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

Lời giải:
$(2x+3)^2+(2x+5)^2=4x^2+12x+9+4x^2+20x+25$

$=8x^2+32x+34$

\(m=\frac{2^5.15^3}{6^3.10^2}=\frac{2^5.\left(3.5\right)^3}{\left(2.3\right)^3.\left(5.2\right)^2}=\frac{2^5.3^3.5^3}{2^5.3^3.5^2}=5\)\(5\)

\(m=\frac{2^5.15^3}{6^3.10^2}=\frac{2^5.\left(3.5\right)^3}{\left(2.3\right)^3.\left(2.5\right)^2}=\frac{2^5.3^3.5^3}{2^3.3^3.2^2.5^2}=5\)

a, Với x ≠ 0,x ≠ ± 5 và x ≠ 5/2 thì 
P = [x/(x^2 - 25)  -  (x - 5)/(x^2 + 5x)]  : (2x - 5)/(x^2 + 5x) + x/(x - 5)
<=>P = [x/(x - 5)(x + 5)  -  (x - 5)/x(x+5)] . x(x + 5)/(2x - 5) + x/(x - 5)
=> P = [x^2 - (x - 5)^2]/x(x - 5)(x + 5) . x(x + 5)/(2x - 5) + x/(x - 5)
<=> P = (x - x + 5)(x + x - 5)/(x - 5)(2x - 5) + x/(x - 5)
<=> P = 5(2x - 5)/(x - 5)(2x - 5) + x/(x - 5)
<=> P = 5/(x - 5) + x/(x - 5)
<=> P = (5 + x)/(x - 5)
b, Với x ≠ 0,x ≠ ± 5 và x ≠ 5/2 (x ∈ Z) thì P ∈ Z <=> (5 + x)/(x - 5) ∈ Z
<=> (x - 5 + 10)/(x - 5) ∈ Z
<=> 1 + 10/(x - 5) ∈ Z
<=> 10/(x - 5) ∈ Z
<=> (x - 5) ∈ Ư(10)
<=> x - 5 = 10  <=> x = 15 (TM)
hoặc x - 5 = -10 <=> x = -5  (TM)
hoặc x - 5 = 5  <=> x = 10  (TM)
hoặc x - 5 = -5 <=> x = 0  (TM)
hoặc x - 5 = 2  <=> x = 7  (TM)
hoặc x - 5 = -2  <=> x = 3  (TM)
hoặc x - 5 = -1  <=> x = 4  (TM)
hoặc x - 5 = 1  <=> x = 6  (TM)
Vậy x ∈ {-5,0,3,4,6,7,10,15} thì P ∈ Z

đk: x>=0; x khác 3

a) \(P=\frac{\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-3}=\frac{\sqrt{x}-3-5+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{x+\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)

\(P=\frac{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+4}{\sqrt{x}+2}\)

b) \(P=\frac{\sqrt{x}+2+2}{\sqrt{x}+2}=1+\frac{2}{\sqrt{x}+2}\)

ta có: \(x\ge0\Rightarrow\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\frac{2}{\sqrt{x}+2}\le1\Leftrightarrow1+\frac{2}{\sqrt{x}+2}\le2\Rightarrow MaxP=2\Rightarrow x=0\)

toán lớp 9 mik mới học lớp 6 thôi

=\(\dfrac{5}{11}\left(\dfrac{5}{7}+\dfrac{2}{7}\right)+\dfrac{6}{11}\)

=\(\dfrac{5}{11}\times1+\dfrac{6}{11}\)

=\(\dfrac{11}{11}\)=1

\(\dfrac{5}{11}\cdot\dfrac{5}{7}+\dfrac{5}{11}\cdot\dfrac{2}{7}+\dfrac{6}{11}=\dfrac{5}{11}\cdot\left(\dfrac{5}{7}+\dfrac{2}{7}\right)+\dfrac{6}{11}=\dfrac{5}{11}\cdot1+\dfrac{6}{11}=\dfrac{5}{11}+\dfrac{6}{11}=1\)