Hỏi đáp môn Toán

Gọi x(h) là thời gian vòi 1 chảy 1 mình đầy bể

y(h) là thời gian vòi 2 chảy 1 mình đầy bể

đk: x,y>16

\(\dfrac{1}{x}\)(bể) là phần bể mà vòi 1 chảy 1 mình trong 1h

\(\dfrac{1}{y}\) (bể) là phần bể mà vòi 2 chảy 1 mình trong 1h

Vì hai vòi cùng chảy thì 16h sẽ đầy bể nên ta có phương trình:

\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{16}\) (1)

\(\dfrac{3}{x}\) (bể) là phần bể mà vòi 1 chảy 1 mình trong 3h

\(\dfrac{6}{y}\) (bể) là phần bể mà vòi 2 chảy 1 mình trong 6h

Vì nếu vòi 1 chảy 3h, vòi 2 chảy 6h thì được 25%\(\left(\dfrac{1}{4}\right)\) bể nên ta có phương trình:

\(\dfrac{3}{x}+\dfrac{6}{y}=\dfrac{1}{4}\left(2\right)\)

Từ (1) và (2) ta có hệ phương trình:

\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{16}\\\dfrac{3}{x}+\dfrac{6}{y}=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}+\dfrac{3}{y}=\dfrac{3}{16}\\\dfrac{3}{x}+\dfrac{6}{y}=\dfrac{1}{4}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-3}{y}=\dfrac{-1}{16}\\\dfrac{3}{x}+\dfrac{6}{y}=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=48\\\dfrac{3}{x}+\dfrac{6}{48}=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=48\\x=24\end{matrix}\right.\left(tmđk\right)\)

Vậy vòi 1 chảy 1 mình trong 24h thì đầy bể; vòi 2 chảy 1 mình trong 48h thì đầy bể

Ta có:

\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)

=>\(\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\)

=> \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{xz}{ac}\right)=1\)

=>\(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{cxy}{abc}+\dfrac{ayz}{abc}+\dfrac{bxz}{abc}\right)=1\) (1)

Lại có:

\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)

=> \(\dfrac{a}{x}.\dfrac{yz}{yz}+\dfrac{b}{y}.\dfrac{xz}{xz}+\dfrac{c}{z}.\dfrac{xy}{xy}=0\)

=>\(\dfrac{ayz}{xuy}+\dfrac{bxz}{xyz}+\dfrac{cxy}{xyz}=0\) (2)

Thay (2) vào (1) ta được

\(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+0=1\)

=> \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)

* Ta có:

\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)

\(\Leftrightarrow\dfrac{axy}{xyz}+\dfrac{bxz}{xyz}+\dfrac{cxy}{xyz}=0\)

\(\Leftrightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)

\(\Leftrightarrow ayz+bxz+cxy=0\)

* Ta có:

\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)

\(\Leftrightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\)

\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\dfrac{xy}{ab}+2\dfrac{xz}{ac}+2\dfrac{yz}{bc}=1\)\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{xz}{ac}+\dfrac{yz}{bc}\right)=1\)\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{b^2}+2\left(\dfrac{cxy}{abc}+\dfrac{bxz}{abc}+\dfrac{ayz}{abc}\right)=1\)\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{cxy+bxz+ayz}{abc}\right)=1\)Mà \(cxy+bxz+ayz=0\)

\(\Rightarrow2\left(\dfrac{cxy+bxz+ayz}{abc}\right)=0\)

\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)

Vậy.........................

9x² + y² + 2z² - 18x + 4z - 6y + 20 = 0

<=> 9x² - 18x + 9 + y² - 6y + 9 + 2x² + 4z + 2 = 0

<=> 9(x² - 2x + 1) + (y - 3)² + 2(z² + 2z + 1) = 0

<=> 9(x - 1)² + (y - 3)² + 2(z + 1)² = 0

<=> \(\left\{\begin{matrix}x-1=0\\y-3=0\\z+1=0\end{matrix}\right.\)

<=> \(\left\{\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)

..

bài 3;

áp dujgn C-S kind Engel \(VT\ge\dfrac{\left(1+1+1\right)^2}{3+xy+yz+xz}\ge\dfrac{9}{6}=\dfrac{3}{2}\)

xảy ra khi x=y=z=1

Bài 1:

\(VT\ge\sqrt{x+3-x-2}=1=VP\)

xảy ra khi x=6

Bài 2 :từ gt=>a=b=0 thay vào

\(\left(85\dfrac{7}{30}-83\dfrac{5}{18}\right):2\dfrac{2}{3}=0,01x:4\)

