Giải BPT sau:
\(\sqrt{4x-11}\) - \(\sqrt{3x-9}\) \(\le\) \(\sqrt{2x-5}\)
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Giải BPT sau:
\(\sqrt{4x-11}\) - \(\sqrt{3x-9}\) \(\ge\) \(\sqrt{2x-5}\)
giải bpt: \(4x^2\le\left(2x+9\right)\left(1-\sqrt{1+2x}\right)^2\)
ĐKXĐ: \(x\ge-\frac{1}{2}\)
\(\Leftrightarrow4x^2\le\left(2x+9\right)\left(2x+2-2\sqrt{1+2x}\right)\)
\(\Leftrightarrow4x^2\le4x^2+22x+18-2\left(2x+9\right)\sqrt{2x+1}\)
\(\Leftrightarrow22x+18-2\left(2x+9\right)\sqrt{2x+1}\ge0\)
Đặt \(\sqrt{2x+1}=t\ge0\Rightarrow2x=t^2-1\)
\(11\left(t^2-1\right)+18-2\left(t^2+8\right)t\ge0\)
\(\Leftrightarrow2t^3-11t^2+16t-7\le0\)
\(\Leftrightarrow\left(t-1\right)^2\left(2t-7\right)\le0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=1\\t\le\frac{7}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{2x+1}=1\\\sqrt{2x+1}\le\frac{7}{2}\end{matrix}\right.\)
\(\Rightarrow-\frac{1}{2}\le x\le\frac{45}{8}\)
Giải phương trình:a)sqrt{sqrt{5}-sqrt{3x}}sqrt{8+2sqrt{15}}b)sqrt{4x-20}-3sqrt{dfrac{x-5}{9}}sqrt{1-x}c) sqrt{4x+8}+2sqrt{x+2}-sqrt{9x+18}1d) sqrt{x^2-6x+9}+x11e) sqrt{3x^2-4x+3}1-2xf) sqrt{16left(x+1right)}-sqrt{9left(x+1right)}4g) sqrt{9x+9}+sqrt{4x+4}sqrt{x+1}
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Giải phương trình:
a)\(\sqrt{\sqrt{5}-\sqrt{3x}}=\sqrt{8+2\sqrt{15}}\)
b)\(\sqrt{4x-20}-3\sqrt{\dfrac{x-5}{9}}=\sqrt{1-x}\)
c) \(\sqrt{4x+8}+2\sqrt{x+2}-\sqrt{9x+18}=1\)
d) \(\sqrt{x^2-6x+9}+x=11\)
e) \(\sqrt{3x^2-4x+3}=1-2x\)
f) \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)
g) \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)
f) Ta có: \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)
\(\Leftrightarrow4\left|x+1\right|-3\left|x+1\right|=4\)
\(\Leftrightarrow\left|x+1\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
g) Ta có: \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)
\(\Leftrightarrow5\sqrt{x+1}-\sqrt{x+1}=0\)
\(\Leftrightarrow x+1=0\)
hay x=-1
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giải bpt:
1. \(\frac{\sqrt{-3x^2+x+4}+2}{x}< 2\)
2. \(\sqrt{x^2-3x+2}+\sqrt{x^2-4x+3}\ge2\sqrt{x^2-5x+4}\)
3. \(\sqrt{x^2-8x+15}+\sqrt{x^2+2x-15}\le\sqrt{4x^2-18x=18}\)
4. 4(x+1)2 \(\ge\) (2x +10)( 1- \(\sqrt{3+2x}\))2
5. \(\sqrt{1+x}-\sqrt{1-x}\ge x\)
