tim x, y thuoc Q: (2x-7)^2018+(3y+8)^2020<=8
cho a/7=b/8=c/9=10. tinh P=6a+3b-7c/8a+9b-10c
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a) \(\dfrac{5}{8}+\dfrac{4}{9}=\dfrac{45}{72}+\dfrac{32}{72}=\dfrac{77}{72}\)
b) \(\dfrac{10}{14}-\dfrac{3}{7}=\dfrac{10}{14}-\dfrac{6}{14}=\dfrac{4}{14}=\dfrac{2}{7}\)
c) \(\dfrac{7}{10}\times\dfrac{25}{24}=\dfrac{175}{240}=\dfrac{35}{48}\)
d) \(\dfrac{3}{4}\div\dfrac{9}{6}=\dfrac{3}{4}\times\dfrac{6}{9}=\dfrac{18}{36}=\dfrac{1}{2}\)
a) \(\frac{x}{3}=\frac{y}{2};\frac{y}{7}=\frac{z}{5}\Rightarrow\frac{x}{21}=\frac{y}{14}=\frac{z}{10}=\frac{3x}{63}=\frac{y}{14}=\frac{4z}{40}=\frac{3x-y+4z}{63-14+40}=\frac{-10}{89}\)
\(\Rightarrow\frac{x}{21}=\frac{-10}{89}\Rightarrow x=\frac{-210}{89};\frac{y}{14}=\frac{-10}{89}\Rightarrow y=\frac{-140}{89};\frac{z}{10}=\frac{-10}{89}\Rightarrow z=\frac{-100}{89}\)
b)\(\frac{x-7+7}{8+7}=\frac{y-8+8}{9+8}=\frac{z-9+9}{10+9}=\frac{x}{15}=\frac{y}{17}=\frac{z}{19}=\frac{2x}{30}=\frac{y}{17}=\frac{3z}{57}=\frac{20}{70}=\frac{2}{7}\)
\(\Rightarrow\frac{x}{15}=\frac{2}{7}\Rightarrow x=\frac{30}{7};\frac{y}{17}=\frac{2}{7}\Rightarrow y=\frac{34}{7};\frac{z}{19}=\frac{2}{7}\Rightarrow z=\frac{38}{7}\)
c)\(a=1+7+7^2+...+7^{100}\)
\(a=\left(1+7\right)+\left(7^2+7^3\right)+...+\left(7^{99}+7^{100}\right)\)
\(a=8+7^2\left(1+7\right)+...+7^{99}\left(1+7\right)\)
\(a=8+7^2.8+...+7^{99}.8\)
\(a=8\left(1+7^2+...+7^{99}\right)\) chia hết cho 8
=> a chia 8 dư 0
a)\(a=\left(1+7+7^2+...+7^{100}\right)\)
=>\(7a=7\left(1+7+7^2+...+7^{100}\right)=7+7^2+7^3+...+7^{101}\)
=>\(7a-a=\left(7+7^2+7^3+...+7^{101}\right)\)\(-\left(1+7+7^2+...+7^{100}\right)\)
=>\(6a=7^{101}-1\Rightarrow a=\frac{7^{101}-1}{6}\)
b)\(6a+1=7^x\Rightarrow6.\frac{7^{101}-1}{6}+1=7^x\Rightarrow7^{101}-1+1=7^x\Rightarrow7^{101}=7^x\)=>x=101
\(a,=6y\left(2x^2-3xy-5y^2\right)\\ =6y\left(2x^2+2xy-5xy-5y^2\right)\\ =6y\left(x+y\right)\left(2x-5y\right)\\ b,=5x\left(x-y\right)-10\left(x-y\right)=5\left(x-2\right)\left(x-y\right)\\ c,=\left(a-b\right)\left(a^2+ab+b^2\right)-3\left(a-b\right)\\ =\left(a-b\right)\left(a^2+ab+b^2-3\right)\\ d,=\left(a^2+3b\right)^2-1=\left(a^2+3b+1\right)\left(a^2+3b-1\right)\\ e,=\left(2x-5\right)\left(2x+5\right)-\left(2x+7\right)\left(2x-5\right)\\ =\left(2x-5\right)\left(2x+5-2x-7\right)\\ =-2\left(2x-5\right)\\ f,=x^2+5x-3x-15=\left(x+5\right)\left(x-3\right)\\ g,=x^3-x-6x-6\\ =x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)\\ =\left(x+1\right)\left(x^2-x-6\right)\\ =\left(x+1\right)\left(x^2-3x+2x-6\right)\\ =\left(x+1\right)\left(x-3\right)\left(x+2\right)\\ l,=x^4+4x^2+4-4x^2\\ =\left(x^2+2\right)^2-4x^2=\left(x^2+2x+2\right)\left(x^2-2x+2\right)\\ h,=y\left(x^2+2x+1\right)=y\left(x+1\right)^2\)