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\(A=\frac{63^2-47^2}{215^2-105^2}=\frac{\left(63+47\right)\left(63-47\right)}{\left(215+105\right)\left(215-105\right)}=\frac{110\cdot16}{320\cdot110}=\frac{1}{20}\)
\(B=\frac{437^2-363^2}{537^2-463^2}=\frac{\left(473-363\right)\left(473+363\right)}{\left(573-463\right)\left(573+463\right)}=\frac{110\cdot836}{110\cdot1036}=\frac{836}{1036}=\frac{4\cdot209}{4\cdot234}=\frac{209}{234}\)
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=\(\frac{4}{5}\)
\(\frac{63^2-47^2}{215^2-105^2}=\) \(\frac{\left(63-47\right)\left(63+47\right)}{\left(215-105\right)\left(215+105\right)}\)
\(=\frac{16.110}{110.320}=\frac{16}{320}\)\(=\frac{1}{20}\)
các câu kia làm tương tự nha
2) a) Ta có B = \(\frac{x+2}{x-2}-\frac{x-2}{x+2}-\frac{16}{4-x^2}=\frac{\left(x+2\right)^2-\left(x-2\right)^2+16}{\left(x-2\right)\left(x+2\right)}=\frac{8\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{8}{x-2}\)
Khi |x - 1| = 2
=> \(\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
Khi x = 3 (thỏa mãn) => A = \(\frac{3^2-2.3}{3+1}=\frac{3}{4}\)
Khi x = - 1 (không thỏa mãn) => Không tìm được A
b) Ta có P = \(A.B=\frac{x^2-2x}{x+1}.\frac{8}{x-2}=\frac{8x\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}=\frac{8x}{x+1}\)
Đẻ P < 8
=> \(\frac{8x}{x+1}< 8\Leftrightarrow\frac{x}{x+1}< 1\)
=> \(\orbr{\begin{cases}x< x+1\left(x>-1\right)\\x>x+1\left(x< -1\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}0x< 1\left(tm\right)\\0x>1\left(\text{loại}\right)\end{cases}}\)
Vậy x > - 1 thì P < 8
Do : \(4x^2=1\)
\(< =>\orbr{\begin{cases}2x=1\\2x=-1\end{cases}}\)
\(< =>\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)
Ta thấy điều kiện xác định của B là \(x\ne-\frac{1}{2}\)
Suy ra \(x=\frac{1}{2}\)
Ta có : \(B=\frac{x^2-x}{2x+1}=\frac{\frac{1}{4}-\frac{1}{2}}{\frac{1}{2}.2+1}=\frac{\frac{-1}{4}}{2}=-\frac{1}{8}\)
Vậy ......
Ta có : \(A=\frac{1}{x-1}+\frac{x}{x^2-1}=\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{2x+1}{x^2-1}\)
Suy ra \(M=\frac{2x+1}{x^2-1}.\frac{x^2-x}{2x+1}=\frac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{x}{x+1}\)
3/
a/ \(A=\left(x-y\right)^2+\left(x+y\right)^2.\)
\(A=\left(x^2-2xy+y^2\right)+\left(x^2+2xy+y^2\right)\)
\(A=x^2-2xy+y^2+x^2+2xy+y^2\)
\(A=2x^2+2y^2\)
b/ \(B=\left(2a+b\right)^2-\left(2a-b\right)^2\)
\(B=\left(4a^2+4ab+b^2\right)-\left(4a^2-4ab+b^2\right)\)
\(B=4a^2+4ab+b^2-4a^2+4ab-b^2\)
\(B=8ab\)
c/ \(C=\left(x+y\right)^2-\left(x-y\right)^2\)
\(C=\left(x^2+2xy+y^2\right)-\left(x^2-2xy+y^2\right)\)
\(C=x^2+2xy+y^2-x^2+2xy-y^2\)
\(C=4xy\)
d/ \(D=\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(D=\left(4x^2-4x+1\right)-2\left(4x^2-12x+9\right)+4\)
\(D=4x^2-4x+1-8x^2+24x-18+4\)
\(D=-4x^2+20x-13\)
\(63^2-47^2\)
= \(\left(63-47\right)\left(63+47\right)\)
= \(16.110\)
= \(1760\)