\(\left(3+\frac{a-2\sqrt{a}}{\sqrt{a}-2}\right)\left(3-\frac{3a+\sqrt{a}}{3\sqrt{a}+1}\right).Rut\:gon\:bieu\:thuc\:nay\)
IQ vô cực mà , bn tự làm đc cái biểu thức dễ ợt này mà
Cho bieu thuc: \(p=\left(\frac{1}{\sqrt{x}-\sqrt{x-1}}-\frac{x-3}{\sqrt{x-1}-\sqrt{2}}\right)\left(\frac{2}{\sqrt{2}-\sqrt{x}}-\frac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}\right)\)
a) Tim DKXD cua bieu thuc p
b) Rut gon bieu thuc p
rut gon bieu thuc: \(\frac{\sqrt{\sqrt{\frac{x-1}{x+1}}+\sqrt{\frac{x+1}{x-1}}-2}\left(2x+\sqrt{x^2+1}\right)}{\sqrt{\left(x+1\right)^3}-\sqrt{\left(x-1\right)^2}}\)
\(\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\)\(\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
Rut gon bieu thuc
tum x de bieu thuc = \(\frac{-1}{7}\)
\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(A=\)\(\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}.\)
\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\)\(\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)\(=\frac{-5x+5\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-5\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)\(=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
\(A=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
\(\Rightarrow\frac{-5\sqrt{x}+2}{\sqrt{x}+3}=-\frac{1}{7}\Rightarrow-7\left(-5\sqrt{x}+2\right)=\sqrt{x}+3\)
\(\Rightarrow35\sqrt{x}-14=\sqrt{x}+3\)
\(\Rightarrow34\sqrt{x}=17\)
\(\Rightarrow\sqrt{x}=\frac{1}{2}\Rightarrow x=\frac{1}{4}\left(tm\right)\)
Vậy với \(x=\frac{1}{4}\)thì \(A=-\frac{1}{7}\)
Rut gon bieu thuc
A=\(\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\)\(\frac{3}{\sqrt{x}+3}\)
\(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)+3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6+3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-3x+11\sqrt{x}-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)\(=\frac{-3x+3\sqrt{x}+8\sqrt{x}-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{-3\sqrt{x}\left(\sqrt{x}-1\right)+8\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(8-3\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}=\frac{8-3\sqrt{x}}{\sqrt{x}-3}\)
cho biểu thức p=\(\frac{\sqrt{a}\left(1-a\right)^2}{1+a}:\left[\left(\frac{1-\sqrt{a^3}}{1-\sqrt{a}}+\sqrt{a}\right).\left(\frac{1+\sqrt{a^3}}{1+\sqrt{a}}-\sqrt{a}\right)\right]\)
a)rut gon p
b) xet dau cua bieu thuc M = a. \(\left(P-\frac{1}{2}\right)\)
1/ Rut gon bieu thuc sau:
a) \(\sqrt{12-2\sqrt{35}}+\sqrt{7-2\sqrt{10}}-\sqrt{\sqrt{49}}\)
b) \(\frac{\sqrt{7}-5}{2}-\frac{6}{\sqrt{7}-2}+\frac{1}{3+\sqrt{7}}+\frac{3}{5+2\sqrt{7}}\)
Rut gon bieu thuc
B=\(\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\)\(\left(\frac{\sqrt{x}+1}{\sqrt{x}-3}-\frac{\sqrt{x}+3}{\sqrt{x}-1}\right)\)
\(B=\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-3}-\frac{\sqrt{x}+3}{\sqrt{x}-1}\right)\)
\(=\frac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)\(:\left(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\)\(\frac{x-1-x+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{2}\)
\(=\frac{\sqrt{x}+3}{2\sqrt{x}}\)
1.Tính: \(\sqrt[3]{26+15\sqrt{3}}\) - \(\sqrt[3]{26-15\sqrt{3}}\)
2.Rút gọn: P=\(\left(\frac{\sqrt{a-2}+2}{3}\right)\).\(\left(\frac{\sqrt{a-2}}{3+\sqrt{a-2}}+\frac{a+7}{11-a}\right)\):\(\left(\frac{3\sqrt{a-2}+1}{a-3\sqrt{a-2}-2}-\frac{1}{\sqrt{a-2}}\right)\)
1. Câu hỏi của Nữ hoàng sến súa là ta - Toán lớp 9 - Học toán với OnlineMath
