\(x-5\)

\(=\left(\sqrt{x}\right)^2-\left(\sqrt{5}\right)^2\)

\(=\left(\sqrt{x}-\sqrt{5}\right)\left(\sqrt{x}+\sqrt{5}\right)\)

Sửa đề: Đưa thừa số vào trong dấu căn

a: \(3\sqrt{x^2}=\sqrt{3^2\cdot x^2}=\sqrt{9x^2}\)

b: \(-5\sqrt{y^4}=-\sqrt{5^2\cdot y^4}=-\sqrt{25y^4}\)

c: \(3\sqrt{5x}=\sqrt{3^2\cdot5x}=\sqrt{45x}\)

d: \(x\sqrt{7}=\sqrt{x^2\cdot7}=\sqrt{7x^2}\)

\(\sqrt{48\cdot45}=12\sqrt{15}\\ \sqrt{225\cdot17}=15\sqrt{17}\\ \sqrt{a^3b^7}=\left|ab^3\right|\sqrt{ab}=ab^3\sqrt{ab}\\ \sqrt{x^5\left(x-3\right)^2}=\left|x^2\left(x-3\right)\right|\sqrt{x}=x^2\left(x-3\right)\sqrt{x}\)

\(\sqrt{48\cdot45}=4\sqrt{3}\cdot3\sqrt{5}=12\sqrt{15}\)

\(\sqrt{225\cdot17}=15\sqrt{17}\)

Bài làm:

a) \(x^2-7=\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\)

b) \(4x^2-5=\left(2x-\sqrt{5}\right)\left(2x+\sqrt{5}\right)\)

c) \(3x^2-1=\left(x\sqrt{3}-1\right)\left(x\sqrt{3}+1\right)\)

d) \(x-1=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)

e) \(x-4=\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)

f) \(9x-4=\left(3\sqrt{x}-2\right)\left(3\sqrt{x}+2\right)\)

\(A=4^4\cdot9^5=\left(2^2\right)^4\cdot\left(3^2\right)^5=2^8\cdot3^{10}\)

\(A=4^4\cdot9^5\)

\(A=\left(2^2\right)^4\cdot\left(3^2\right)^5\)

\(A=2^{2\cdot4}\cdot3^{2\cdot5}\)

\(A=2^8\cdot3^{10}\)

=36 mũ 9 nhé

a: =2^100*117*5^2024

b: =7^100(1+7+7^2)=57*7^100

Ta có:

P= (x+1)(x+3)(x+5)(x+7)+15

=((x+1)(x+7))((x+3)(x+5))+15

=(x^2+8x+7)(x^2+8x+15)+15

Đặt t=x^2+8x+11, ta có:

P=(t-4)(t+4)+15

P=t^2-16+15

P=t^2-1=(t-1)(t+1)

Vậy: P=(x^2+8x+10)(x^2+8x+12)

          =(x^2+8x+10)(x+6)(x+2)

a)x-9=\(\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\)