40 + 40 + 40

=   40 x 5

=    200

 Chúc bé học tốt

a.

19 + 18 + 17 + 16 + 14 + 21 + 22 + 23 + 24 + 26

= (19 + 21) + (18 + 22) + (17 + 23) + ( 16 + 24) + (14 + 26)

= 40 + 40 + 40 + 40 + 40

= 40 x 5

= 200

câu b mk đã lm ở dưới r nhé

c) 64 x 4 + 18 x 4 + 9 x 8

= 64 x 4 + 18 x 4 + 9 x 2 x 4

= 4 x ( 64 + 18 + 9 x 2)

=4 x (82 + 18)

= 4 x 100

= 400

\(15,2-x+12=93\)

\(15,2-x=93-12\)

\(15,2-x=81\)

\(x=15,2-81\)

\(x=-65,8\)

Em có nhầm ko nhỉ, lớp 4 sao đã có số âm được

lớp 4 chưa học số thập phân và số âm, hình như em viết sai đề bài.

x.16 - x.4 - x = 2

x. (16 - 4 - 1) = 2

x.       11        = 2

x                    = 2. 11

                  x = 22

           Vậy x = 22

x.16 - x.4 - x = 2

x.(16-4-1) = 2

x.11 = 2

x = 2:11

x = \(\dfrac{2}{11}\)

3 x 9 + 18 x 2 + 2 x 9 + 9 

= 3 x  9 + 9 x 2 x 2 + 2 x 9 + 9

= 9 x (3 + 2 + 2 + 2 + 1)

= 9 x 10

= 90

\(=3\times9+9\times2\times2+2\times9+9\)

\(=3\times9+4\times9+2\times9+9\)

\(=9\times\left(3+4+2+1\right)\)

\(=9\times10\)

\(=90\)

A = {x l x ≤ 12 l x ϵ N}

A = {0; 1; 2; 3; 4; 5; 6; 7; 8; 9; 10; 11; 12}

C1:

\(A=\left\{0;1;2;3;4;5;6;7;8;9;10;11;12\right\}\)

\(C2:A=\left\{x\inℕ|x\le12\right\}\)

a.

\(\sqrt{x^2-4x+1}=x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2-4x+1=x^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\-4x+1=0\end{matrix}\right.\)

\(\Rightarrow x=\dfrac{1}{4}\)

b.

\(\sqrt{5x^2-2x+2}=x+1\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1\ge0\\5x^2-2x+2=\left(x+1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\4x^2-4x+1=0\end{matrix}\right.\)

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

\(\Rightarrow x=\dfrac{1}{2}\)

c.

\(\sqrt{x^2-8x+16}=4-x\)

\(\Leftrightarrow\sqrt{\left(4-x\right)^2}=4-x\)

\(\Leftrightarrow\left|4-x\right|=4-x\)

\(\Leftrightarrow4-x\ge0\)

\(\Rightarrow x\le4\)

d.

\(\sqrt{3x+1}=\sqrt{4x-3}\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x-3\ge0\\3x+1=4x-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\end{matrix}\right.\)

\(\Rightarrow x=4\)

\(\dfrac{28}{36}=\dfrac{28\div4}{36\div4}=\dfrac{7}{9}\)

\(\dfrac{21}{81}=\dfrac{21\div3}{81\div3}=\dfrac{7}{27}\)

\(\dfrac{150}{100}=\dfrac{150\div50}{100\div50}=\dfrac{3}{2}\)

\(\dfrac{70}{80}=\dfrac{70\div10}{80\div10}=\dfrac{7}{8}\)