C

c

Bài 5 là quá kiểu hiển nhiên roài phá ra là xong mà :))))))

Bài 6:

\(A=\left(x-y\right)\left(x+y\right)=\left(87-13\right)\left(87+13\right)=74.100=7400\) 

\(B=\left(5x-3\right)^2=\left(5.2-3\right)^2=7^2=49\)

\(C=\left(2x-7\right)^2=\left(2.2-7\right)^2=\left(4-7\right)^2=\left(-3\right)^2=9\)

Bài 1:

a) \(\left(a+b\right)^2+\left(a-b\right)^2=a^2+2ab+b^2+a^2-2ab+b^2\)

\(=a^2+b^2+a^2+b^2=2a^2+2b^2=2\left(a^2+b^2\right)\)(Đpcm)

b) \(\left(a+b+c\right)^2=\left[\left(a+b\right)+c\right]^2=\left(a+b\right)^2+2\left(a+b\right)c+c^2\)

\(=a^2+2ab+b^2+2ac+2bc+c^2\)

\(=a^2+b^2+c^2+2ab+2bc+2ca\)(Đpcm)

Bài 2:

a) \(x^2-y^2=\left(x-y\right)\left(x+y\right)=\left(87-13\right)\left(87+13\right)=74.100=7400\)

b)\(25x^2-30x+9=\left(5x\right)^2-2.5.3x+3^2=\left(5x-3\right)^2=\left(5.2-3\right)^2=7^2=49\)

c)\(4x^2-28x+49=\left(2x\right)^2-2.2.7x+7^2=\left(2x-7\right)^2=\left(2.4-7\right)^2=1^2\)

B, C và D

mấy cái đó là đúng hả bạn

b) Ta có : a\(^2\)+ b\(^2\)+ c\(^2\) =ab+bc+ca

=> 2(a\(^2\)+b\(^2\)+c\(^2\))= 2(ab+bc+ca)

<=>2a\(^2\)+2b\(^2\)+2c\(^2\)=2ab+2bc+2ca

<=> 2a\(^2\)+2b\(^2\)+2c\(^2\)-2ab-2bc-2ca=0

<=> a\(^2\)+a\(^2\)+b\(^2\)+b\(^2\)+c\(^2\)+c\(^2\)-2ab-2bc=2ca=0

<=> (a\(^2\)-2ab+b\(^2\))+(b\(^2\)-2bc+b\(^2\))+(a\(^2\)-2ca+c\(^2\))

<=> (a-b)\(^2\)+(b-c)\(^2\)+(a-c)\(^2\) =a

<=> hoặc a-b=0 hoặc b-c=o hoặc a-c=o <=>a=b hoặc b=c hoặc a=c

=>a=b=c (đpcm)

a) Theo đề bài: \(a^2+b^2=ab\)

=>\(a^2+b^2-ab=0\)

=>\(a^2-2ab+b^2+ab=0\)

=>\(\left(a-b\right)^2+ab=0\)

Vì \(\left(a-b\right)^2\ge0\)  để \(\left(a-b\right)^2+ab=0\) <=> \(\left(a-b\right)^2=ab=0\)

(a-b)²=0 <=> a-b=0 <=> a=b (đpcm)

b)\(a^2+b^2+c^2=ab+bc+ca\)

=>\(2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ac\right)\)

=>\(2a^2+2b^2+2c^2=2ab+2bc+2ac\)

=>\(2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)

=>\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)

=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

Vì \(\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(a-c\right)^2\ge0\end{cases}\) để \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

<=>\(\left(a-b\right)^2=\left(b-c\right)^2=\left(a-c\right)^2=0\)

<=>a-b=b-c=a-c=0

<=>a=b=c (đpcm)

(a^2+b^2)/2>=ab

<=>(a^2+b^2)>=2ab

 <=> a^2+2ab+b^2>=2ab 

<=>a^2+b^2>=0(luôn đúng)

=> điều phải chứng minh.

Xét hiệu:  \(a^2+b^2-2ab=\left(a-b\right)^2\ge0\)

=>  \(a^2+b^2\ge2ab\)

Dấu "=" xra  <=>  a = b

Áp dụng ta có:

a)  \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge2a.2b.2c=8abc\)

dấu "=" xra  <=>  a = b = c = 1

b)  \(\left(a^2+4\right)\left(b^2+4\right)\left(c^2+4\right)\left(d^2+4\right)\ge4a.4b.4c.4d=256abcd\)

Dấu "=" xra  <=>  a = b= c = d = 2

b)

\(-x^2+2x-6=-\left(x^2-2x+6\right)\)

\(=-\left(x^2-2x+1+5\right)=-\left(x+1\right)^2-6\)

vì \(\left(x-1\right)^2\ge0\)với mọi \(x\in R\)

nên \(-\left(x-1\right)^2\le0\)với mọi \(x\in R\)

do đó \(-\left(x-1\right)-5< 0\)với mọi \(x\in R\)

vậy \(-x^2+2x-6< 0\)với mọi \(x\in R\)

a) \(x^2+2x+7=x^2+2x+1+6\)

                            \(=\left(x+1\right)^2+6\)

vì \(\left(x+1\right)^2\ge0\)với mọi \(x\in R\)

nên \(\left(x+1\right)^2+6>0\)với mọi \(x\in R\)

vậy \(x^2+2x+7>0\)với mọi \(x\in R\)

a) \(\left(a^2+b+c\right)^2\)

\(=\left(a^2+b\right)^2+2\left(a^2+b\right)c+c^2\)

\(=a^4+2a^2b+b^2+2a^2c+2bc+c^2\)

b) \(\left(a+b+c\right)^2\)

\(=\left(a+b\right)^2+2\left(a+b\right)c+c^2\)

\(=a^2+2ab+b^2+2ca+2bc+c^2\)

a) (a^2+b+c)^2(a^2+b+c)^2

=(a^2+b)^2+2(a^2+b)c+c^2

=a^4+2a2b+b^2+2a2c+2bc+c^2

b) (a+b+c)^2(a+b+c)^2

=(a+b)^2+2(a+b)c+c^2

=a^2+2ab+b^2+2ca+2bc+c^2

=\(\left(2-1\right)\left(2+1\right)\left(2^2-1\right)....\left(2^{20}-1\right)\) +1

=\(\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{20}+1\right)+1\)

=\(\left(2^4-1\right)\left(2^4+1\right)....\left(2^{20}+1\right)+1\)

=.....

=\(\left(2^{20}-1\right)\left(2^{20}+1\right)+1\)

=\(2^{40}-1+1\)

=\(2^{40}\)

Chuc ban hoc tot

Sai rồi, nếu mũ là 32 thì bài này làm thế đc chứ mũ 20 thì ko làm như này được