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Ta có :
\(\left(x+\frac{1}{x}\right)\cdot\left(y+\frac{1}{y}\right)=3.5\)
\(\Leftrightarrow xy+\frac{x}{x}+\frac{y}{y}+\frac{1}{xy}=15\)
\(\Leftrightarrow xy+\frac{1}{xy}=15-2\)
\(\Leftrightarrow xy+\frac{1}{xy}=13\)
Hay A = 13
\(M=\frac{x^3}{x^2-4}-\frac{x}{x-2}-\frac{2}{x+2}\)
\(M=\frac{x^3}{\left(x-2\right)\left(x+2\right)}-\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(M=\frac{x^3-x^2-2x-2x+4}{\left(x-2\right)\left(x+2\right)}\)
\(M=\frac{x^3-x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\)
\(M=\frac{x^2\left(x-1\right)-4\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}\)
\(M=\frac{\left(x-1\right)\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(M=x-1\)
\(M=\frac{x^3}{x^2-4}-\frac{x}{x-2}-\frac{2}{x+2}\)
\(=\frac{x^3}{\left(x-2\right)\left(x+2\right)}-\frac{x}{x-2}-\frac{2}{x+2}\)
\(=\frac{x^3-x\left(x+2\right)-2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^3-x^2-2x-2x+4}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^3-x^2-4x-4}{\left(x-2\right)\left(x+2\right)}\)
ĐK: -1 <= x <= 1
Đặt y = \(\sqrt{1-x^2}\)
=> y2 = 1 - x2 (y >= 0)
=> x = \(\sqrt{1-y^2}\)
<=>
x3 + y3 = 2xy
x2 + y2 = 1
<=>
(x + y)3 - 3x2y - 3xy2 = 2xy
(x + y) - 2xy = 1
<=>
(x + y)3 - 3xy(x + y) = 2xy
(x + y) - 2xy = 1
Đặt S = x + y, P = xy
=>
S3 - 3SP = 2P
S - 2P = 1
\(A=\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)
\(\sqrt{2}A=\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)
\(\sqrt{2}A=\sqrt{5}+\sqrt{5}+1-1\)
\(\sqrt{2}A=2\sqrt{5}\)
\(A=\sqrt{10}\)
P/s tham khảo nha
Do \(\sqrt{18}< \sqrt{20};\sqrt{19}< \sqrt{20}\)nên
\(\sqrt{18}+\sqrt{19}< 2\sqrt{20}=\sqrt{80}< \sqrt{81}=9\)
Vậy
28 con số 1vaf 1 số 0
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