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1) Ta có: \(\sqrt{21-x}+1=x\)
\(\Leftrightarrow21-x=\left(x-1\right)^2\)
\(\Leftrightarrow x^2-2x+1-21+x=0\)
\(\Leftrightarrow x^2-3x-20=0\)
\(\text{Δ}=\left(-3\right)^2-4\cdot1\cdot\left(-20\right)=9+80=89\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{3+\sqrt{89}}{2}\\x_2=\dfrac{3-\sqrt{89}}{2}\end{matrix}\right.\)
1)\(\sqrt{21-x}+1=x\)
\(\Leftrightarrow21-x=\left(x-1\right)^2\)
\(\Leftrightarrow21-x=x^2-2x+1\)
\(\Leftrightarrow x^2-x-20=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-4\end{matrix}\right.\)
2)\(\sqrt{8-x}+2=x\)
\(\Leftrightarrow8-x=\left(x-2\right)^2\)
\(\Leftrightarrow8-x=x^2-4x+4\)
\(\Leftrightarrow x^2-3x-4=0\Leftrightarrow\left(x-4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)
\(pt\Rightarrow\sqrt{x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}}}=2-x\\ \Leftrightarrow x+\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}=\left(2-x\right)^2\\ \Leftrightarrow x+\dfrac{1}{4}+\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{4}=\left(x-2\right)^2\\ \Leftrightarrow\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2=\left(x-2\right)^2\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}=x-2\left(1\right)\\\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}=2-x\left(2\right)\end{matrix}\right.\)
Tới đây giải \(pt\left(1\right)\left(2\right)\), sau đó thế lại vào cái pt ban đầu, từ đó nhận hoặc loại nghiệm tìm được
( Không giải được 2 cái kia thì cmt nhắc nha )
ĐKXĐ: \(x\ge-\dfrac{1}{4}\)
Ta có: \(x+\sqrt{x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}}}=2\)
\(\Leftrightarrow x+\sqrt{x+\dfrac{1}{4}+2\cdot\sqrt{x+\dfrac{1}{4}}\cdot\dfrac{1}{2}+\dfrac{1}{4}}=2\)
\(\Leftrightarrow x+\sqrt{\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2}=2\)
\(\Leftrightarrow x+\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}=2\)
\(\Leftrightarrow x+\dfrac{1}{4}+2\cdot\sqrt{x+\dfrac{1}{4}}\cdot\dfrac{1}{2}+\dfrac{1}{4}=2\)
\(\Leftrightarrow\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2=2\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}=-2\\\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+\dfrac{1}{4}}=-\dfrac{5}{2}\left(loại\right)\\\sqrt{x+\dfrac{1}{4}}=\dfrac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow x+\dfrac{1}{4}=\dfrac{9}{4}\)
hay x=2(thỏa ĐK)
Vậy: x=2
Bài 2
b, `\sqrt{3x^2}=x+2` ĐKXĐ : `x>=0`
`=>(\sqrt{3x^2})^2=(x+2)^2`
`=>3x^2=x^2+4x+4`
`=>3x^2-x^2-4x-4=0`
`=>2x^2-4x-4=0`
`=>x^2-2x-2=0`
`=>(x^2-2x+1)-3=0`
`=>(x-1)^2=3`
`=>(x-1)^2=(\pm \sqrt{3})^2`
`=>` $\left[\begin{matrix} x-1=\sqrt{3}\\ x-1=-\sqrt{3}\end{matrix}\right.$
`=>` $\left[\begin{matrix} x=1+\sqrt{3}\\ x=1-\sqrt{3}\end{matrix}\right.$
Vậy `S={1+\sqrt{3};1-\sqrt{3}}`
Lời giải:
ĐKXĐ: $x\geq 5$
$2x^2-8x-6=2\sqrt{x-5}\leq (x-5)+1$ theo BĐT Cô-si
$\Leftrightarrow 2x^2-9x-2\leq 0$
$\Leftrightarrow 2x(x-5)+(x-2)\leq 0$
Điều này vô lý do $2x(x-5)\geq 0; x-2\geq 3>0$ với mọi $x\geq 5$
Vậy pt vô nghiệm nên không có đáp án nào đúng.
