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\(a.\left(4+\sqrt{7}\right)\left(\sqrt{14}-\sqrt{2}\right)\sqrt{4-\sqrt{7}}=\left(4+\sqrt{7}\right)\left(\sqrt{7}-1\right)\sqrt{7-2\sqrt{7}+1}=\left(4+\sqrt{7}\right)\left(\sqrt{7}-1\right)^2=2\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)=2\left(16-7\right)=18\) \(b.\dfrac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\dfrac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}=\dfrac{4\sqrt{2}+\sqrt{14}}{6+\sqrt{7+2\sqrt{7}+1}}+\dfrac{4\sqrt{2}-\sqrt{14}}{6-\sqrt{7-2\sqrt{7}+1}}=\dfrac{4\sqrt{2}+\sqrt{14}}{7+\sqrt{7}}+\dfrac{4\sqrt{2}-\sqrt{14}}{7-\sqrt{7}}=\dfrac{\left(4\sqrt{2}+\sqrt{14}\right)\left(7-\sqrt{7}\right)+\left(4\sqrt{2}-\sqrt{14}\right)\left(7+\sqrt{7}\right)}{49-7}=\dfrac{28\sqrt{2}-4\sqrt{14}+7\sqrt{14}-7\sqrt{2}+28\sqrt{2}+4\sqrt{14}-7\sqrt{14}-7\sqrt{2}}{42}=\dfrac{42\sqrt{2}}{42}=\sqrt{2}\)
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 \(\sqrt{9+4\sqrt{2}}=\sqrt{8+2\sqrt{8}+1}=\sqrt{\left(\sqrt{8}+1\right)^2}=\sqrt{8}+1\)\(\Rightarrow6\sqrt{2+\sqrt{9+4\sqrt{2}}}=6\sqrt{2+\sqrt{8}+1}=6\sqrt{2+2\sqrt{2}+1}\)
=\(6\sqrt{\left(\sqrt{2}+1\right)^2}=6\left(\sqrt{2}+1\right)=6\sqrt{2}+6\)\(\Rightarrow D=\sqrt{17-6\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{17-6\sqrt{2}-6}=\sqrt{11-6\sqrt{2}}=\sqrt{9-6\sqrt{2}+2}\)
=\(\sqrt{\left(3-\sqrt{2}\right)^2}=3-\sqrt{2}\)
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 \(B=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
\(B^2=4+\sqrt{7}-2\sqrt{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}+4-\sqrt{7}\)
\(=8-2\sqrt{16-7}=8-2\sqrt{9}=8-2.3=8-6=2\)
\(\Rightarrow B=\sqrt{2}\)
B4
a) \(\frac{9}{\sqrt{3}}=\frac{9\cdot\sqrt{3}}{\sqrt{3}\cdot\sqrt{3}}=\frac{9\sqrt{3}}{3}=3\sqrt{3}\)
b)\(\frac{3}{\sqrt{5}-\sqrt{2}}=\frac{3\left(\sqrt{5}+\sqrt{2}\right)}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}=\frac{3\left(\sqrt{5}+\sqrt{2}\right)}{3}=\sqrt{5}+\sqrt{2}\)
c)\(\frac{\sqrt{2}+1}{\sqrt{2}-1}=\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\frac{\left(\sqrt{2}+1\right)^2}{1}=\left(\sqrt{2}+1\right)^2\)
d)\(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{1}=14\)
B3
a)\(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\) \(đk:x\ge1\)
\(\frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\sqrt{x-1}\cdot\left(\frac{1}{2}-\frac{9}{2}+3\right)=-17\)
\(\sqrt{x-1}\cdot\left(-1\right)=-17\)
\(\sqrt{x-1}=17\)
\(\left[{}\begin{matrix}x-1=289\left(tm\right)\\x-1=-289\left(ktm\right)\end{matrix}\right.\)
\(x=290\left(tm\right)\)
Bài làm:
a) \(\sqrt{4-\sqrt{7}}=\frac{\sqrt{2\left(4-\sqrt{7}\right)}}{\sqrt{2}}=\sqrt{\frac{8-2\sqrt{7}}{2}}=\sqrt{\frac{7-2\sqrt{7}+1}{2}}\)
