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a;(18 - 9 x 2)x (2+ 4 + 6 + 8 + 10)
= (18 - 18)x (2+ 4 + 6+ 8 +10)
= 0 x (2+ 4+ 6+ 8 + 10)
= 0
b; (7 x 8 -56): (2+ 4 + 6 +8 + 112)
= (56 - 56) : (2 +4 +6+ 8+ 112)
= 0 : (2 + 4+ 6 + 8+ 112)
= 0
c; (2 + 125 + 6 + 145 + 112) x (42 - 6 x 7)
= (2+ 125+ 6 + 145+ 112) x (42 - 42)
= (2 + 125+ 6 + 145+ 112) x 0
= 0
Câu d:
(12 x 6 - 12 x 4 - 12 x 2) x (347 + 125)
= 12x(6 - 4 -2) x (347 + 125)
= 12 x (2 - 2) x (347 + 125)
= 12 x 0 x (347 + 125)
= 0 x (347 + 125)
= 0
a: Ta có: \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-\dfrac{3}{\sqrt{x}+2}+\dfrac{12}{x-4}\)
\(=\dfrac{x+4\sqrt{x}+4-3\sqrt{x}+6+12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+\sqrt{x}+22}{x-4}\)
d: Ta có: \(D=\dfrac{1}{\sqrt{x}+3}-\dfrac{\sqrt{x}}{3-\sqrt{x}}+\dfrac{2\sqrt{x}-12}{x-9}\)
\(=\dfrac{\sqrt{x}-3+x+3\sqrt{x}+2\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x+6\sqrt{x}-15}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
a, \(\left(\dfrac{7}{2}-2x\right).\dfrac{10}{3}=\dfrac{22}{3}\Leftrightarrow\dfrac{7}{2}-2x=\dfrac{22}{10}=\dfrac{11}{5}\)
\(\Leftrightarrow2x=\dfrac{13}{10}\Leftrightarrow x=\dfrac{13}{20}\)
b, \(\dfrac{4x}{9}=\dfrac{9}{8}-\dfrac{125}{1000}=1\Leftrightarrow x=\dfrac{9}{4}\)
c, \(-\dfrac{x}{21}=\dfrac{60}{21}\Rightarrow x=-60\)
a) \(6.8^{x-1}+8^{x+1}=6.8^{19}+8^{21}\)
\(\Rightarrow x-1+x+1=19+21\)
\(=2x=40\)
\(\Rightarrow x=20\)
b) \(4.3^{x-1}+2.3^{x+2}=4.3^6+2.3^9\)
\(\Rightarrow x-1+x+2=6+9\)
\(\Rightarrow2x+1=15\)
\(\Rightarrow2x=14\)
\(\Rightarrow x=7\)
Ta có: \(B=\left(\dfrac{21}{x^2-9}-\dfrac{x-4}{3-x}-\dfrac{x-1}{3+x}\right):\left(1-\dfrac{1}{x+3}\right)\)
\(=\left(\dfrac{21}{\left(x-3\right)\left(x+3\right)}+\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\right):\left(\dfrac{x+3}{x+3}-\dfrac{1}{x+3}\right)\)
\(=\dfrac{21+x^2+3x-4x-12-\left(x^2-4x+3\right)}{\left(x+3\right)\left(x-3\right)}:\dfrac{x+3-1}{x+3}\)
\(=\dfrac{x^2-x+9-x^2+4x-3}{\left(x+3\right)\left(x-3\right)}:\dfrac{x+2}{x+3}\)
\(=\dfrac{3x+6}{\left(x+3\right)\left(x-3\right)}:\dfrac{x+2}{x+3}\)
\(=\dfrac{3\left(x+2\right)}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{x+3}{x+2}\)
\(=\dfrac{3}{x-3}\)