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Bài 1 :
Đặt f(x) = \(\sqrt{x}-\sqrt{x-1}\) tập xác định [1;+∞)
Dễ thấy f(x) > 0
f(x) = \(\left(\sqrt{x}-1\right)-\sqrt{x-1}+1=\dfrac{x-1}{\sqrt{x}+1}-\sqrt{x-1}+1\)
= \(\sqrt{x-1}\left(\dfrac{\sqrt{x-1}}{\sqrt{x+1}}-1\right)+1\le\sqrt{x-1}\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)+1=\dfrac{-\sqrt{x-1}}{\sqrt{x+1}}+1\le1\)
Và f(1) = 1
Vậy f(x) có tập giá trị là (0;1]
* Nếu m \(\ge1\) thì bpt vô nghiệm
* Nếu m < 1 thì bpt có nghiệm
Vậy tập hợp m thỏa mãn là (0;1)
(0;1)
ei ~ atr ăn cắp ảnh nka , chưa xin phép eg , atr lấy ảnh eg từ khi nào vậy , khai mau
\( 1)\sqrt[3]{{12 - x}} + \sqrt[3]{{14 + x}} = 2\\ \Leftrightarrow 12 - x + 3\sqrt[3]{{{{\left( {12 - x} \right)}^2}.\left( {14 + x} \right)}} + 3\sqrt[3]{{\left( {12 - x} \right){{\left( {14 + x} \right)}^2}}} + 14 + x = 8\\ \Leftrightarrow 3\sqrt[3]{{\left( {12 - x} \right)\left( {14 + x} \right)}}\left( {\sqrt[3]{{12 - x}} + \sqrt[3]{{14 + x}}} \right) = - 18\\ \Leftrightarrow 3\sqrt[3]{{\left( {12 - x} \right)\left( {14 + x} \right)}}.2 = - 18\\ \Leftrightarrow \sqrt[3]{{\left( {12 - x} \right)\left( {14 + x} \right)}} = - 3\\ \Leftrightarrow \left( {12 - x} \right)\left( {14 + x} \right) = {\left( { - 3} \right)^3}\\ \Leftrightarrow 168 - 2x - {x^2} = - 27\\ \Leftrightarrow {x^2} + 2x - 195 = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = - 15\\ x = 13 \end{array} \right. \)
Vậy...
1.
Đặt\(\left\{{}\begin{matrix}u=\sqrt[3]{12-x}\\v=\sqrt[3]{14+x}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u^3=12-x\\v^3=14+x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}u^3+v^3=26\\u+v=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(u+v\right)\left(u^2-uv+v^2\right)=26\\u+v=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u^2-uv+v^2=13\\v=2-u\end{matrix}\right.\)
\(\Rightarrow u^2-u\left(2-u\right)+\left(2-u\right)^2=13\) \(\Leftrightarrow3u^2-6u-9=0\) \(\Rightarrow\left[{}\begin{matrix}u=3\Rightarrow v=-1\\u=-1\Rightarrow v=3\end{matrix}\right.\) Tìm x.
2.ĐK: \(-40\le x\le57\)
Đặt \(\left\{{}\begin{matrix}\sqrt[4]{57-x}=u\\\sqrt[4]{x+40}=v\end{matrix}\right.\) \(\left(u,v\ge0\right)\) \(\Rightarrow\left\{{}\begin{matrix}u^4=57-x\\v^4=x+40\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u+v=5\\u^4+v^4=97\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}u^2+v^2=25-2uv\\\left(u^2+v^2\right)^2-2u^2v^2=97\end{matrix}\right.\) \(\Rightarrow\left(25-2uv\right)^2-2u^2v^2=97\)
\(\Leftrightarrow2u^2v^2-100uv+528=0\) \(\Rightarrow\left[{}\begin{matrix}uv=44\\uv=6\end{matrix}\right.\) Kết hợp \(u+v=5\) giải 2 trường hợp.
3.
ĐK: \(-\sqrt{17}\le x\le\sqrt{17}\)
Đặt \(x+\sqrt{17-x^2}=t\) \(\Rightarrow\frac{t^2-17}{2}=x\sqrt{17-x^2}\)
\(PT\Leftrightarrow t+\frac{t^2-17}{2}=9\) \(\Leftrightarrow t^2+2t-35=0\Rightarrow\left[{}\begin{matrix}t=5\\t=-7\end{matrix}\right.\) Giải tiếp.
