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=-27/5x -4/15 + -4/15x2010 - -4/15x-27/5 =-4/15x2010 =-52
\(x^2=\frac{4}{5}.\frac{15}{27}=\frac{4.3.5}{5.3^3}=\frac{4}{3^2}=\frac{2^2}{3^2}=\left(\frac{2}{3}\right)^2\)
=> \(x=\pm\frac{2}{3}\)
Qui đồng rồi khử mẫu, ta được:
\(4x+12.\left(27-x\right)=15x+5.\left(27-x\right)\)
\(\Leftrightarrow4x+324-12x=15x+135-5x\)
\(\Leftrightarrow4x-12x-15x+5x=135-324\)
\(\Leftrightarrow-18x=-189\Leftrightarrow x=\frac{21}{2}=10,5\)
Vậy x = 10,5
\(\frac{x}{15}+\frac{27-x}{5}=\frac{x}{4}+\frac{27-x}{12}\)
\(\frac{x}{15}+\frac{3\left(27-x\right)}{15}=\frac{3x}{12}+\frac{27-x}{12}\)
\(\frac{x}{15}+\frac{81-3x}{15}=\frac{3x}{12}+\frac{27-x}{12}\)
\(\frac{x+81-3x}{15}=\frac{3x+27-x}{12}\)
\(\frac{-2x+81}{15}=\frac{2x+27}{12}\)
\(12\left(-2x+81\right)=15\left(2x+27\right)\)
\(-24x+972=30x+405\)
\(972-405=30x+24x\)
\(567=54x\)
\(x=567:54\)
\(x=10,5\)
a) \(\frac{4}{9}\)và \(\frac{7}{15}\)
Ta có: \(9 = 3^2 ; 15 = 3.5\) nên \(BCNN (9,15) = 3^2. 5 = 45\). Do đó ta có thể chọn mẫu chung là 45.
\(\frac{4}{9}=\frac{4.5}{9.5}=\frac{20}{45}\)
\(\frac{7}{15}=\frac{7.3}{15.3}=\frac{21}{45}\)
b) \(\frac{5}{12}; \frac{7}{15}\) và \(\frac{4}{27}\)
Ta có: \(12=2^2.3\); \(15 = 3.5\) ; \(27=3^3\) nên BCNN(12, 15, 27) =\(2^2.3^3.5=540\). Do đó ta có thể chọn mẫu chung là 540.
\(\frac{5}{12}=\frac{5.45}{12.45}=\frac{225}{540}\)
\(\frac{7}{15}=\frac{7.36}{15.36}=\frac{252}{540}\)
\(\frac{4}{27}=\frac{4.20}{27.20}=\frac{80}{540}\)
a) (x + 5) / 95 + (x +10)/90 + (x + 15)/85 + (x + 20)/80 = -4
<=> (x + 5)/95 + (x + 5)/90 + 5/90 + (x + 5)/85 + 10/85+ (x + 5)/80 + 15/80 = -4
<=> (x + 5)(1/95+1/90+1/85+1/80) =-4 -5/90-10/85-15/85
<=> (x + 5)(1/95+1/90+1/85+1/80)= -1-(1 + 5/90 )-(1 + 10/85) - (1 + 15/80)
<=>(x + 5)(1/95+1/90+1/85+1/80) = -1 - 95/90 - 95/85 - 95/80
<=>(x + 5)(1/95+1/90+1/85+1/80) = -95 (1/95+1/90+1/85+1/80)
<=> x + 5 = -95 => x = -100
\(\frac{5-\frac{5}{3}+\frac{5}{9}-\frac{5}{27}}{8-\frac{8}{3}+\frac{8}{9}-\frac{8}{27}}:\frac{15-\frac{15}{11}+\frac{15}{121}}{16-\frac{16}{11}+\frac{16}{121}}\)
\(=\frac{5\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}{8\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}:\frac{15\left(1-\frac{1}{11}+\frac{1}{121}\right)}{16\left(1-\frac{1}{11}+\frac{1}{121}\right)}\)
\(=\frac{5}{8}:\frac{15}{16}\)
\(=\frac{2}{3}\)