Phân tích thành nhân tử biểu thức :
ab+\(b\sqrt{a}+\sqrt{a}+1\) với a≥0
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\(a\sqrt{b}+\sqrt{ab}+\sqrt{a}+1\)
\(=\sqrt{ab}\cdot\sqrt{a}+\sqrt{ab}+\sqrt{a}+1\)
\(=\left(\sqrt{ab}\cdot\sqrt{a}+\sqrt{ab}\right)+\left(\sqrt{a}+1\right)\)
\(=\sqrt{ab}\cdot\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)\)
\(=\left(\sqrt{ab}+1\right)\left(\sqrt{a}+1\right)\)
a√b + √(ab) + √a + 1
= [a√b + √(ab)] + (√a + 1)
= √(ab)(√a + 1) + (√a + 1)
= (√a + 1)[√(ab) + 1]
a) Ta có: \(-7xy\cdot\sqrt{\dfrac{3}{xy}}\)
\(=\dfrac{-7xy\cdot\sqrt{3xy}}{xy}\)
\(=-7\sqrt{3}\cdot\sqrt{xy}\)
b) Ta có: \(ab+b\sqrt{a}+\sqrt{a}+1\)
\(=b\sqrt{a}\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)\)
\(=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)
$a)-7xy.\sqrt{\dfrac{3}{xy}}$
$=-7.\sqrt{x^2y^2.\dfrac{3}{xy}}(do \,x,y>0a\to xy>0)$
$=-7.\sqrt{\dfrac{xy}{3}}$
$b)ab+b\sqrt{a}+\sqrt{a}+1(a \ge 0)$
$=b\sqrt{a}(\sqrt{a}+1)+\sqrt{a}+1$
$=(\sqrt{a}+1)(b\sqrt{a}+1)$
a) \(-7xy.\sqrt{\dfrac{3}{xy}}=-7xy.\dfrac{\sqrt{3xy}}{xy}=-7\sqrt{3xy}\)
b) \(ab+b\sqrt{a}+\sqrt{a}+1=b\sqrt{a}\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)
a: \(-7xy\cdot\sqrt{\dfrac{3}{xy}}=-7xy\cdot\dfrac{\sqrt{3}}{\sqrt{xy}}=-7\sqrt{3xy}\)
b: \(ab+b\sqrt{a}+\sqrt{a}+1\)
\(=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)
d: \(=-\left(x+\sqrt{x}-12\right)=-\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)\)
a) \(\sqrt{a^3}-\sqrt{b^3}+\sqrt{a^2b}-\sqrt{ab^2}\)
\(=a\sqrt{a}-b\sqrt{b}+a\sqrt{b}-b\sqrt{a}\)
\(=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)-\left(\sqrt{a}-\sqrt{b}\right)\sqrt{ab}\)
\(=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b-\sqrt{ab}\right)\)
\(=\left(\sqrt{a}-\sqrt{b}\right)\left(a+b\right)\)
b) \(x-y+\sqrt{xy^2}-\sqrt{y^3}\)
\(=\left(x-y\right)+\left(y\sqrt{x}-y\sqrt{y}\right)\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)+y\left(\sqrt{x}-\sqrt{y}\right)\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}+y\right)\)
\(\sqrt{ab}-\sqrt{a}-\sqrt{b}+1\)
\(=\sqrt{a}\left(\sqrt{b}-1\right)-\left(\sqrt{b}-1\right)\)
\(=\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\)
\(\sqrt{ab}-\sqrt{a}-\sqrt{b}+1=\sqrt{a}\left(\sqrt{b}-1\right)-\left(\sqrt{b}-1\right)=\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\)
\(=\sqrt{a}\left(\sqrt{a}+1\right)+2\sqrt{b}\left(\sqrt{a}+1\right)=\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\sqrt{b}\right)\)
a) \(A=\left(\sqrt{x}+3\right)^2-4\sqrt{x}-6\)
\(A=x+6\sqrt{x}+9-4\sqrt{x}-6\)
\(A=x+2\sqrt{x}-3\)
b) \(A=x+2\sqrt{x}-3\)
\(A=x+3\sqrt{x}-\sqrt{x}-3\)
\(A=\sqrt{x}\left(\sqrt{x}+3\right)-\left(\sqrt{x}+3\right)\)
\(A=\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)\)
a: A=x+6căn x+9-4căn x-6
=x+2căn x+3
b: A ko phân tích được nha bạn
\(ab+b\sqrt{a}+\sqrt{a}+1=b\sqrt{a}\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)\)
\(=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)
\(=b\sqrt{a}\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)\)
\(=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)