a. C1: \(2^{10}\):\(2^8\)=1024:256=4

C2:\(2^{10}\):\(2^8\)=\(2^2\)=4

b C1:.\(4^6\):\(4^3\)=4096:64=64

C2: \(4^6\):\(4^3\)=\(4^3\)=64

c.C1:\(8^5\):\(8^4\)=32768:4096=8

C2: \(8^5\):\(8^4\)=\(8^1\)=8

d. C1: \(7^4\):\(7^4\)=2401:2401=1

C2;\(7^4\):\(7^4\)=\(7^0\)=1

b.\(3^{x+2}\)+4.\(3^{x+1}\)=7.\(3^6\)

\(3^2\).\(3^x\)+4.3.\(3^x\)=7.\(3^6\)

9.\(3^x\)+12.\(3^x\)=7.\(3^6\)

\(3^x\)(9+12)=7.\(3^6\)

\(3^x\).21=7.\(3^6\)

\(3^x\)=7.\(3^6\):21

\(3^x\)=\(3^6\):3

\(3^x\)=\(3^5\)

x=5

b, vế trái = (a-b)-(c-d)= a-b-c+d=(a+d)-(b+c)= vế phải

a. ta có Vế trái = (a-b)+(c-d)=a-b+c-d=(a+c)-(b+d)= vế trái

thay 11=x+1 ta có:

f(x)= \(x^{10}\)-11\(x^9\)+11\(x^8\)-11\(x^7\)+....+11\(x^2\)-11x+100

     =\(x^{10}\)-(x+1)\(x^9\)+(x+1)\(x^8\)-(x+1)\(x^7\)+...+(x+1)\(x^2\)-(x+1)x+100

     =\(x^{10}\)-\(x^{10}\)-\(x^9\)+\(x^9\)+\(x^8\)-\(x^8\)-\(x^7\)+......+\(x^3\)+\(x^2\)-\(x^2\)-x+100

     =-x+100

=> f(10)=-10+100=90      

\(\left(x+y+z\right)^3\)=\(\left[\left(x+y\right)+z\right]^3\)=\(\left(x+y\right)^3\)+3.\(\left(x+y\right)^2\).z+3.\(\left(x+y\right)^{ }\).\(z^2\)+\(z^3\)

                    =\(x^3\)+3\(x^2y\)+3x\(y^{^{ }2}\)+\(y^3\)+3.(\(x^2\)+2xy+\(y^2\)).z+ 3x\(z^2\)+3y\(z^2\)+\(z^3\)

                    =\(x^3\)+3\(x^2y\)+3x\(y^{^{ }2}\)+\(y^3\)+3\(x^2\)z+6xyz+3\(y^2\)z+3x\(z^2\)+3y\(z^2\)+\(z^3\)

\(\left(x+y+x\right)^3\)=\(\left(2x+y\right)^3\)=\(\left(2x\right)^3\)+3.\(\left(2x\right)^2\).y+3.2x.\(y^2\)+\(y^3\)

                    =8\(x^3\)+12\(x^2\)y+6x\(y^2\)+\(y^3\)

a,Q(2) = 4a+2b+c

Q(-1)=a-b+c

Ta có: Q(2)+Q(-1)= 4a+2b+c+a-b+c=5a+b+2c

mà 5a+b+2c=0 => Q(2)=-Q(-1)

Nên Q(2).Q(-1)\(\le\)0

A=\(\frac{2xy-x^2+z^2-y^2}{x^2+z^2-y^2+2xz}\)=\(\frac{z^2-\left(x^2-2xy+y^2\right)}{\left(x^2+2xz+z^2\right)-y^2}\)=\(\frac{z^2-\left(x-y\right)^2}{\left(x+z\right)^2-y^2}\)=\(\frac{\left(z+x-y\right)\left(z-x+y\right)}{\left(x+z-y\right)\left(x+z+y\right)}\)=\(\frac{\left(z-x+y\right)}{\left(x+z+y\right)}\)

Vì \(a^2\)\(\ge\)0; \(b^2\)\(\ge\)0; 1>0 nên ta áp dụng bất đẳng thức Cosi cho từng cặp ta có:

\(a^2\)+\(b^2\)\(\ge\)2\(\sqrt{a^2b^2}\)=2ab    (1)

\(a^2\)+1\(\ge\)2\(\sqrt{a^21}\)=2a          (2)

\(b^2\)+1\(\ge\)2\(\sqrt{b^2.1}\)=2b         (3)

Cộng vế với vế của (1); (2) và (3) ta có:

2\(a^2\)+2\(b^2\)+2\(\ge\)2ab+2a+2b

\(a^2\)+\(b^2\)+1\(\ge\)ab+a+b( chia cả 2 vế của Bất phương trình cho 2)

Dấu = xảy ra khi a=b=1