\(8\sqrt{x^2+\frac{3}{4}+\sqrt{4x^2-1}}=10\sqrt{x^2+\frac{3}{4}-\sqrt{x^2}+\frac{1}{2}}+7\)
K
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13 tháng 11 2016
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
AH
Akai Haruma
Giáo viên
16 tháng 7 2020
k) ĐK: $x^2\geq 5$
PT $\Leftrightarrow 2\sqrt{x^2-5}-\frac{1}{3}\sqrt{x^2-5}+\frac{3}{4}\sqrt{x^2-5}-\frac{5}{12}\sqrt{x^2-5}=4$
$\Leftrightarrow 2\sqrt{x^2-5}=4$
$\Leftrightarrow \sqrt{x^2-5}=2$
$\Rightarrow x^2-5=4$
$\Leftrightarrow x^2=9\Rightarrow x=\pm 3$ (đều thỏa mãn)
l) ĐKXĐ: $x\geq -1$
PT $\Leftrightarrow 2\sqrt{x+1}+3\sqrt{x+1}-\sqrt{x+1}=4$
$\Leftrightarrow 4\sqrt{x+1}=4$
$\Leftrightarrow \sqrt{x+1}=1$
$\Rightarrow x+1=1$
$\Rightarrow x=0$
m)
ĐKXĐ: $x\geq -1$
PT $\Leftrightarrow 4\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}+3\sqrt{x+1}$
$\Leftrightarrow 6\sqrt{x+1}=16+2\sqrt{x+1}$
$\Leftrightarrow 4\sqrt{x+1}=16$
$\Leftrightarrow \sqrt{x+1}=4$
$\Rightarrow x=15$ (thỏa mãn)
AH
Akai Haruma
Giáo viên
16 tháng 7 2020
h)
ĐKXĐ: $x\geq -5$
PT $\Leftrightarrow \sqrt{x+5}=6$
$\Rightarrow x+5=36\Rightarrow x=31$ (thỏa mãn)
i) ĐKXĐ: $x\geq 5$
PT \(\Leftrightarrow \sqrt{x-5}+4\sqrt{x-5}-\sqrt{x-5}=12\)
\(\Leftrightarrow 4\sqrt{x-5}=12\Leftrightarrow \sqrt{x-5}=3\Rightarrow x-5=9\Rightarrow x=14\) (thỏa mãn)
j)
ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow 3\sqrt{2x}+\sqrt{2x}-6\sqrt{2x}+4=0$
$\Leftrightarrow -2\sqrt{2x}+4=0$
$\Leftrightarrow \sqrt{2x}=2$
$\Rightarrow x=2$ (thỏa mãn)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 11 2019
a/ ĐKXĐ: \(-\frac{3}{2}\le x\le4\)
\(\sqrt{2x+3}+\sqrt{4-x}=6x-3\left(x+7-2\sqrt{\left(2x+3\right)\left(4-x\right)}\right)-10\)
\(\Leftrightarrow\sqrt{2x+3}+\sqrt{4-x}=3\left(x+7+2\sqrt{\left(2x+3\right)\left(4-x\right)}\right)-52\)
Đặt \(\sqrt{2x+3}+\sqrt{4-x}=a>0\Rightarrow a^2=x+7+2\sqrt{\left(2x+3\right)\left(4-x\right)}\)
Phương trình trở thành:
\(a=3a^2-52\Leftrightarrow3a^2-a-52=0\Rightarrow\left[{}\begin{matrix}a=-4\left(l\right)\\a=\frac{13}{3}\end{matrix}\right.\)
\(\sqrt{2x+3}+\sqrt{4-x}=\frac{13}{3}\)
Phương trình này vô nghiệm nên ko muốn giải tiếp, bạn bình phương lên và chuyển vế thôi :(
b/ ĐKXĐ: \(-\frac{1}{4}\le x\le1\)
Đặt \(\sqrt{4x+1}+2\sqrt{1-x}=a>0\Rightarrow a^2=5+4\sqrt{-4x^2+3x+1}\)
\(\Rightarrow\sqrt{-4x^2+3x+1}=\frac{a^2-5}{4}\)
Pt trở thành:
\(a+10\left(\frac{a^2-5}{4}\right)=13\)
\(\Leftrightarrow5a^2+2a-51=0\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{17}{5}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{-4x^2+3x+1}=\frac{a^2-5}{4}=1\)
