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\(A=\frac{\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}+\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}}{\sqrt{\left(\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}\right)}-\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}}\)=\(\frac{\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}}{\sqrt{\left(\sqrt{x-1}+1\right)^2}-\sqrt{\left(\sqrt{x-1}-1\right)^2}}\)Vì x>/2
=\(\frac{\sqrt{x-1}+1+\sqrt{x-1}-1}{\sqrt{x-1}+1-\sqrt{x-1}+1}=\frac{2\sqrt{x-1}}{2}=\sqrt{x-1}\)
(14,78-a)/(2,87+a)=4/1
14,78+2,87=17,65
Tổng số phần bằng nhau là 4+1=5
Mỗi phần có giá trị bằng 17,65/5=3,53
=>2,87+a=3,53
=>a=0,66.
a,\(\sqrt{x-4+4\sqrt{x-4}+4}\) +\(\sqrt{x-4-4\sqrt{x-4}+4}\)
=\(\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\) (vi x>=8)
=\(\sqrt{x-4}+2+\sqrt{x-4}-2=2\sqrt{x-4}\)
b, \(\sqrt{x-1+2\sqrt{x\left(x-1\right)}+x}+\sqrt{x-1-2\sqrt{x\left(x-1\right)}+x}\)
=\(\sqrt{x-1}+\sqrt{x}+\left|\sqrt{x-1}-\sqrt{x}\right|\)
=\(\sqrt{x}+\sqrt{x-1}+\sqrt{x}-\sqrt{x-1}\) =\(2\sqrt{x}\)
c,d sai dau bai hay sao y
a) \(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-\left(x-2\sqrt{xy}+y\right)\)
\(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y=\sqrt{xy}\)
b) \(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\frac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
c) \(4x-\sqrt{8}+\frac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}=4x-\sqrt{8}+\frac{\sqrt{x^2\left(x+2\right)}}{x+2}=4x-\sqrt{8}+x=5x-\sqrt{8}\)
\(A=\left(a-1\right)\sqrt{\frac{a}{a-1}}+\sqrt{a\left(a-1\right)}-a\sqrt{\frac{a-1}{a}}\)
\(A=\sqrt{\left(a-1\right)^2.\frac{a}{a-1}}+\sqrt{a\left(a-1\right)}-\sqrt{a^2.\frac{a-1}{a}}\)
\(A=\sqrt{\left(a-1\right)a}+\sqrt{a\left(a-1\right)}-\sqrt{a\left(a-1\right)}\)
\(A=\sqrt{a\left(a-1\right)}\)
\(\frac{A}{\sqrt{2}}\)=\(\frac{\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}}{\sqrt{2x-1+2\sqrt{2x-1}+1}-\sqrt{2x-1-2\sqrt{2x-1}+1}}\) (DK \(x\ge1\)
\(=\frac{\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|}{\sqrt{2x-1}+1-\left|\sqrt{2x-1}-1\right|}\)
vs \(x\ge2\) \(\frac{\sqrt{x-1}+1+\sqrt{x-1}-1}{\sqrt{2x-1}+1-\sqrt{2x-1}+1}=\frac{2\sqrt{x-1}}{2}=\sqrt{x-1}\) \(\Rightarrow A=\sqrt{2x-2}\)
vs \(1\le x< 2\) \(\frac{\sqrt{x-1}+1+1-\sqrt{x-1}}{\sqrt{2x-1}+1-1+\sqrt{2x-1}}=\frac{1}{\sqrt{2x-1}}\) \(\Rightarrow A=\frac{\sqrt{2}}{\sqrt{2x-1}}\)
\(\sqrt{2X-1}\ge1\Leftrightarrow X\ge1\)NEN SUY RA THEO CACH LAM CUA TO
THOI U AM BUSY SEE YOU AGAIN
\(\frac{1}{\sqrt{2}}.A=\frac{\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}+\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}}{\sqrt{\left(2x-1\right)+2\sqrt{2x-1}+1}-\sqrt{\left(2x-1\right)-2\sqrt{2x-1}+1}}\)
\(=\frac{\sqrt{\left[\left(\sqrt{x-1}+1\right)\right]^2}+\sqrt{\left[\left(\sqrt{x-1}-1\right)^2\right]}}{\sqrt{\left[\sqrt{2x-1}+1\right]^2}-\sqrt{\left[\left(\sqrt{2x-1}\right)-1\right]^2}}\)
\(=\frac{\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|}{\left|\sqrt{2x-1}+1\right|-\left|\sqrt{2x-1}-1\right|}\)
DO X>2 NÊN TOÀN BỘ BIỂU THỨC TRONG TRỊ TUYỆT ĐỐI ĐỀU DƯƠNG
\(\frac{1}{\sqrt{2}}.A=\frac{2\sqrt{x-1}}{2}=\sqrt{x-1}\)
=>\(A=\frac{\sqrt{x-1}}{\sqrt{2}}\)