$\color{#ADD8E6}{\text{19 + 6 - 7 + 23}}$

$\color{#ADD8E6}{\text{= 25 - 7 + 23}}$

$\color{#ADD8E6}{\text{= 22 + 23}}$

$\color{#ADD8E6}{\text{= 45}}$

$\color{#ADD8E6}{\text{26 - 4 + 3 + 10 - 8}}$

$\color{#ADD8E6}{\text{= 22 + 3 + 10 - 8}}$

$\color{#ADD8E6}{\text{= 25 + 10 - 8}}$

$\color{#ADD8E6}{\text{= 35 - 8}}$

$\color{#ADD8E6}{\text{= 27}}$

$\color{#6495ED}{\text{4}}$$\color{#87CEFA}{\text{5}}$$\color{#ADD8E6}{\text{6}}$

màu mê nhoa =)

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

\(x^2-xy+x-y\\ =x\left(x-y\right)+\left(x-y\right)\\ =\left(x+1\right)\left(x-y\right)\)

:|

a) 

\(A\left(x\right)=3\left(x^2+2-4x\right)-2x\left(x-2\right)+17\\ =3x^2+6-12x-2x^2+4x+17\\ =\left(3x^2-2x^2\right)+\left(-12x+4x\right)+\left(6+17\right)\\ =x^2-8x+23\)

\(B\left(x\right)=3x^2-7x+3-3\left(x^2-2x+4\right)\\ =3x^2-7x+3-3x^2+6x-12\\ =\left(3x^2-3x^2\right)+\left(-7x+6x\right)+\left(3-12\right)\\ =-x-9\) 

b) 

\(N\left(x\right)-B\left(x\right)=A\left(x\right)\\ \Rightarrow N\left(x\right)=A\left(x\right)+B\left(x\right)\\ =\left(x^2-8x+23\right)+\left(-x-9\right)\\ =x^2-8x+23-x-9\\ =x^2-9x+14\)

\(A\left(x\right)-M\left(x\right)=B\left(x\right)\\ \Rightarrow M\left(x\right)=A\left(x\right)-B\left(x\right)\\ =\left(x^2-8x+23\right)-\left(-x-9\right)\\ =x^2-8x+23+x+9\\ =x^2-7x+32\) 

c) Thay `x=2` vào N(x) ta có:

\(N\left(2\right)=2^2-9\cdot2+14=0\)

=> x=2 là nghiệm của N(x) 

\(N\left(x\right)=x^2-9x+14\\ =x^2-2x-7x+14\\ =x\left(x-2\right)-7\left(x-2\right)\\ =\left(x-7\right)\left(x-2\right)\)

=> Nghiệm còn lại là: x - 7 = 0 => x = 7 

d) Thay `x=2/3` vào A(x) ta có:

\(A\left(x\right)=\left(\dfrac{2}{3}\right)^2-8\cdot\left(\dfrac{2}{3}\right)+23=\dfrac{4}{9}-\dfrac{16}{3}+23=\dfrac{4}{9}-\dfrac{48}{9}+\dfrac{207}{9}=\dfrac{163}{9}\)

\(\dfrac{3}{x+4}=\dfrac{2}{2x+1}\left(x\ne-4;x\ne-\dfrac{1}{2}\right)\\ \Leftrightarrow3\left(2x+1\right)=2\left(x+4\right)\\ \Leftrightarrow6x+3=2x+8\\ \Leftrightarrow6x-2x=8-3\\ \Leftrightarrow4x=5\\ \Leftrightarrow x=\dfrac{5}{4}\left(tm\right)\)

Gọi n là số người

5 x n+3=7 x n - 3

2 x n = 6

n=3

Vậy có 3 người và 5x3+3 = 18 quả táo

\(25< 5^{2x-1}< 5^5\)

\(5^2< 5^{2x-1}< 5^5\)

\(2< 2x-1< 5\)

\(2+1< 2x< 5+1\)

\(3< 2x< 6\)

\(\dfrac{3}{2}< x< 3\)

x=2

\(A=\dfrac{6n-1}{2n+3}=\dfrac{3\left(2n+3\right)-10}{2n+3}\\ =3-\dfrac{10}{2n+3}\)

Để A nguyên thì: \(\dfrac{10}{2n+3}\) nguyên

\(\Rightarrow10⋮\left(2n+3\right)\)

\(\Rightarrow2n+3\inƯ\left(10\right)=\left\{1;-1;2;-2;5;-5;10;-10\right\}\\ \Rightarrow2n\in\left\{-2;-4;-1;-5;2;-8;7;-13\right\}\\ \Rightarrow n\in\left\{-1;-2;-\dfrac{1}{2};-\dfrac{5}{2};1;-4;\dfrac{7}{2};-\dfrac{13}{2}\right\}\)

help me

\(\dfrac{5}{\sqrt{x}+3}\left(x\ge0;x\ne9\right)\\ =\dfrac{5\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{5\left(\sqrt{x}-3\right)}{\left(\sqrt{x}\right)^2-3^2}\\ =\dfrac{5\sqrt{x}-15}{x-9}\)