ĐK: \(x\ge5\)
Chuyển vế, bình phương ta đc:
\(\sqrt{5x^2+14x+9}=5\sqrt{\left(x^2-x-20\right)\left(x+1\right)}\)
Nhận xét:
Không tồn tại số \(\alpha,\beta\) để: \(2x^2-5x+2=\alpha\left(x^2-x-20\right)+\beta\left(x+1\right)\)
Ta có: \(\left(x^2-x-20\right)\left(x+1\right)=\left(x+4\right)\left(x-5\right)\left(x+1\right)=\left(x+4\right)\left(x^2-4x-5\right)\)
PT đc vt lại là: \(2\left(x^2-4x-5\right)+3\left(x+4\right)=5\sqrt{\left(x^2-4x-5\right)\left(x+4\right)}\)
Đặt: \(\left\{{}\begin{matrix}u=x^2-4x-5\\v=x+4\end{matrix}\right.\)
Khi đó PT trở thành:
\(2u+3v=5\sqrt{uv}\Leftrightarrow\left[{}\begin{matrix}u=v\\u=\frac{9}{4}v\end{matrix}\right.\)
Xét \(u=v\) ta có PT:
\(x^2-4x-5=x+4\Leftrightarrow x^2-5x+9=0\Leftrightarrow\left[{}\begin{matrix}x=\frac{5+\sqrt{61}}{2}\\x=\frac{5-\sqrt{61}}{2}\left(loại\right)\end{matrix}\right.\)
Xét \(u=\frac{9}{4}v\) ta có PT:
\(x^2-4x-5=\frac{9}{4}\left(x+4\right)\Leftrightarrow4x^2-25x-56=0\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-\frac{7}{4}\left(loại\right)\end{matrix}\right.\)
Vậy PT có 2 nghiệm là \(x=8;x=\frac{5+\sqrt{61}}{2}\)
`x=(\sqrt{x}+2004)(1-\sqrt{1-\sqrt{x}})^2(1>=x>=0)`
`<=>x=((\sqrt{x}+2004)(1-\sqrt{1-\sqrt{x}})^2(1+\sqrt{1-\sqrt{x}}))/(1+\sqrt{1-\sqrt{x}})`
`<=>x=(\sqrt{x}+2004)(1-\sqrt{1-\sqrt{x})(1-1+\sqrt{x}))/(1+\sqrt{1-\sqrt{x}})`
`<=>x=\sqrt{x}.(\sqrt{x}+2004)(1-\sqrt{1-\sqrt{x}))/(1+\sqrt{1-\sqrt{x}})`
`<=>\sqrt{x}((\sqrt{x}+2004)(1-\sqrt{1-\sqrt{x}))/(1+\sqrt{1-\sqrt{x}})-1)=0`
Có `x>=0`
`=>1-\sqrt{x}<=1`
`=>1+\sqrt{1-\sqrt{x}}<=2`
`=>1/(1+\sqrt{1-\sqrt{x}})>=1/2`
Mà `(\sqrt{x}+2004)>=2004`
`=>(\sqrt{x}+2004)(1-\sqrt{1-\sqrt{x})>=2004`
`=>(\sqrt{x}+2004)(1-\sqrt{1-\sqrt{x}))/(1+\sqrt{1-\sqrt{x}})>=1002>0`
`=>\sqrt{x}=0`
`=>x=0`
Vậy `S={0}`
ĐKXĐ: \(x\ge0\)
\(\Leftrightarrow x=\left(2004+\sqrt{x}\right)\left(\dfrac{\sqrt{x}}{1+\sqrt{1-\sqrt{x}}}\right)^2\)
\(\Leftrightarrow x=\dfrac{x\left(2004+\sqrt{x}\right)}{2-\sqrt{x}+2\sqrt{1-\sqrt{x}}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{2004+\sqrt{x}}{2-\sqrt{x}+2\sqrt{1-\sqrt{x}}}=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2004+\sqrt{x}=2-\sqrt{x}+2\sqrt{1-\sqrt{x}}\)
\(\Leftrightarrow1001+\sqrt{x}=\sqrt{1-\sqrt{x}}\)
\(VT\ge1001\) ; \(VP\le1\) nên (1) vô nghiệm
1/ ĐKXĐ:...
