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a,Ta có: \(\frac{a^2+b^2}{a^2+c^2}=\frac{bc+b^2}{bc+c^2}=\frac{b\left(c+b\right)}{c\left(c+b\right)}=\frac{b}{c}\)
b, \(\frac{a+b}{a-b}=\frac{c+a}{c-a}\Rightarrow\frac{a+b}{c+a}=\frac{a-b}{c-a}=\frac{\left(a+b\right)+\left(a-b\right)}{\left(c+a\right)+\left(c-a\right)}=\frac{2a}{2c}=\frac{a}{c}\)(1)
Mặt khác: \(\frac{a+b}{c+a}=\frac{a-b}{c-a}=\frac{\left(a+b\right)-\left(a-b\right)}{\left(c+a\right)-\left(c-a\right)}=\frac{2b}{2a}=\frac{b}{a}\)(2)
Từ (1);(2)\(\Rightarrow\frac{a}{c}=\frac{b}{a}\Leftrightarrow a^2=bc\)
c, Áp dụng tính chất dãy tỉ số bằng nhau
\(\frac{a}{b}=\frac{c}{d}=\frac{m}{n}=\frac{a+c+m}{b+d+n}\)
Ta có : \(a^2=bc\)
\(\Rightarrow\frac{a^2+b^2}{a^2+c^2}=\frac{bc+b^2}{bc+c^2}=\frac{b\left(b+c\right)}{c\left(b+c\right)}=\frac{b}{c}\)(đpcm)
a) \(\frac{-28}{4}\le x\le\frac{-21}{7}\)
\(\Rightarrow-7\le x\le-3\)
\(\Rightarrow x=\left\{-3;-4;-5;-6;-7\right\}\)
b) \(\frac{-5}{12}=\frac{7}{72}\)
\(\Rightarrow-5.72=y.12\)
\(\Rightarrow y=\frac{-5.72}{12}\)
\(\Rightarrow y=-30\)
c) \(\frac{x}{19}=4\)
\(\Rightarrow x\div19=4\)
\(\Rightarrow x=4.19\)
\(\Rightarrow x=76\)
d) \(\frac{z+3}{15}=\frac{-1}{3}\)
\(\Rightarrow\left(z+3\right).3=-1.15\)
\(\Rightarrow z+3=\frac{-1.15}{3}\)
\(\Rightarrow z+3=-5\)
\(\Rightarrow z=-5-3\)
\(\Rightarrow z=-8\)
a) ta có : 3/4 = -x/4
=> -x = 3×4/4
=> -x =3
=> x = -3
Mặt khác: -x/4 =21/y
Với x = -3, ta có :
-3/4 = 21/y
=> y = 21×4/-3 = -28
Lại có : 21/y = z/-80
Với y = -28, ta có:
22/-28 = z/-80
=> z = 21×-80/-28 = 60
Vậy x= -3; y = -28; z = 60
b) Ta có: y-2/2 = 18/-2
=> y -2 = 2×18/-2
=> y-2 = -18 => y = -16
Lại có : x/3 = y-2/2
Với y = -16, ta có:
x/3 = -16-2/2
=> x/3 = -18/2
=> x = 3×-18/2 => x = -27
Vậy x = -27; y = -16
Bài 1:
\(S=4\left(\dfrac{1}{1\cdot7}+\dfrac{1}{7\cdot13}+...+\dfrac{1}{43\cdot49}\right)\)
\(=\dfrac{4}{6}\left(\dfrac{6}{1\cdot7}+\dfrac{6}{7\cdot13}+...+\dfrac{6}{43\cdot49}\right)\)
\(=\dfrac{2}{3}\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+...+\dfrac{1}{43}-\dfrac{1}{49}\right)\)
\(=\dfrac{2}{3}\cdot\dfrac{48}{49}=\dfrac{96}{147}=\dfrac{32}{49}\)
Bài 3:
Theo đề, ta có:
\(\dfrac{a}{b}=\dfrac{a+10}{b+10}\)
=>ab+10a=ab+10b
=>10a=10b
=>a/b=1
Bài 1:
a)\(\frac{x}{5}=\frac{-12}{20}\Rightarrow20x=5.\left(-12\right)=-60\Rightarrow x=-3\)
b)\(\frac{2}{y}=\frac{11}{-66}\Rightarrow2.\left(-66\right)=11y\Rightarrow11y=-132\Rightarrow y=-12\)
c)\(\frac{-3}{6}=\frac{x}{-2}=\frac{-18}{y}=\frac{-z}{24}\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{-3}{6}=\frac{x}{-2}\Rightarrow x=\frac{\left(-3\right)\left(-2\right)}{6}=1\\\frac{-3}{6}=\frac{-18}{y}\Rightarrow y=\frac{\left(-18\right).6}{-3}=36\\\frac{-3}{6}=\frac{-z}{24}\Rightarrow-z=\frac{\left(-3\right).24}{6}=-12\Rightarrow z=12\end{matrix}\right.\)
Bài 2:
\(\frac{-2}{x}=\frac{y}{3}\Rightarrow xy=\left(-2\right).3=-6\)
Mà \(x< 0< y\) nên ta có bảng sau:
| \(x\) | \(-6\) | \(-3\) | \(-2\) | \(-1\) |
| \(y\) | 1 | 2 | 3 | 6 |
1.a.ta có:\(\frac{2017+2018}{2018+2019}=\frac{2017}{2018+2019}+\frac{2018}{2018+2019}\)
mà \(\frac{2017}{2018}>\frac{2017}{2018+2019};\frac{2018}{2019}>\frac{2018}{2018+2019}\)
\(\Rightarrow M>N\)
b.ta thấy:
\(\frac{n+1}{n+2}>\frac{n+1}{n+3}>\frac{n}{n+3}\Rightarrow\frac{n+1}{n+2}>\frac{n}{n+3}\)
=> A>B