rút gọn hộ với mình cảm ơn trước
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( a + 2 )3 - a( a - 3 )2
= a3 + 6a2 + 12a + 8 - a( a2 - 6a + 9 )
= a3 + 6a2 + 12a + 8 - a3 + 6a2 - 9a
= 12a2 + 3a + 8
cách của symbolab:
\(\left(a+2\right)^3-a\left(a-3\right)^2\)
\(=a^3+6a^2+12a+8-a\left(a-3\right)^2\)
\(=a^3+6a^2+12a+8-a\left(a^2-6a+9\right)\)
\(=a^3+6a^2+12a+8-a^3+6a^2-9a\)
\(=12a^2+3a+8\)
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\(\dfrac{2.3.5.\left(-7\right).12}{3.\left(-5\right).7.14}=\dfrac{2.3.5.\left(-7\right).12}{3.5.\left(-7\right).2.7}=\dfrac{12}{7}\)
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Ta có : \(\frac{a}{b}=\frac{3}{4}\)(1)
\(\frac{a+15}{b}=\frac{7}{6}\)
\(\Rightarrow\frac{a}{b}+\frac{15}{b}=\frac{7}{6}\)(2)
Thay (1) vào (2) ta có :
\(\frac{3}{4}+\frac{15}{b}=\frac{7}{6}\)
\(\Rightarrow\frac{15}{b}=\frac{7}{6}-\frac{3}{4}\)
\(\Rightarrow\frac{15}{b}=\frac{5}{12}\)
\(\Rightarrow5b=12.15\)
\(\Rightarrow b=12.15:5\)
\(\Rightarrow b=36\)
Thay b\(\Rightarrow a=27\) vào (1) ta có :
\(\frac{a}{b}=\frac{a}{36}=\frac{3}{4}\)
\(\Rightarrow4a=36.3\)
\(\Rightarrow a=36.3:4\)
\(\Rightarrow a=27\)
\(\Rightarrow\frac{a}{b}=\frac{27}{36}\)
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a) \(A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2x+10}{\left(x+5\right)\left(x-5\right)}\)
\(A=\dfrac{x-5+2x+10-2x-10}{\left(x+5\right)\left(x-5\right)}=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}=\dfrac{1}{x+5}\)
b) \(A=-3\Rightarrow\dfrac{1}{x+5}=-3\)
\(\Leftrightarrow x+5=-\dfrac{1}{3}\Leftrightarrow x=-\dfrac{1}{3}-5=\dfrac{-16}{3}\)
\(9x^2-42x+49=\left(3x-7\right)^2=\left(3.\dfrac{-16}{3}-7\right)^2=\left(-23\right)^2=529\) \(\left(x=\dfrac{-16}{3}\right)\)
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a,\(\left(\sqrt{3}-\sqrt{2}\right)+\sqrt{2}=\sqrt{3}\) (vi \(\sqrt{3}>\sqrt{2}\) )
b,\(3\sqrt{5}-\left(\sqrt{5}-1\right)\) =\(3\sqrt{5}-\sqrt{5}+1=2\sqrt{5}+1\)
c,\(\sqrt{\left(\sqrt{3}\right)^2-2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
\(B=\dfrac{2\sqrt{x}-6\sqrt{x}+x+3\sqrt{x}-3x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{-3}{\sqrt{x}-3}\)
\(B=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ B=\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3}{\sqrt{x}+3}\)