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Bài 1:
a) \(\frac{4}{9}x^2-y^2=\left(\frac{2}{3}x-y\right)\left(\frac{2}{3}x+y\right)\)
b) \(x^2-5=\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\)
c) \(4x^2+6x+9=\left(2x+2\right)^2+5\)ko hiểu ???
d) \(\frac{1}{9}x^2-\frac{4}{3}xy+4=\left(\frac{1}{3}x\right)^2-2.\frac{1}{3}x.2+2^2=\left(\frac{1}{3}x-2\right)^2\)
Bài 2:
a) \(\left(\frac{1}{2}x-\frac{1}{3}y\right)\left(\frac{1}{2}x+\frac{1}{3}y\right)=\frac{1}{4}x^2-\frac{1}{9}y^2\)
b) \(\left(2x-\frac{1}{3}y\right)\left(4x^2+\frac{2}{3}xy+\frac{1}{9}x^2\right)=8x^3-\frac{1}{27}y^3\)
c) \(\left(3x-5y\right)\left(9x^2+15xy+\frac{1}{9}x^2\right)=27x^3-125y^3\)
Bài 1:
\(a,27x^3+27x^2+9x+1\)
\(=\left(3x\right)^3+3.\left(3x\right)^2.1+3.3x.1^2+1^3\)
\(=\left(3x+1\right)^3\)
\(b,x^3+3\sqrt{2}x^2y+6xy^2+2\sqrt{2}y^3\)
\(=x^3+3.x^2.\sqrt{2}y+3.x.\left(\sqrt{2}y\right)^2+\left(\sqrt{2}y\right)^3\)
\(=\left(x+\sqrt{2}y\right)^3\)
Bài 2:
\(a,x^3+9x^2+27x+27=0\)
\(\Leftrightarrow\left(x+3\right)^3=0\)
\(\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
\(b,\left(x+1\right)^3-x\left(x-2\right)^2+x-1=0\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3-4x^2+4x+x-1=0\)
\(\Leftrightarrow-x^2+8x=0\)
\(\Leftrightarrow-x\left(x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)
1)
a) = (3x+1)3
b) (x+\(\sqrt{2}\) )3
2)
a)\(x^3+9x^2+27x+27=0\\ \left(x+3\right)^3=0\\ =>x=-3\)
b) Bài cuối bạn tự làm nhé! Mình mắc học bài
# Chúc bạn học tốt !
b1:
câu a,f áp dụng a2-b2=(a-b)(a+b)
câu b,c áp dụng a3-b3=(a-b)(a2+ab+b2)
câu d: \(x^2+2xy+x+2y=x\left(x+2y\right)+\left(x+2y\right)=\left(x+1\right)\left(x+2y\right)\)
câu e: \(7x^2-7xy-5x+5y=7x\left(x-y\right)-5\left(x-y\right)=\left(7x-5\right)\left(x-y\right)\)
câu g xem lại đề
\(P=\left(x-y\right)^2+\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)-4x^2=\left(x-y-x-y\right)^2-\left(2x\right)^2=\left(-2y\right)^2-\left(2x\right)^2\)
\(=\left(2y-2x\right)\left(2y+2x\right)=2\left(y-x\right)2\left(y+x\right)=4\left(x+y\right)\left(y-x\right)\)
\(x^3-x^2y+3x-3y=x^2\left(x-y\right)+3\left(x-y\right)=\left(x-y\right)\left(x^2+3\right)\)
\(x^3-2x^2-4xy^2+x=x\left(x^2-2x+1-4y^2\right)=x\left[\left(x-1\right)^2-\left(2y\right)^2\right]=x\left(x+2y-1\right)\left(x-2y-1\right)\)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)
Đặt \(x^2+7x+10=t\), ta có:
\(t\left(t+2\right)-8=t^2+2t-8=t^2-2t+4t-8=t\left(t-2\right)+4\left(t-2\right)=\left(t-2\right)\left(t+4\right)\)
\(=\left(x^2+7x+10+4\right)\left(x^2+7x+10-2\right)=\left(x^2+7x+14\right)\left(x^2+7x-8\right)\)
a)\(\left(3x\right)^3-\left(\dfrac{1}{3}\right)^3=\left(3x-\dfrac{1}{3}\right)\left(9x^2+x+\dfrac{1}{9}\right)\)
b)\(x^3-x^2+5x^2-5x-24x+24\)
\(=x^2\left(x-1\right)+5x\left(x-1\right)-24\left(x-1\right)\)
\(=\left(x^2+5x-24\right)\left(x-1\right)\)
\(=\left(x^2+8x-3x-24\right)\left(x-1\right)\)
\(=\left[x\left(x+8\right)-3\left(x+8\right)\right]\left(x-1\right)\)
\(=\left(x-3\right)\left(x+8\right)\left(x-1\right)\)
c)\(x^2+7x+12=x^2+3x+4x+12\)
\(=x\left(x+3\right)+4\left(x+3\right)=\left(x+3\right)\left(x+4\right)\)
d)\(3x^2-3xy-5x+5y\)
\(=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)
a )x2+2y2-2xy+2x-4y+2=0
<=>x2-2x(y-1)+y2-2y+1+y2-2y+1=0
<=>x2-2x(y-1)+(y-1)2+(y-1)2=0
<=>(x-y+1)2+(y-1)2=0
<=>x-y+1=0 va y-1=0
<=>x=y-1 y=1
<=>x=1-1=0 y=1