a) \(x^2+10x+26+y^2+2y\)

= \(x^2+10x+25+y^2+2y+1\)

= \(\left(x+5\right)^2+\left(y+1\right)^2\)

b) \(x^2-2xy+2y^2+2y+1\)

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

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

c) \(z^2-6z+5-t^2-4t\)

= \(z^2-6z+9-\left(t^2+4t+4\right)\)

= \(\left(z-3\right)^2-\left(t+2\right)^2\)

d) \(4x^2-12x-y^2+2y+1\)

Hình như câu này sai đề -_-

ko có quả này vì quả này có ai biết đâu

a) \(\dfrac{15^{30}}{45^{15}}=\dfrac{15^{30}}{3^{15}.15^{15}}=\dfrac{15^{15}}{3^{15}}=5^{15}\)

b) \(\dfrac{2^{15}.9^4}{6^6.8^3}=\dfrac{8^5.3^8}{2^6.3^6.8^3}=\dfrac{8^2.3^2}{2^6}=\dfrac{2^6.3^2}{2^6}=3^2=9\)

c) \(\dfrac{14^{10}.21^{32}.35^{48}}{10^{10}.15^{32}.7^{96}}=\dfrac{2^{10}.7^{10}.3^{32}.7^{32}.5^{48}.7^{48}}{2^{10}.5^{10}.3^{32}.5^{32}.7^{96}}\)

= \(\dfrac{2^{10}.7^{58}.3^{32}.5^{48}}{2^{10}.5^{42}.3^{32}.7^{96}}=\dfrac{5^6}{7^{38}}\) ( Câu này làm bừa, có lẽ sai đấy :)) )

2. So sánh

a) 3²⁰⁰ = 9¹⁰⁰

2³⁰⁰ = 8¹⁰⁰

Vì 9¹⁰⁰ > 8¹⁰⁰ nên 3²⁰⁰ < 2³⁰⁰

b) 9¹² = 729⁴

26⁸ = 676⁴

Vì 729⁴ > 676⁴ nên 9¹² > 26⁸

c) 2²⁴ = 8⁸

3¹⁶ = 9⁸

Vì 8⁸ < 9⁸ nên 2²⁴ < 3¹⁶

1 nha Phạm Ngọc Huyền

x + y + z = 0

x + y = -z

( x + y )³ = ( -z )³

x³ + 3x²y +3xy² + y³ = -z³

x³ + y³ + z³ = 3x²y - 3xy²

x³ + y³ + z³ = - 3xy ( x + y )

x³ + y³ + z³ = -3xy. ( -z )

x³ + y³ + z³ = 3xyz ( đpcm )

B1:

a) \(\left(10x+9\right)x-\left(5x-1\right)\left(2x+3\right)=8\)

\(10x^2+9x-10x^2-15x+2x+3-8=0\)

\(-4x-5=0\)

\(-4x=5\Leftrightarrow x=-\dfrac{5}{4}\)

b) \(\left(3x-5\right)\left(7-5x\right)+\left(5x+2\right)\left(3x-2\right)-2=0\)

\(21x-15x^2-35+25x+15x^2-10x+6x-4-2=0\)

\(42x-41=0\)

\(x=\dfrac{41}{42}\)

\(x^3-4x^2+2x^2-1\)

= \(x^3-2x^2+2x-1\)

= \(x^3-x^2-x^2+x+x-1\)

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

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

Ta thấy \(x^2-x+1=\left(x-\dfrac{1}{4}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Vậy x=1