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Bài 1:
a)
\(\sqrt{13-2\sqrt{42}}=\sqrt{6+7-2\sqrt{6.7}}=\sqrt{(\sqrt{7}-\sqrt{6})^2}=|\sqrt{7}-\sqrt{6}|=\sqrt{7}-\sqrt{6}\)
b)
\(\sqrt{46+6\sqrt{5}}=\sqrt{46+2\sqrt{45}}=\sqrt{45+1+2\sqrt{45.1}}=\sqrt{(\sqrt{45}+1)^2}=\sqrt{45}+1\)
\(=3\sqrt{5}+1\)
c)
\(\sqrt{12-3\sqrt{15}}=\sqrt{\frac{24-6\sqrt{15}}{2}}=\sqrt{\frac{24-2\sqrt{135}}{2}}=\sqrt{\frac{15+9-2\sqrt{15.9}}{2}}\)
\(=\sqrt{\frac{(\sqrt{15}-\sqrt{9})^2}{2}}=\frac{\sqrt{15}-\sqrt{9}}{\sqrt{2}}=\frac{\sqrt{15}-3}{\sqrt{2}}\)
d)
\(\sqrt{11+\sqrt{96}}=\sqrt{11+2\sqrt{24}}=\sqrt{8+3+2\sqrt{8.3}}\)
\(=\sqrt{(\sqrt{8}+\sqrt{3})^2}=\sqrt{8}+\sqrt{3}\)
Bài 1:
a)
\(\sqrt{13-2\sqrt{42}}=\sqrt{6+7-2\sqrt{6.7}}=\sqrt{(\sqrt{7}-\sqrt{6})^2}=|\sqrt{7}-\sqrt{6}|=\sqrt{7}-\sqrt{6}\)
b)
\(\sqrt{46+6\sqrt{5}}=\sqrt{46+2\sqrt{45}}=\sqrt{45+1+2\sqrt{45.1}}=\sqrt{(\sqrt{45}+1)^2}=\sqrt{45}+1\)
\(=3\sqrt{5}+1\)
c)
\(\sqrt{12-3\sqrt{15}}=\sqrt{\frac{24-6\sqrt{15}}{2}}=\sqrt{\frac{24-2\sqrt{135}}{2}}=\sqrt{\frac{15+9-2\sqrt{15.9}}{2}}\)
\(=\sqrt{\frac{(\sqrt{15}-\sqrt{9})^2}{2}}=\frac{\sqrt{15}-\sqrt{9}}{\sqrt{2}}=\frac{\sqrt{15}-3}{\sqrt{2}}\)
d)
\(\sqrt{11+\sqrt{96}}=\sqrt{11+2\sqrt{24}}=\sqrt{8+3+2\sqrt{8.3}}\)
\(=\sqrt{(\sqrt{8}+\sqrt{3})^2}=\sqrt{8}+\sqrt{3}\)
Bài 4 :
\(a,\sqrt{x-1}=2\)
=> \(x-1=2^2=4\)
=>\(x=4+1=5\)
Vậy \(x\in\left\{5\right\}\)
\(b,\sqrt{x^2-3x+2}=2\)
=> \(x^2-3x+2=2\)
=> \(x^2-3x=2-2=0\)
=>\(x.\left(x-3\right)=0\)( phân tích đa thức thanh nhân tử )
=> \(\left[{}\begin{matrix}x=0\\x-3=0=>x=0+3=3\end{matrix}\right.\)
Vậy \(x\in\left\{0;3\right\}\)
MÌNH Biết vậy thôi ,
Bài 4 :
c) \(\sqrt{4x+1}=x+1\)ĐK : \(x\ge-1\)
\(\Leftrightarrow4x+1=\left(x+1\right)^2\)
\(\Leftrightarrow x^2+2x+1-4x-1=0\)
\(\Leftrightarrow x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)( thỏa )
d) \(\sqrt{x+2\sqrt{x-1}}-\sqrt{x-2\sqrt{x-1}}=2\)
\(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}-\sqrt{x-1-2\sqrt{x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}-\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\)
\(\Leftrightarrow\left|\sqrt{x-1}+1\right|-\left|\sqrt{x-1}-1\right|=2\)
+) Xét \(x\ge2\)
\(pt\Leftrightarrow\sqrt{x-1}+1-\sqrt{x-1}+1=2\)
\(\Leftrightarrow2=2\)( luôn đúng )
+) Xét \(1\le x< 2\):
\(pt\Leftrightarrow\sqrt{x-1}+1-1+\sqrt{x-1}=2\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\)( loại )
Vậy \(x\ge2\)
a: \(=\sqrt{4+2+\sqrt{3}}=\sqrt{6+\sqrt{3}}\)
c: \(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)
\(=\sqrt{13+30\left(\sqrt{2}+1\right)}\)
