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a, \(A=x^2-x\sqrt{y}-2x\sqrt{y}+2y\)
\(=x\left(x-\sqrt{y}\right)-2\sqrt{y}\left(x-\sqrt{y}\right)\)
\(=\left(x-2\sqrt{y}\right)\left(x-\sqrt{y}\right)\)
\(a,\)\(A=x^2-3x\sqrt{y}+2y\)
\(=x^2-2x\sqrt{y}-x\sqrt{y}+2y\)
\(=x\left(x-2\sqrt{y}\right)-\sqrt{y}\left(x-2\sqrt{y}\right)\)
\(=\left(x-\sqrt{y}\right)\left(x-2\sqrt{y}\right)\)
\(b,\)Ta có : \(x=\frac{1}{\sqrt{5}-2}=\frac{\sqrt{5}+2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}=\frac{\sqrt{5}+2}{5-4}=\sqrt{5}+2\)
\(y=\frac{1}{9+4\sqrt{5}}=\frac{9-4\sqrt{5}}{\left(9+4\sqrt{5}\right)\left(9-4\sqrt{5}\right)}=\frac{9-4\sqrt{5}}{81-80}=9-4\sqrt{5}=\left(\sqrt{5}-2\right)^2\)
\(\Rightarrow A=\left[\sqrt{5}+2-\sqrt{\left(\sqrt{5}-2\right)^2}\right]\left[\sqrt{5}+2-2\sqrt{\left(\sqrt{5}-2\right)^2}\right]\)
\(=\left(\sqrt{5}+2-\sqrt{5}-2\right)\left(\sqrt{5}+2-2\sqrt{5}+4\right)\)
\(=4\left(6-\sqrt{5}\right)\)
\(=24-4\sqrt{5}\)
2
\(M=2y-3x\sqrt{y}+x^2=y-2x\sqrt{y}+x^2+y-x\sqrt{y}\\ =\left(\sqrt{y}-x\right)^2+\sqrt{y}\left(\sqrt{y}-x\right)\\ =\left(\sqrt{y}-x\right)\left(\sqrt{y}-x+\sqrt{y}\right)\\ =\left(\sqrt{y}-x\right)\left(2\sqrt{y}-x\right)\)
b
\(y=\dfrac{18}{4+\sqrt{7}}=\dfrac{18\left(4-\sqrt{7}\right)}{16-7}=\dfrac{72-18\sqrt{7}}{9}=\dfrac{72}{9}-\dfrac{18\sqrt{7}}{9}=8-2\sqrt{7}\\ =7-2\sqrt{7}.1+1=\left(\sqrt{7}-1\right)^2\)
Thế x = 2 và y = \(\left(\sqrt{7}-1\right)^2\) vào M được:
\(M=2\left(\sqrt{7}-1\right)^2-3.2.\sqrt{\left(\sqrt{7}-1\right)^2}+2^2\\ =2\left(8-2\sqrt{7}\right)-6.\left(\sqrt{7}-1\right)+4\\ =16-4\sqrt{7}-6\sqrt{7}+6+4\\ =26-10\sqrt{7}\)
1:
a: =>2x-2căn x+3căn x-3-5=2x-4
=>căn x-8=-4
=>căn x=4
=>x=16
b: \(\Leftrightarrow\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)-3\sqrt{x}\left(\sqrt{x}-2\right)=0\)
=>(căn x-2)(x-căn x+4)=0
=>căn x-2=0
=>x=4
a: (x+y+z)^3-x^3-y^3-z^3
=(x+y+z-x)(x^2+2xy+y^2-x^2-xy-xz+z^2)-(y+z)(y^2-yz+z^2)
=(x+y)(y+z)(x+z)
b: x^3+y^3+z^3=1
x+y+z=1
=>x+y=1-z
x^3+y^3+z^3=1
=>(x+y)^3+z^3-3xy(x+y)=1
=>(1-z)^3+z^3-3xy(1-z)=1
=>1-3z-3z^2-z^3+z^3-3xy(1-z)=1
=>1-3z+3z^2-3xy(1-z)=1
=>-3z+3z^2-3xy(1-z)=0
=>-3z(1-z)-3xy(1-z)=0
=>(z-1)(z+xy)=0
=>z=1 và xy=0
=>z=1 và x=0; y=0
A=1+0+0=1
1.
\(y^2+y\left(x^3+x^2+x\right)+x^5-x^4+2x^3-2x^2\)
\(\Delta=\left(x^3+x^2+x\right)^2-4\left(x^5-x^4+2x^3-2x^2\right)\)
\(=\left(x^3-x^2+3x\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}y=\dfrac{-x^3-x^2-x+x^3-x^2+3x}{2}=-x^2+x\\y=\dfrac{-x^3-x^2-x-x^3+x^2-3x}{2}=-x^3-2x\end{matrix}\right.\)
Hay đa thức trên có thể phân tích thành:
\(\left(x^2-x+y\right)\left(x^3+2x+y\right)\)
Dựa vào đó em tự tách cho phù hợp
Lời giải:
$4x^2-y^2+4x+1=(4x^2+4x+1)-y^2=(2x+1)^2-y^2$
$=(2x+1-y)(2x+1+y)=(2.10+1-5)(2.10+1+5)$
$=16.26=416$
a, \(=\left(2x+1\right)^2-y^2=\left(2x+1-y\right)\left(2x+1+y\right)\)
Thay x = 10 ; y = 5
\(\left(20+1-5\right)\left(20+1+5\right)=16.26=416\)
b, \(=x^2-\left(y^2+2y+1\right)=x^2-\left(y+1\right)^2=\left(x-y-1\right)\left(x+y+1\right)\)
Thay x = 93 ; y = 6
\(\left(93-6-1\right)\left(93+6+1\right)=8600\)