Lần sau bạn chú ý viết đầy đủ đề.

1.

\(\sqrt{9+4\sqrt{5}-\sqrt{9-4\sqrt{5}}}=\sqrt{9+4\sqrt{5}-\sqrt{5-2\sqrt{4.5}+4}}\)

\(=\sqrt{9+4\sqrt{5}-\sqrt{(\sqrt{5}-\sqrt{4})^2}}=\sqrt{9+4\sqrt{5}-(\sqrt{5}-\sqrt{4})}\)

\(=\sqrt{9+4\sqrt{5}-\sqrt{5}+2}=\sqrt{11+3\sqrt{5}}\)

2.

\(\sqrt{8-2\sqrt{7}-\sqrt{8+2\sqrt{7}}}=\sqrt{8-2\sqrt{7}-\sqrt{7+2\sqrt{7}+1}}\)

\(=\sqrt{8-2\sqrt{7}-\sqrt{(\sqrt{7}+1)^2}}\)

\(=\sqrt{8-2\sqrt{7}-\sqrt{7}-1}=\sqrt{7-3\sqrt{7}}\)

Phạm Mạnh Kiên: sửa lại theo ý bạn thì làm như sau:

1.

\(\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}=\sqrt{5+2\sqrt{5}.\sqrt{4}+4}-\sqrt{5-2\sqrt{5}.\sqrt{4}+4}\)

\(=\sqrt{(\sqrt{5}+\sqrt{4})^2}-\sqrt{(\sqrt{5}-\sqrt{4})^2}=|\sqrt{5}+2|-|\sqrt{5}-2|\)

\(=\sqrt{5}+2-(\sqrt{5}-2)=4\)

2.

\(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)

\(=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}=|\sqrt{7}-1|-|\sqrt{7}+1|\)

\(=-2\)

\(\sqrt{4\sqrt{2}+4\sqrt{10-8\sqrt{3-2\sqrt{2}}}}=\sqrt{4\sqrt{2}+4\sqrt{10-8\sqrt{\left(\sqrt{2}-1\right)^2}}}=\sqrt{4\sqrt{2}+4\sqrt{10-8\sqrt{2}+8}}=\sqrt{4\sqrt{2}+4\sqrt{18-8\sqrt{2}}}=\sqrt{4\sqrt{2}+4\sqrt{\left(4-\sqrt{2}\right)^2}}=\sqrt{4\sqrt{2}+16-4\sqrt{2}}=\sqrt{16}=4\)

\(\sqrt{4\sqrt{2}+4\sqrt{10-8\sqrt{3-2\sqrt{2}}}}\)

\(=\sqrt{4\sqrt{2}+4\sqrt{10-8\left(\sqrt{2}-1\right)}}\)

\(=\sqrt{4\sqrt{2}+4\cdot\sqrt{18-8\sqrt{2}}}\)

\(=\sqrt{4\sqrt{2}+4\left(4-\sqrt{2}\right)}\)

=4

Đề không rõ ràng. Bạn xem lại.

1) \(\left(\sqrt{19}-3\right)\left(\sqrt{19}+3\right)=\left(\sqrt{19}\right)^2-3^2=19-9=10\)

2) \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\sqrt{\dfrac{8+2\sqrt{7}}{2}}-\sqrt{\dfrac{8-2\sqrt{7}}{2}}\)

\(=\sqrt{\dfrac{\left(\sqrt{7}\right)^2+2.\sqrt{7}.1+1^2}{2}}-\sqrt{\dfrac{\left(\sqrt{7}\right)^2-2.\sqrt{7}.1+1^2}{2}}\)

\(=\sqrt{\dfrac{\left(\sqrt{7}+1\right)^2}{2}}-\sqrt{\dfrac{\left(\sqrt{7}-1\right)^2}{2}}=\dfrac{\left|\sqrt{7}+1\right|}{\sqrt{2}}-\dfrac{\left|\sqrt{7}-1\right|}{\sqrt{2}}\)

\(=\dfrac{\sqrt{7}+1}{\sqrt{2}}-\dfrac{\sqrt{7}-1}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)

3) \(\sqrt{8+\sqrt{60}}+\sqrt{45}-\sqrt{12}=\sqrt{8+\sqrt{4.15}}+\sqrt{9.5}-\sqrt{4.3}\)

\(=\sqrt{8+2\sqrt{15}}+3\sqrt{5}-2\sqrt{3}\)

\(=\sqrt{\left(\sqrt{5}\right)^2+2.\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2}+3\sqrt{5}-2\sqrt{3}\)

\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}+3\sqrt{5}-2\sqrt{3}=\left|\sqrt{5}+\sqrt{3}\right|+3\sqrt{5}-2\sqrt{3}\)

