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a, \(\sqrt{\left(2-\sqrt{5}\right)^2}=\sqrt{5}-2\left(\sqrt{5}>2\right)\)
b, \(\sqrt{\left(3-\sqrt{2}\right)^2}=3-\sqrt{2}\left(3>\sqrt{2}\right)\)
c, Với a < 3
\(\sqrt{\left(a-3\right)^2}+\left(a-9\right)=3-a+a-9=-6\)
d, \(A=\sqrt{\left(2a+5\right)^2}-\left(2a-7\right)\)
\(=\left|2a+5\right|-2a+7\)
+) Xét \(x\ge\dfrac{-5}{2}\) có:
\(A=2a+5-2a+7=12\)
+) Xét \(x< \dfrac{-5}{2}\) có:
\(A=-2a-5-2a+7=-4a+2\)
Vậy...
A) ĐKXĐ : \(x\ge0\) và \(x\ne4\)
Rút gọn :\(A=\frac{2}{2+\sqrt{x}}+\frac{1}{2-\sqrt{x}}+\frac{4\sqrt{x}}{4-x}\)
\(A=\frac{2\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}+\frac{2+\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}+\frac{4\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
\(A=\frac{4-2\sqrt{x}+2+\sqrt{x}+4\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
\(A=\frac{6+3\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
\(A=\frac{3\left(2+\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
\(A=\frac{3}{2-\sqrt{x}}\)
b) thay \(x=7+4\sqrt{3}\) vào A
ta được :\(A=\frac{3}{2-\sqrt{7+4\sqrt{3}}}=\frac{3}{2-2+\sqrt{3}}=\frac{3}{\sqrt{3}}\)
vậy vói \(x=7+4\sqrt{3}\) thì \(A=\frac{3}{\sqrt{3}}\)
c)với\(x\ge0\) và \(x\ne4\)
Để \(A=-\frac{3}{7}\Leftrightarrow\frac{3}{2-\sqrt{x}}=-\frac{3}{7}\)
\(\Leftrightarrow3.7=-3\left(2-\sqrt{x}\right)\)
\(\Leftrightarrow21=-6+3\sqrt{x}\)
\(\Leftrightarrow21+6=3\sqrt{x}\)
\(\Leftrightarrow27=3\sqrt{x}\)
\(\Leftrightarrow\sqrt{x}=9\)
\(\Leftrightarrow x=81\)
Vậy để\(A=-\frac{3}{7}\Leftrightarrow x=81\)
a/ \(A=\frac{30\left(\sqrt{6}-1\right)}{5}+\frac{2\left(\sqrt{6}+2\right)}{2}-\frac{6\left(3+\sqrt{6}\right)}{3}=6\sqrt{6}-6+\sqrt{6}+2-6-2\sqrt{6}\)
\(A=5\sqrt{6}-10\)
\(B=\sqrt{17-6\sqrt{2}+\sqrt{8+4\sqrt{2}+1}}\)
\(B=\sqrt{17-6\sqrt{2}+\sqrt{\left(2\sqrt{2}+1\right)^2}}=\sqrt{18-4\sqrt{2}}\)
Đến đây ko rút gọn được nữa, nhưng nếu đề là:
\(B=\sqrt{17+6\sqrt{2}+\sqrt{8+4\sqrt{2}+1}}=\sqrt{18+8\sqrt{2}}=4+\sqrt{2}\)
c/
\(C=\sqrt{8-2\sqrt{7}}+\sqrt{8+2\sqrt{7}}=\sqrt{\left(\sqrt{7}-1\right)^2}+\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(C=\sqrt{7}-1+\sqrt{7}+1=2\sqrt{7}\)
\(D=\sqrt{a-2\sqrt{a}+1}-\sqrt{a-8\sqrt{a}+16}\)
\(D=\sqrt{\left(\sqrt{a}-1\right)^2}-\sqrt{\left(4-\sqrt{a}\right)^2}=\sqrt{a}-1-\left(4-\sqrt{a}\right)=2\sqrt{a}-5\)
\(E=\sqrt{a-2+2\sqrt{a-2}+1}+\sqrt{a-2-2\sqrt{a-2}+1}\) (\(a\ge2\))
\(E=\sqrt{\left(\sqrt{a-2}+1\right)^2}+\sqrt{\left(\sqrt{a-2}-1\right)^2}\)
\(E=\sqrt{a-2}+1+\left|\sqrt{a-2}-1\right|\)
\(\Rightarrow\left[{}\begin{matrix}E=2\sqrt{a-2}\left(a\ge3\right)\\E=2\left(2\le a\le3\right)\end{matrix}\right.\)
