\(\sqrt{9a}\left(5-3\sqrt{a}\right)-5\left(3\sqrt{a}-5\right)\\ =\sqrt{9a}\left(5-3\sqrt{a}\right)+5\left(5-3\sqrt{a}\right)\\ =\left(5-3\sqrt{a}\right)\left(\sqrt{9a}+5\right)\)

\(\sqrt{9a}\left(5-3\sqrt{a}\right)-5\left(3\sqrt{a}-5\right)\\ =3\sqrt{a}\left(5-3\sqrt{a}\right)+5\left(5-3\sqrt{a}\right)\\ =\left(5-3\sqrt{a}\right)\left(5+3\sqrt{a}\right)\\ =5^2-\left(3\sqrt{a}\right)^2\\ =25-9a\)

hình như đề cứ sao sao á

sửa lại rồi đấy

\(7ab\cdot\sqrt{\dfrac{36a^4}{49b^2}}\\ =>7ab\cdot\sqrt{\left(\dfrac{6a^2}{7b}\right)^2}\\ =>7ab\cdot\dfrac{6a^2}{7b}\\ =>\dfrac{7ab\cdot6a^2}{7b}\\ =>6a^3\)

\(A=7ab.\sqrt{\dfrac{36a^4}{49b^2}}\)

\(=7ab.\dfrac{6a^2}{\left|7b\right|}\)

\(=7ab.\dfrac{6a^2}{7b}\left(vib>0\right)\)

\(=6a^3\)

Tham khảo:

AE là phân giác

=>EB/EC=AB/AC

=>EB/EC=6/9=2/3

=>EB/2=EC/3=(EC-EB)/(3-2)=10/1

=>EB=20cm

AB đâu 

\(a)ĐK:\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

\(A=\dfrac{x-2\sqrt{x}+1}{-\left(\sqrt{x}-1\right)}.\left(\dfrac{2}{\sqrt{x}-1}-\dfrac{3}{\sqrt{x}+1}\right)\)

\(=\dfrac{(\sqrt{x}-1)^2}{-\left(\sqrt{x}-1\right)}.\left(\dfrac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)

\(=-\left(\sqrt{x}-1\right).\dfrac{2\sqrt{x}+2-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=-\dfrac{\left(-\sqrt{x}+5\right)}{\sqrt{x}+1}\)

\(=\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\)

\(b)A=\dfrac{\sqrt{x}-5}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1-6}{\sqrt{x}+1}=1-\dfrac{6}{\sqrt{x}+1}\)

\(Vì\) \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1\Rightarrow\dfrac{6}{\sqrt{x}+1}\le6\Rightarrow A=1-\dfrac{6}{\sqrt{x}+1}\ge1-6=-5\)

Dấu "=" xảy ra \(\Leftrightarrow x=0\)

\(Vậy,A\min\limits=-5\Leftrightarrow x=0_{ }\)

\(c)A=1-\dfrac{6}{\sqrt{x}+1}\in Z\Leftrightarrow\dfrac{6}{\sqrt{x}+1}\in Z\Leftrightarrow\sqrt{x}+1\inƯ\left(6\right)=\left\{1;2;3;6\right\}\)(Vì \(x\ge0;x\in Z\))

\(TH1:\sqrt{x}+1=1\)

\(\Leftrightarrow\sqrt{x}=0\)

\(\Leftrightarrow x=0\left(tm\right)\)

\(TH2:\sqrt{x}+1=2\)

\(\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\left(loại\right)\)

\(TH3:\sqrt{x}+1=3\)

\(\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)

\(TH4:\sqrt{x}+1=6\)

\(\Leftrightarrow\sqrt{x}=5\Leftrightarrow x=25\left(tm\right)\)

\(Vậy,A\in Z\Leftrightarrow x\in\left\{0;4;25\right\}\)

ΔABC vuông tại A có AB=4m; góc B=48 độ. Tính AC

AC=AB*tan48

=4*tan48

\(\simeq4,44\left(m\right)\)

Đề là sắp xếp nha em! 

cos75^o < cos56^o < sin65^o < tan72^o

cos56° ≃ 0,5591 = a

sin 65° ≃ 0,9063 = b 

tan 72° ≃ 3,0776 = c 

cos75° ≃ 0,2588 = d

Dễ thấy d < a < b < c

Vậy dãy tăng dần là cos75°; cos56°; sin 65°; tan 72°

\(\sqrt{10+2\sqrt{21}}+4\sqrt{\left(\sqrt{3}-\sqrt{7}\right)^2}\)

\(=\sqrt{(\sqrt{3})^2+2\sqrt{3}.\sqrt{7}+\left(\sqrt{7}\right)^2}+4\left|\sqrt{3}-\sqrt{7}\right|\)

\(=\sqrt{\left(\sqrt{3}+\sqrt{7}\right)^2}+4\left(\sqrt{7}-\sqrt{3}\right)\)

\(=\sqrt{3}+\sqrt{7}+4\left(\sqrt{7}-\sqrt{3}\right)\)

\(=5\sqrt{7}-3\sqrt{3}\)

\(\sqrt[]{9-4\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)

\(=\sqrt{(2-\sqrt{5})^2}-\sqrt{\left(1-\sqrt{5}\right)^2}\)

\(=\left|2-\sqrt{5}\right|-\left|1-\sqrt{5}\right|\)

\(=\sqrt{5}-2-\left(\sqrt{5}-1\right)=\sqrt{5}-2-\sqrt{5}+1\)

\(=-1\)