Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(c,=\left(31,8-21,8\right)^2=10^2=100\\ 12,\\ a,\left(n+2\right)^2-\left(n-2\right)^2\\ =\left(n+2-n+2\right)\left(n+2+n-2\right)\\ =4\cdot2n=8n⋮8\\ b,\left(n+7\right)^2-\left(n-5\right)^2\\ =\left(n+7-n+5\right)\left(n+7+n-5\right)\\ =12\left(2n+2\right)=24\left(n+1\right)⋮24\)
\(=\frac{\left(a-b\right)^3-c^3+3ab\left(a-b\right)-3abc}{a^2+2ab+b^2+b^2-2bc+c^2+c^2+2ca+a^2}\)
\(=\frac{\left(a-b-c\right)\left(a^2-2ab+b^2+ac-bc+c^2\right)+3ab\left(a-b-c\right)}{\left(a-b-c\right)^2+a^2+b^2+c^2}\)
\(=\frac{\left(\cdot a-b-c\right)\left(a^2+b^2+c^2+ac+ab-bc\right)}{4+a^2+b^2+c^2}\)
\(=\frac{2a^2+2b^2+2c^2+2ab-2bc+2ca}{4+a^2+b^2+c^2}\)
\(=\frac{\left(a-b-c\right)^2+a^2+b^2+c^2}{4+a^2+b^2+c^2}=1\)
k mk nha
\(M=\dfrac{\left(a-b\right)^3-c^3+3ab\left(a-b\right)-3abc}{\left(a+b\right)^2+\left(b-c\right)^2+\left(c+a\right)^2}\)
\(=\dfrac{\left(a-b-c\right)\left(a^2-2ab+b^2+ac-bc+c^2+3ab\right)}{2a^2+2b^2+2c^2+2ab-2bc+2ac}\)
\(=\dfrac{\left(a-b-c\right)\cdot\left(a^2+b^2+c^2-ab-bc+ac\right)}{2\cdot\left(a^2+b^2+c^2+ab-bc+ac\right)}=\dfrac{2}{2}=1\)
e) Ta có: \(2\left|x-\dfrac{1}{2}\right|\ge0\forall x\)
\(\Leftrightarrow2\left|x-\dfrac{1}{2}\right|+2021\ge2021\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
\(x\left(x+1\right)^4+x\left(x+1\right)^3+x\left(x+1\right)^2+\left(x+1\right)^2\)
\(=\left(x+1\right)^2\left[x\left(x+1\right)^2+x\left(x+1\right)+x+1\right]\)
\(=\left(x+1\right)^2\left[x\left(x+1\right)\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\right]\)
\(=\left(x+1\right)^2\left\{\left(x+1\right)\left[x\left(x+1\right)+x+1\right]\right\}\)
\(=\left(x+1\right)^2\left\{\left(x+1\right)\left[x^2+x+x+1\right]\right\}\)
\(=\left(x+1\right)^2\left[\left(x+1\right)\left(x^2+2x+1\right)\right]\)
\(=\left(x+1\right)^2\cdot\left(x+1\right)^3\)
\(=\left(x+1\right)^5\left(đpcm\right)\)
Mình năm nay lớp 7 nên chưa chắc đúng đâu nha :
\(a\left(b+1\right)+b\left(a+1\right)=\left(a+b\right)\left(b+1\right)\left(1\right)\)
=) \(ab+a+ab+b=\left(a+b\right)\left(b+1\right)\)
=) \(1+a+1+b=\left(a+b\right)\left(b+1\right)\)
=) \(2+a+b=\left(a+b\right)\left(b+1\right)\)
=) \(2=\left(a+b\right)\left(b+1\right)-\left(a+b\right)\)
=) \(2=\left(a+b\right).\left(b+1-1\right)\)=) \(2=\left(a+b\right).b=ab+b^2\)
=) \(2=1+b^2\)=) \(b^2=2-1=1\)=) \(b=1\)
=) \(a=1:b=1:1=1\)
Thay vào \(\left(1\right)\):
\(1.\left(1+1\right)+1.\left(1+1\right)=\left(1+1\right).\left(1+1\right)\)
=) \(1.2+1.2=2.2\)
=) \(4=4\)( Đúng )
Vậy nếu \(ab=1\Leftrightarrow a\left(b+1\right)+b\left(a+1\right)=\left(a+b\right)\left(b+1\right)\left(ĐPCM\right)\)
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