3.phan tich da thuc thanh nhan tu:
a.\(1+\sqrt{a}+\sqrt{b}+\sqrt{ab}\)
b.\(\sqrt{x}+\sqrt{y}+\sqrt{x^2y}+\sqrt{xy^2}\)
phan tich da thuc thanh nhan tu
5x+ 7$\sqrt xy $ -6y+$\sqrt x $ - 2$\sqrt y $
\(7\sqrt{ab}+7b-\sqrt{a-}\sqrt{b}\)
phan tich da thuc thanh nhan tu
\(7\sqrt{ab}+7b-\sqrt{a}-\sqrt{b}\) =\(7\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)-\left(\sqrt{a}+\sqrt{b}\right)\) =\(\left(\sqrt{a}+\sqrt{b}\right)\left(7\sqrt{b}-1\right)\)
Phan tich da thuc thanh nhan tu
\(x+3\sqrt{x}+2\)
\(2x+\sqrt{x}-3\)
\(x+\sqrt{x}+2\sqrt{x}+2\)
= \(\sqrt{x}\left(\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)
= \(\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)\)
\(2x-2\sqrt{x}+3\sqrt{x}-3\)
= \(2\sqrt{x}\left(\sqrt{x}-1\right)+3\left(\sqrt{x}-1\right)\)
= \(\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)\)
5x + \(7\sqrt{xy}-6y+\sqrt{x}+2\sqrt{y}
phan tich thanh nhan tu
b2. phan tich da thuc thanh nhan tu
1.x+2\(\sqrt{x-1}\)
2.x-2\(\sqrt{x-1}\)
3.x -4\(\sqrt{x-4}\)
4. x+4+4\(\sqrt{x}\)
a/ \(=x-1+2\sqrt{x-1}+1=\left(\sqrt{x-1}+1\right)^2\)
b/ \(=x-1-2\sqrt{x-1}+1=\left(\sqrt{x-1}-1\right)^2\)
c/ \(=x-4-4\sqrt{x-4}+4=\left(\sqrt{x-4}-2\right)^2\)
d/ \(=\left(\sqrt{x}+2\right)^2\)
\(Timx,biet:\sqrt{X-5}=1+\sqrt{X}\) \(b,\sqrt{3-x}+\sqrt{x-5}=10\) \(Phan-tich-thanh-nhan-tu\) : \(x-2\sqrt{xy}+xy,,,\)
phan tich da thuc thanh nhan tu
\(3-\sqrt{3}+15-3\sqrt{5}\)
\(3-\sqrt{3}+15-3\sqrt{5}=18-\sqrt{3}-3\sqrt{5}=\sqrt{3}\left(6\sqrt{3}-1-\sqrt{15}\right)\)
1,tim x de bieu thuc sau co nghia \(\sqrt{x+\dfrac{3}{x}}+\sqrt{-3x}\)
b,\(\sqrt{x^2+4x+5}\)
c,\(\sqrt{2x^2+4x+5}\)
2, phan tich thanh nhan tu
a,\(x+5\sqrt{x}+6\) b,\(x+4\sqrt{x}+3\)
GIUP MINH VS MINH CAN GAP MINH CAM ON TRUOC NHA
\(1a.\) Để : \(\sqrt{x+\dfrac{3}{x}}+\sqrt{-3x}\) xác định thì :
\(x+\dfrac{3}{x}\) ≥ 0 và \(-3x\) ≥ 0
⇔ \(\dfrac{x^2+3}{x}\) ≥ 0 và : x ≤ 0 ⇔ x > 0 và : x ≤ 0 ( Vô lý )
⇔ x ∈ ∅
b. Để : \(\sqrt{x^2+4x+5}\) xác định thì :
\(x^2+4x+5\) ≥ 0
Mà : \(x^2+4x+5=\left(x+2\right)^2+1>0\)
Vậy , ........
c. Để : \(\sqrt{2x^2+4x+5}\) xác định thì :
\(2x^2+4x+5\) ≥ 0
Mà : \(2\left(x^2+2x+1\right)+3=2\left(x+1\right)^2+3>0\)
Vậy ,.........
Bài 2. \(a.x+5\sqrt{x}+6=x+2.\dfrac{5}{2}\sqrt{x}+\dfrac{25}{4}+6-\dfrac{25}{4}=\left(\sqrt{x}+\dfrac{5}{2}\right)^2-\dfrac{1}{4}=\left(\sqrt{x}+\dfrac{5}{2}-\dfrac{1}{2}\right)\left(\sqrt{x}+\dfrac{5}{2}+\dfrac{1}{2}\right)=\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)\)
\(b.x+4\sqrt{x}+3=x+\sqrt{x}+3\sqrt{x}+3=\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}+1\right)=\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)\)
\(\sqrt[3]{x+6}+\sqrt{x-1}+1-x^2\)Phan tich thanh nhan tu