tính nhanh
\(a,\left(31,8\right)^2-2.31,8.21,8+\left(21,8\right)^2\)
\(b,\left(7,5\right)^2+2.7,5.2,5+\left(2,5\right)^2\)
giúp mình nhé mình đang cần gấp
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\(c,=\left(31,8-21,8\right)^2=10^2=100\\ 12,\\ a,\left(n+2\right)^2-\left(n-2\right)^2\\ =\left(n+2-n+2\right)\left(n+2+n-2\right)\\ =4\cdot2n=8n⋮8\\ b,\left(n+7\right)^2-\left(n-5\right)^2\\ =\left(n+7-n+5\right)\left(n+7+n-5\right)\\ =12\left(2n+2\right)=24\left(n+1\right)⋮24\)
\(=\frac{\left(a-b\right)^3-c^3+3ab\left(a-b\right)-3abc}{a^2+2ab+b^2+b^2-2bc+c^2+c^2+2ca+a^2}\)
\(=\frac{\left(a-b-c\right)\left(a^2-2ab+b^2+ac-bc+c^2\right)+3ab\left(a-b-c\right)}{\left(a-b-c\right)^2+a^2+b^2+c^2}\)
\(=\frac{\left(\cdot a-b-c\right)\left(a^2+b^2+c^2+ac+ab-bc\right)}{4+a^2+b^2+c^2}\)
\(=\frac{2a^2+2b^2+2c^2+2ab-2bc+2ca}{4+a^2+b^2+c^2}\)
\(=\frac{\left(a-b-c\right)^2+a^2+b^2+c^2}{4+a^2+b^2+c^2}=1\)
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 !
(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!
\(M=\dfrac{\left(a-b\right)^3-c^3+3ab\left(a-b\right)-3abc}{\left(a+b\right)^2+\left(b-c\right)^2+\left(c+a\right)^2}\)
\(=\dfrac{\left(a-b-c\right)\left(a^2-2ab+b^2+ac-bc+c^2+3ab\right)}{2a^2+2b^2+2c^2+2ab-2bc+2ac}\)
\(=\dfrac{\left(a-b-c\right)\cdot\left(a^2+b^2+c^2-ab-bc+ac\right)}{2\cdot\left(a^2+b^2+c^2+ab-bc+ac\right)}=\dfrac{2}{2}=1\)
\(a,-7-\left[\left(-19\right)+\left(21\right)\right].\left(-3\right)-\left[\left(32\right)+\left(-7\right)\right]\)
\(=-7-\left[21-19\right].\left(-3\right)-\left[32-7\right]\)
\(=-7-2.\left(-3\right)-25\)
\(=-7+6-25=-26\)
\(b,\left(-2\right)^2.3-\left(1^{10}+8\right):\left(-3\right)^2\)
\(=4.3-9:9\)
\(=12-1=11\)
Giải:
a) \(\left(31,8\right)^2-2.31,8.21,8+\left(21,8\right)^2\)
\(=\left(31,8-21,8\right)^2\)
\(=10^2=100\)
Vậy ...
b) \(\left(7,5\right)^2+2.7,5.2,5+\left(2,5\right)^2\)
\(=\left(7,5+2,5\right)^2\)
\(=10^2=100\)
Vậy ...
HĐT: \(\left(A\pm B\right)^2=A^2\pm2.A.B+B^2\)
a, (31,8)2 - 2.31,8.21,8+(21,8)2
=(31,8+21,8)2 =102 =100
b,(7,5)2 + 2.7,5.2,5 +(2,5)2
=(7,5 +2,5)2 =102 =100