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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) \(\sqrt{x-2\sqrt{x-1}}\)
\(=\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
\(=\left|\sqrt{x-1}-1\right|\)
điều kiện -4<=x<=4x<=4
\(a,\sqrt{\left(x+4\right)^2}+\sqrt{\left(x-4\right)^2}\)
\(A=\left|x+4\right|+\left|x-4\right|\)
KẾT HỢP ĐIỀU KIỆN
\(A=x+4+4-x\)
\(A=8\)
\(B=\sqrt{\left(3x\right)^2-6x+1}+\sqrt{\left(2x\right)^2-12x+3^2}\)
\(B=\sqrt{\left(3x-1\right)^2}+\sqrt{\left(2x-3\right)^2}\)
\(B=\left|3x-1\right|+\left|2x-3\right|\)
\(TH1:x>=\frac{3}{2}\)
\(B=3x-1+2x-3\)
\(B=5x-4\)
\(TH2:\frac{1}{3}< =x< \frac{3}{2}\)
\(B=3x-1-2x+3\)
\(B=x+2\)
\(TH3:x< \frac{1}{3}\)
\(B=-3x+1-2x+3\)
\(B=4-5x\)
câu c và câu d tương tự
câu c tách ra: \(C=\sqrt{\left(\sqrt{x}-3\right)^2}-\sqrt{\left(2\sqrt{x}+1\right)^2}\)
còn câu d tách ra :\(D=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)
\(D=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
bạn tự làm nốt câu c, d nha
a)\(4\sqrt{x}-5\sqrt{4x}-\sqrt{25x}-3\sqrt{x}-5\)
=\(4\sqrt{x}-10\sqrt{x}-5\sqrt{x}-3\sqrt{x}-5\)
=\(-14\sqrt{x}-5\)
b)\(\sqrt{16x}-5\left(\sqrt{x}-2\right)\sqrt{79x}-5\)
=\(4\sqrt{x}-\left(5\sqrt{x}-10\right)\sqrt{79x}-5\)
=\(4\sqrt{x}-\left(5\sqrt{79}x-10\sqrt{79}x\right)-5\)
=\(4\sqrt{x}+5\sqrt{79}x-5\)
\(a.\sqrt{49-56x+16x^2}=\sqrt{16x^2-2.4x.7+49}=\sqrt{\left(4x-7\right)^2}=\text{|}4x-7\text{|}=4x-7\)\(b.\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1-2\sqrt{x-1}+1}=\sqrt{\left(\sqrt{x-1}-1\right)^2}=\text{|}\sqrt{x-1}-1\text{|}=\sqrt{x-1}-1\)\(c.\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}=\sqrt{x-4+2.2\sqrt{x-4}+4}+\sqrt{x-4-2.2\sqrt{x-4}+4}=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}=\text{|}\sqrt{x-4}+2\text{|}+\text{|}\sqrt{x-4}-2\text{|}=2\sqrt{x-4}\)