Chứng minh \(\sqrt{x^2-2x+2}+\sqrt{x^2-4x+8}\ge\sqrt{10}\)
Những câu hỏi liên quan
giải bpt:
1. \(\frac{\sqrt{-3x^2+x+4}+2}{x}< 2\)
2. \(\sqrt{x^2-3x+2}+\sqrt{x^2-4x+3}\ge2\sqrt{x^2-5x+4}\)
3. \(\sqrt{x^2-8x+15}+\sqrt{x^2+2x-15}\le\sqrt{4x^2-18x=18}\)
4. 4(x+1)2 \(\ge\) (2x +10)( 1- \(\sqrt{3+2x}\))2
5. \(\sqrt{1+x}-\sqrt{1-x}\ge x\)
Tìm điều kiện có nghĩa:1) sqrt{2x^2}2) sqrt{-x}3) sqrt{-x^2-3}4) sqrt{x^2+2x+3}5) sqrt{-a^2+8a-16}6) sqrt[]{16x^2-25}7) sqrt{4x^2-49}8) sqrt{8-x^2}9) sqrt{x^2-12}10) sqrt{x^2+2x-3}11) sqrt{2x^2+5x+3}12) sqrt{dfrac{4}{x-1}}13) sqrt{dfrac{-1}{x-3}}14) sqrt{dfrac{-3}{x+2}}15) sqrt{dfrac{1}{2a-1}}16) sqrt{dfrac{2}{3-2a}}17) sqrt{dfrac{-1}{2a-5}}18) sqrt{dfrac{-2}{3-5a}}19) sqrt{dfrac{-a}{5}}20) dfrac{1}{sqrt{-3a}}
Đọc tiếp
Tìm điều kiện có nghĩa:
1) \(\sqrt{2x^2}\)
2) \(\sqrt{-x}\)
3) \(\sqrt{-x^2-3}\)
4) \(\sqrt{x^2+2x+3}\)
5) \(\sqrt{-a^2+8a-16}\)
6) \(\sqrt[]{16x^2-25}\)
7) \(\sqrt{4x^2-49}\)
8) \(\sqrt{8-x^2}\)
9) \(\sqrt{x^2-12}\)
10) \(\sqrt{x^2+2x-3}\)
11) \(\sqrt{2x^2+5x+3}\)
12) \(\sqrt{\dfrac{4}{x-1}}\)
13) \(\sqrt{\dfrac{-1}{x-3}}\)
14) \(\sqrt{\dfrac{-3}{x+2}}\)
15) \(\sqrt{\dfrac{1}{2a-1}}\)
16) \(\sqrt{\dfrac{2}{3-2a}}\)
17) \(\sqrt{\dfrac{-1}{2a-5}}\)
18) \(\sqrt{\dfrac{-2}{3-5a}}\)
19) \(\sqrt{\dfrac{-a}{5}}\)
20) \(\dfrac{1}{\sqrt{-3a}}\)
1) \(ĐK:x\in R\)
2) \(ĐK:x< 0\)
3) \(ĐK:x\in\varnothing\)
4) \(=\sqrt{\left(x+1\right)^2+2}\)
\(ĐK:x\in R\)
5) \(=\sqrt{-\left(a-4\right)^2}\)
\(ĐK:x\in\varnothing\)
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Chứng minh
a) \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\)
b) \(\sqrt{2x+\sqrt{4x-1}}+\sqrt{2x-\sqrt{4x-1}}=\sqrt{6}\)
Trả lời:
a,\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2.\sqrt{x-1}\)
Đặt \(\sqrt{x-1}=t\)\(\Rightarrow x=t^2+1\)
Đẳng thức đã cho trở thành:
\(VT=\)\(\sqrt{t^2+1+2t}+\sqrt{t^2+1-2t}\)
\(=\sqrt{t^2+2t+1}+\sqrt{t^2-2t+1}\)
\(=\sqrt{\left(t+1\right)^2}+\sqrt{\left(t-1\right)^2}\)
\(=t+1+t-1\)
\(=2t\)
\(=2.\sqrt{x-1}=VP\)
Vậy \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2.\sqrt{x-1}\)
b, \(\sqrt{2x+\sqrt{4x-1}}+\sqrt{2x-\sqrt{4x-1}}=\sqrt{6}\)
Đặt \(\sqrt{4x-1}=t\)\(\Rightarrow2x=\frac{t^2+1}{2}\)
Đẳng thức đã cho trở thành:
\(VT=\sqrt{\frac{t^2+1}{2}+t}+\sqrt{\frac{t^2+1}{2}-t}\)
\(=\sqrt{\frac{t^2+2t+1}{2}}+\sqrt{\frac{t^2-2t+1}{2}}\)
\(=\sqrt{\frac{\left(t+1\right)^2}{2}}+\sqrt{\frac{\left(t-1\right)^2}{2}}\)
\(=\frac{t+1}{\sqrt{2}}+\frac{t-1}{\sqrt{2}}\)
\(=\frac{2t}{\sqrt{2}}\)
\(=\frac{2.\sqrt{4x-1}}{\sqrt{2}}\)
GIẢI CÁC PT SAU:
\(\sqrt{5x+10}=8-x\)
\(\sqrt{4x^2+x-12}=3x-5\)
\(\sqrt{x^2-2x+6}=2x-3\)
\(\sqrt{3x^2-2x+6}+3-2x=0\)
\(\sqrt{\sqrt{ }x+\dfrac{\sqrt{x^2}-4}{\sqrt{x}}}+\sqrt{\sqrt{ }x-\dfrac{\sqrt{x^2-4}}{\sqrt{x}}}=\sqrt{\dfrac{2x+4}{\sqrt{x}}}\)
chứng minh đẳng thức trên với x\(_{\ge}\)2
\(\sqrt{x}+2\sqrt{y}=10\) chứng minh x+y \(\ge\) 20
help
ta có: \(\sqrt{x}+2\sqrt{y}=10=>\left(\sqrt{x}+2\sqrt{y}\right)^2=100\)
áp dụng BDT Bunhia
\(\sqrt{x}+2\sqrt{y}\le\sqrt{\left(1+2^2\right)\left(x+y\right)}\)
\(=>100\le5\left(x+y\right)=>x+y\ge\dfrac{100}{5}=20\)
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Chứng minh với mọi giá trị của x ta có:\(\frac{x^2+4x+9}{\sqrt{x^2+4x+8}}\ge\frac{5}{2}\)
giải pt:
a,\(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
b,\(\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4\)
c, \(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)
Cho x, y, z > 0 và x + y + z = 1. Chứng minh rằng: \(\sqrt{2x^2+xy+2y^2}+\sqrt{2y^2+yz+2z^2}+\sqrt{2z^2+zx+2x^2}\ge\sqrt{5}\)
\(VT=\sum\sqrt{\frac{1}{2}\left(x^2+2xy+y^2\right)+\frac{3}{2}\left(x^2+y^2\right)}\)
\(VT\ge\sum\sqrt{\frac{1}{2}\left(x+y\right)^2+\frac{3}{4}\left(x+y\right)^2}=\sum\sqrt{\frac{5}{4}\left(x+y\right)^2}\)
\(VT\ge\frac{\sqrt{5}}{2}\left(x+y\right)+\frac{\sqrt{5}}{2}\left(y+z\right)+\frac{\sqrt{5}}{2}\left(z+x\right)\)
\(VT\ge\sqrt{5}\left(x+y+z\right)=\sqrt{5}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)
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