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Bài 1:
a, \(\dfrac{x+5}{x}=\dfrac{4}{3}\)
\(\Rightarrow3x+15=4x\\ \Rightarrow4x-3x=15\\ \Rightarrow x=15\)
b, \(\dfrac{x-20}{x-10}=\dfrac{x+40}{x+70}\)
\(\Rightarrow\left(x-20\right).\left(x+70\right)=\left(x+40\right)\left(x-10\right)\)
\(\Rightarrow x^2+70x-20x-1400=x^2-10x+40x-400\)
\(\Rightarrow x^2-x^2+70x-20x+10x-40x=-400+1400\)
\(\Rightarrow20x=1000\Rightarrow x=50\)
c, \(4^x=\dfrac{1.2.3.....31}{4.6.8.....64}\)
\(\Rightarrow4^x=\dfrac{1}{2.2.2.2.....2.2.64}\) (có 30 số 2)
\(\Rightarrow4^x=\dfrac{1}{2^{30}.4^3}\Rightarrow4^x=\dfrac{1}{4^{15}.4^3}\)
\(\Rightarrow4^x=\dfrac{1}{4^{18}}\)
\(\Rightarrow4^x=4^{-18}\)
Vì \(4\ne-1;4\ne0;4\ne1\) nên \(x=-18\)
Chúc bạn học tốt!!!
a , \(\dfrac{x+5}{x}=\dfrac{4}{3}\Leftrightarrow3\left(x+5\right)=4x\)
<=> 3x+15=4x
<=> x= 15
b , \(\dfrac{x-20}{x-10}=\dfrac{x+40}{x+70}\)
<=> \(\dfrac{x-10}{x-10}-\dfrac{10}{x-10}=\dfrac{x+70}{x+70}-\dfrac{30}{x+70}\)
<=> \(1-\dfrac{10}{x-10}=1-\dfrac{30}{x+70}\)
<=> \(\dfrac{10}{x-10}=\dfrac{30}{x+70}\Leftrightarrow\dfrac{1}{x-10}=\dfrac{3}{x+70}\)
<=> (x+70)=3(x-10)
<=> x+70 = 3x-30
<=> 100=2x
<=> x= 50
\(a,A=\dfrac{\dfrac{3}{4}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}-\dfrac{5}{6}+\dfrac{5}{8}}\\ A=\dfrac{\dfrac{405}{572}}{\dfrac{645}{1001}}+\dfrac{\dfrac{5}{12}}{\dfrac{25}{24}}\\ A=\dfrac{189}{172}+\dfrac{2}{5}\\ A=\dfrac{1289}{860}\)
Câu 2:
Ta có: \(x^2=1\)
=>x=1 hoặc x=-1
=>x là số hữu tỉ
Bài 1:
|\(x\)| = 1 ⇒ \(x\) \(\in\) {-\(\dfrac{1}{3}\); \(\dfrac{1}{3}\)}
A(-1) = 2(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)) + 5
A(-1) = \(\dfrac{2}{9}\) + 1 + 5
A (-1) = \(\dfrac{56}{9}\)
A(1) = 2.(\(\dfrac{1}{3}\) )2- \(\dfrac{1}{3}\).3 + 5
A(1) = \(\dfrac{2}{9}\) - 1 + 5
A(1) = \(\dfrac{38}{9}\)
|y| = 1 ⇒ y \(\in\) {-1; 1}
⇒ (\(x;y\)) = (-\(\dfrac{1}{3}\); -1); (-\(\dfrac{1}{3}\); 1); (\(\dfrac{1}{3};-1\)); (\(\dfrac{1}{3};1\))
B(-\(\dfrac{1}{3}\);-1) = 2.(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)).(-1) + (-1)2
B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) - 1 + 1
B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\)
B(-\(\dfrac{1}{3}\); 1) = 2.(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)).1 + 12
B(-\(\dfrac{1}{3};1\)) = \(\dfrac{2}{9}\) + 1 + 1
B(-\(\dfrac{1}{3}\); 1) = \(\dfrac{20}{9}\)
B(\(\dfrac{1}{3};-1\)) = 2.(\(\dfrac{1}{3}\))2 - 3.(\(\dfrac{1}{3}\)).(-1) + (-1)2
B(\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) + 1 + 1
B(\(\dfrac{1}{3}\); -1) = \(\dfrac{20}{9}\)
B(\(\dfrac{1}{3}\); 1) = 2.(\(\dfrac{1}{3}\))2 - 3.(\(\dfrac{1}{3}\)).1 + (1)2
B(\(\dfrac{1}{3}\); 1) = \(\dfrac{2}{9}\) - 1 + 1
B(\(\dfrac{1}{3}\);1) = \(\dfrac{2}{9}\)
a: \(=\left(\dfrac{-1}{3}:\dfrac{-2}{3}\right)^3+\left(\dfrac{4}{21}\cdot\dfrac{21}{4}\right)^{50}+0.01\)
\(=\left(\dfrac{1}{2}\right)^3+1^{50}+0.01=0.125+1+0.01=1.135\)
b: \(=x:y+\left(\dfrac{2x}{y}\right)^2-11x+12x-12y\)
\(=\dfrac{x}{y}+\dfrac{4x^2}{y^2}+x-12y\)
\(=\dfrac{x^2+4x^2+xy^2-12y^3}{y^2}=\dfrac{5x^2+xy^2-12y^3}{y^2}\)
b: \(3x^2y^3=\dfrac{1}{9}\)
\(\Leftrightarrow3x^2=\dfrac{1}{9}:\dfrac{1}{27}=3\)
=>x=1 hoặc x=-1
a: \(A=\dfrac{-2}{3}\cdot\left(-27\right)\cdot4\cdot\dfrac{1}{2}=18\cdot2=36\)