\(\Leftrightarrow-4x^2+3x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{3}{4}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 11 2019
c/ \(\Leftrightarrow x^2\left(x^2+2\right)=12-x\sqrt{2x^2+4}\)
\(\Leftrightarrow x^2\left(2x^2+4\right)=24-2x\sqrt{2x^2+4}\)
Đặt \(x\sqrt{2x^2+4}=a\) ta được:
\(a^2=24-2a\Leftrightarrow a^2+2a-24=0\Leftrightarrow\left[{}\begin{matrix}a=4\\a=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\sqrt{2x^2+4}=4\left(x>0\right)\\x\sqrt{2x^2+4}=-6\left(x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2\left(2x^2+4\right)=16\\x^2\left(2x^2+4\right)=36\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^4+2x^2-8=0\\x^4+2x^2-18=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2=2\\x^2=-4\left(l\right)\\x^2=\sqrt{19}-1\\x^2=-\sqrt{19}-1\left(l\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}< 0\left(l\right)\\x=-\sqrt{\sqrt{19}-1}\\x=\sqrt{\sqrt{19}-1}>0\left(l\right)\end{matrix}\right.\)
MC
15 tháng 3 2020
a) \(\lim\limits_{x\rightarrow0}\frac{\sqrt{1+2x}-1}{2x}=\lim\limits_{x\rightarrow0}\frac{2x}{2x\left(\sqrt{1+2x}+1\right)}=\lim\limits_{x\rightarrow0}\frac{1}{\sqrt{1+2x}+1}=\frac{1}{2}\)
b) \(\lim\limits_{x\rightarrow0}\frac{4x}{\sqrt{9+x}-3}=\lim\limits_{x\rightarrow0}\frac{4x\left(\sqrt{9+x}+3\right)}{x}=\lim\limits_{x\rightarrow0}[4\left(\sqrt{9+x}+3\right)=24\)
c) \(\lim\limits_{x\rightarrow2}\frac{\sqrt{x+7}-3}{x-2}=\lim\limits_{x\rightarrow2}\frac{x-2}{\left(x-2\right)\left(\sqrt{x+7}+3\right)}=\lim\limits_{x\rightarrow2}\frac{1}{\sqrt{x+7}+3}=\frac{1}{6}\)
d) \(\lim\limits_{x\rightarrow1}\frac{3x-2-\sqrt{4x^2-x-2}}{x^2-3x+2}=\lim\limits_{x\rightarrow1}\frac{\left(3x-2\right)^2-\left(4x^2-4x-2\right)}{(x^2-3x+2)\left(3x-2+\sqrt{4x^2-x-2}\right)}=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(5x-6\right)}{\left(x-1\right)\left(x-2\right)\left(3x-2+\sqrt{4x^2-x-2}\right)}=\frac{1}{2}\\ \\\\ \\ \\ \\ \)
e)\(\lim\limits_{x\rightarrow1}\frac{\sqrt{2x+7}+x-4}{x^3-4x^2+3}=\lim\limits_{x\rightarrow1}\frac{2x+7-\left(x^2-8x+16\right)}{\left(x-1\right)\left(x^2-3x-3\right)\left(\sqrt{2x+7}-x+4\right)}=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x-9\right)}{\left(x-1\right)\left(x^2-3x-3\right)\left(\sqrt{2x+7}-x+4\right)}=\lim\limits_{x\rightarrow1}\frac{x-9}{\left(x^2-3x-3\right)\left(\sqrt{2x+7}-x+4\right)}=-8\)
f) \(\lim\limits_{x\rightarrow1}\frac{\sqrt{2x+7}-3}{2-\sqrt{x+3}}=\lim\limits_{x\rightarrow1}\frac{(2x-2)\left(2+\sqrt{x+3}\right)}{\left(1-x\right)\left(\sqrt{2x+7}+3\right)}=\lim\limits_{x\rightarrow1}\frac{-2\left(2+\sqrt{x+3}\right)}{\sqrt{2x+7}+3}=\frac{-4}{3}\)
