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1.ĐK: \(x\ge\dfrac{1}{4}\)
bpt\(\Leftrightarrow5x+1+4x-1-2\sqrt{20x^2-x-1}< 9x\)
\(\Leftrightarrow2\sqrt{20x^2-x-1}>0\)
\(\Leftrightarrow20x^2-x-1>0\)
\(\Leftrightarrow\left[{}\begin{matrix}x< \dfrac{-1}{5}\\x>\dfrac{1}{4}\end{matrix}\right.\)
2.ĐK: \(-2\le x\le\dfrac{5}{2}\)
bpt\(\Leftrightarrow x+2+3-x-2\sqrt{-x^2+x+6}< 5-2x\)
\(\Leftrightarrow2x< 2\sqrt{-x^2+x+6}\)
\(\Leftrightarrow x^2< -x^2+x+6\)
\(\Leftrightarrow-2x^2+x+6>0\)
\(\Leftrightarrow\dfrac{-3}{2}< x< 2\)
3. ĐK: \(\left\{{}\begin{matrix}12+x-x^2\ge0\\x\ne11\\x\ne\dfrac{9}{2}\end{matrix}\right.\)
.bpt\(\Leftrightarrow\sqrt{12+x-x^2}\left(\dfrac{1}{x-11}-\dfrac{1}{2x-9}\right)\ge0\)
\(\Leftrightarrow\sqrt{-x^2+x+12}.\dfrac{x+2}{\left(x-11\right)\left(2x-9\right)}\ge0\)
\(\Rightarrow\dfrac{x+2}{\left(x-11\right)\left(2x-9\right)}\ge0\)
\(\Leftrightarrow\dfrac{x+2}{2x^2-31x+99}\ge0\)
*Xét TH1: \(\left\{{}\begin{matrix}x+2\ge0\\2x^2-31x+99>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\\left[{}\begin{matrix}x< \dfrac{9}{2}\\x>11\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}-2\le x< \dfrac{9}{2}\\x>11\end{matrix}\right.\)
*Xét TH2: \(\left\{{}\begin{matrix}x+2\le0\\2x^2-31x+99< 0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le-2\\\dfrac{9}{2}< x< 11\end{matrix}\right.\)\(\Rightarrow\dfrac{9}{2}< x< 11\)
a/ \(x\le8\)
\(\Leftrightarrow x^2+x+12=\left(8-x\right)^2\)
\(\Leftrightarrow x^2+x+12=x^2-16x+64\)
\(\Leftrightarrow17x=52\Rightarrow x=\frac{52}{17}\)
b/ \(x\le4\)
\(\Leftrightarrow x^2+3x-1=\left(4-x\right)^2\)
\(\Leftrightarrow x^2+3x-1=x^2-8x+16\)
\(\Leftrightarrow11x=17\Rightarrow x=\frac{17}{11}\)
c/ \(\left\{{}\begin{matrix}x^2-3x\ge0\\2x-1\ge0\end{matrix}\right.\) \(\Rightarrow x\ge3\)
\(x^2-3x=2x-1\)
\(\Leftrightarrow x^2-5x+1=0\Rightarrow\left[{}\begin{matrix}x=\frac{5+\sqrt{21}}{2}\\x=\frac{5-\sqrt{21}}{2}\left(l\right)\end{matrix}\right.\)
d/ \(2-x\ge0\Rightarrow x\le2\)
\(x^2+2x+4=2-x\)
\(\Leftrightarrow x^2+3x+2=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)
e/ \(2x^2-x\ge0\Rightarrow\left[{}\begin{matrix}x\le0\\x\ge\frac{1}{2}\end{matrix}\right.\)
\(x^2+2x+4=2x^2-x\)
\(\Leftrightarrow x^2-3x-4=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)
f/ \(x\ge2\)
\(2x-1=\left(x-2\right)^2\)
\(\Leftrightarrow x^2-6x+5=0\Rightarrow\left[{}\begin{matrix}x=1\left(l\right)\\x=5\end{matrix}\right.\)
ĐKXĐ: x – 6 ≥ 0 ⇔ x > 6. Bình phương hai vế thì được 5x + 6 = (x – 6)2 ⇔ x2 = 2 (loại), x2 = 15 (nhận).
b) ĐKXĐ: – 2 ≤ x ≤ 3. Bình phương hai vế thì được 3 - x = x + 3 + 2
⇔ -2x = 2.
Điều kiện x ≤ 0. Bình phương tiếp ta được:
x2 = x + 2 => x1 = -1 (nhận); x2 = 2 (loại).
Kết luận: Tập nghiệm S {-1}.
c) ĐKXĐ: x ≥ -2.
=> 2x2 + 5 = (x + 2)2 => x2 - 4x + 1 = 0
=> x1 =2 – (nhận), x2 = 2 +
(nhận).
d) ĐK: x ≥ .
=> 4x2 + 2x + 10 = (3x + 1)2 => x1 = (loại), x2 = 1 (nhận).
