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b) Ta có: \(x+\sqrt{3}=2\Leftrightarrow x-2=-\sqrt{3}\Leftrightarrow\left(x-2\right)^2=3\Leftrightarrow x^2-4x+1=0\)
\(B=x^5-3x^4-3x^3+6x^2-20x+2021\)
\(B=\left(x^5-4x^4+x^3\right)+\left(x^4-4x^3+x^2\right)+\left(5x^2-20x+5\right)+2016\)
\(B=x^3\left(x^2-4x+1\right)+x^2\left(x^2-4x+1\right)+5\left(x^2-4x+1\right)+2016\)
Thế \(x^2-4x+1=0\)\(\Rightarrow B=2016.\)
a) \(\left\{{}\begin{matrix}x\ge0\\-\sqrt{x+7}< 0\\-5x-4\ne0\\-3x+2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x+7>0\\-5x\ne4\\-3x\ne-2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x>-7\\x\ne\frac{-4}{5}\\x\ne\frac{2}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne\frac{2}{3}\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x\ge0\\x+4\ne0\\x-2\ge0\\-2x-10\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne-4\\x\ge2\\-2x\ne10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ne-5\end{matrix}\right.\Leftrightarrow x\ge2\)
c) \(\left\{{}\begin{matrix}x\ge0\\-x-3\ne0\\2x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne-3\\x\ne-\frac{3}{2}\end{matrix}\right.\Leftrightarrow x\ge0\)
d) \(\left\{{}\begin{matrix}2x-7\ge0\\x\ge0\\3x-4\ne0\\x-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{7}{2}\\x\ge0\\x\ne\frac{4}{3}\\x\ne3\end{matrix}\right.\Leftrightarrow x\ge\frac{7}{2}\)
ai nay dung kinh nghiem la chinh
cau a)
ta thay \(10+6\sqrt{3}=\left(1+\sqrt{3}\right)^3\)
\(6+2\sqrt{5}=\left(1+\sqrt{5}\right)^2\)
khi do \(x=\frac{\sqrt[3]{\left(\sqrt{3}+1\right)^3}\left(\sqrt{3}-1\right)}{\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{5}}\)
\(x=\frac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{1+\sqrt{5}-\sqrt{5}}\)
\(x=\frac{3-1}{1}=2\)
suy ra
x^3-4x+1=1
A=1^2018
A=1
b)
ta thay
\(7+5\sqrt{2}=\left(1+\sqrt{2}\right)^3\)
khi do
\(x=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\frac{1}{\sqrt[3]{\left(1+\sqrt{2}\right)^3}}\)
\(x=1+\sqrt{2}-\frac{1}{1+\sqrt{2}}=\frac{\left(1+\sqrt{2}\right)^2-1}{1+\sqrt{2}}=\frac{2+2\sqrt{2}}{1+\sqrt{2}}\)
x=2
thay vao
x^3+3x-14=0
B=0^2018
B=0
Bài 1 : Ta có :
\(A=\sqrt{3x+\sqrt{6x-1}}+\sqrt{3x-\sqrt{6x-1}}\)
\(A\sqrt{2}=\sqrt{6x+2\sqrt{6x-1}}+\sqrt{6x-2\sqrt{6x-1}}\)
\(=\sqrt{6x-1+2\sqrt{6x-1}+1}+\sqrt{6x-1-2\sqrt{6x-1}+1}\)
\(=\sqrt{\left(\sqrt{6x-1}+1\right)^2}+\sqrt{\left(\sqrt{6x-1}-1\right)^2}\)
\(=\left|\sqrt{6x-1}+1\right|+\left|\sqrt{6x-1}-1\right|\)
\(=\sqrt{6x-1}+1+\sqrt{6x-1}-1\)
\(=2\sqrt{6x-1}\)
\(\Rightarrow A=\sqrt{2}\left(\sqrt{6x-1}\right)\)
Thay \(x=4+\sqrt{10}\) vào A ta được :
\(A=\sqrt{2}.\sqrt{6\left(4+\sqrt{10}\right)-1}=\sqrt{2}.\sqrt{24+6\sqrt{10}-1}\)
\(=\sqrt{2}.\sqrt{23+6\sqrt{10}}=\sqrt{46+12\sqrt{10}}\)
\(=\sqrt{36+12\sqrt{10}+10}=\sqrt{\left(6+\sqrt{10}\right)^2}=6+\sqrt{10}\)
Vậy \(A=6+\sqrt{10}\) tại \(x=4+\sqrt{10}\)
Em xin phép làm bài EZ nhất :)
4,ĐK :\(\forall x\in R\)
Đặt \(x^2+x+2=t\) (\(t\ge\dfrac{7}{4}\))
\(PT\Leftrightarrow\sqrt{t+5}+\sqrt{t}=\sqrt{3t+13}\)
\(\Leftrightarrow2t+5+2\sqrt{t\left(t+5\right)}=3t+13\)
\(\Leftrightarrow t+8=2\sqrt{t^2+5t}\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge-8\\\left(t+8\right)^2=4t^2+20t\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\3t^2+4t-64=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left(t-4\right)\left(3t+16\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left[{}\begin{matrix}t=4\left(tm\right)\\t=-\dfrac{16}{3}\left(l\right)\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow x^2+x+2=4\)\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy ....
Đk: x = \(5+2\sqrt{7}\)> 5
Đặt A = \(\sqrt{3x+\sqrt{6x-1}}-\sqrt{3x-\sqrt{6x-1}}\)
A2 = \(\left(\sqrt{3x+\sqrt{6x-1}}-\sqrt{3x-\sqrt{6x-1}}\right)^2\)
A2 = \(3x+\sqrt{6x-1}+3x-\sqrt{6x-1}-2\sqrt{\left(3x+\sqrt{6x-1}\right)\left(3x-\sqrt{6x-1}\right)}\)
A2 = \(6x-2\sqrt{9x^2-6x+1}\)
A2 = \(6x-2\sqrt{\left(3x-1\right)^2}\) (vì x > \(\frac{1}{3}\))
A2 = \(6x-2\left(3x-1\right)\)
A2 = \(6x-6x+2\)
A2 = 2
=> A = \(\sqrt{2}\)
Vậy ....
Đặt: \(A=\sqrt{3x+\sqrt{6x-1}}-\sqrt{3x-\sqrt{6x-1}}\)
=> \(A^2=3x+\sqrt{6x-1}+3x-\sqrt{6x-1}-2\sqrt{\left(3x+\sqrt{6x-1}\right)\left(3x-\sqrt{6x-1}\right)}\)
=> \(A^2=6x-2\sqrt{9x^2-6x+1}\)
=> \(A^2=6x-2\sqrt{\left(3x-1\right)^2}\)
Mà: \(x=5+2\sqrt{7}\Rightarrow x>\frac{1}{3}\Rightarrow3x>1\Rightarrow3x-1>0\)
=> \(A^2=6x-2\left(3x-1\right)\)
=> \(A^2=6x-6x+2=2\)
Mà: \(\sqrt{3x+\sqrt{6x-1}}>\sqrt{3x-\sqrt{6x-1}}\Rightarrow A>0\)
=> \(A=\sqrt{2}\)
VẬY \(A=\sqrt{2}\)