g) \(\lim\limits_{x\rightarrow0}\frac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4}=\lim\limits_{x\rightarrow0}\frac{x^2\left(\sqrt{x^2+16}+4\right)}{x^2\left(\sqrt{x^2+1}+1\right)}=4\)
h)
\(\lim\limits_{x\rightarrow4}\frac{\sqrt{x+5}-\sqrt{2x+1}}{x-4}=\lim\limits_{x\rightarrow4}\frac{\sqrt{x+5}-3}{x-4}+\lim\limits_{x\rightarrow4}\frac{3-\sqrt{2x+1}}{x-4}=\lim\limits_{x\rightarrow4}\frac{1}{\sqrt{x+5}+4}+\lim\limits_{x\rightarrow4}\frac{8-2x}{\left(x-4\right)\left(3+\sqrt{2x+1}\right)}=\frac{1}{7}-\frac{1}{3}=\frac{-4}{21}\)
k) \(\lim\limits_{x\rightarrow0}\frac{\sqrt{x+1}+\sqrt{x+4}-3}{x}=\lim\limits_{x\rightarrow0}\frac{\sqrt{x+1}-1}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt{x+4}-2}{x}=\lim\limits_{x\rightarrow0}\frac{1}{\sqrt{x+1}+1}+\lim\limits_{x\rightarrow0}\frac{1}{\sqrt{x+4}+2}=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 10 2019
a/ ĐKXĐ: ...
\(\Leftrightarrow3\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)-7\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow a^2=x+\frac{1}{4x}+1\)
\(\Rightarrow x+\frac{1}{4x}=a^2-1\)
Pt trở thành:
\(3a=2\left(a^2-1\right)-7\)
\(\Leftrightarrow2a^2-3a-9=9\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=3\)
\(\Leftrightarrow2x-6\sqrt{x}+1=0\)
\(\Rightarrow\sqrt{x}=\frac{3+\sqrt{7}}{2}\Rightarrow x=\frac{8+3\sqrt{7}}{2}\)
b/ ĐKXĐ:
\(\Leftrightarrow5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow x+\frac{1}{4x}=a^2-1\)
\(\Rightarrow5a=2\left(a^2-1\right)+4\Leftrightarrow2a^2-5a+2=0\)
\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{2\sqrt{x}}=2\\\sqrt{x}+\frac{1}{2\sqrt{x}}=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-4\sqrt{x}+1=0\\2x-\sqrt{x}+1=0\left(vn\right)\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 10 2019
c/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)
\(\Leftrightarrow\frac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)
\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\frac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)
\(\Leftrightarrow2x^2-8x+5=0\)
d/ ĐKXĐ: ...
\(\Leftrightarrow x+1-\frac{15}{6}\sqrt{x}+\sqrt{x^2-4x+1}-\frac{1}{2}\sqrt{x}=0\)
\(\Leftrightarrow\frac{x^2-\frac{17}{4}x+1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{x^2-\frac{17}{4}x+1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}=0\)
\(\Leftrightarrow\left(x^2-\frac{17}{4}x+1\right)\left(\frac{1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}\right)=0\)
\(\Leftrightarrow x^2-\frac{17}{4}x+1=0\)
\(\Leftrightarrow4x^2-17x+4=0\)