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NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 12 2020
a.
ĐKXĐ: \(x\ge1\)
\(\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x-1\right)\left(x^3+x^2+x+1\right)}\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x^3+x^2+x+1}-1\right)-\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x^3+x^2+x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x^3+x^2+x=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 12 2020
b.
ĐKXĐ: \(x\ge-1\)
\(x^2-6x+9+x+1-4\sqrt{x+1}+4=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(\sqrt{x+1}-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\\sqrt{x+1}-2=0\end{matrix}\right.\)
\(\Leftrightarrow x=3\)
c.
ĐKXĐ: \(-2\le x\le\dfrac{4}{5}\)
\(VT=2x+3\sqrt{4-5x}+1.\sqrt{x+2}\)
\(VT\le2x+\dfrac{1}{2}\left(9+4-5x\right)+\dfrac{1}{2}\left(1+x+2\right)=8\)
Dấu "=" xảy ra khi và chỉ khi \(x=-1\)
NQ
giải pt
a)\(\dfrac{1}{x+1}+\dfrac{3}{2x+1}=\dfrac{8}{x-2}\)
b)\(\sqrt{2x+1}+\sqrt{3-x}=\sqrt{3x+5}\)
0
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 8 2021
ĐKXĐ: \(x\ne-1\)
\(\dfrac{6x^2+4x+8}{x+1}=5\sqrt{2x^2+3}\)
\(\Rightarrow6x^2+4x+8=5\left(x+1\right)\sqrt{2x^2+3}\)
\(\Leftrightarrow2\left(2x^2+3\right)-5\left(x+1\right)\sqrt{2x^2+3}+2\left(x+1\right)^2=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{2x^2+3}=a\\x+1=b\end{matrix}\right.\)
\(\Rightarrow2a^2-5ab+2b^2=0\)
\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{2x^2+3}=2\left(x+1\right)\\2\sqrt{2x^2+3}=x+1\end{matrix}\right.\) (\(x\ge-1\))
\(\Rightarrow\left[{}\begin{matrix}2x^2+3=4\left(x+1\right)^2\\4\left(2x^2+3\right)=\left(x+1\right)^2\end{matrix}\right.\) (\(x\ge-1\))
\(\Leftrightarrow...\)
13 tháng 2 2022
\(\left(x\ne-y;x>\dfrac{y}{2}\right)\Rightarrow\left\{{}\begin{matrix}\dfrac{4}{\sqrt{2x-y}}-\dfrac{21}{x+y}=\dfrac{1}{2}\\\dfrac{3}{\sqrt{2x-y}}+\dfrac{7-\left(x+y\right)}{x+y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{\sqrt{2x-y}}-\dfrac{21}{x+y}=\dfrac{1}{2}\\\dfrac{3}{\sqrt{2x-y}}+\dfrac{7}{x+y}=2\end{matrix}\right.\)
\(đặt:\dfrac{1}{\sqrt{2x-y}}=a>0;\dfrac{1}{x+y}=b\)
\(\Rightarrow\left\{{}\begin{matrix}4a-21b=\dfrac{1}{2}\\3a+7b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\left(tm\right)\\b=\dfrac{1}{14}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{2x-y}}=\dfrac{1}{2}\\\dfrac{1}{x+y}=\dfrac{1}{14}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x-y=4\\x+y=14\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=8\end{matrix}\right.\)(thỏa)
PA
18 tháng 8 2017
\(\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}=m\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(2-\sqrt{x-4}\right)^2}=m\)
\(\Leftrightarrow\left|\sqrt{x-4}+2\right|+\left|2-\sqrt{x-4}\right|=m\)
mà \(\left|\sqrt{x-4}+2\right|+\left|2-\sqrt{x-4}\right|\)
\(\ge\left|\sqrt{x-4}+2+2-\sqrt{x-4}\right|=4\)
\(\Rightarrow m\ge4\) thì pt trên có no
NA
12 tháng 8 2021
a,ĐK: x\(\ge\)1
⇔\(\sqrt{x-1-2\sqrt{x-1}+1}\)=\(\sqrt{2}\)
⇔\(\sqrt{\left(\sqrt{x-1}-1\right)^2}\)=\(\sqrt{2}\)
⇔\(\left|\sqrt{x-1}-1\right|\)=\(\sqrt{2}\)
TH1:\(\sqrt{x-1}\)-1≥0⇒\(\left|\sqrt{x-1}-1\right|\)=\(\sqrt{x-1}\)-1 bn tự giải ra nha
TH2:\(\sqrt{x-1}\)-1<0⇒\(\left|\sqrt{x-1}-1\right|\)=1-\(\sqrt{x-1}\) bn tự lm nha
PA
28 tháng 9 2017
Đặt \(\left\{{}\begin{matrix}\sqrt{x+\dfrac{1}{y}}=a\left(a\ge0\right)\\x+y=b\left(b\ge3\right)\end{matrix}\right.\), ta có hpt:
\(\left\{{}\begin{matrix}a+\sqrt{b-3}=3\left(1\right)\\a^2+b=8\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{b-3}=3-a\)
\(\Leftrightarrow\left\{{}\begin{matrix}3-a\ge0\\b-3=9-6a+a^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}0\le a\le3\\b=a^2-6a+12\left(3\right)\end{matrix}\right.\). Thay (3) vào (2)
\(\Rightarrow a^2+a^2-6a+12=8\)
\(\Leftrightarrow2\left(a-1\right)\left(a-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a=2\end{matrix}\right.\left(n\right)\)
TH1: \(a=1;b=7\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x+\dfrac{1}{y}}=1\left(4\right)\\x+y=7\end{matrix}\right.\). Thay \(x=7-y\) vào (4)
\(\Rightarrow7-y+\dfrac{1}{y}=1\)
\(\Leftrightarrow7y-y^2+1=y\)
\(\Leftrightarrow\left(y-3\right)^2-10=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=3+\sqrt{10}\\y=3-\sqrt{10}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4-\sqrt{10}\\x=4+\sqrt{10}\end{matrix}\right.\)
TH2: \(a=2;b=4\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x+\dfrac{1}{y}}=2\left(5\right)\\x+y=4\end{matrix}\right.\). Thay \(x=4-y\) vào (5)
\(\Rightarrow4-y+\dfrac{1}{y}=4\)
\(\Leftrightarrow4y-y^2+1=4y\)
\(\Leftrightarrow\left(1-y\right)\left(1+y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=5\end{matrix}\right.\)
Vậy . . .
\(\sqrt[4]{x}=\dfrac{3}{8}+2x\)
<=> \(x=\left(\dfrac{3}{8}+2x\right)^4\)
<=> \(x=\left[\left(\dfrac{3}{8}+2x\right)^2\right]^2\)
<=> \(x=\left(\dfrac{9}{64}+\dfrac{3}{2}x+4x^2\right)^2\)
<=> \(x=\dfrac{1}{16}\)