\(F=\sqrt[3]{10+6\sqrt{3}}-\sqrt{3}=\sqrt[3]{1+3.1.\sqrt{3}+3.1.\sqrt{3}^2+\sqrt{3}^3}-\sqrt{3}\)
\(F=\sqrt[3]{\left(1+\sqrt{3}\right)^3}-\sqrt{3}=1+\sqrt{3}-\sqrt{3}=1\)
\(G=\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\Rightarrow G^3=\left(\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\right)^3\)
\(\Rightarrow G^3=14+3\left(\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\right)\left(\sqrt[3]{49-50}\right)\)
\(\Rightarrow G^3=14-3G\Rightarrow G^3+3G-14=0\)
\(\Rightarrow G=2\)
\(A=\frac{a\left(\sqrt{a}+2\right)-4\left(\sqrt{a}+2\right)}{a-4}=\frac{\left(a-4\right)\left(\sqrt{a}+2\right)}{a-4}=\sqrt{a}+2\)
\(B=\frac{12\sqrt{6}}{\sqrt{\sqrt{\left(\sqrt{6}+1\right)^2}-\sqrt{\left(\sqrt{6}-1\right)^2}}}=\frac{12\sqrt{6}}{\sqrt{2}}=12\sqrt{3}\)
C k thấy đề
\(A=\frac{a\sqrt{a}-8+2a-4\sqrt{a}}{a-4}=\frac{\left(a\sqrt{a}-4\sqrt{a}\right)+\left(2a-8\right)}{a-4}=\frac{\left(a-4\right)\left(\sqrt{a}+2\right)}{a-4}=\sqrt{a}+2\)
\(B=\frac{12\sqrt{6}}{\sqrt{7+2\sqrt{6}}-\sqrt{7-2\sqrt{6}}}=\frac{12\sqrt{6}}{\sqrt{1+6+2\sqrt{6}}-\sqrt{1+6-2\sqrt{6}}}\)
\(=\frac{12\sqrt{6}}{\sqrt{\left(1+\sqrt{6}\right)^2}-\sqrt{\left(\sqrt{6}-1\right)^2}}=\frac{12\sqrt{6}}{1+\sqrt{6}-\sqrt{6}+1}=6\sqrt{6}\)
\(C=\frac{\sqrt{c^2+2c+1}}{\left|c\right|-1}=\frac{\left|c+1\right|}{\left|c\right|-1}\)
a) \(\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}=\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=\sqrt{4-3}=\sqrt{1}=1\)
b)
Đặt \(B=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
\(B^2=4+\sqrt{7}-2\sqrt{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}+4-\sqrt{7}\)
\(=8-2\sqrt{16-7}=8-2\sqrt{9}=8-2.3=8-6=2\)
\(\Rightarrow B=\sqrt{2}\)
câu đầu bạn xem lại đề đi nha
các phần còn lại
b)B=\(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)=\(\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)=\(\sqrt{7}-1-\left(\sqrt{7}+1\right)=-2\)
c)tính từng căn nha
\(\sqrt{13-4\sqrt{3}}=\sqrt{12-2\sqrt{12}+1}=\sqrt{\left(\sqrt{12}-1\right)^2}=\sqrt{12}-1=2\sqrt{3}-1\)
\(\sqrt{22-12\sqrt{2}}=\sqrt{18-4\sqrt{18}+4}=\sqrt{\left(\sqrt{18}-2\right)^2}=\sqrt{18}-2=3\sqrt{2}-3\)
\(\sqrt{\left(2\sqrt{3}-3\sqrt{2}\right)^2}=3\sqrt{2}-2\sqrt{3}\)
thay vào tính C đc C=2
d)có \(\sqrt{9+4\sqrt{2}}=\sqrt{8+2\sqrt{8}+1}=\sqrt{\left(\sqrt{8}+1\right)^2}=\sqrt{8}+1\)\(\Rightarrow6\sqrt{2+\sqrt{9+4\sqrt{2}}}=6\sqrt{2+\sqrt{8}+1}=6\sqrt{2+2\sqrt{2}+1}\)
=\(6\sqrt{\left(\sqrt{2}+1\right)^2}=6\left(\sqrt{2}+1\right)=6\sqrt{2}+6\)\(\Rightarrow D=\sqrt{17-6\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{17-6\sqrt{2}-6}=\sqrt{11-6\sqrt{2}}=\sqrt{9-6\sqrt{2}+2}\)
=\(\sqrt{\left(3-\sqrt{2}\right)^2}=3-\sqrt{2}\)