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
a) ĐKXĐ: \(\left[{}\begin{matrix}x\ge1\\0>x\ge-1\end{matrix}\right.\). Để pt có nghiệm => x>0=> \(x\ge1\) pt<=> \(x-\sqrt{1-\dfrac{1}{x}}=\sqrt{x-\dfrac{1}{x}}.Bìnhphương2vetaco\left(x-\sqrt{1-\dfrac{1}{x}}\right)^2=x-\dfrac{1}{x}\)\(\Leftrightarrow x^2+1-\dfrac{1}{x}-2x\sqrt{1-\dfrac{1}{x}}=x-\dfrac{1}{x}\Leftrightarrow x^2-x+1=2\sqrt{x^2-x}\Leftrightarrow\left(\sqrt{x^2-x}-1\right)^2=0\Leftrightarrow x^2-x=1\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{5}{4}\)
b) ĐKXĐ\(0\le x\le1\) pt \(\Leftrightarrow\left(\sqrt{x^2+x}+\sqrt{x-x^2}\right)^2=\left(x+1\right)^2\Leftrightarrow2x+2x.\sqrt{1-x^2}=x^2+2x+1\Leftrightarrow x^2-2x\sqrt{1-x^2}+1-x^2+x^2=0\Leftrightarrow\left(x-\sqrt{1-x^2}\right)^2+x^2=0\)
Chứng minh:\(\left(\dfrac{x\sqrt{x}-3\sqrt{3x}}{x-27}+\dfrac{x^3-x^2+x}{3\sqrt{3x}+x\sqrt{x}}\right)\div\dfrac{x^2+1}{\sqrt{x}+3\sqrt{3}}\)= 1
Biến đổi vế trái ta được:
VT=\(\left(\dfrac{(x\sqrt{x}-3\sqrt{3x)}\times\left(3\sqrt{3x}+x\sqrt{x}\right)}{(x-27)\times\left(3\sqrt{3x}+x\sqrt{x}\right)}+\dfrac{\left(x-27\right)\times(x^3-x^2+x)}{\left(x-27\right)\times\left(3\sqrt{3x}+x\sqrt{x}\right)}\right)\div\dfrac{x^2+1}{\sqrt{x}+3\sqrt{3}}\)
=\(\left(\dfrac{x^3-27x^2}{\left(x-27\right)\times\left(3\sqrt{3x}+x\sqrt{x}\right)}+\dfrac{\left(x-27\right)\times\left(x^3-x^2+x\right)}{\left(x-27\right)\times\left(3\sqrt{3x}+x\sqrt{x}\right)}\right)\div\dfrac{x^2+1}{\sqrt{x}+3\sqrt{3}}\)
=\(\left(\dfrac{x^2}{3\sqrt{3x}+x\sqrt{x}}+\dfrac{x^3-x^2+x}{3\sqrt{3x}+x\sqrt{x}}\right)\div\dfrac{x^2+1}{\sqrt{x}+3\sqrt{3}}\)
=\(\dfrac{x^3+x}{3\sqrt{3x}+x\sqrt{x}}\div\dfrac{x^2+1}{\sqrt{x}+3\sqrt{3}}\)
=\(\dfrac{(x^3+x)\times\left(\sqrt{x}+3\sqrt{3}\right)}{(3\sqrt{3x}+x\sqrt{x})\times(x^2+1)}\)
=\(\dfrac{x\times\left(x^2+1\right)\times\left(\sqrt{x}+3\sqrt{3}\right)}{\left(x^2+1\right)\times\left(3\sqrt{3x}+x\sqrt{x}\right)}\)
=\(\dfrac{x\times\left(\sqrt{x}+3\sqrt{3}\right)}{\sqrt{x}\times\left(x+3\sqrt{3}\right)}\)
=\(\dfrac{x\times\left(\sqrt{x}+3\sqrt{3}\right)}{x\times\left(\sqrt{x}+3\sqrt{3}\right)}\)= 1 =VP
Vậy đẳng thức được chứng minh