Giải các phương trình sau:
a) \(x^3-x^2+2x=\sqrt{2x-1}+\sqrt{4x-3}\)
b) \(x^3-x^2+3x+13=4\left(\sqrt{x+3}+\sqrt{3x+1}\right)\)
c) \(x^3-4x^2+6x-1=\sqrt{2x-3}+2\sqrt{x-1}\)
d) \(x^3+4x^2+9x+9=2\sqrt{3x+4}+\sqrt{2x+3}\)
e) \(2x^2-4x+11=2\sqrt{3x-5}+3\sqrt{2x+5}\)
Giải các phương trình sau:
a) \(x^3-x^2+2x=\sqrt{2x-1}+\sqrt{4x-3}\)
b) \(x^3-x^2+3x+13=4\left(\sqrt{x+3}+\sqrt{3x+1}\right)\)
c) \(x^3-4x^2+6x-1=\sqrt{2x-3}+2\sqrt{x-1}\)
d) \(x^3+4x^2+9x+9=2\sqrt{3x+4}+\sqrt{2x+3}\)
e) \(2x^2-4x+11=2\sqrt{3x-5}+3\sqrt{2x+5}\)
26 tháng 2 2021 lúc 16:06
Giải bpt
\(\dfrac{1}{\sqrt{x^2+1}}+\dfrac{1}{\sqrt{3x^2+5}}\le\dfrac{2}{\sqrt{x^2-2}+1}\)
ĐKXĐ: \(x^2\ge2\)
Đặt \(\sqrt{x^2-2}=a\ge0\)
BPT tương đương: \(\dfrac{1}{\sqrt{a^2+3}}+\dfrac{1}{\sqrt{3a^2+11}}\le\dfrac{2}{a+1}\)
Ta có: \(VT^2\le2\left(\dfrac{1}{a^2+3}+\dfrac{1}{3a^2+11}\right)< 2\left(\dfrac{1}{a^2+3}+\dfrac{1}{3a^2+1}\right)=\dfrac{8\left(a^2+1\right)}{\left(3a^2+1\right)\left(a^2+3\right)}\)
Mặt khác ta có: \(\left(a-1\right)^4\ge0\Leftrightarrow a^4-4a^3+6a^2-4a+1\ge0\)
\(\Leftrightarrow3a^4+10a^2+3\ge2a^4+4a^3+4a^2+4a+2\)
\(\Leftrightarrow\left(3a^2+1\right)\left(a^2+3\right)\ge2\left(a^2+1\right)\left(a+1\right)^2\)
\(\Rightarrow\dfrac{8\left(a^2+1\right)}{\left(3a^2+1\right)\left(a^2+3\right)}\le\dfrac{4}{\left(a+1\right)^2}\)
\(\Rightarrow VT^2< \dfrac{4}{\left(a+1\right)^2}\Rightarrow VT< \dfrac{2}{a+1}\)
\(\Rightarrow\) BPT đã cho đúng với mọi \(a\ge0\) hay nghiệm của BPT là \(x^2\ge2\)
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giải phương trình:
1,\(\sqrt{3x-8}\)-\(\sqrt{x+1}\)=\(\dfrac{2x-11}{5}\)
2,3x2-3x+18=10\(\sqrt{x^3+8}\)
3,\(\sqrt{5+2x}\)+\(\sqrt{5-2x}\)+5=3\(\sqrt{25-4x^2}\)
Giải các bất phương trình sau:
1. \(\sqrt{5x+1}-\sqrt{4x-1}< 3\sqrt{x}\)
2. \(\sqrt{x+2}-\sqrt{3-x}< \sqrt{5-2x}\)
3 \(\dfrac{\sqrt{12+x-x^2}}{x-11}\ge\dfrac{\sqrt{12+x-x^2}}{2x-9}\)
4.\(\sqrt{x^2-8x+15}+\sqrt{x^2+2x-15}\le\sqrt{4x^2-18x+18}\).
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: \(\left\{{}\begin{matrix}x+2\le0\\2x^2-31x+99< 0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le-2\\\dfrac{9}{2}< x< 11\end{matrix}\right.\)\(\Rightarrow\dfrac{9}{2}< x< 11\)
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