PTĐTTNT : 2x-1-x^2
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Ta có:
\(A=\dfrac{\sqrt{x}-x}{\sqrt{x}-1}=\dfrac{-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=-\sqrt{x}\)
Vậy \(A=-\sqrt{x}\)
\(A=\dfrac{-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=-\sqrt{x}\)
( x + 2,7 ) : 4,9 = 25,3
( x + 2,7 ) = 25,3 x 4,9
x + 2,7 = 123,97
x = 123,97 - 2,7
x = 121,27
(x+2,7):4,9=25,3
x+2,7 =25,3x4,9
x+2,7 =123,97
x =123,97-2,7
x =121,27
\(a,\dfrac{2}{7}+\dfrac{1}{4}=\dfrac{8}{28}+\dfrac{7}{28}=\dfrac{15}{28}\\ b,\dfrac{3}{5}+\dfrac{3}{8}=\dfrac{3.8}{5.8}+\dfrac{3.5}{5.8}=\dfrac{6}{5}\\ c,=\dfrac{4.3}{9.3}+\dfrac{10}{27}=\dfrac{12+10}{27}=\dfrac{22}{27}\\ d,=\dfrac{2.3}{3.3}+\dfrac{7}{9}=\dfrac{6+7}{9}=\dfrac{13}{9}\\ e,=\dfrac{5.5}{12.5}+\dfrac{7.4}{15.4}=\dfrac{25+28}{60}=\dfrac{53}{60}\)
\(\dfrac{2}{7}+\dfrac{1}{4}=\dfrac{8}{28}+\dfrac{7}{28}=\dfrac{15}{28};\dfrac{3}{5}+\dfrac{3}{8}=\dfrac{24}{40}+\dfrac{15}{40};\dfrac{4}{9}+\dfrac{10}{27}=\dfrac{12}{27}+\dfrac{10}{27}=\dfrac{22}{27};\dfrac{2}{3}+\dfrac{7}{9}=\dfrac{18}{27}+\dfrac{21}{27}=\dfrac{39}{27};\dfrac{5}{12}+\dfrac{7}{15}=\dfrac{75}{180}+\dfrac{84}{180}=\dfrac{159}{180}\)
\(g,4=2^2;6=2.3\\ \Rightarrow BCNN\left(4,6\right)=2^2.3=12\\ \Rightarrow x\in BC\left(4,6\right)=B\left(12\right)=\left\{0;12;24;36;48;60;...\right\}\\ \text{Mà }0< x< 50\\ \Rightarrow x\in\left\{12;24;36;48\right\}\\ h,12=2^2.3;18=2.3^2\\ \Rightarrow BCNN\left(12,18\right)=2^2.3^2=36\\ \Rightarrow x\in BC\left(12,18\right)=B\left(36\right)=\left\{0;36;72;108;144;180;216;252;...\right\}\\ \text{Mà }x< 250\\ \Rightarrow x\in\left\{0;36;72;108;144;180;216\right\}\)
g,\(x⋮4,x⋮6\Rightarrow x\in BC\left(4,6\right)=\left\{\pm0;\pm12;\pm24;\pm36;\pm48;\pm60;...\right\}\)
Mà \(0< x< 50\Rightarrow x\in\left\{12;36;48\right\}\)
h,\(x⋮12,x⋮18\Rightarrow x\in BC\left(12,18\right)=\left\{0;\pm36;\pm72;\pm108;\pm144;\pm180;\pm216;\pm252;...\right\}\)
Mà \(x< 50\Rightarrow x\in\left\{0;\pm36;\pm72;\pm108;\pm144;\pm180;\pm216\right\}\)
a) \(\dfrac{x+9}{x^2-9}\)-\(\dfrac{3}{x^2+3x}\) = \(\dfrac{x+9}{\left(x-3\right)\left(x+3\right)}\)-\(\dfrac{3}{x\left(x+3\right)}\)
= \(\dfrac{x^2+9x-3x+9}{x\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{x^2+6x+9}{x\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{\left(x+3\right)^2}{x\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{x+3}{x\left(x-3\right)}\)
\(2x-1-x^2\\ =x+x-1-x^2\\ =\left(x-x^2\right)+\left(x-1\right)\\ =-x\left(x-1\right)+\left(x-1\right)\\ =\left(x-1\right)\left(1-x\right)\)
2x - 1 - x²
= -x² + 2x - 1
= -(x² - 2x + 1)
= -(x - 1)²