\(\Leftrightarrow[\left(85-83\right)+\left(\dfrac{7}{30}-\dfrac{5}{18}\right)]:2\dfrac{2}{3}=0,01x:4\)

\(\Leftrightarrow1\dfrac{43}{45}:2\dfrac{2}{3}=0,01x:4\)

\(\Leftrightarrow\dfrac{88}{45}:\dfrac{8}{3}=0,01x:4\)

\(\Leftrightarrow\dfrac{88}{45}.\dfrac{3}{8}=0,01x:4\)

\(\Leftrightarrow\dfrac{11}{15}=0,01x:4\)

\(\Leftrightarrow0,01x:4=\dfrac{11}{15}\)

\(\Leftrightarrow x=\dfrac{11}{15}.4:0,01\)

\(\Leftrightarrow x=\dfrac{880}{3}\)

Vậy x = \(\dfrac{880}{3}\)

\(\dfrac{1}{3}-\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)

\(=\left(\dfrac{1}{3}-\dfrac{2}{9}-\dfrac{1}{36}\right)+(-\dfrac{3}{4}+\dfrac{1}{64})+(-\dfrac{3}{5}+\dfrac{1}{15})\)

\(=\left(\dfrac{12}{36}-\dfrac{8}{36}-\dfrac{1}{36}\right)+\left(-\dfrac{48}{64}+\dfrac{1}{64}\right)+(-\dfrac{9}{15}+\dfrac{1}{15})\)

\(=\dfrac{1}{12}+\dfrac{-47}{64}-\dfrac{8}{15}=\dfrac{80}{960}-\dfrac{705}{960}-\dfrac{512}{960}\)

\(\dfrac{80-705-512}{960}=-\dfrac{1137}{960}=-\dfrac{379}{320}\)

Ta có:

\(VT=\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}\)

\(=\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)-2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)}\)

\(=\sqrt{\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2-2\left(\dfrac{c}{abc}+\dfrac{a}{abc}+\dfrac{b}{bca}\right)}\)

\(=\sqrt{\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2-2\left(\dfrac{a+b+c}{abc}\right)}\)

\(=\sqrt{\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}\)

\(=\left|\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right|\)

\(\Rightarrow VT=VP\)

Vậy \(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}=\left|\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right|\) (Đpcm)

Ta có: \(\dfrac{5n+2}{n-1}=\dfrac{5n-5+7}{n-1}=5+\dfrac{7}{n-1}\)

Mà 5 là số tự nhiên nên để bt trên là số tự nhiên nên:

\(n-1\inƯ\left(7\right)\)

\(\Leftrightarrow n-1\in\left\{1,7\right\}\)

\(\Rightarrow n=7\left(chọn\right)\)

Vậy nếu n =7 thì bt trên là số tự nhiên

Để:

\(n\in N\)

\(\Rightarrow5n+2⋮n-1\)

\(5n-5+7⋮n-1\)

\(5\left(n-1\right)+7⋮n-1\)

\(7⋮n-1\)

\(\Rightarrow n-1\inƯ\left(7\right)\)

\(Ư\left(7\right)=\left\{\pm1;\pm7\right\}\)

\(\Leftrightarrow n-1=1\Rightarrow n=2\)

\(\Leftrightarrow n-1=-1\Rightarrow n=0\)

\(\Leftrightarrow n-1=7\Rightarrow n=8\)

\(\Leftrightarrow n-1=-7\Rightarrow n=-8\)(loại)

\(\Rightarrow n\in\left\{0;2;8\right\}\)

\(B=4+4^2+4^3+.....+4^{2016}\)

\(4B=4\left(4+4^2+4^3+.....+4^{2016}\right)\)

\(4B=4^2+4^3+4^4+.....+4^{2017}\)

\(4B-B=\left(4^2+4^3+4^4+......+4^{2017}\right)-\left(4+4^2+4^3+.....+4^{2016}\right)\)

\(3B=4^{2017}-4\)

\(B=\dfrac{4^{2017}-4}{3}\)

c, Đặt x²-3x+2=0

=> x²-2x-x+2=0

=> x²-x-2x+2=0

=>x(x-1)-2(x-1)=0

=>(x-2).(x-1)=0

=>\(\left\{{}\begin{matrix}x-2=0\\x-1=0\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}x=0+2\\x=0+1\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)

Vậy x\(\in\){2;1}là nghiệm của đa thức x²-3x+2

b,Đặt 2x²-3x =0

=> x.(2x-3)=0

=>\(\left\{{}\begin{matrix}x=0\\2x-3=0\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}x=0\\2x=3\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy x\(\in\){0;\(\dfrac{3}{2}\)} là nghiệm của đa thức 2x²-3x