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a: A=x+3+|x-3|
=x+3+3-x(x<=3)
=6
b:\(B=\sqrt{x^2+4x+4}-\sqrt{x^2}\)
\(=\left|x+2\right|-\left|x\right|\)
=x+2-x=2
c: \(C=\dfrac{\sqrt{x^2-2x+1}}{x-1}\)
\(=\dfrac{\left|x-1\right|}{x-1}=\dfrac{x-1}{x-1}=1\)
Bài 1:
Ta có: \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\left(a^3+3a^2b+3ab^2+b^3\right)+c^3-3a^2b-3ab^2-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0\left(do.a+b+c\ne0\right)\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(a-c\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow a=b=c\)
\(M=\dfrac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\dfrac{3a^2}{\left(3a\right)^2}=\dfrac{3a^2}{9a^2}=\dfrac{1}{3}\)
Bài 2:
a) \(=\dfrac{x\left(x^2+x-6\right)}{x\left(x^2-4\right)}=\dfrac{x\left(x-2\right)\left(x+3\right)}{x\left(x-2\right)\left(x+2\right)}=\dfrac{x+3}{x+2}\)
b) \(=\dfrac{x\left(x+1\right)+7\left(x+1\right)}{x\left(x^2+2x+1\right)}=\dfrac{\left(x+1\right)\left(x+7\right)}{x\left(x+1\right)^2}=\dfrac{x+7}{x\left(x+1\right)}=\dfrac{x+7}{x^2+x}\)
\(a.x+3+\sqrt{x^2-6x+9}=x+3+\text{ |}x-3\text{ |}=x+3+3-x=6\) \(b.\sqrt{x^2+4x+4}-\sqrt{x^2}=\text{ |}x+2\text{ |}-\text{ |}x\text{ |}=x+2-\left(-x\right)=x+2+x=2x+2\) \(c.\dfrac{\sqrt{x^2-2x+1}}{x-1}=\dfrac{x-1}{x-1}=1\)
\(d.\text{ |}x-2\text{ |}+\dfrac{\sqrt{x^2-4x+4}}{x-2}=\text{ |}x-2\text{ |}+\dfrac{\text{ |}x-2\text{ |}}{x-2}=2-x+\dfrac{-\left(x-2\right)}{x-2}=2-x-1=1-x\)
a/ ĐKXĐ:...
\(\Leftrightarrow4x^2-4x\sqrt{2x-1}-3x^2+6x-3=0\)
\(\Leftrightarrow4x\left(x-\sqrt{2x-1}\right)-3\left(x-1\right)^2=0\)
\(\Leftrightarrow\frac{4x\left(x-1\right)^2}{x+\sqrt{2x-1}}-3\left(x-1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\frac{4x}{x+\sqrt{2x-1}}=3\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow4x=3x+3\sqrt{2x-1}\)
\(\Leftrightarrow x=3\sqrt{2x-1}\)
\(\Leftrightarrow x^2-18x+9=0\) \(\Rightarrow9\pm6\sqrt{2}\)
Vậy pt có 3 nghiệm....
b/ ĐKXĐ:...
\(\Leftrightarrow4x^2-4x\sqrt{4x-3}-x^2+4x-3=0\)
\(\Leftrightarrow4x\left(x-\sqrt{4x-3}\right)-\left(x^2-4x+3\right)=0\)
\(\Leftrightarrow\frac{4x\left(x^2-4x+3\right)}{x+\sqrt{4x-3}}-\left(x^2-4x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x+3=0\Rightarrow x=...\\\frac{4x}{x+\sqrt{4x-3}}=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow4x=x+\sqrt{4x-3}\)
\(\Leftrightarrow3x=\sqrt{4x-3}\)
\(\Leftrightarrow9x^2-4x+3=0\) (vô nghiệm)
Vậy...
a: =>|2x-1|=3
=>2x-1=3 hoặc 2x-1=-3
=>2x=-2 hoặc 2x=4
=>x=2 hoặc x=-1
c: \(\Leftrightarrow\left|x-3\right|=11-x\)
=>x<=11 và (x-3)^2=(11-x)^2
=>x<=11 và x^2-6x+9=x^2-22x+121
=>x<=11 và 16x=112
=>x=7
d:
ĐKXĐ: 3x+19>=0
=>x>=-19/3
PT =>x>=-3 và (3x+19)=(x+3)^2=x^2+6x+9
=>x>=-3 và x^2+6x+9-3x-19=0
=>x>=-3 và (x+5)(x-2)=0
=>x=2
e: =>\(\sqrt{x^2+x+5}=x+1\)
=>x>=-1 và x^2+x+5=x^2+2x+1
=>x>=-1 và 2x+1=x+5
=>x=4
1.
$x+3+\sqrt{x^2-6x+9}=x+3+\sqrt{(x-3)^2}=x+3+|x-3|$
$=x+3+(3-x)=6$
2.
