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\(x^2+6x+9=\left(x+3\right)^2\)
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\(x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2\)
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\(x^3+12x^2+48x+64=\left(x+4\right)^3\)
1) \(\dfrac{\left(x+5\right)^2+\left(x-5\right)^2}{x^2+25}\)
\(=\dfrac{x^2+10x+25+x^2-10x+25}{x^2+25}\)
\(=\dfrac{2x^2+50}{x^2+25}\)
\(=\dfrac{2\left(x^2+25\right)}{x^2+25}=2\)
2) \(\left(x+3\right)\left(x^2-3x+9\right)-\left(54+x^3\right)\)
\(=x^3+3^3-54-x^3\)
\(=27-54=-27\)
3) \(\left(2x+y\right)^2-\left(y+3x\right)^2\)
\(=4x^2+4xy+y^2-y^2-6xy-9x^2\)
\(=-5x^2-2xy\)
4) \(\left(2x+1\right)^3-\left(2x-1\right)^3-24x^2\)
\(=8x^3+12x^2+6x+1-8x^3+12x^2-6x+1-24x^2\)
\(=2\)
a) \(9x^2+6x+1=\left(3x+1\right)^2\)
b)\(x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2\)
c)\(x^2y^4-2xy^2+1=\left(xy^2-1\right)^2\)
d) \(x^2+\frac{2}{3}x+\frac{1}{9}=\left(x+\frac{1}{3}\right)^2\)
a) 9x2 + 6x + 1 = ( 3x + 1 )2
b) x2 - x + 1/4 = ( x - 1/2)2
c) x2 . y4 - 2xy2 + 1 = ( xy2 - 1 ) 2
d) x2 + 2/3x + 1/9 = (x+1/3)2
1)\(25x^2y^4+30xy^2z+9z^2=\left(5xy^2+3z\right)^2\)
\(\dfrac{16}{9}x^2+4xyz^2+\dfrac{9}{4}y^2z^4=\left(\dfrac{4}{3}x+\dfrac{3}{2}yz^2\right)^2\)
2)
a)\(\dfrac{9}{25}x^2+\dfrac{12}{35}xy+\dfrac{4}{49}y^2=\left(\dfrac{3}{5}x+\dfrac{2}{7}y\right)^2=\left(\dfrac{3}{5}.5+\dfrac{2}{7}.\left(-7\right)\right)^2=\left(3-2\right)^2=1\)b)\(\dfrac{25}{16}u^4v^2+\dfrac{1}{5}u^2v^3+\dfrac{4}{625}v^4\)
\(=\left(\dfrac{5}{4}u^2v+\dfrac{2}{25}v^2\right)^2=\left(\dfrac{5}{4}.\dfrac{4}{25}.\left(-5\right)+\dfrac{2}{25}.\left(-5\right)^2\right)^2\)
\(=\left(-1+2\right)^2=1\)
a) \(9x^2+6x+1=\left(3x\right)^2+2.3x.1+1^2=\left(3x+1\right)^2\)
b) \(x^2-x+\dfrac{1}{4}=x^2-2.\dfrac{1}{2}x+\left(\dfrac{1}{2}\right)^2=\left(x-0,5\right)^2\)
c) \(x^2y^4-2xy^2+1=\left(xy^2\right)^2-2.xy^2.1+1^2=\left(xy^2-1\right)^2\)
d) \(x^2+\dfrac{2}{3}x+\dfrac{1}{9}=x^2+2.x.\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2=\left(x+\dfrac{1}{3}\right)^2\)
a) \(9x^2+6x+1\)
\(=\left(3x\right)^2+2.3x.1+1^2\)
\(=\left(3x+1\right)^2\)
a) x2 +4x+4 = ( x + 2 )2
b) 16x2 - 8xy + y2 = ( 4x - y )2
c)9a2 +16b2 - 24ab = ( 3a - 4y ) 2
d) x2 - x + \(\dfrac{1}{4}\)= ( x - \(\dfrac{1}{2}\))2
e) y2 + \(\dfrac{1}{2}y\) + \(\dfrac{1}{16}\) = ( y + \(\dfrac{1}{4}\))2
a, ĐKXĐ : \(x-1\ne0\)
=> \(x\ne1\)
TH1 : \(x-2\ge0\left(x\ge2\right)\)
=> \(\left|x-2\right|=x-2=1\)
=> \(x=3\left(TM\right)\)
- Thay x = 3 vào biểu thức P ta được :
\(P=\frac{3+2}{3-1}=\frac{5}{2}\)
TH2 : \(x-2< 0\left(x< 2\right)\)
=> \(\left|x-2\right|=2-x=1\)
=> \(x=1\left(KTM\right)\)
Vậy giá trị của P là \(\frac{5}{2}\) .
a) \(P=\frac{x+2}{x-1}\) \(\left(ĐKXĐ:x\ne1\right)\)
Ta có: \(\left|x-2\right|=1\text{⇔}\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\text{⇔}\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\) (loại x = 1 vì x ≠ 1)
Thay \(x=3\) vào P, ta có:
\(P=\frac{3+2}{3-2}=\frac{5}{1}=5\)
Vậy P = 5 tại x = 3.
b) \(Q=\frac{x-1}{x}+\frac{2x+1}{x^2+x}=\frac{x-1}{x}+\frac{2x+1}{x\left(x+1\right)}=\frac{x^2-1}{x\left(x+1\right)}+\frac{2x+1}{x\left(x+1\right)}\) (ĐKXĐ: x ≠ 0, x ≠ -1)
\(=\frac{x^2+2x}{x\left(x+1\right)}=\frac{x\left(x+2\right)}{x\left(x+1\right)}=\frac{x+2}{x+1}\)
\(25x^2y^4+30xy^2z+9z^2=\left(5xy^2\right)^2+2.5xy^2.3z+\left(3z\right)^2=\left(5xy^2+3z\right)^2\)
\(\frac{16}{9}x^2+4xyz^2+\frac{9}{4}y^2z^4=\left(\frac{4}{3}x\right)^2+2.\frac{4}{3}x.\frac{3}{2}yz^2+\left(\frac{3}{2}yz^2\right)^2=\left(\frac{4}{3}x+\frac{3}{2}yz^2\right)^2\)
\(\frac{9}{25}x^2+\frac{12}{35}xy+\frac{4}{49}y^2=\left(\frac{3}{5}x\right)^2+2.\frac{3}{5}x.\frac{2}{7}y+\left(\frac{2}{7}y\right)^2=\left(\frac{3}{5}x+\frac{2}{7}y\right)^2\)( tự thay vào tính nhé )
\(\frac{25}{16}u^4y^2+\frac{1}{5}u^2+y^3+\frac{4}{625}y^4=\left(\frac{5}{4}u^2y\right)^2+2.\frac{5}{4}u^2y.\frac{2}{25}.y^2+\left(\frac{2}{25}y^2\right)^2=\left(\frac{5}{4}u^2y+\frac{2}{25}y^2\right)^2\)( tự thay vào tính nhé )
Tham khảo nhé~