\(TXĐ:D=R\)
\(pt\Leftrightarrow\sqrt{\left(2x-1\right)^2+1^2}+\sqrt{\left(\sqrt{3}x+1\right)^2+\left(x+1\right)^2}\)
\(+\sqrt{\left(\sqrt{3}x-1\right)^2+\left(x+1\right)^2}=3\sqrt{2}\left(1\right)\)
Chọn \(\hept{\begin{cases}\overrightarrow{u}=\left(1;1-2x\right)\\\overrightarrow{v}=\left(\sqrt{3}x+1;x+1\right)\\\overrightarrow{w}=\left(1-\sqrt{3}x;x+1\right)\end{cases}}\)\(\Rightarrow\overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}=\left(3;3\right)\)
\(\Rightarrow\left|\overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}\right|=3\sqrt{2}\)(2)
Ta có: \(\left|\overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}\right|\le\left|\overrightarrow{u}\right|+\left|\overrightarrow{v}\right|+\left|\overrightarrow{w}\right|\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)^2+1^2}+\sqrt{\left(\sqrt{3}x+1\right)^2+\left(x+1\right)^2}\)
\(+\sqrt{\left(\sqrt{3}x-1\right)^2+\left(x+1\right)^2}\ge3\sqrt{2}\)
Dấu "=" xảy ra khi \(\overrightarrow{u};\overrightarrow{v};\overrightarrow{w}\)cùng hướng
Từ (1) và (2) suy ra \(\overrightarrow{u};\overrightarrow{v};\overrightarrow{w}\)cùng hướng
\(\Leftrightarrow\exists k,l>0\hept{\begin{cases}\overrightarrow{v}=k.\overrightarrow{u}\\\overrightarrow{v}=l.\overrightarrow{w}\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{3}x+1=k.1;x+1=k\left(1-2x\right)\\\sqrt{3}x+1=l\left(1-\sqrt{3}x\right);x+1=l\left(x+1\right)\end{cases}}\)
Vậy x = 0
a)
Pt\(\Leftrightarrow\left\{{}\begin{matrix}3x-4=\left(x-3\right)^2\\x-3\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x-4=x^2-6x+9\\x\ge3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-9x+13=0\\x\ge3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x_1=\dfrac{9+\sqrt{29}}{2}\\x_2=\dfrac{9-\sqrt{29}}{2}\end{matrix}\right.\\x\ge3\end{matrix}\right.\)\(\Leftrightarrow x=\dfrac{9+\sqrt{29}}{2}\)
Vậy \(x=\dfrac{9+\sqrt{29}}{2}\) là nghiệm của phương trình.
b) Pt \(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x+3=\left(2x-1\right)^2\\2x-1\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3x^2-2x-2=0\\x\ge\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x_1=\dfrac{1+\sqrt{7}}{3}\\x_2=\dfrac{1-\sqrt{7}}{3}\end{matrix}\right.\\x\ge\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow x=\dfrac{1+\sqrt{7}}{3}\)
Vậy phương trình có duy nhất nghiệm là: \(x=\dfrac{1+\sqrt{7}}{3}\)
a) \(x+1+\dfrac{2}{x+3}=\dfrac{x+5}{x+3}\)
\(\Leftrightarrow x+\dfrac{x+5}{x+3}=\dfrac{x+5}{x+3}\)
\(\Leftrightarrow x=0\)
b) \(2x+\dfrac{3}{x-1}=\dfrac{3x}{x-1}\)
\(\Leftrightarrow x+x+\dfrac{3}{x-1}=\dfrac{3x}{x-1}\)
\(\Leftrightarrow x+\dfrac{x\left(x-1\right)+3}{x-1}=\dfrac{3x}{x-1}\)
\(\Leftrightarrow x+\dfrac{x^2-x+3}{x-1}=\dfrac{3x}{x-1}\)
\(\Leftrightarrow\dfrac{x^2-x+3}{x-1}=\dfrac{3x}{x-1}-x\)
\(\Leftrightarrow\dfrac{x^2-x+3}{x-1}=\dfrac{3x-x\left(x-1\right)}{x-1}\)
\(\Leftrightarrow\dfrac{x^2-x+3}{x-1}=\dfrac{3x-x^2+x}{x-1}\)
\(\Leftrightarrow x^2-x+3=3x-x^2+x\) ( điều kiện \(x\ne1\) )
\(\Leftrightarrow2x^2-5x+3=0\)
\(\Delta=b^2-4ac\)
\(\Delta=1\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{3}{2}\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=1\left(loại\right)\end{matrix}\right.\)
Vậy \(x=\dfrac{3}{2}\)
c) \(\dfrac{x^2-4x-2}{\sqrt{x-2}}=\sqrt{x-2}\)
\(\Leftrightarrow x^2-4x-2=\sqrt{\left(x-2\right)^2}\) ( điều kiện \(x>2\) )
\(\Leftrightarrow x^2-4x-2=x-2\)
\(\Leftrightarrow x^2-5x=0\)
\(\Leftrightarrow x\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=5\end{matrix}\right.\)
Vậy \(x=5\)
d) \(\dfrac{2x^2-x-3}{\sqrt{2x-3}}=\sqrt{2x-3}\)
\(\Leftrightarrow2x^2-x-3=\sqrt{\left(2x-3\right)^2}\) ( điều kiện \(x>\dfrac{3}{2}\) )
\(\Leftrightarrow2x^2-x-3=2x-3\)
\(\Leftrightarrow2x^2-3x=0\)
\(\Leftrightarrow x\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=\dfrac{3}{2}\left(loại\right)\end{matrix}\right.\)
Vậy phương trình vô nghiệm