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P= (\(\frac{3\sqrt{a}}{\sqrt{a}+4}+\frac{\sqrt{a}}{\sqrt{a}-4}+\frac{4\left(a+2\right)}{16-a}\)):\(\left(1-\frac{2\sqrt{a}+5}{\sqrt{a}-4}\right)\)
=\(\left(\frac{3\sqrt{a}\left(\sqrt{a}-4\right)}{a-16}+\frac{\sqrt{a}\left(\sqrt{a}+4\right)}{a-16}-\frac{4a+8}{a-16}\right):\left(\frac{\sqrt{a}-4-2\sqrt{a}-5}{\sqrt{a}-4}\right)\)
= \(\left(\frac{3a-12\sqrt{a}+a+4\sqrt{a}-4a-8}{a-16}\right):\left(\frac{-\sqrt{a}-9}{\sqrt{a}-4}\right)\)
=\(\left(\frac{-8\sqrt{a}-8}{a-16}\right).\left(\frac{\sqrt{a}-4}{-\sqrt{a}-9}\right)=\frac{8\sqrt{a}+8}{\left(\sqrt{a}+4\right).\left(\sqrt{a}+9\right)}=\frac{8\sqrt{a}+8}{a+13\sqrt{a}+36}\)
\(A=4\sqrt{x}-\frac{x+6\sqrt{x}+9}{x-9}\)
\(=4\sqrt{x}-\frac{\left(\sqrt{x}+3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=4\sqrt{x}-\frac{\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)}\)
\(=\frac{4\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}-3}-\frac{\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)}\)
\(=\frac{4x-12\sqrt{x}-\sqrt{x}-3}{\sqrt{x}-3}\)
\(=\frac{4x-13\sqrt{x}-3}{\sqrt{x}-3}\)
C.Tham khảo ở dây:Câu hỏi của Đặng Phương Thảo - Toán lớp 9 - Học toán với OnlineMath
\(B=\frac{5\sqrt{x}-\left(x-10\sqrt{x}+25\right)\left(\sqrt{x}+5\right)}{x-25}\)
\(=\frac{5\sqrt{x}-\left(\sqrt{x}-5\right)^2\left(\sqrt{x}+5\right)}{x-25}\)
\(=\frac{5\sqrt{x}-\left(\sqrt{x}-5\right)\left(x-25\right)}{x-25}\)
\(=\frac{5\sqrt{x}-\left(x\sqrt{x}-25\sqrt{x}-5x+125\right)}{x-25}\)
\(=\frac{5\sqrt{x}-x\sqrt{x}+25\sqrt{x}+5x-125}{x-25}\)
\(=\frac{-x\sqrt{x}+30\sqrt{x}+5x-125}{x-25}\)
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\(A=2-x\sqrt{\frac{x\left(x-2\right)}{\left(x-2\right)^2}+\frac{1}{\left(x-2\right)^2}}=2-x\sqrt{\frac{\left(x-1\right)^2}{\left(x-2\right)^2}}\)
\(=2-x\cdot\frac{x-1}{x-2}=\frac{2x-4}{x-2}-\frac{x^2-x}{x-2}=\frac{-x^2+3x-4}{x-2}\)
\(B=\frac{2\sqrt{5}x}{x-2}\cdot\left|x-2\right|+\frac{3\sqrt{5}x^2}{x}=\frac{2\sqrt{5}x}{x-2}\cdot\left|x-2\right|+3\sqrt{5}x\)
Với 0 < x < 2 \(B=-2\sqrt{5}x+3\sqrt{5}x=\sqrt{5}x\)
Với x > 2 \(B=2\sqrt{5}x+3\sqrt{5}x=5\sqrt{5}x\)
\(C=\frac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\sqrt{x}\left(\sqrt{x}+5\right)}+\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-5\right)^2}}=\frac{\sqrt{x}-5}{\sqrt{x}}+\left|\frac{\sqrt{x}-1}{\sqrt{x}-5}\right|\)
Với 0 < x < 1 \(C=\frac{\sqrt{x}-5}{\sqrt{x}}+\frac{\sqrt{x}-1}{\sqrt{x}-5}=\frac{x-10\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}+\frac{x-\sqrt{x}}{x\left(\sqrt{x}-5\right)}=\frac{2x-11\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}\)
