3*4 = mấy ạ
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\(\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}\right)^2-2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}-1\right)^2}=|\sqrt{2}-1|=\sqrt{2}-1\)
Tương tự \(\sqrt{4-2\sqrt{3}}=\sqrt{3}-1\); \(\sqrt{7-4\sqrt{3}}=2-\sqrt{3}\)
\(\Rightarrow BTT=\sqrt{2}-1+\sqrt{3}-1+2-\sqrt{3}=\sqrt{2}\)
\(\sqrt{3-2\sqrt{2}}+\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{2-2\sqrt{2}+1}+\sqrt{3-2\sqrt{3}+1}-\sqrt{4-4\sqrt{3}+3}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\sqrt{2}-1+\sqrt{3}-1-2+\sqrt{3}\)
\(=2\sqrt{3}+\sqrt{2}-4\)
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\(x^3-4x^2+4x-xy^2\)
\(=x\left(x^2-4x+4-y^2\right)\)
\(=x\left[\left(x^2-4x+4\right)-y^2\right]\)
\(=x\left[\left(x-2\right)^2-y^2\right]\)
\(=x\left(x-2-y\right)\left(x-2+y\right)\)
x3 - 4x2 + 4x - xy2
= x ( x2 - 4x + 4 - y2 )
= x [ ( x2 - 4x + 22 ) - y2 ]
= x [ ( x - 2 )2 - y2 ]
= x ( x - 2 - y ) ( x - 2 + y )
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\(\left|\frac{13}{4}-2x\right|=\frac{2}{5}+\frac{5}{2}=\frac{29}{10}\left(1\right)\)
+ Nếu \(\frac{13}{4}-2x\ge0\Leftrightarrow x\le\frac{13}{8}\)
\(\Rightarrow\left(1\right)\Leftrightarrow\frac{13}{4}-2x=\frac{29}{10}\Rightarrow x=\frac{7}{40}\) so với điều kiện \(x\le\frac{13}{8}\) nên thoả mãn
+ Nếu \(\frac{13}{4}-2x< 0\Leftrightarrow x>\frac{13}{8}\)
\(\Rightarrow\left(1\right)\Leftrightarrow2x-\frac{13}{4}=\frac{29}{10}\Leftrightarrow x=\frac{123}{40}\) so với điều kiện \(x>\frac{13}{8}=\frac{65}{40}\) nên thoả mãn
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn sai ở dấu bằng thứ 4. Mình làm lại nhé.
\(\left(x+y\right)^4+x^4+y^4\)
\(=\left[\left(x+y\right)^2\right]^2+x^4+y^4\)
\(=\left(x^2+2xy+y^2\right)^2+x^4+y^4\)
\(=x^4+4x^2y^2+y^4+4x^3y+4xy^3+2x^2y^2+x^4+y^4\)
\(=2x^4+4x^3y+6x^2y^2+4xy^3+2y^4\)
\(=2\left(x^4+2x^3y+3x^2y^2+2xy^3+y^4\right)\)
\(=2.\left[\left(x^4+2x^3y+x^2y^2\right)+\left(2x^2y^2+2xy^3\right)+y^4\right]\)
\(=2.\left[\left(x^2+xy\right)^2+2.\left(x^2+xy\right).y^2+\left(y^2\right)^2\right]\)
\(=2.\left(x^2+xy+y^2\right)^2\)
Học tốt nhe.
TL :
3*4 = 3 . 4 = 12
_HT_
3 x 4 = 12