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đề rút gọn \(A=x+2y-\sqrt{x^2-4xy+4y^2}\) biết \(x\ge2y\)
\(A=x+2y-\sqrt{x^2-4xy+4y^2}\)
\(A=x+2y-\sqrt{\left(x-2y\right)^2}\)
\(A=x+2y-x+2y=4y\) (do \(x\ge2y\))
Lời giải:
Sửa đề: Rút gọn \(x+2y-\sqrt{x^2-4xy+4y^2}\) \((x\geq 2y)\)
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Ta có:
\(x+2y-\sqrt{x^2-4xy+4y^2}=x+2y-\sqrt{x^2-2.x.2y+(2y)^2}\)
\(=x+2y-\sqrt{(x-2y)^2}\)
\(=x+2y-|x-2y|=x+2y-(x-2y)=4y\)
(do \(x\geq 2y\Rightarrow |x-2y|=x-2y\) )
1) \(\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\)
= \(\frac{ \left(\sqrt{7}+\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}+\frac{\left(\sqrt{7}-\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}\)
= \(\frac{\left(\sqrt{7}+\sqrt{5}\right)^2+\left(\sqrt{7}-\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}\) = \(\frac{\left(\sqrt{7}\right)^2+2\sqrt{7}.\sqrt{5}+\left(\sqrt{5}\right)^2+\left(\sqrt{7}\right)^2-2\sqrt{7}.\sqrt{5}+\left(\sqrt{5}\right)^2}{\left(\sqrt{7}\right)^2-\left(\sqrt{5}\right)^2}\)
= \(\frac{7+2\sqrt{35}+5+7-2\sqrt{35}+5}{7-5}\) = \(\frac{24}{2}=12\)
2) \(x+2y-\sqrt{\left(x^2-4xy+4y^2\right)^2}\left(x\ge2y\right)\)
= \(x+2y-\sqrt{\left(x-2y\right)^4}\) = \(x+2y-|x-2y|\)
= \(x+2y-\left(x-2y\right)\) = \(x+2y-x+2y=4y\)
3)\(4x+\sqrt{\left(x-12\right)^2}\left(x\ge2\right)\)
= \(4x+x-12=5x-12\)
1,Sửa lại điều kiện,mình nghĩ là: \(x \geq 12\)(chắc bạn ghi nhầm)
Vì \(x \geq 12\) \(\Rightarrow\) \(x-12 \geq 0\) \(\Rightarrow\) \(\sqrt{\left(x-12\right)^2}=x-12\)
Ta có \(4x+\sqrt{\left(x-12\right)^2}\) = \(4x+x-12\) = 5x-12
2, Dư bình phương ở phần căn
Vì \(x \geq 2y\) \(\Rightarrow\) \(x-2y \geq 0\)
Ta có : \(x+2y-\sqrt{\left(x^2-4xy+4y^2\right)}=x+2y-\sqrt{\left(x-2y\right)^2}=x+2y-\left(x-2y\right)=x+2y-x+2y=4y\)
a)=1-4a
b) = 2x - 4y
c) = 2x - 2 (nếu x>5)
=2x(nếu x<5)
\(a,\sqrt{1-4a+4a^2}-2a\)
\(=\sqrt{\left(1-2a\right)^2}-2a\)
\(=1-2a-2a\)
\(=1-4a\)
\(b,x-2y-\sqrt{x^2-4xy+4y^2}\)
\(=x-2y-\sqrt{\left(x-2y\right)^2}\)
\(=x-2y-\left(x-2y\right)\)
\(=x-2y-x+2y\)
\(=0\)
\(c,x^2+\sqrt{x^4-8x^2+16}\)
\(=x^2+\sqrt{\left(x^2-4\right)^2}\)
\(=x^2+x^2-4\)
\(=2x^2-4\)
Các câu còn lại tương tự nha
\(a,\sqrt{1-4a+4a^2}-2a\)
\(=\sqrt{\left(1-2a\right)^2}-2a\)
\(=\left(1-2a\right)-2a\)
\(=1-4a\)
\(b,x-2y-\sqrt{x^2-4xy+4y^2}\)
\(=x-2y-\sqrt{\left(x-2y\right)^2}\)
\(=x-2y-\left(x-2y\right)\)
\(=x-2y-x+2y\)
\(=0\)
\(c,x^2+\sqrt{x^4-8x^2+16}\)
\(=x^2+\sqrt{\left(x^2-2^2\right)^2}\)
\(=x^2+\left(x^2-4\right)\)
\(=x^2+x^2-4\)
\(=2x^2-4\)
\(d,2x-1-\frac{\sqrt{x^2-10x+25}}{x-5}\)
\(=2x-1-\frac{\sqrt{\left(x-5\right)^2}}{x-5}\)
\(=2x-1-\frac{x-5}{x-5}\)
\(=2x-1-1\)
\(=2x-2\)
\(=2\left(x-1\right)\)
vế 2 là hằng đẳng thức đầu tiên ý