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6/ Đặt \(\hept{\begin{cases}\sqrt[4]{x}=a\\\sqrt[4]{2-x}=b\end{cases}}\)
\(\Rightarrow b^4+a^4=2\)
Từ đó ta có: a + b = 2
Ta có: \(a^4+b^2\ge\frac{\left(a^2+b^2\right)^2}{2}\ge\frac{\left(a+b\right)^4}{8}=\frac{16}{8}=2\)
Dấu = xảy ra khi a = b = 1
=> x = 1
Giải PT
a) \(3\sqrt{9x}+\sqrt{25x}-\sqrt{4x} = 3\)
\(\Leftrightarrow\) \(3.3\sqrt{x} +5\sqrt{x} - 2\sqrt{x} = 3 \)
\(\Leftrightarrow\) \(9\sqrt{x}+5\sqrt{x}-2\sqrt{x} = 3 \)
\(\Leftrightarrow\) \(12\sqrt{x} = 3\)
\(\Leftrightarrow\) \(\sqrt{x} = 4 \)
\(\Leftrightarrow\) \(\sqrt{x^2} = 4^2\)
\(\Leftrightarrow\) \(x=16\)
b) \(\sqrt{x^2-2x-1} - 3 =0\)
\(\Leftrightarrow\) \(\sqrt{(x-1)^2} -3=0\)
\(\Leftrightarrow\) \(|x-1|=3\)
* \(x-1=3\)
\(\Leftrightarrow\) \(x=4\)
* \(-x-1=3\)
\(\Leftrightarrow\) \(-x=4\)
\(\Leftrightarrow\) \(x=-4\)
c) \(\sqrt{4x^2+4x+1} - x = 3\)
<=> \(\sqrt{(2x+1)^2} = 3+x\)
<=> \(|2x+1|=3+x\)
* \(2x+1=3+x\)
<=> \(2x-x=3-1\)
<=> \(x=2\)
* \(-2x+1=3+x\)
<=> \(-2x-x = 3-1\)
<=> \(-3x=2\)
<=> \(x=\dfrac{-2}{3}\)
d) \(\sqrt{x-1} = x-3\)
<=> \(\sqrt{(x-1)^2} = (x-3)^2\)
<=> \(|x-1| = x^2-2.x.3+3^2\)
<=> \(|x-1| = x-6x+9\)
<=> \(|x-1| = -5x+9\)
* \(x-1= -5x+9\)
<=> \(x+5x = 9+1\)
<=> \(6x=10\)
<=> \(x= \dfrac{10}{6} =\dfrac{5}{3}\)
* \(-x-1 = -5x+9\)
<=> \(-x+5x = 9+1\)
<=> \(4x = 10\)
<=> \(x= \dfrac{10}{4} = \dfrac{5}{2}\)
Bài 1
a) √81a - √36a - √144a = 9√a - 6√a - 12√a = -9√a
b) √75 - √48 - √300 = 5√3 - 4√3 - 10√3 = -9√3
Bài 2
a) √2x-3 = 7
⇒ 2x-3 = 49 ⇔ 2x = 52 ⇔ x =26
c) √16x - √9x = 2
⇔ 4√x - 3√x = 2 ⇔ √x = 2 ⇔ x = 4
Bài 3
a) √(2-√5)2 = l 2-√5 l = √5-2
b) (a - 3)2 + (a - 9)
= a2 - 6a + 9 + a - 9 = a2 - 5a
c) A=\(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}:\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
=\(\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
=\(\left(\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\right)\)
=\(\left(\dfrac{-3\sqrt{x}-3}{x-9}\right).\left(\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\right)\)
=\(\left(\dfrac{-3\left(\sqrt{x}+1\right)}{x-9}\right).\left(\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\right)\)
=\(\dfrac{-3\sqrt{x}+9}{x-9}\)
Lời giải:
a) ĐKXĐ: $x\in\mathbb{R}$
\(\sqrt{x^2-2x+4}=2x-2\Leftrightarrow \left\{\begin{matrix} 2x-2\geq 0\\ x^2-2x+4=(2x-2)^2\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ 3x^2-6x=0\end{matrix}\right.\Rightarrow x=2\)
b) ĐKXĐ: $-x^2+x+4\geq 0$
\(\sqrt{-x^2+x+4}=x-3\Leftrightarrow \left\{\begin{matrix} x-3\geq 0\\ -x^2+x+4=(x-3)^2\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq 3\\ 2x^2-7x+5=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 3\\ (2x-5)(x-1)=0\end{matrix}\right.\) (không thỏa mãn)
Vậy pt vô nghiệm
c) ĐK: $x\leq 0$
PT $\Rightarrow x^2-2x=2-3x$
$\Leftrightarrow x^2+x-2=0$
$\Leftrightarrow (x+2)(x-1)=0$
Vì $x\leq 0$ nên $x=-2$ là nghiệm duy nhất của pt.
