Tìm MinA: \(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}\)
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KC
2
7 tháng 11 2017
A= \(\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}\)
=\(\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)^2}\)
=\(\left|\sqrt{x-1}-1\right|+\left|\sqrt{x-1}+1\right|\)
\(=\left|1-\sqrt{x-1}\right|+\left|\sqrt{x-1}+1\right|\)
\(\ge\left|\sqrt{x-1}+1+1-\sqrt{x-1}\right|\)
=2.
dấu = khi và chỉ khi \(\left(\sqrt{x-1}+1\right).\left(1-\sqrt{x-1}\right)=0\)
KC
6
11 tháng 8 2017
\(A=\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}\)
\(=\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)^2}\)
\(=\left|\sqrt{x-1}-1\right|+\left|\sqrt{x-1}+1\right|\)
\(=\left|1-\sqrt{x-1}\right|+\left|\sqrt{x-1}+1\right|\)
\(\ge\left|1-\sqrt{x-1}+\sqrt{x-1}+1\right|=2\)
Dấu "=" xảy ra \(\Leftrightarrow\left(1-\sqrt{x-1}\right)\left(\sqrt{x-1}+1\right)\ge0\Leftrightarrow0\le x\le2\)
Vậy \(A_{min}=2\) tại \(0\le x\le2\)
AS
1
28 tháng 6 2019
a) \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)
\(=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
\(=\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\)
b) \(\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\)
\(=\left|\sqrt{x-1}+1\right|+\left|1-\sqrt{x-1}\right|\)
\(\ge\left|\sqrt{x-1}+1+1-\sqrt{x-1}\right|=\left|2\right|=2\)
Dấu "=" xảy ra \(\Leftrightarrow1\le x\le2\)
28 tháng 8 2023
1: \(\Leftrightarrow\dfrac{3x-1}{x+2}=4\)
=>4x+8=3x-1
=>x=-9
2: \(\Leftrightarrow\dfrac{5x-7}{2x-1}=4\)
=>8x-4=5x-7
=>3x=-3
=>x=-1
3: ĐKXD: x>=0
\(PT\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)
=>\(x+\sqrt{x}-6=x-1\)
=>căn x=-1+6=5
=>x=25
4: ĐKXĐ: x>=0
PT =>\(\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)=\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)
=>x-2*căn x-3=x-4
=>-2căn x-3=-4
=>2căn x+3=4
=>2căn x=1
=>căn x=1/2
=>x=1/4
XO
27 tháng 12 2021
a) ĐKXĐ : \(3\le x\le7\)
Ta có \(A=1.\sqrt{x-3}+1.\sqrt{7-x}\)
\(\le\sqrt{\left(1+1\right)\left(x-3+7-x\right)}=\sqrt{8}\)(BĐT Bunyacovski)
Dấu "=" xảy ra <=> \(\dfrac{1}{\sqrt{x-3}}=\dfrac{1}{\sqrt{7-x}}\Leftrightarrow x=5\)
ĐK: \(x\ge1\)
\(A=\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}+\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)^2}\)
\(=\left|1-\sqrt{x-1}\right|+\left|\sqrt{x-1}+1\right|\)
\(\ge\left|1-\sqrt{x-1}+\sqrt{x-1}+1\right|=2\)
Đẳng thức xảy ra \(\Leftrightarrow\left(1-\sqrt{x-1}\right)\left(\sqrt{x-1}+1\right)\ge0\)
\(\Leftrightarrow1\le x\le2\)