Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{63^2-47^2}{215^2-105^2}=\frac{\left(63+47\right)\left(63-47\right)}{\left(215+105\right)\left(215-105\right)}=\frac{110\cdot16}{320\cdot110}=\frac{1}{20}\)
\(B=\frac{437^2-363^2}{537^2-463^2}=\frac{\left(473-363\right)\left(473+363\right)}{\left(573-463\right)\left(573+463\right)}=\frac{110\cdot836}{110\cdot1036}=\frac{836}{1036}=\frac{4\cdot209}{4\cdot234}=\frac{209}{234}\)
Trả lời:
\(A=\frac{63^2-47^2}{215^2-105^2}=\frac{\left(63-47\right).\left(63+47\right)}{\left(215-105\right).\left(215+105\right)}=\frac{16.110}{110.320}=\frac{1}{20}\)
\(B=\frac{437^2-363^2}{537^2-463^2}=\frac{\left(437-363\right).\left(437+363\right)}{\left(537-463\right).\left(537+463\right)}=\frac{74.800}{74.1000}=\frac{4}{5}\)
Học tốt
A=\(\frac{63^2-47^2}{215^2-105^2}\)
A=\(\frac{\left(63-47\right).\left(63+47\right)}{\left(215-105\right).\left(215+105\right)}\)
A=\(\frac{16.110}{110.320}\)
A=\(\frac{1760}{35200}\)
\(A=\frac{1}{20}\)
B=\(\frac{437^2-363^2}{537^2-463^2}\)
B=\(\frac{\left(437-363\right).\left(437+363\right)}{\left(537-463\right).\left(537+463\right)}\)
B=\(\frac{74.800}{74.1000}\)
B=\(\frac{4}{5}\)
b) \(263^2+74.263+37^2\)
\(=\left(263+37\right)^2\)
\(=300^2\)
\(=90000\)
c) \(136^2-92.136+46^2\)
\(=\left(136-46\right)^2\)
\(=90^2\)
\(=8100\)
b: Sửa đề: \(B=263^2+54\cdot263+27^2\)
\(=263^2+2\cdot263\cdot27+27^2\)
\(=\left(263+27\right)^2=290^2=84100\)
c: \(C=136^2-2\cdot46\cdot136+46^2\)
\(=\left(136-46\right)^2=90^2=8100\)
d: \(D=\left(50^2-49^2\right)+\left(48^2-47^2\right)+...+\left(2^2-1^2\right)\)
\(=50+49+...+2+1\)
Số số hạng là (50-1):1+1=50(số)
Tổng là;
\(D=\dfrac{\left(50+1\right)\cdot50}{2}=51\cdot25=1225\)
a) \(\dfrac{63^2-47^2}{215^2-105^2}\)
= \(\dfrac{\left(63-47\right)\left(63+47\right)}{\left(215-105\right)\left(215+105\right)}\)
= \(\dfrac{16.110}{110.320}=\dfrac{16}{320}=\dfrac{1}{20}\)
b) \(\dfrac{437^2-363^2}{537^2-463^2}\)
= \(\dfrac{\left(437-363\right)\left(437+363\right)}{\left(537-463\right)\left(537+463\right)}\)
= \(\dfrac{74.800}{74.1000}=\dfrac{800}{1000}=\dfrac{4}{5}\)
2)
A = \(26^2-24^2=\left(26-24\right)\left(26+24\right)=2.50=100\)
B = \(27^2-25^2=\left(27-25\right)\left(27+25\right)=2.52=104\)
Từ đó suy ra A < B
1.
\(a.\: \dfrac{63^2-47^2}{215^2-105^2}=\dfrac{\left(63-47\right)\left(63+47\right)}{\left(215-105\right)\left(215+105\right)}\\ =\dfrac{16.110}{110.320}=\dfrac{16}{320}=\dfrac{1}{20}\)
\(b.\dfrac{437^2-363^2}{537^2-463^2}=\dfrac{\left(437-363\right)\left(437+363\right)}{\left(537-463\right)\left(537+463\right)}\\ =\dfrac{74.800}{74.1000}=\dfrac{800}{1000}=\dfrac{4}{5}\)
2.