d) ĐK: $x\geq 3$
PT $\Leftrightarrow \sqrt{x-3}-2\sqrt{(x-3)(x+3)}=0$
$\Leftrightarrow \sqrt{x-3}(1-2\sqrt{x+3})=0$
$\Rightarrow \sqrt{x-3}=0$ hoặc $1-2\sqrt{x+3}=0$
Nếu $\sqrt{x-3}=0\Rightarrow x=3$ (thỏa mãn)
Nếu $1-2\sqrt{x+3}=0\Rightarrow x=\frac{-11}{4}< 3$ (không thỏa ĐKXĐ)
Vậy ...........
Bài 1:
a) Để căn thức \(\sqrt{\frac{2}{9-x}}\) có nghĩa thì \(\left\{{}\begin{matrix}\frac{2}{9-x}\ge0\\9-x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9-x>0\\x\ne9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 9\\x\ne9\end{matrix}\right.\Leftrightarrow x< 9\)
b) Ta có: \(x^2+2x+1\)
\(=\left(x+1\right)^2\)
mà \(\left(x+1\right)^2\ge0\forall x\)
nên \(x^2+2x+1\ge0\forall x\)
Do đó: Căn thức \(\sqrt{x^2+2x+1}\) xác được với mọi x
c) Để căn thức \(\sqrt{x^2-4x}\) có nghĩa thì \(x^2-4x\ge0\)
\(\Leftrightarrow x\left(x-4\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x-4\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x-4< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x\ge4\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x< 4\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge4\\x< 0\end{matrix}\right.\)
Bài 3:
a) Ta có: \(\sqrt{\left(3-\sqrt{10}\right)^2}\)
\(=\left|3-\sqrt{10}\right|\)
\(=\sqrt{10}-3\)(Vì \(3< \sqrt{10}\))
b) Ta có: \(\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(=\left|\sqrt{5}-2\right|\)
\(=\sqrt{5}-2\)(Vì \(\sqrt{5}>2\))
c) Ta có: \(3x-\sqrt{x^2-2x+1}\)
\(=3x-\sqrt{\left(x-1\right)^2}\)
\(=3x-\left|x-1\right|\)
\(=\left[{}\begin{matrix}3x-\left(x-1\right)\left(x\ge1\right)\\3x-\left(1-x\right)\left(x< 1\right)\end{matrix}\right.\)
\(=\left[{}\begin{matrix}3x-x+1\\3x-1+x\end{matrix}\right.=\left[{}\begin{matrix}2x+1\\4x-1\end{matrix}\right.\)
a: =>\(x^2\cdot2\sqrt{2}+x\left(2+2\sqrt{2}\right)+4=0\)
\(\text{Δ}=\left(2\sqrt{2}+2\right)^2-4\cdot2\sqrt{2}\cdot4=12-24\sqrt{2}< 0\)
=>PTVN
b:
\(\Leftrightarrow2x^2+2x+\sqrt{3}-x^2+2\sqrt{3}x+\sqrt{3}=0\)
=>\(x^2+x\left(2\sqrt{3}+2\right)+2\sqrt{3}=0\)
\(\text{Δ}=\left(2\sqrt{3}+2\right)^2-4\cdot2\sqrt{3}=16>0\)
PT có hai nghiệm là;
\(\left\{{}\begin{matrix}x_1=\dfrac{-2\sqrt{3}-2-4}{2}=-\sqrt{3}-3\\x=\dfrac{-2\sqrt{3}-2+4}{2}=-\sqrt{3}+1\end{matrix}\right.\)