Với 1 < x < 5 \(C=\frac{\sqrt{x}-5}{\sqrt{x}}-\frac{\sqrt{x}-1}{\sqrt{x}-5}=\frac{x-10\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}-\frac{x-\sqrt{x}}{x\left(\sqrt{x}-5\right)}=\frac{-9\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}\)
Với x > 5 \(C=\frac{\sqrt{x}-5}{\sqrt{x}}+\frac{\sqrt{x}-1}{\sqrt{x}-5}=\frac{x-10\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}+\frac{x-\sqrt{x}}{x\left(\sqrt{x}-5\right)}=\frac{2x-11\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}\)
\(A=\left(1+\frac{5}{\sqrt{x}-2}\right).\left(\sqrt{x}-\frac{x+2\sqrt{x}+4}{\sqrt{x}+3}\right).\)
\(=\frac{\sqrt{x}-2+5}{\sqrt{x}-2}.\frac{x+3\sqrt{x}-x-2\sqrt{x}-4}{\sqrt{x}+3}\)
\(=\frac{\sqrt{x}+3}{\sqrt{x}-2}.\frac{\sqrt{x}-4}{\sqrt{x}+3}=\frac{\sqrt{x}-4}{\sqrt{x}-2}\)
Đặt: \(a=\sqrt{2+x};b=\sqrt{2-x}\left(a,b\ge0\right)\)
\(\Rightarrow\hept{\begin{cases}a^2+b^2=4\\a^2-b^2=2x\end{cases}}\)
\(\Rightarrow A=\frac{\sqrt{2+ab}\left(a^3-b^3\right)}{4+ab}=\frac{\sqrt{2+ab}\left(a-b\right)\left(a^2+b^2+ab\right)}{4+ab}\)
\(\Rightarrow A=\frac{\sqrt{2+ab}\left(a-b\right)\left(4+ab\right)}{4+ab}=\sqrt{2+ab}\left(a-b\right)\)
\(\Rightarrow A\sqrt{2}=\sqrt{4+2ab}\left(a-b\right)\)
\(\Rightarrow A\sqrt{2}=\sqrt{\left(a^2+b^2+2ab\right)}\left(a-b\right)=\left(a+b\right)\left(a-b\right)\)
\(\Rightarrow A\sqrt{2}=a^2-b^2=2x\)
\(\Rightarrow A=x\sqrt{2}\)
A = \(\frac{8}{\sqrt{5}-1}\) - (\(2\sqrt{5}-1\) ) ( chúng ta cần trục căn thức lên để khử mẫu )
= \(\frac{8\left(\sqrt{5}+1\right)}{5-1}\)- \(\left(2\sqrt{5}-1\right)\)
= \(2\sqrt{5}\)+ 2 - \(2\sqrt{5}\)+1
= 3
B = \(\frac{\left(1-\sqrt{x}\right)^2+4\sqrt{x}}{1+\sqrt{x}}\)( x \(\ge\)0 )
= \(\frac{1-2\sqrt{x}+x+4\sqrt{x}}{1+\sqrt{x}}\)
= \(\frac{1+2\sqrt{x}+x}{1+\sqrt{x}}\)
= \(\frac{\left(1+\sqrt{x}\right)^2}{1+\sqrt{x}}\)
= 1 +\(\sqrt{x}\)
#mã mã#
b) \(\sqrt{\left(7-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=7-\sqrt{3}+\sqrt{3}+1\)
\(=8\)
Đặt \(B=4^{1993}+4^{1992}+.......+4^2+1\)
\(\Rightarrow4B=4^{1994}+4^{1993}+....+4^3+4\)
\(\Rightarrow3B=4^{1994}-1\)
Mà: \(A=75B+25=25\left(3B+1\right)=25\left(4^{1994}-1+1\right)=25.4^{1994}\)
\(A=75\left(4^{1993}+5^{1992}+...+4^2+5\right)+25=75B+25\)
Xét \(B=4^{1993}+4^{1992}+...+4^2+5=4^{1993}+4^{1992}+...+4^2+4+1\)
\(\Rightarrow4B=4^{1994}+4^{1993}+...+4^2+4\)
\(\Rightarrow4B+1-4^{1994}=4^{1993}+4^{1992}+...+4^2+4+1=B\)
\(\Rightarrow3B=4^{1994}-1\Rightarrow B=\dfrac{4^{1994}-1}{3}\)
Vậy \(A=75.\dfrac{\left(4^{1994}-1\right)}{3}+25=25.4^{1994}-25+25\)
\(\Rightarrow A=25.4^{1994}\)