\(=\sqrt{\frac{\left(\sqrt{7}-1\right)^2}{2}}=\frac{\left(\sqrt{7}-1\right)\sqrt{2}}{2}=\frac{\sqrt{14}-\sqrt{2}}{2}\)
b) \(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\) (đề vậy chứ)
\(=\sqrt{3+2\sqrt{3}+1}-\sqrt{3-2\sqrt{3}+1}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\sqrt{3}+1-\sqrt{3}+1\)
\(=2\)
c) \(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)
\(=\sqrt{5-4\sqrt{5}+4}-\sqrt{5+4\sqrt{5}+4}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)
\(=\sqrt{5}-2-\sqrt{5}-2\)
\(=-4\)
d) \(\sqrt{x-2\sqrt{x-1}}\)
\(=\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
\(=\left|\sqrt{x-1}-1\right|\)
a) Đặt A=\(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
<=> \(\sqrt{2}\cdot A=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}\)=\(\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}\)
= \(\sqrt{7}+1-\sqrt{7}+1=2\)
=> \(A=\frac{2}{\sqrt{2}}\sqrt{2}\)
b) Ta đặt \(B=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
=> \(B^2=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}\)
= \(8+2\sqrt{6-2\sqrt{5}}=8+2\sqrt{5-2\sqrt{5}+1}\)=\(8+2\sqrt{\left(\sqrt{5}-1\right)^2}=8+2\sqrt{5}-2=6+2\sqrt{5}\)
= \(5+2\sqrt{5}+1=\left(\sqrt{5}+1\right)^2\)
=> B=\(\sqrt{5}+1\)
c) Ta xét \(A=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}\)
=> \(\sqrt{2}\cdot A=\sqrt{8+2\sqrt{3}\cdot\sqrt{5}}+\sqrt{8-2\sqrt{3}\cdot\sqrt{5}}\)
= \(\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
= \(\sqrt{3}+\sqrt{5}+\sqrt{5}-\sqrt{3}\)= \(2\sqrt{5}\)
=> A=\(\sqrt{5}\)
Ta có : \(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
= \(A-\sqrt{6-2\sqrt{5}}\)
= \(\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-\sqrt{5}+1\)=1
\(a.\sqrt{23+8\sqrt{7}}-\sqrt{7}=\sqrt{16+2.4\sqrt{7}+7}-\sqrt{7}=4+\sqrt{7}-\sqrt{7}=4\) \(b.\sqrt{4+2\sqrt{3}}-\sqrt{3}=\sqrt{3+2\sqrt{3}+1}-\sqrt{3}=\sqrt{3}+1-\sqrt{3}=1\) \(c.\sqrt{13-4\sqrt{3}}-\left(1+\sqrt{3}\right)^2=\sqrt{12-2.2\sqrt{3}+1}-\left(1+\sqrt{3}\right)^2=2\sqrt{3}-1+4+2\sqrt{3}=4\sqrt{3}+3\)
Dòng cuối nhầm : \(2\sqrt{3}-1-4-2\sqrt{3}=-5\)Lê Vương Kim Anh
Với bài này bạn áp dụng công thức : \(\sqrt{x^2}= \left|x\right|\); Nếu \(x\ge0\) thì \(\left|x\right|=x\)
Nếu \(x< 0\) thì \(\left|x\right|=-x\)
Áp dụng :
\(A=\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}=\sqrt{\left(2-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}=\left(2-\sqrt{3}\right)-\left(2+\sqrt{3}\right)=-2\sqrt{3}\)
điều kiện :a<=0
\(A^2=7-4\sqrt{3}-2\sqrt{\left(7-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)}+7+4\sqrt{3}\)
\(=14-2\sqrt{49-48}=12\)
\(\Rightarrow A=\sqrt{12}\left(LOẠI\right)HAYA=-\sqrt{12}\left(NHẬN\right)\)