\(A=26^2-24^2=\left(26-24\right)\left(26+24\right)=2.50=100\)
\(B=27^2-25^2=\left(27-25\right)\left(27+25\right)=2.52=104\)
\(vì\:100< 104\:nên\:26^2-24^2< 27^2-25^2\\ hay\:A< B\)
B1: a) \(\left|x-2\right|+9y^2+12xy+4x^2=0\)
=> \(\left|x-2\right|+\left(3y+2x\right)^2=0\)
Ta có: \(\left|x-2\right|\ge0\forall x\)
\(\left(3y+2x\right)^2\ge0\forall x;y\)
=> \(\left|x-2\right|+\left(3y+2x\right)^2\ge0\forall x;y\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}x-2=0\\3y+2x=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\3y=-2x\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\3y=-2.2=-4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=-\frac{4}{3}\end{cases}}\)
Vậy ...
Bài 3:
a) \(\left|x-2\right|+9y^2+12xy+4x^2=0\)
\(\Leftrightarrow\left|x-2\right|+\left(3y+2x\right)^2=0\)
Dễ thấy \(VT\ge0\forall x;y\)
\(\Rightarrow\left\{{}\begin{matrix}x-2=0\\3y+2x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\frac{-4}{3}\end{matrix}\right.\)
Vậy...
b) \(3x^2+y^2+10x-2xy+26=0\)
\(\Leftrightarrow x^2-2xy+y^2+2x^2+10x+26=0\)
\(\Leftrightarrow\left(x-y\right)^2+2\left(x^2+5x+\frac{25}{4}\right)+\frac{27}{2}=0\)
\(\Leftrightarrow\left(x-y\right)^2+2\left(x+\frac{5}{2}\right)^2=\frac{-27}{2}\)
Dễ thấy \(VT\ge0\forall x;y\) mặt khác \(VP< 0\)
Do đó pt vô nghiệm
Bài 2:
\(A=263^2+74\cdot263+37^2\)
\(A=263^2+2\cdot263\cdot37+37^2\)
\(A=\left(263+37\right)^2\)
\(A=300^2\)
\(A=90000\)
b) tương tự
\(C=-1^2+2^2-3^2+...-99^2+100^2\)
\(C=\left(2^2-1^2\right)+\left(4^2-3^2\right)+...+\left(100^2-99^2\right)\)
\(C=\left(2-1\right)\left(1+2\right)+\left(4-3\right)\left(3+4\right)+...+\left(100-99\right)\left(99+100\right)\)
\(C=1+2+3+4+...+99+100\)
\(C=\frac{\left(100+1\right)\cdot100}{2}=5050\)
\(D=\left(3+1\right)\left(3^2+1\right)...\left(3^{32}+1\right)\)
\(2D=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)...\left(3^{32}+1\right)\)
\(2D=\left(3^2-1\right)\left(3^2+1\right)...\left(3^{32}+1\right)\)
\(2D=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{32}+1\right)\)
\(2D=\left(3^8-1\right)\left(3^8+1\right)...\left(3^{32}+1\right)\)
\(2D=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2D=\left(3^{32}-1\right)\left(3^{32}+1\right)\)
\(2D=3^{64}-1\)
\(D=\frac{3^{64}-1}{2}\)
\(\frac{63^2-47^2}{215^2-105^2}=\) \(\frac{\left(63-47\right)\left(63+47\right)}{\left(215-105\right)\left(215+105\right)}\)
\(=\frac{16.110}{110.320}=\frac{16}{320}\)\(=\frac{1}{20}\)
các câu kia làm tương tự nha
Thanks bạn nhiều nhiều nha!