Phân tích thành nhân tử :
a) \(1+x\sqrt{x}\)
b) \(x^2\sqrt{x}+x^2-\sqrt{x}-1\)
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d: \(=-\left(x+\sqrt{x}-12\right)=-\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)\)
\(x+\sqrt{x}=\sqrt{x}\left(\sqrt{x}+1\right)\\ x-\sqrt{x}=\sqrt{x}\left(\sqrt{x}-1\right)\\ a+3\sqrt{a}-10=a+5\sqrt{a}-2\sqrt{a}-10=\sqrt{a}\left(\sqrt{a}+5\right)-2\left(\sqrt{a}+5\right)=\left(\sqrt{a}-2\right)\left(\sqrt{a}+5\right)\)
\(x\sqrt{x}+\sqrt{x}-x-1=\left(x\sqrt{x}-x\right)+\left(\sqrt{x}-1\right)=x\left(\sqrt{x}-1\right)+\sqrt{x}-1=\left(\sqrt{x}-1\right)\left(x+1\right)\\ x+\sqrt{x}-2=x+2\sqrt{x}-\sqrt{x}-2=\sqrt{x}\left(\sqrt{x}+2\right)-\left(\sqrt{x}+2\right)=\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)\\ x-5\sqrt{x}+6=x-2\sqrt{x}-3\sqrt{x}-6=\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)=\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)\)
Mấy bạn còn lại tương tự những bài trên nhé. Nếu còn thắc mắc ở chỗ nào bạn có thể liên hệ mình nhé. Nhớ lần sau bạn tách ra nha, chứ nhiều câu quá.
1. $x+\sqrt{x}=\sqrt{x}(\sqrt{x}+1)$
2. $x-\sqrt{x}=\sqrt{x}(\sqrt{x}-1)$
3. $a+3\sqrt{a}-10=(a-2\sqrt{a})+(5\sqrt{a}-10)$
$=\sqrt{a}(\sqrt{a}-2)+5(\sqrt{a}-2)=(\sqrt{a}+5)(\sqrt{a}-2)$
4. $x\sqrt{x}+\sqrt{x}-x-1=(x\sqrt{x}+\sqrt{x})-(x+1)=\sqrt{x}(x+1)-(x+1)$
$=(x+1)(\sqrt{x}-1)$
5. $x+\sqrt{x}-2=(x-\sqrt{x})+(2\sqrt{x}-2)$
$=\sqrt{x}(\sqrt{x}-1)+2(\sqrt{x}-1)=(\sqrt{x}-1)(\sqrt{x}+2)$
\(x+2\sqrt{x-1}=\left(x-1\right)+2\sqrt{x-1}+1=\left(\sqrt{x-1}+1\right)^2\)
\(x-4\sqrt{x-2}+2=\left(x-2\right)-4\sqrt{x-2}+4=\left(\sqrt{x-2}-2\right)^2\)
\(x+2\sqrt{x-1}=\left(\sqrt{x-1}+1\right)^2\)
\(x-4\sqrt{x-2}+2=\left(\sqrt{x-2}+4\right)^2\)
a, \(ab+b\sqrt{a}+\sqrt{a}+1=\sqrt{a}b\left(\sqrt{a}+1\right)+\sqrt{a}+1\)
\(=\left(b\sqrt{a}+1\right)\left(\sqrt{a}+1\right)\)
b, \(\sqrt{x^3}-\sqrt{y^3}+\sqrt{x^2y}-\sqrt{xy^2}\)
\(=\sqrt{x^2}\left(\sqrt{x}+\sqrt{y}\right)-\sqrt{y^2}\left(\sqrt{y}+\sqrt{x}\right)=\left(\left|x\right|-\left|y\right|\right)\left(\sqrt{x}+\sqrt{y}\right)\)
a) \(x\sqrt{x}+\sqrt{x}-x-1\)
\(=\left(x\sqrt{x}-x\right)+\left(\sqrt{x}-1\right)\)
\(=x\left(\sqrt{x}-1\right)+\left(\sqrt{x}-1\right)\)
\(=\left(\sqrt{x}-1\right)\left(x+1\right)\)
b) \(\sqrt{ab}+2\sqrt{a}+3\sqrt{b}+6\)
\(=\sqrt{a}\left(\sqrt{b}+2\right)+3\left(\sqrt{b}+2\right)\)
\(=\left(\sqrt{b}+2\right)\left(\sqrt{a}+3\right)\)
a: =(căn a-3)^2-b^2
=(căn a-3-b)(căn a-3+b)
b: \(x-9=\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\)
c: \(x-7\sqrt{x}+12=x-3\sqrt{x}-4\sqrt{x}+12=\left(\sqrt{x}-3\right)\left(\sqrt{x}-4\right)\)
d: x*căn x-64
=(căn x)^3-4^3
=(căn x-4)(x+4căn x+16)
\(a-6\sqrt{a}+9-b^2\\ =\left(\sqrt{a}+3\right)^2-b^2\\ =\left(\sqrt{a}+3-b\right)\left(\sqrt{a}+3+b\right)\)
\(x-9=\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\)
\(x-7\sqrt{x}+12\\ =x-4\sqrt{x}-3\sqrt{x}+12\\ =\sqrt{x}\left(\sqrt{x}-4\right)-3\left(\sqrt{x}-4\right)\\ =\left(\sqrt{x}-4\right)\left(\sqrt{x}-3\right)\)
\(x\sqrt{x}+64\\ =\sqrt{x^3}+4^3\\ =\left(\sqrt{x}\right)^3+4^3\\ =\left(\sqrt{x}+4\right)\left(x-4\sqrt{x}+16\right)\)
với a,b,x,y không âm ta có
a,\(ab+b\sqrt{a}+\sqrt{a}+1\)
\(=b\sqrt{a}\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)\)
\(=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)
b, \(\sqrt{x^3}-\sqrt{y^3}+\sqrt{x^2y}-\sqrt{xy^2}\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)+\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)^2\)
a) \(\sqrt{x}\left(\frac{1}{\sqrt{x}}+1\right)\)
b)\(x^2\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)\)
=\(\left(\sqrt{x}+1\right)\left(x^2-1\right)\)
=\(\left(\sqrt{x}+1\right)\left(x-1\right)\left(x+1\right)\)
k mình nha