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HT
15 tháng 12 2016
1) ĐK: \(x\ge\frac{3}{2}\)
pt \(\Leftrightarrow\frac{2x-2-\left(6x-9\right)}{\sqrt{2x-2}+\sqrt{6x-9}}=16x^2-28x-20x+35\)
\(\Leftrightarrow\frac{-4x+7}{\sqrt{2x-2}+\sqrt{6x-9}}=4x\left(4x-7\right)-5\left(4x-7\right)\)
\(\Leftrightarrow-\frac{4x-7}{\sqrt{2x-2}+\sqrt{6x-9}}=\left(4x-7\right)\left(4x-5\right)\)
\(\Leftrightarrow\left(4x-7\right)\left(\frac{1}{\sqrt{2x-2}+\sqrt{6x-9}}+4x-5\right)=0\)
\(\Leftrightarrow4x-7=0\Leftrightarrow x=\frac{7}{4}\) (nhận)
HT
15 tháng 12 2016
2) ĐK: \(2\le x\le4\)
pt \(\Leftrightarrow\sqrt{x-2}+\sqrt{a-x}=2\left(x^2-6x+9\right)+7x-19\)
\(\Leftrightarrow\sqrt{x-2}-\left(7x-20\right)+\sqrt{4-x}-1=2\left(x-3\right)^2\)
\(\Leftrightarrow\frac{x-2-\left(7x-20\right)^2}{\sqrt{x-2}+7x-20}+\frac{4-x-1}{\sqrt{4-x}+1}=2\left(x-3\right)^2\)
\(\Leftrightarrow\frac{\left(x-3\right)\left(134-49x\right)}{\sqrt{x-2}+\left(7x-20\right)}+\frac{3-x}{\sqrt{4-x}+1}=2\left(x-3\right)^2\)
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\) (nhận)
BT
3 tháng 5 2017
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.
BT
3 tháng 5 2017
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: \(x=\dfrac{1+\sqrt{7}}{3}\)
4 tháng 12 2019
1.
ĐK: \(-1\le x\le4\)
Đặt \(\sqrt{x+1}+\sqrt{4-x}=t\left(t\ge0\right)\)
\(\Leftrightarrow\sqrt{\left(x+1\right)\left(4-x\right)}=\frac{t^2-5}{2}\)
\(PT\Leftrightarrow t+\frac{t^2-5}{2}=5\Rightarrow t^2+2t-15=0\) \(\Rightarrow\left[{}\begin{matrix}t=3\\t=-5\left(l\right)\end{matrix}\right.\)
\(t=3\Rightarrow\sqrt{-x^2+3x+4}=2\) \(\Leftrightarrow-x^2+3x+4=4\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\) (tm)
2.
ĐK:\(x\ge4\)
Đặt \(\sqrt{x+4}+\sqrt{x-4}=t\left(t\ge0\right)\)
\(\Rightarrow2\sqrt{x^2-16}=t^2-2x\)
\(PT\Leftrightarrow t=2x-12+t^2-2x\)
\(\Leftrightarrow t^2-t-12=0\Rightarrow\left[{}\begin{matrix}t=4\\t=-3\left(l\right)\end{matrix}\right.\) Giải tiếp như trên.
HM
0
N
3 tháng 3 2019
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: \(\left\{{}\begin{matrix}x+2\le0\\2x^2-31x+99< 0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le-2\\\dfrac{9}{2}< x< 11\end{matrix}\right.\)\(\Rightarrow\dfrac{9}{2}< x< 11\)
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 11 2019
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/ \(\left\{{}\begin{matrix}x^2-3x\ge0\\2x-1\ge0\end{matrix}\right.\) \(\Rightarrow x\ge3\)
\(x^2-3x=2x-1\)
\(\Leftrightarrow x^2-5x+1=0\Rightarrow\left[{}\begin{matrix}x=\frac{5+\sqrt{21}}{2}\\x=\frac{5-\sqrt{21}}{2}\left(l\right)\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 11 2019
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/ \(x\ge2\)
\(2x-1=\left(x-2\right)^2\)
\(\Leftrightarrow x^2-6x+5=0\Rightarrow\left[{}\begin{matrix}x=1\left(l\right)\\x=5\end{matrix}\right.\)
tttttttttttttttttttt
\(ĐK:X\ge-1\)
xét x = -1 ta có \(\sqrt{\left(-1\right)^2+1+9}=2\left(-1\right)-4+\sqrt{-1+1}\)
\(\Leftrightarrow\sqrt{11}=-6\left(voli\right)\)
=> x = - 1 không là nghiệm của pt
=> x > -1
=> x + 1 > 0
chia cả 2 vế cho \(\sqrt{x+1}\) ta được :
\(\sqrt{\frac{x^2+x+9}{x+1}}=\frac{2\left(x-2\right)}{\sqrt{x+1}}+1\)
\(\Leftrightarrow\sqrt{\frac{x^2-4x+4+5x+5}{x+1}}=2\cdot\frac{x-2}{\sqrt{x+1}}+1\)
\(\Leftrightarrow\sqrt{\frac{\left(x-2\right)^2}{x+1}+5}=2\cdot\frac{x-2}{\sqrt{x+1}}+1\)
\(\Leftrightarrow\sqrt{\left(\frac{x-2}{\sqrt{x+1}}\right)^2+5}=2\cdot\frac{x-2}{\sqrt{x+1}}+1\)
đặt \(\frac{x-2}{\sqrt{x+1}}=t\) ta có
\(\sqrt{t^2+5}=2t+1\left(đk:t\ge-\frac{1}{2}\right)\)
\(\Leftrightarrow t^2+5=4t^2+4t+1\)
\(\Leftrightarrow3t^2+4t-4=0\)
\(\Leftrightarrow\orbr{\begin{cases}t=\frac{2}{3}\left(tm\right)\\t=-2\left(loai\right)\end{cases}}\)
t = 2/3 => \(\frac{x-2}{\sqrt{x+1}}=\frac{2}{3}\Leftrightarrow3x-6=2\sqrt{x+1}\) (đk x >= 2)
\(\Rightarrow9x^2-36x+36=4x+4\)
\(\Leftrightarrow9x^2-40x+32=0\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{20-4\sqrt{7}}{9}\left(loai\right)\\x=\frac{20+4\sqrt{7}}{9}\left(tm\right)\end{cases}}\)
vậy x = ...