Cho 3x-4y=0 . Tim gia tri nho nhat cua bieu thuc : \(M=x^2+y^2\)
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x3 +27 - (x3 +9x2 +3x+27)=0
=>x3 +27 -x3 -9x2-3x-27=0
=>-9x2 -3x =0
=> -3x(3x+1)=0
=>x=0 ; x =-1/3
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\(\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2.\frac{xy}{ab}+2.\frac{xz}{ac}+2.\frac{yz}{bc}=1\)
Ta có: \(2.\frac{xy}{ab}+2.\frac{xz}{ac}+2.\frac{yz}{bc}=2.\left(\frac{xy}{ab}+\frac{xz}{ac}+\frac{yz}{bc}\right)\)
Mặt khác, \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\) => \(\frac{ayz+bxz+cxy}{xyz}=0\)=> ayz + bxz + cxy = 0
=> \(\frac{ayz+bxz+cxy}{abc}=0\) => \(\frac{yz}{bc}+\frac{xz}{ac}+\frac{xy}{ab}=0\)
Do đó, \(2.\frac{xy}{ab}+2.\frac{xz}{ac}+2.\frac{yz}{bc}=2.\left(\frac{xy}{ab}+\frac{xz}{ac}+\frac{yz}{bc}\right)=0\)
=> đpcm
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x>/5
Hai vế không âm bình phương 2 vế
x+1 = x2 -10x +25
x2 -11x +24 =0
(x-3)(x-8) =0
=> x = 3 (loại) hoặc x =8 (TM)
Vậy x =8
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x2+y2=5
=>x2+2xy+y2=5+2xy
=>(x+y)2=5+2xy
=>32=5+2xy
=>4=2xy =>xy=2
Suy ra: x3+y3=(x+y)(x2-xy+y2)=3.(5-2)=9
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\(\frac{x^2+x+1}{\left(x+1\right)^2}=\frac{x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}}{\left(x+1\right)^2}=\frac{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}{\left(x+1\right)^2}\)
\(\text{Đặt }y=x+1\Rightarrow y-\frac{1}{2}=x+\frac{1}{2}\text{ ta được:}\)
\(\frac{\left(y-\frac{1}{2}\right)^2+\frac{3}{4}}{y^2}=\frac{y^2-y+\frac{1}{4}+\frac{3}{4}}{y^2}\)
\(=\frac{y^2-y+1}{y^2}=\frac{\left(y^2-y+1\right):y^2}{y^2:y^2}=1-\frac{1}{y}+\frac{1}{y^2}\)
\(=\frac{1}{4}-2.\frac{1}{y}.\frac{1}{2}+\frac{1}{y^2}+\frac{3}{4}\)
\(=\left(\frac{1}{y}-\frac{1}{2}\right)^2+\frac{3}{4}\)
Vậy GTNN của \(\frac{x^2+x+1}{\left(x+1\right)^2}\) là 3/4 tại:
\(\frac{1}{y}=\frac{1}{2}\Rightarrow y=2\Rightarrow x=y-1=2-1=1\)
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=x^2-4x+3x-12
=x[x-4]+3[x-4]
=[x+3][x-4]
C2:=x^2-16-[x-4]
=[x-4][x+4]-[x-4]
=[x-4][x+4-1]
=[x-4][x+3]
Cách nữa nè !
x^2-x-12
=(x^2-9)-(x+3)
=(x-3)(x+3)-(x+3)
=(x+3)(x-4)
25
cho mìn ****