\(\Leftrightarrow\sqrt{x+1+2\sqrt{x+1}+1}+\sqrt{x+1-2\sqrt{x+1}+1}=\frac{x+5}{2}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x+1}+1\right)^2}+\sqrt{\left(1-\sqrt{x+1}\right)^2}=\frac{x+5}{2}\)
\(\Leftrightarrow\sqrt{x+1}+1+\left|1-\sqrt{x+1}\right|=\frac{x+5}{2}\)
Nếu \(0\ge x\ge-1\Rightarrow\left|1-\sqrt{x+1}\right|=1-\sqrt{x+1}\)
\(\Rightarrow2=\frac{x+5}{2}\Leftrightarrow x=-1\left(tm\right)\)
Nếu \(x>0\Rightarrow\left|1-\sqrt{x+1}\right|=\sqrt{x+1}-1\)
\(\Rightarrow2\sqrt{x+1}=\frac{x+5}{2}\Leftrightarrow16x+16=x^2+10x+25\)
\(\Leftrightarrow x^2-6x+9=0\Leftrightarrow x=3\left(tm\right)\)
Vậy...
Câu dưới tương tự
\(x^3=3+2\sqrt{2}+3-2\sqrt{2}+3\cdot\sqrt[3]{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)\\ \Leftrightarrow x^3=6+3x\sqrt[3]{1}\\ \Leftrightarrow x^3-3x=6\)
\(y^3=17+12\sqrt{2}+17-12\sqrt{2}+3\sqrt[3]{\left(17-12\sqrt{2}\right)\left(17+12\sqrt{2}\right)}\left(\sqrt[3]{17-12\sqrt{2}}+\sqrt[3]{17+12\sqrt{2}}\right)\\ \Leftrightarrow y^3=34+3x\sqrt[3]{1}\\ \Leftrightarrow y^3-3y=34\)
Thay vào P, ta được
\(P=x^3+y^3-3x-3y+1979\\ P=\left(x^3-3x\right)+\left(y^3-3y\right)+1979\\ P=6+34+1979=2019\)
\(x^3=6+3\sqrt[3]{\left(3+2\sqrt[]{2}\right)\left(3-2\sqrt[]{2}\right)}\left(\sqrt[3]{3+2\sqrt[]{2}}+\sqrt[3]{3-2\sqrt[]{2}}\right)\)
\(\Rightarrow x^3=6+3x\)
\(\Rightarrow x^3-3x=6\)
Tương tự:
\(y^3=34+3\sqrt[3]{\left(17+12\sqrt[]{2}\right)\left(17-12\sqrt[]{2}\right)}\left(\sqrt[3]{17+12\sqrt[]{2}}+\sqrt[3]{17-12\sqrt[]{2}}\right)\)
\(\Rightarrow y^3=34+3y\)
\(\Rightarrow y^3-3y=34\)
Do đó:
\(P=\left(x^3-3x\right)+\left(y^3-3y\right)+1979=6+34+1979=...\)
Nhận thấy \(x=0\) không phải nghiệm, pt tương đương:
\(x+\sqrt[3]{x-\dfrac{1}{x}}=2+\dfrac{1}{x}\)
\(\Leftrightarrow x-\dfrac{1}{x}+\sqrt[3]{x-\dfrac{1}{x}}-2=0\)
Đặt \(\sqrt[3]{x-\dfrac{1}{x}}=t\)
\(\Rightarrow t^3+t-2=0\Leftrightarrow\left(t-1\right)\left(t^2+t+2\right)=0\)
\(\Leftrightarrow t=1\Rightarrow x-\dfrac{1}{x}=1\)
\(\Leftrightarrow x^2-x-1=0\Leftrightarrow...\)
ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}u=v\\v=\sqrt{17-x^2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}u+v+uv=9\\u^2+v^2=17\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}uv=9-\left(u+v\right)\\\left(u+v\right)^2-2uv=17\end{matrix}\right.\)
\(\Rightarrow\left(u+v\right)^2+2\left(u+v\right)-35=0\)
\(\Rightarrow\left[{}\begin{matrix}u+v=5\Rightarrow uv=4\\u+v=-7\Rightarrow uv=16\end{matrix}\right.\)
\(\Rightarrow...\)