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![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 4:
a: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{x+y}{4+3}=\dfrac{14}{7}=2\)
Do đó: x=8; y=6
b: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{2x+3y}{2\cdot8+3\cdot12}=\dfrac{13}{52}=\dfrac{1}{4}\)
Do đó: x=2; y=3
a) \(\dfrac{x}{5}=\dfrac{2}{5}.\Rightarrow x=2.\)
b) \(\dfrac{3}{x-5}=\dfrac{-4}{x+2.}.\left(x\ne5;x\ne-2\right).\)
\(\Leftrightarrow\dfrac{3}{x-5}+\dfrac{4}{x+2}=0.\Leftrightarrow\dfrac{3x+6+4x-20}{\left(x-5\right)\left(x+2\right)}=0.\)
\(\Rightarrow7x-14=0.\Leftrightarrow x=2\left(TM\right).\)
c) \(\dfrac{x}{-2}=\dfrac{-8}{x}.\left(x\ne0\right).\Leftrightarrow\dfrac{-x}{2}+\dfrac{8}{x}=0.\Leftrightarrow\dfrac{-x^2+16}{2x}=0.\Rightarrow-x^2+16=0.\Leftrightarrow x^2=16.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=4^2.\\x^2=\left(-4\right)^2.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4.\\x=-4.\end{matrix}\right.\)\(\left(TM\right).\)
d) \(\dfrac{x+2}{5}=\dfrac{45}{x+2}.\left(x\ne-2\right).\)
\(\Leftrightarrow\dfrac{x+2}{5}-\dfrac{45}{x+2}=0.\Leftrightarrow\dfrac{x^2+4x+4-225}{5x+10}=0.\Rightarrow x^2+4x-221=0.\)
\(\Leftrightarrow\left(x-13\right)\left(x+17\right)=0.\Leftrightarrow\left[{}\begin{matrix}x-13=0.\\x+17=0.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=13.\\x=-17.\end{matrix}\right.\) \(\left(TM\right).\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
a: =>1<x<=4
hay \(x\in\left\{2;3;4\right\}\)
b: =>-104<x<-99
hay \(x\in\left\{-103;-102;-101;-100\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
b: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)
hay \(x\in\left\{3;-3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
c: ⇔n+2∈{1;−1;5;−5}⇔n+2∈{1;−1;5;−5}
hay n∈{−1;−3;3;−7}n∈{−1;−3;3;−7}
d: ⇔n+2∈{1;−1;2;−2;4;−4}⇔n+2∈{1;−1;2;−2;4;−4}
hay n∈{−1;−3;0;−4;2;−6}n∈{−1;−3;0;−4;2;−6}
a: ⇔n−1∈{1;−1;5;−5}⇔n−1∈{1;−1;5;−5}
hay n∈{2;0;6;−4}
b: \(\Leftrightarrow n+1\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(n\in\left\{0;-2;1;-3;3;-5\right\}\)
c: \(\Leftrightarrow n+2\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{-1;-3;3;-7\right\}\)
d: \(\Leftrightarrow n+2\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(n\in\left\{-1;-3;0;-4;2;-6\right\}\)
a: \(\Leftrightarrow n-1\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{2;0;6;-4\right\}\)
d: =-76x10=-760
senpai trả lời đầy đủ ấy