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mong bn thông cảm !$$%

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mik ko biết

mong bn thông cảm 

nha ................

kho lam a?

\(\left(3x-2y\right)^2+4\left(3x-2y\right)+4\\ =\left(3x-2y\right)^2+2.2\left(3x-2y\right)+2^2\\ =\left(3x-2y+2\right)^2\)

Áp dụng HĐT số 1 : \(A^2+2AB+B^2=\left(A+B\right)^2\)

   (3\(x\) - 2y)² + 4.(3\(x\) - 2y) + 4

=(3\(x\) - 2y)² + 2.2 (3\(x-2y\)) + 2²

= (3\(x\) - 2y + 2)²

(3x - 1)² - (2x - 5)²

= (3x - 1 - 2x + 5)(3x - 1 + 2x - 5)

= (x + 4)(5x - 6)

\(\left(3x-1\right)^2-\left(2x-5\right)^2\)

\(=\left(3x-1-2x+5\right)\left(3x-1+2x-5\right)\)

\(=\left(x+4\right)\left(5x-6\right)\)

\(17-\dfrac{12}{\sqrt[]{2}}=17-6\sqrt[]{2}=18-6\sqrt[]{2}+1-2\)

\(=\left(\sqrt[]{18}-1\right)^2-2=\left(3\sqrt[]{2}-1\right)^2-2\)

\(x^2+y^2-2x+2y-2=0\)

\(\left(x^2-2x+1\right)+\left(y^2+2y+1\right)-4=0\)

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

Đặt \(A=\sqrt{2+\sqrt{3}}\)

    \(\Rightarrow A\sqrt{2}=\sqrt{4+2\sqrt{3}}\)

                       \(=\sqrt{3+2\sqrt{3}+1}\)

                       \(=\sqrt{\left(\sqrt{3}+1\right)^2}\)

                        \(=\sqrt{3}+1\)

\(\Rightarrow A=\frac{\sqrt{3}+1}{2}hay\sqrt{2+\sqrt{3}}=\frac{\sqrt{3}+1}{2}\)

TK nha!

\(\sqrt{2+\sqrt{3}}=\sqrt{\frac{4+2\sqrt{3}}{2}}\)

                           \(=\sqrt{\frac{\left(\sqrt{3}+1\right)^2}{2}}\)

                           \(=\frac{\sqrt{3}+1}{\sqrt{2}}\)

                              \(=\frac{\sqrt{6}+\sqrt{2}}{2}\)

a)  \(x^2-4x+5+y^2+2y=\left(x^2-4x+4\right)+\left(y^2+2y+1\right)\)

\(=\left(x-2\right)^2+\left(y+1\right)^2\)

b)  \(2x^2+y^2-2xy+10x+25=\left(x^2+10x+25\right)+\left(x^2-2xy+y^2\right)\)

\(=\left(x+5\right)^2+\left(x-y\right)^2\)

c)  \(2x^2+2y^2=\left(x^2-2xy+y^2\right)+\left(x^2+2xy+y^2\right)=\left(x-y\right)^2+\left(x+y\right)^2\)

bạn ơi , bạn lấy bài này ở đâu vậy bạn

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