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a) Giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk,c=dk\)
Ta có:
\(\frac{ab}{cd}=\frac{bkb}{dkd}=\frac{b^2}{d^2}\) (1)
\(\frac{a^2-b^2}{c^2-d^2}=\frac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\frac{b^2.k^2-b^2}{d^2.k^2-d^2}=\frac{b^2.\left(k^2-1\right)}{d^2.\left(k^2-1\right)}=\frac{b^2}{d^2}\) (2)
Từ (1) và (2) suy ra \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\left(đpcm\right)\)
b) Giải:
Để \(P\in Z\Rightarrow2x-3⋮x+1\)
Ta có:
\(2x-3⋮x+1\)
\(\Rightarrow\left(2x+2\right)-5⋮x+1\)
\(\Rightarrow5⋮x+1\)
\(\Rightarrow x+1\in\left\{1;-1;5;-5\right\}\)
+) \(x+1=1\Rightarrow x=0\)
+) \(x+1=-1\Rightarrow x=-2\)
+) \(x+1=5\Rightarrow x=4\)
+) \(x+1=-5\Rightarrow x=-6\)
Vậy \(x\in\left\{0;-2;4;-6\right\}\)
\(\Rightarrow5⋮x+1\)
1)Ta có:\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{a}{c}\cdot\frac{b}{d}=\frac{ab}{cd}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\)(tính chất dãy tỉ số bằng nhau)
\(\Rightarrow\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\left(đpcm\right)\)
2)\(P=\frac{2x-3}{x+1}=\frac{2x+2-5}{x+1}=\frac{2\left(x+1\right)-5}{x+1}=2-\frac{5}{x+1}\)
\(\Rightarrow P\in Z\Leftrightarrow2-\frac{5}{x+1}\in Z\Leftrightarrow\frac{5}{x+1}\in Z\Leftrightarrow5⋮x+1\Leftrightarrow x+1\inƯ\left(5\right)\)
\(\Rightarrow x+1\in\left\{-1;-5;1;5\right\}\)
\(\Rightarrow x\in\left\{-2;-6;0;4\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
bạn đăg tách ra cho m.n cùng giúp nhé
Bài 2 :
a, \(A=\left|2x-4\right|+2\ge2\)
Dấu ''='' xảy ra khi x = 2
Vậy GTNN A là 2 khi x = 2
b, \(B=\left|x+2\right|-3\ge-3\)
Dấu ''='' xảy ra khi x = -2
Vậy GTNN B là -3 khi x = -2
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(\left(a+1\right)\left(b+1\right)=a\left(b+1\right)+\left(b+1\right)=ab+a+b+1\)
\(\Rightarrow D=\left(a+b\right)\left(ab+a+b+1\right)=\left(a+b\right)\left[ab+\left(a+b\right)+1\right]=3.\left(-5+3+1\right)=-3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(D=\left(a+b\right)\left(a+1\right)\left(b+1\right)\)
\(\Rightarrow D=3\left(a+1\right)\left(b+1\right)\)
\(\Rightarrow D=\left(3a+3\right)\left(b+1\right)\)
\(\Rightarrow D=3ab+3b+3a+3\)
\(\Rightarrow D=-15+3\left(a+b\right)+3\)
\(\Rightarrow D=-15+9+3\)
\(\Rightarrow D=-3\)
Vậy D = -3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{25.5^3.1}{625.5^2}=\frac{5^2.5^3}{5^4.5^2}=\frac{1}{5}\)
các câu sau mk k hiểu bn viết j
`D=(a+b)(a+1)(b+1)`
`=3[ab+(a+b)+1]`
`=3(5+3+1)`
`=27`.
D=10
(mk tính ra là như thế )