\(\sqrt{5}+\sqrt{3}+3\sqrt{5}-2\sqrt{3}=4\sqrt{5}-\sqrt{3}\)

4) \(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)

\(=\sqrt{\left(\sqrt{5}\right)^2-2.2.\sqrt{5}+2^2}-\sqrt{\left(\sqrt{5}\right)^2+2.2.\sqrt{5}+2^2}\)

\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}=\left|\sqrt{5}-2\right|-\left|\sqrt{5}+2\right|\)

\(=\sqrt{5}-2-\sqrt{5}-2=-4\)

1) \(\left(\sqrt{19}-3\right)\left(\sqrt{19}+3\right)=19-9=10\)

4) \(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}=\sqrt{5}-2-\sqrt{5}-2=-4\)

\(\sqrt{3x+7}-\sqrt{x-1}=3\)

Đkxđ:\(\left\{{}\begin{matrix}3x+7\ge0\\x+1\ge0\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}x\ge-\frac{7}{3}\\x\ge-1\end{matrix}\right.\rightarrow x\ge-1\)

\(PT\rightarrow\sqrt{3x+7}=2+\sqrt{x+1}\)

\(\Rightarrow3x+7=\left(2+\sqrt{x+1}\right)^2\)

\(\Rightarrow3x+7=4+4\sqrt{x+1}+x+1\)

\(\Rightarrow2x+2=4\sqrt{x+1}\)

\(\Rightarrow x+1=2\sqrt{x+1}\)

\(\Rightarrow x^2+2x+1=4\left(x+1\right)\)

\(\Rightarrow x^2-2x-3=0\)

\(\Rightarrow x^2-3x+x-3=0\)

\(\Rightarrow\left(x-3\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=3\left(TM\right)\\x=-1\left(TM\right)\end{matrix}\right.\)

Vậy ....

đkxđ:....

Rút gọn:

\(\dfrac{1}{2\left(\sqrt{x}-1\right)}\cdot\left(\dfrac{x^2-8\sqrt{x}}{x+2\sqrt{x}+4}+1\right)-\dfrac{x-\sqrt{x}-1}{2\sqrt{x}}\)

\(=\dfrac{1}{2\left(\sqrt{x}-1\right)}\cdot\left[\dfrac{\sqrt{x}\left(\sqrt{x}^3-8\right)}{x+2\sqrt{x}+4}+1\right]-\dfrac{x-\sqrt{x}-1}{2\sqrt{x}}\)

\(=\dfrac{1}{2\left(\sqrt{x}-1\right)}\cdot\left[\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}{x+2\sqrt{x}+4}+1\right]-\dfrac{x-\sqrt{x}-1}{2\sqrt{x}}\)

\(=\dfrac{1}{2\left(\sqrt{x}-1\right)}\cdot\left[\sqrt{x}\left(\sqrt{x}-2\right)+1\right]-\dfrac{x-\sqrt{x}-1}{2\sqrt{x}}\)

\(=\dfrac{x-2\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}-\dfrac{x-\sqrt{x}-1}{2\sqrt{x}}\)

\(=\dfrac{\left(\sqrt{x}-1\right)^2}{2\left(\sqrt{x}-1\right)}-\dfrac{x-\sqrt{x}-1}{2\sqrt{x}}\)

\(=\dfrac{\sqrt{x}-1}{2}-\dfrac{x-\sqrt{x}-1}{2\sqrt{x}}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-x+\sqrt{x}+1}{2\sqrt{x}}\)

\(=\dfrac{x-\sqrt{x}-x+\sqrt{x}+1}{2\sqrt{x}}=\dfrac{1}{2\sqrt{x}}\)

\(=\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-\dfrac{8\left(\sqrt{5}+1\right)}{4}\\ =2\sqrt{5}-2\sqrt{5}-2=-2\)

ta làm, dù là

Điệp ngữ "ta làm" lặp đi lặp lại diễn tả khát vọng muốn được hòa nhập, muốn được cống hiến vào cuộc đời chung, sự nghiệp chung của đất nước.

Điệp ngữ : dù là" khẳng định ước muốn, tâm nguyện thường trực, thường xuyên, bền bỉ vượt thời gian, cả khi trẻ lẫn khi về già, cả khi khỏe mạnh hay bệnh tật. Đặt bài thơ vào cảnh ngộ lúc này của tác giả, ta thấy càng trân trọng hơn ước nguyện cao quý đó của nhà thơ.

a,

\(pt\Leftrightarrow\left(x-1-2\sqrt{x-1}+1\right)+\left(y-2-4\sqrt{y-2}+4\right)+\left(z-3-6\sqrt{z-3}+9\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=6\\z=12\end{cases}}\)