$\sqrt{x^2+4x+4}-\sqrt{x^2}=\sqrt{(x+2)^2}-\sqrt{x^2}$
$=|x+2|-|x|=x+2-(-x)=2x+2$
3.
$\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}$
$=\sqrt{(\sqrt{x^2-1}+1)^2}-\sqrt{(\sqrt{x^2-1}-1)^2}$
$=|\sqrt{x^2-1}+1|+|\sqrt{x^2-1}-1|$
$=\sqrt{x^2-1}+1+|\sqrt{x^2-1}-1|$
4.
$\frac{\sqrt{x^2-2x+1}}{x-1}=\frac{\sqrt{(x-1)^2}}{x-1}$
$=\frac{|x-1|}{x-1}=\frac{x-1}{x-1}=1$
5.
$|x-2|+\frac{\sqrt{x^2-4x+4}}{x-2}=2-x+\frac{\sqrt{(x-2)^2}}{x-2}$
$=2-x+\frac{|x-2|}{x-2}|=2-x+\frac{2-x}{x-2}=2-x+(-1)=1-x$
6.
$2x-1-\frac{\sqrt{x^2-10x+25}}{x-5}=2x-1-\frac{\sqrt{(x-5)^2}}{x-5}$
$=2x-1-\frac{|x-5|}{x-5}$
a, \(A=\left(\sqrt{12}-2\sqrt{5}\right)\sqrt{3}+\sqrt{60}\)
\(=\left(2\sqrt{3}-2\sqrt{5}\right)\sqrt{3}+2\sqrt{15}\)
\(=2\sqrt{9}-2\sqrt{15}+2\sqrt{15}=2\sqrt{9}\)
b, \(B=\frac{\sqrt{4x}}{x-3}\sqrt{\frac{x^2-6x+9}{x}}=\frac{2\sqrt{x}}{x-3}.\sqrt{\frac{\left(x-3\right)^2}{x}}\)
\(=\frac{2\sqrt{x}}{x-3}.\frac{x-3}{\sqrt{x}}=2\)
`A=-x^2+2x+10`
`=-(x^2-2x)+10`
`=-(x-1)^2+11<=11`
Dấu "=" xảy ra khi `x=1`.
`B=4x-2x^2+8`
`=-2(x^2-2x)+8`
`=-2(x^2-2x+1)+10`
`=-2(x-1)^2+10<=10`
Dấu "=" xảy ra khi `x=1`
`C=-x^2-x+1`
`=-(x^2+x)+1`
`=-(x^2+x+1/4)+1+1/4`
`=-(x+1/2)^2+5/4<=5/4`
Dấu "=" xảy ra khi `x=-1/2`
`D=-4x^2+6x+3`
`=-(4x^2-6x)+3`
`=-(4x^2-6x+9/4)+21/4`
`=-(2x-3/2)^2+21/4<=21/4`
Dấu "=' xảy ra khi `2x=3/2<=>x=3/4`
\(a,A=-x^2+2x+10=-x^2+2x-1+11=-\left(x^2-2x+1\right)+11\)
\(=11-\left(x-1\right)^2\)
- Thấy : \(\left(x-1\right)^2\ge0\forall x\in R\)
\(\Rightarrow A=11-\left(x-1\right)^2\le11\)
Vậy MaxA = 11 <=> x = 1 .
\(b,B=-2x^2+4x-2+10=-2\left(x^2-2x+1\right)+10=10-2\left(x-1\right)^2\)
- Thấy : \(\left(x-1\right)^2\ge0\forall x\in R\)
\(\Rightarrow B=10-2\left(x-1\right)^2\le10\)
Vậy MaxB = 10 <=> x = 1 .
\(c,C=-x^2-\dfrac{1}{2}.2.x-\dfrac{1}{4}+\dfrac{5}{4}=\dfrac{5}{4}-\left(x+\dfrac{1}{2}\right)^2\)
- Thấy : \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\in R\)
\(\Rightarrow C=\dfrac{5}{4}-\left(x+\dfrac{1}{2}\right)^2\le\dfrac{5}{4}\)
Vậy MaxC = 5/4 <=> x = -1/2 .
\(d,D=-4x^2+6x+3=-4x^2+2x.2.\dfrac{6}{4}-\dfrac{9}{4}+\dfrac{21}{4}=-\left(4x^2-6x+\dfrac{9}{4}\right)+\dfrac{21}{4}\)
\(=\dfrac{21}{4}-\left(2x-\dfrac{3}{2}\right)^2\)
- Thấy : \(\left(2x-\dfrac{3}{2}\right)^2\ge0\forall x\in R\)
\(\Rightarrow A=\dfrac{21}{4}-\left(2x-\dfrac{3}{2}\right)^2\le\dfrac{21}{4}\)
Vậy MaxD=21/4 <=> x = 3/4 .
