\(c,=\left(31,8-21,8\right)^2=10^2=100\\ 12,\\ a,\left(n+2\right)^2-\left(n-2\right)^2\\ =\left(n+2-n+2\right)\left(n+2+n-2\right)\\ =4\cdot2n=8n⋮8\\ b,\left(n+7\right)^2-\left(n-5\right)^2\\ =\left(n+7-n+5\right)\left(n+7+n-5\right)\\ =12\left(2n+2\right)=24\left(n+1\right)⋮24\)

tui chiuj

\(=\frac{\left(a-b\right)^3-c^3+3ab\left(a-b\right)-3abc}{a^2+2ab+b^2+b^2-2bc+c^2+c^2+2ca+a^2}\)

\(=\frac{\left(a-b-c\right)\left(a^2-2ab+b^2+ac-bc+c^2\right)+3ab\left(a-b-c\right)}{\left(a-b-c\right)^2+a^2+b^2+c^2}\)

\(=\frac{\left(\cdot a-b-c\right)\left(a^2+b^2+c^2+ac+ab-bc\right)}{4+a^2+b^2+c^2}\)

\(=\frac{2a^2+2b^2+2c^2+2ab-2bc+2ca}{4+a^2+b^2+c^2}\)

\(=\frac{\left(a-b-c\right)^2+a^2+b^2+c^2}{4+a^2+b^2+c^2}=1\)

k mk nha

\(a,\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)=4-a\)

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

a) \(\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)=2^2-\left(\sqrt{a}\right)^2=4-a\)

b) \(\left(3+\sqrt{a}\right)\left(3-\sqrt{a}\right)=3^2-\left(\sqrt{a}\right)^2=9-a\)

a,40.213+163.29+87.40-63.29

=40.7.29+163.29+3.29.40-63.29

=(40.7+163+3.40-63)29

=(280+163+120-63).29

=500.29

=14500

\(a.\)\(40.213+163.29+87.40-63.29\)

\(=40.7.20+163.29+3.29.40-63.29\)

\(=29\left(40.7+163+3.40-63\right)\)

\(=29\left(280+163+120-63\right)\)

\(=29.500\)

\(=14500\)

Đây là tính nha, ko thích thì tính theo thứ tự !

\(b.\)\(140-\left\{\left[50-\left(8-1\right)^2\right]+4^2\right\}\)

\(=140-\left\{\left[50-7^2\right]+16\right\}\)

\(=140-\left\{\left[50-49\right]+16\right\}\)

\(=140-\left\{1+16\right\}\)

\(=140-17\)

\(=123\)

(a² + b²)(c² + d²) ≥ (ac + bd)² \(\forall a,b,c,d\)

↔ (ac)² + (ad)² + (bc)² + (bd)² ≥ (ac)² + 2abcd + (bd)² \(\forall a,b,c,d\)

↔ (ad)² + (bc)² ≥ 2abcd \(\forall a,b,c,d\) 

↔ (ad)² - 2abcd + (bc)² ≥ 0 \(\forall a,b,c,d\) 

↔ (ad - bc)² ≥ 0 \(\forall a,b,c,d\) 

=> luôn đúng

Vậy.....

!Chúc Bạn Học Tốt!

\(M=\dfrac{\left(a-b\right)^3-c^3+3ab\left(a-b\right)-3abc}{\left(a+b\right)^2+\left(b-c\right)^2+\left(c+a\right)^2}\)

\(=\dfrac{\left(a-b-c\right)\left(a^2-2ab+b^2+ac-bc+c^2+3ab\right)}{2a^2+2b^2+2c^2+2ab-2bc+2ac}\)

\(=\dfrac{\left(a-b-c\right)\cdot\left(a^2+b^2+c^2-ab-bc+ac\right)}{2\cdot\left(a^2+b^2+c^2+ab-bc+ac\right)}=\dfrac{2}{2}=1\)

\(a,-7-\left[\left(-19\right)+\left(21\right)\right].\left(-3\right)-\left[\left(32\right)+\left(-7\right)\right]\)

\(=-7-\left[21-19\right].\left(-3\right)-\left[32-7\right]\)

\(=-7-2.\left(-3\right)-25\)

\(=-7+6-25=-26\)

\(b,\left(-2\right)^2.3-\left(1^{10}+8\right):\left(-3\right)^2\)

\(=4.3-9:9\)

\(=12-1=11\)