Bài 1:

a. \(=[(3x+(4y-5z)][3x-(4y-5z)]=(3x)^2-(4y-5z)^2\)

\(=9x^2-(16y^2-40yz+25z^2)=9x^2-16y^2+40yz-25z^2\)

b.

\(=(3a-1)^2+2(3a-1)(3a+1)+(3a+1)^2=[(3a-1)+(3a+1)]^2=(6a)^2=36a^2\)

Bài 2:

\((x+y+z)^3=[(x+y)+z]^3=(x+y)^3+3(x+y)^2z+3(x+y)z^2+z^3\)

\(=[x^3+y^3+3xy(x+y)]+3(x+y)z(x+y+z)+z^3\)

\(=x^3+y^3+z^3+3xy(x+y)+3(x+y)z(x+y+z)\)

\(=x^3+y^3+z^3+3(x+y)(xy+zx+zy+z^2)\)

\(=x^3+y^3+z^3+3(x+y)(z+x)(z+y)\) (đpcm)

a) \(=6a-3+15-5a=a+12\)

b) \(=25x-12x+4+35-14x=-x+39\)

d) \(=2ab+8a^2-b^2-4ab+2ab-6a^2=2a^2-b^2\)

e) \(=x+x^2-x^3+x^4-x^5+1+x-x^2+x^3-x^4=-x^5+2x+1\)

f) \(=6y^3-3y^2+y-y+y^2-y^3-y^2+y=5y^3-3y^2+y\)

a) 3( 2a -1) +5( 3-a)

   = 3. 2a -3.1 +5. 3- 5.a

   = 6a -3+ 15-5a

   =(6a -5a )+ (-3+ 15)

b) 25x - 4(3x - 1) +7(5 - 2x)

   = 25x -4.3x + 4.1 + 7.5 - 7.2

   =25x - 12x + 4 +35 - 14x

   = (25x-12x-14x)+(4+35)

   = -x=39

c) -12x3 -x1-2x-18x2

   = -36x-x-2x-36x

   = -75x

d) (2a-b)(b+4a)+2a(b-3a)

   = 2ab+2a4a-bb-b4a+2ab-2a3b

   = 2ab+8a²-b²-4ab+2ab-6a²

   =(2ab-4ab+2ab)+(8a²-6a²)-b²

   = 2a²-b²

e) (x+1)(2+x-x2+x3-x4)

   = (x+1)(2-2x)

   = x2-x2x+1.2-1.2x

   =(2x-2x)-2x²+2

   = -2x²+2

a) A = ( 6 a   +   2 ) 2 .             b) B = 1 4 ( 3 x + 1 ) 2 .  

a) Gợi ý: a 2 − 5 a + 4 = ( a − 1 ) ( a − 4 ) ; a 2 + 3 a − 4 = ( a − 1 ) ( a + 4 )  

Ta rút gọn được A = a + 1 a − 4  

b) Thay a = 5 vào biểu thức A tìm được A = 6

c) Ta biến đổi A = a + 1 a − 4 = 1 + 5 a − 4  

⇒ A ∈ ℤ ⇒ a ∈ − 1 ;   3 ;   5 ;   9

\(log_{a^3}b.log_ba=\dfrac{1}{3}.log_ab.log_ba=\dfrac{1}{3}\)

\(log_{a^{10}}b^5.log_{b^3}a^9=\dfrac{1}{10}.5.log_ab.\dfrac{1}{3}.9.log_ba=\dfrac{3}{2}\)

\(log_{a^{107}}b^{101}.log_{b^{303}}a^{428}=\dfrac{1}{107}.101.log_ab.\dfrac{1}{303}.428.log_ba=\dfrac{4}{3}.log_ab.log_ba=\dfrac{4}{3}\)

a: \(log_{a^3}b\cdot log_ba=\dfrac{1}{3}\cdot log_ab\cdot log_ba=\dfrac{1}{3}\)

b: \(log_{a^{10}}b^5\cdot log_{b^3}a^9\)

\(=\dfrac{1}{10}\cdot log_ab^5\cdot\dfrac{1}{3}\cdot log_ba^9\)

\(=\dfrac{1}{30}\cdot5\cdot log_ab\cdot9\cdot log_ba=\dfrac{45}{30}=\dfrac{3}{2}\)

c: \(log_{a^{107}}b^{101}\cdot log_{b^{303}}a^{428}\)

\(=\dfrac{1}{107}\cdot log_ab^{101}\cdot\dfrac{1}{303}\cdot log_ba^{428}\)

\(=\dfrac{1}{107}\cdot101\cdot log_ab\cdot\dfrac{1}{303}\cdot428\cdot log_ba\)

\(=4\cdot\dfrac{1}{3}=\dfrac{4}{3}\)

a)\(\sqrt{\left(2-\sqrt{3}\right)^2}=2-\sqrt{3}\)

b)\(\sqrt{\left(2-\sqrt{11}\right)^2}=2-\sqrt{11}\)

c)\(2\sqrt{a^2}=2a\) vì a≥0

a: \(=2x^2-6x+x-3-20x+8x^2\)

\(=10x^2-25x-3\)

b: \(=x^2+4x+4-2\left(x^2-9\right)+10\)

\(=x^2+4x+14-2x^2+18\)

\(=-x^2+4x+32\)

\(A=\left(x+2\right)^3-\left(x-2\right)^3-12x^2=x^3+6x^2+12x+8-x^3+6x^2-12x+8-12x^2=16\)

\(A=a+2\sqrt{a}-3\sqrt{a}-6-a-2\sqrt{a}-1+3\sqrt{a}\)

\(A=-7\)

Ta có: \(A=\left(\sqrt{a}+2\right)\left(\sqrt{a}-3\right)-\left(\sqrt{a}+1\right)^2+\sqrt{9a}\)

\(=a-3\sqrt{a}+2\sqrt{a}-6-a-2\sqrt{a}-1+3\sqrt{a}\)

\(=-7\)