\(\text{Đặt: }\sqrt{6+\sqrt{6+\sqrt{6+....}}}=a\Rightarrow a^2=6+a\Leftrightarrow a^2-a-6=\left(a-3\right)\left(a+2\right)=0\)

thấy ngay a không thể đạt giá trị âm nên 

a=3 thay vào P=0 (vô lí) -> đề sai.

a) Ta có:

\(A=\frac{\sqrt{x}-3}{x-\sqrt{x}+1}\)

\(A=\frac{\sqrt{4}-3}{4-\sqrt{4}+1}\)

\(A=\frac{2-3}{4-2+1}=-\frac{1}{3}\)

b) đk: \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)

\(B=\left(\frac{3\sqrt{x}+6}{x-9}-\frac{2}{\sqrt{x}-3}\right):\frac{1}{\sqrt{x}+3}\)

\(B=\frac{3\sqrt{x}+6-2\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\left(\sqrt{x}+3\right)\)

\(B=\frac{3\sqrt{x}+6-2\sqrt{x}-6}{\sqrt{x}-3}\)

\(B=\frac{\sqrt{x}}{\sqrt{x}-3}\)

\(BT=\frac{2\sqrt{3}}{3+\sqrt{3}}-\frac{1}{\sqrt{3}-2}+\frac{6}{\sqrt{3}+3}=\frac{6+2\sqrt{3}}{3+\sqrt{3}}-\frac{1}{\sqrt{3}-2}=2-\frac{1}{\sqrt{3}-2}\)

\(=\frac{2\sqrt{3}-4-1}{\sqrt{3}-2}=\frac{2\sqrt{3}-5}{\sqrt{3}-2}=\left(5-2\sqrt{3}\right)\left(2+\sqrt{2}\right)=10+5\sqrt{2}-4\sqrt{3}+2\sqrt{6}\)

Nếu tính không lầm thì như vậy.

sao a ko trục căn thức từng phân thức cho nhanh ? 

\(\frac{2}{\sqrt{3}+1}-\frac{1}{\sqrt{3}-2}+\frac{6}{\sqrt{3}+3}\)

\(=\frac{2\left(\sqrt{3}-1\right)}{2}-\frac{\sqrt{3}+2}{-1}+\frac{6\left(\sqrt{3}-3\right)}{-6}\)

\(=\sqrt{3}-1+\sqrt{3}+2-\sqrt{3}+3=\sqrt{3}+4\)

Ai giúp tui với

Theo đề ra: HB = 1cm

                    HC = 2cm

Ta có: BC = HB + HC

          BC = 1cm + 2cm

          BC = 3cm

Theo đề ra: ΔABC vuông tại A, đường cao AH

\(\rightarrow AB^2=BH.BC=1.3=3\)

\(\rightarrow AB=\sqrt{3}\)

\(\rightarrow AC^2=CH.BC=2.3=6\)

\(\rightarrow AC=\sqrt{6}\)

\(K=\left(\sqrt{35}+5\right)\sqrt{6-\sqrt{35}}\)

\(\Rightarrow\sqrt{2}K=\left(\sqrt{35}+5\right)\sqrt{12-2\sqrt{35}}\)

mà \(\sqrt{12-2\sqrt{35}}=\sqrt{\left(\sqrt{7}\right)^2-2\sqrt{7.5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}\)

\(=\sqrt{7}-\sqrt{5}\)vì \(\sqrt{7}-\sqrt{5}>0\)

\(\Leftrightarrow\sqrt{2}K=\left(\sqrt{35}+5\right)\left(\sqrt{7}-\sqrt{5}\right)\)

\(\Leftrightarrow K=\frac{7\sqrt{5}-5\sqrt{7}+5\sqrt{7}-5\sqrt{7}}{\sqrt{2}}=\frac{\sqrt{2}\left(7\sqrt{5}-5\sqrt{7}\right)}{2}\)

\(=\frac{7\sqrt{10}-5\sqrt{14}}{2}\)

a, \(\sqrt{7-2\sqrt{10}}=\sqrt{7-2\sqrt{5.2}}=\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5.2}+\left(\sqrt{2}\right)^2}\)

\(=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\sqrt{5}-\sqrt{2}\)vì \(\sqrt{5}-\sqrt{2}>0\)

b, \(\frac{4}{\sqrt{7}-\sqrt{3}}+\frac{6}{3+\sqrt{3}}+\frac{\sqrt{7}-7}{\sqrt{7}-1}\)

\(=\frac{4\left(\sqrt{7}+\sqrt{3}\right)}{4}+\frac{6\left(3-\sqrt{3}\right)}{6}+\frac{\sqrt{7}\left(1-\sqrt{7}\right)}{\sqrt{7}-1}\)

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

a, √7−2√10=√7−2√5.2=√(√5)2−2√5.2+(√2)2

=√(√5−√2)2=√5−√2vì √5−√2>0

b, 4√7−√3 +63+√3 +√7−7√7−1 

=4(√7+√3)4 +6(3−√3)6 +√7(1−√7)√7−1 

=√7+√3+3−√3−√7=3