\(=\sqrt{43+30\sqrt{2}}\)
d: \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)
\(=\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\)
\(=\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|\)
TH1: x>=2
\(D=\sqrt{x-1}+1+\sqrt{x-1}-1=2\sqrt{x-1}\)
TH2: 0<=x<2
\(D=\sqrt{x-1}+1+1-\sqrt{x-1}=2\)
chú ý\(x=\sqrt{x}^2\) tương tự với y , và các số tự nhiên dương
\(A=\frac{\sqrt{x}^2+2\sqrt{x}-3}{\sqrt{x}-1}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)}=\sqrt{x}+3\)
\(B=\frac{\left(2\sqrt{y}\right)^2+3\sqrt{y}-7}{4\sqrt{y}+7}=\frac{\left(\sqrt{y}-1\right)\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}=\sqrt{y}-1\)
\(C=\frac{\sqrt{x}^2\sqrt{y}-\sqrt{y}^2\sqrt{x}}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}=\sqrt{xy}\)
\(D=\frac{\sqrt{x}^2-3\sqrt{x}-4}{\sqrt{x}^2-\sqrt{x}-12}=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}=\frac{\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)}\)
\(E=\sqrt{1+2\sqrt{5}+5}+\sqrt{\sqrt{5}-2\sqrt{5}+1}=\sqrt{\left(1+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
=>\(E=1+\sqrt{5}+\sqrt{5}-1=2\sqrt{5}\)
CÂU CUỐI chưa làm đc
ý cuối cùng này :
\(D=\sqrt{13-4\sqrt{10}}+\sqrt{13+4\sqrt{10}}\)lấy bình phương 2 vế ta có
\(D^2=13-4\sqrt{10}+13+4\sqrt{10}+2\sqrt{13-4\sqrt{10}}\sqrt{13+4\sqrt{10}}\)
\(D^2=26+2\sqrt{13^2-16\sqrt{10}^2}\Leftrightarrow D^2=26+2\sqrt{9}\)
\(D^2=32\Leftrightarrow D=\sqrt{32}=4\sqrt{2}\)
1: \(=\dfrac{\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}+1+\sqrt{7}-1}{\sqrt{2}}=\dfrac{2\sqrt{7}}{\sqrt{2}}=\sqrt{14}\)
3: \(=\sqrt{6+2\sqrt{2\cdot\sqrt{3-\sqrt{3}-1}}}\)
\(=\sqrt{6+2\sqrt{2\cdot\sqrt{2-\sqrt{3}}}}\)
\(=\sqrt{6+2\sqrt{\sqrt{2}\left(\sqrt{3}-1\right)}}\)
\(=\sqrt{6+2\sqrt{\sqrt{6}-\sqrt{2}}}\)
+) ta có : \(A=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\dfrac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}=\dfrac{\sqrt{7}+1-\sqrt{7}+1}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
+) ta có : \(B=\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}=\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}\)
\(=\sqrt{5}-\sqrt{6-2\sqrt{5}}=\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-\sqrt{5}+1=1\)
+) điều kiện : \(x\ge1\)
ta có : \(C=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)
\(=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
\(=\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}C=2\sqrt{x-1}\left(x\ge2\right)\\C=2\left(1\le x< 2\right)\end{matrix}\right.\)
câu c này mk sữa đề nhát
thanks so much !