Cho \(\dfrac{x}{3}=\dfrac{c}{a};\dfrac{y}{d}=\dfrac{3}{b}\) và c + d = 27. Tính \(\sqrt{ax+by}\) (a, b, c, d, x là các số dương).
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b/ \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
\(\Rightarrow\left(\dfrac{a}{b}\right)^3=\dfrac{a}{d}\left(1\right)\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
=> \(\left(\dfrac{a}{b}\right)^3=\left(\dfrac{a+b+c}{c+d+b}\right)^3\) (2)Từ (1) và (2)=>đpcm
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\(1,Q=\dfrac{a^4-2a^2+a^3-2a+a^2-2}{a^4-2a^2+2a^3-4a+a^2-2}\\ Q=\dfrac{\left(a^2-2\right)\left(a^2+a+1\right)}{\left(a^2-2\right)\left(a^2+2a+1\right)}=\dfrac{a^2+a+1}{a^2+2a+1}\)
\(Q=\dfrac{x^2+x+1}{\left(x+1\right)^2}-\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{x^2+x+1-\dfrac{3}{4}x^2-\dfrac{3}{2}x-\dfrac{3}{4}}{\left(x+1\right)^2}+\dfrac{3}{4}\\ Q=\dfrac{\dfrac{1}{4}x^2-\dfrac{1}{2}x+\dfrac{1}{4}}{\left(x+1\right)^2}+\dfrac{3}{4}=\dfrac{\dfrac{1}{4}\left(x-1\right)^2}{\left(x+1\right)^2}+\dfrac{3}{4}\ge\dfrac{3}{4}\\ Q_{min}=\dfrac{3}{4}\Leftrightarrow x=1\)
\(2,\text{Từ GT }\Leftrightarrow\dfrac{ayz+bxz+czy}{xyz}=0\\ \Leftrightarrow ayz+bxz+czy=0\\ \text{Ta có }\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\\ \Leftrightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{zx}{ca}\right)=0\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\cdot\dfrac{cxy+ayz+bzx}{abc}=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\cdot\dfrac{0}{abc}=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)
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\(a,A=\dfrac{5-3}{5+2}=\dfrac{2}{7}\\ b,B=\dfrac{3x-9+2x+6-3x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ c,C=AB=\dfrac{x-3}{x+2}\cdot\dfrac{2}{x-3}=\dfrac{2}{x+2}\\ C=-\dfrac{1}{3}\Leftrightarrow x+2=-6\Leftrightarrow x=-8\left(tm\right)\)
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a: \(B=\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{x+9}{x-9}\)
\(=\dfrac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3}{\sqrt{x}-3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=\left(\dfrac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right)\cdot\dfrac{\sqrt{x}-3}{2\sqrt{x}+4}\)
\(=\dfrac{-3}{2\sqrt{x}+4}\)
Để \(C< -\dfrac{1}{3}\) thì \(\dfrac{-3}{2\sqrt{x}+4}+\dfrac{1}{3}< 0\)
\(\Leftrightarrow-9+2\sqrt{x}+4< 0\)
\(\Leftrightarrow\sqrt{x}< \dfrac{5}{2}\)
hay \(0\le x< \dfrac{25}{4}\)
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2:
a: Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{24}{9}=\dfrac{8}{3}\)
=>x=16/3; y=8; z=32/3
A=3x+2y-6z
=3*16/3+2*8-6*32/3
=16+16-64
=-32
b: Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{x-y+z}{5-6+7}=\dfrac{6\sqrt{2}}{6}=\sqrt{2}\)
=>x=5căn 2; y=6căn 2; y=7căn 2
B=xy-yz
=y(x-z)
=6căn 2(5căn 2-7căn 2)
=-6căn 2*2căn 2
=-24
![](https://rs.olm.vn/images/avt/0.png?1311)
1
Áp dụng tính chất dãy tỉ số bằng nhau
`=>a/(b+c)=c/(a+b)=b/(a+c)=(a+b+c)/(2a+2b+2c)=1/2`
`=>b+c=2a`
`=>a+b+c=3a`
Hoàn toàn tương tự:
`a+b+c=3b`
`a+b+c=3c`
`=>a=b=c`
`=>A=1/2+1/2+1/2=3/2`
2
`A in Z`
`=>x+3 vdots x-2`
`=>x-2+5 vdots x-2`
`=>5 vdots x-2`
`=>x-2 in Ư(5)={1,-1,5,-5}`
`+)x-2=1=>x=3(TM)`
`+)x-2=-1=>x=1(TM)`
`+)x-2=5=>x=7(TM)`
`+)x-2=-5=>x=-3(TM)`
Vậy với `x in {1,3,-3,7}` thì `A in Z`
`A in Z`
`=>1-2x vdots x+3`
`=>-2(x+3)+1+6 vdots x+3`
`=>7 vdots x+3`
`=>x+3 in Ư(7)={1,-1,7,-7}`
`+)x+3=1=>x=-2(TM)`
`+)x+3=-1=>x=-4(TM)`
`+)x+3=-7=>x=-10(TM)`
`+)x+3=7=>x=4(TM)`
Vậy `x in {2,-4,4,10}` thì `A in Z`
![](https://rs.olm.vn/images/avt/0.png?1311)
1, Ta có: \(x+y=9\Rightarrow\left(x+y\right)^2=81\)
\(\Rightarrow x^2+2xy+y^2=81\)
\(\Rightarrow x^2+y^2=45\)
\(\Rightarrow x^2+y^2-2xy=9\)
\(\Rightarrow\left(x-y\right)^2=9\Rightarrow\left[{}\begin{matrix}x-y=3\\x-y=-3\end{matrix}\right.\)
\(A=x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(\Rightarrow\left[{}\begin{matrix}A=3.63=189\\A=-3.63=-189\end{matrix}\right.\)
Vậy...
x/3=c/a=>ax=3c
y/d=3/b=>by=3d
ax+by=3c+3d=3(c+d)=3.27=34
\(\sqrt{ax+by}=\sqrt{3^4}=3^2=9\)