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a. ĐKXĐ: \(x\ge-1\)
\(y=\sqrt{x^3+1+2\sqrt{x^3+1}+1}+\sqrt{x^3+1-2\sqrt{x^3+1}+1}\)
\(=\sqrt{\left(\sqrt{x^3+1}+1\right)^2}+\sqrt{\left(\sqrt{x^3+1}-1\right)^2}\)
\(=\left|\sqrt{x^3+1}+1\right|+\left|1-\sqrt{x^3+1}\right|\ge\left|\sqrt{x^3+1}+1+1-\sqrt{x^3+1}\right|=2\)
b.
\(f\left(x\right)=\dfrac{x-1}{2}+\dfrac{2}{x-1}+\dfrac{1}{2}\ge2\sqrt{\dfrac{2\left(x-1\right)}{2\left(x-1\right)}}+\dfrac{1}{2}=\dfrac{5}{2}\)
c.
\(y=\dfrac{x-2018+1}{\sqrt{x-2018}}=\sqrt{x-2018}+\dfrac{1}{\sqrt{x-2018}}\ge2\sqrt{\dfrac{\sqrt{x-2018}}{\sqrt{x-2018}}}=2\)
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a, ĐKXĐ: ...
\(\sqrt{3x^2-2x+6}+3-2x=0\)
\(\Leftrightarrow\sqrt{3x^2-2x+6}=2x-3\)
\(\Leftrightarrow3x^2-2x+6=4x^2-12x+9\)
\(\Leftrightarrow4x^2-10x+3=0\)
.....
b, ĐKXĐ: ...
\(\sqrt{x+1}+\sqrt{x-1}=4\\ \Leftrightarrow x+1+x-1+2\sqrt{\left(x+1\right)\left(x-1\right)}=16\\ \Leftrightarrow2\sqrt{x^2-1}=16-2x\\ \Leftrightarrow\sqrt{x^2-1}=8-x\\ \Leftrightarrow x^2-1=64-16x+x^2\\ \Leftrightarrow65-16x=0\\ \Leftrightarrow x=\dfrac{65}{16}\)
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`\sqrt{x+3}+\sqrt{6-x}=\sqrt{(x+3)(6-x)}+3(-3<=x<=6)`
`<=>x+3+6-x=(x+3)(6-x)+9+6\sqrt{(x+3)(6-x)}`
`<=>9=9+(x+3)(6-x)+6\sqrt{(x+3)(6-x)}`
`<=>(x+3)(6-x)+6\sqrt{(x+3)(6-x)}=0`
`<=>\sqrt{(x+3)(6-x)}(\sqrt{(x+3)(6-x)}+6)=0`
`<=>\sqrt{(x+3)(6-x)}=0`
`<=>x=-3\or\x=6`
Vậy `S={-3,6}`
![](https://rs.olm.vn/images/avt/0.png?1311)
b) Đặt \(\sqrt{x^2-6x+6}=a\left(a\ge0\right)\)
\(\Rightarrow a^2+3-4a=0\)
=> (a - 3).(a - 1) = 0
=> \(\left[{}\begin{matrix}a=3\\a=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-6x+6}=3\\\sqrt{x^2-6x+6}=1\end{matrix}\right.\)
Bình phương lên giải tiếp nhé!
c) Tương tư câu b nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(\Leftrightarrow\left(x^2-6x-7\right)^2-\left(3x^2-12x-9\right)^2=0\)
\(\Leftrightarrow\left(3x^2-12x-9-x^2+6x+7\right)\left(3x^2-12x-9+x^2-6x-7\right)=0\)
\(\Leftrightarrow\left(2x^2-6x-2\right)\left(4x^2-18x-16\right)=0\)
\(\Leftrightarrow\left(x^2-3x-1\right)\left(2x^2-9x-8\right)=0\)
hay \(x\in\left\{\dfrac{3+\sqrt{13}}{2};\dfrac{3-\sqrt{13}}{2};\dfrac{9+\sqrt{145}}{4};\dfrac{9-\sqrt{145}}{4}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a.
ĐKXĐ: \(1\le x\le7\)
\(\Leftrightarrow x-1-2\sqrt{x-1}+2\sqrt{7-x}-\sqrt{\left(x-1\right)\left(7-x\right)}=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-2\right)-\sqrt{7-x}\left(\sqrt{x-1}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-\sqrt{7-x}\right)\left(\sqrt{x-1}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=\sqrt{7-x}\\\sqrt{x-1}=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=7-x\\x-1=4\end{matrix}\right.\)
\(\Leftrightarrow...\)
b. ĐKXĐ: ...
Biến đổi pt đầu:
\(x\left(y-1\right)-\left(y-1\right)^2=\sqrt{y-1}-\sqrt{x}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\ge0\\\sqrt{y-1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow a^2b^2-b^4=b-a\)
\(\Leftrightarrow b^2\left(a+b\right)\left(a-b\right)+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(b^2\left(a+b\right)+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow\sqrt{x}=\sqrt{y-1}\Rightarrow y=x+1\)
Thế vào pt dưới:
\(3\sqrt{5-x}+3\sqrt{5x-4}=2x+7\)
\(\Leftrightarrow3\left(x-\sqrt{5x-4}\right)+7-x-3\sqrt{5-x}=0\)
\(\Leftrightarrow\dfrac{3\left(x^2-5x+4\right)}{x+\sqrt{5x-4}}+\dfrac{x^2-5x+4}{7-x+3\sqrt{5-x}}=0\)
\(\Leftrightarrow\left(x^2-5x+4\right)\left(\dfrac{3}{x+\sqrt{5x-4}}+\dfrac{1}{7-x+3\sqrt{5-x}}\right)=0\)
\(\Leftrightarrow...\)
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\(\sqrt{x-5}+\sqrt{x-3}-2\sqrt{x^2+2x-8}+4=0\left(1\right)\\ \Leftrightarrow\sqrt{x-5}+\sqrt{x-3}+4=2\sqrt{x^2+2x-8}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x-5\ge0\\x-3\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge5\\x\ge3\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{x-5}+\sqrt{x-3}+4=2\sqrt{x^2+2x-8}\\ \Leftrightarrow\left(\sqrt{x-5}\right)^2+\left(\sqrt{x-3}\right)^2+4^2=\left(2\sqrt{x^2+2x-8}\right)^2\\ \Leftrightarrow x-5+x-3+16=4.\left(x^2+2x-8\right)\\ \Leftrightarrow x-5+x-3+16=4x^2+8x-32\\ \Leftrightarrow x-5+x-3+16-4x^2-8x+32=0\\ \Leftrightarrow-4x^2-6x+40=0\)
Ta có: \(\Delta=b^2-4ac=\left(-6\right)^2-4.\left(-4\right).40=676\)
\(\Rightarrow\left[{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-\left(-6\right)+\sqrt{676}}{2.\left(-4\right)}=-4\left(nhận\right)\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-\left(-6\right)-\sqrt{676}}{2.\left(-4\right)}=\dfrac{5}{2}=2,5\left(loại\right)\end{matrix}\right.\)
Vậy phương trình (1) không có nghiệm thỏa mãn.
Mình nhầm chỗ \(x_1=-4\) là loại mà mình nhấn nhầm là nhận!