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Câu 1:
Ta có: \(\left[\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{65.68}\right]x-\dfrac{7}{34}=\dfrac{19}{68}\)
\(\Rightarrow\left[\dfrac{1}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{65.68}\right)\right]x=\dfrac{33}{68}\)
\(\Rightarrow\left[\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\right]x=\dfrac{33}{68}\)
\(\Rightarrow\left[\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\right]x=\dfrac{33}{68}\)
\(\Rightarrow\dfrac{11}{68}x=\dfrac{33}{68}\)
\(\Rightarrow x=3\)
Vậy \(x=3.\)
3a) A=\(\dfrac{5}{x+xy+xyz}+\dfrac{5}{y+yz+1}+\dfrac{5xyz}{z+xz+xyz}\)
=\(\dfrac{5}{x\left(1+y+yz\right)}+\dfrac{5}{y+yz+1}+\dfrac{5xy}{1+x+xy}\)
=\(\dfrac{5}{x\left(1+y+zy\right)}+\dfrac{5x}{x\left(1+zy+y\right)}+\dfrac{5xy}{x\left(1+y+zy\right)}\)
=\(\dfrac{5+5x+5xy}{x\left(1+yz+y\right)}\)
=\(\dfrac{5x\left(yz+1+y\right)}{x\left(1+yz+y\right)}=5\)
Ta có:
\(b^2=ac\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\left(1\right)\)
\(c^2=bd\Rightarrow\dfrac{b}{c}=\dfrac{c}{d}\left(2\right)\)
Từ (1) và (2), suy ra: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
\(\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}\)
Vậy \(\dfrac{a}{d}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\)(đpcm)
~ Học tốt!~
2) \(\dfrac{x}{y}=\left(\dfrac{x}{y}\right)^2\)
\(\Rightarrow\left(\dfrac{x}{y}\right)^2-\dfrac{x}{y}=0\)
\(\Rightarrow\dfrac{x}{y}\left(\dfrac{x}{y}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{y}=0\Rightarrow x=0;y\in R\\\dfrac{x}{y}-1=0\Rightarrow\dfrac{x}{y}=1\Rightarrow x=y\end{matrix}\right.\)
3) \(16^5+2^{15}=\left(2^4\right)^5+2^{15}=2^{20}+2^{15}=2^{15}.2^5+2^{15}.1=2^{15}.33⋮33\rightarrowđpcm\)
4)\(\left(x-3\right)^2+\left(y+2\right)^2=0\)
\(\left\{{}\begin{matrix}\left(x-3\right)^2\ge0\\\left(y+2\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow\left(x-3\right)^2+\left(y+2\right)^2\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left(x-3\right)^2=0\Rightarrow x-3=0\Rightarrow x=3\\\left(y+2\right)^2=0\Rightarrow y+2=0\Rightarrow y=-2\end{matrix}\right.\)
\(\left(x-12+y\right)^{200}+\left(x-4-y\right)^{200}=0\)
\(\left\{{}\begin{matrix}\left(x-12+y\right)^{200}\ge0\\\left(x-4-y\right)^{200}\ge0\end{matrix}\right.\)
\(\Rightarrow\left(x-12+y\right)^{200}+\left(x-y-4\right)^{200}\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left(x-12+y\right)^{200}=0\\\left(x-y-4\right)^{200}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-12+y=0\Rightarrow x+y=12\\x-y-4=0\Rightarrow x-y=4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x+y\right)+\left(x-y\right)=12+4\Rightarrow x+y+x-y=16\Rightarrow2x=16\Rightarrow x=8\\y=8-4=4\end{matrix}\right.\)
5a.
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+....+\dfrac{1}{19.21}\\ =\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{19}-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}.\dfrac{20}{21}=\dfrac{10}{21}\)
b.
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\\ =\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)< \dfrac{1}{2}.1=\dfrac{1}{2}\)
b/
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}=\dfrac{2b+c-a+2c-b+a+2a+b-c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
* \(\left\{{}\begin{matrix}2b+c-a=2a\\2c-b+a=2b\\2a+b-c=2c\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2b+c=3a\\2c+a=3b\\2a+b=3c\end{matrix}\right.\)
+)\(\Rightarrow\left\{{}\begin{matrix}c=3a-2b\\a=3b-2c\\b=3c-2a\end{matrix}\right.\)
\(\Rightarrow\left(3a-2b\right)\left(3b-2c\right)\left(3c-2a\right)=abc\left(1\right)\)
+) \(\Rightarrow\left\{{}\begin{matrix}2b=3c-a\\2c=3b-a\\2a=3c-b\end{matrix}\right.\)
\(\Rightarrow\left(3a-c\right)\left(3b-a\right)\left(3c-b\right)=8abc\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\dfrac{abc}{8abc}=\dfrac{1}{8}\)
\(\Rightarrow P=\dfrac{1}{8}\)
Bài 1: Vì: 2x^3 - 1 = 15
=> 2x^3 = 16
=> x^3 = 8
=> x = 2 (1)
Ta có:
* (x + 16)/9 = (y - 25)/16
<=> (2 + 16)/9 = (y - 25)/16
<=> 18/9 = (y - 25)/16
<=> 2 = (y - 25)/16
<=> y - 25 = 16.2 = 32
=> y = 32+25 = 57 (2)
* (x + 16)/9 = (z + 9)/25
<=> (2 + 16)/9 = (z + 9)/25
<=> 2 = (z + 9)/25
<=> z + 9 = 25.2 = 50
=> z = 50 - 9 = 41 (3)
Từ (1), (2) và (3) => x + y + z = 2 + 57 + 41 = 100
Bài 2:
c) vì a,b,c là độ dài các cạnh của tam giác:
\(\Rightarrow\left\{{}\begin{matrix}a< b+c\\b< a+c\\c< a+b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{b+c}< 1\\\dfrac{b}{a+c}< 1\\\dfrac{c}{a+b}< 1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{b+c}< \dfrac{2a}{a+b+c}\\\dfrac{b}{a+c}< \dfrac{2b}{a+b+c}\\\dfrac{c}{a+b}< \dfrac{2c}{a+b+c}\end{matrix}\right.\)
\(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}< \dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}\)
\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}< 2\) (đpcm)