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Ở tử đặt nhân tử chung căn x chung rồi lại đặt căn x +1 chung
Ở mẫu tách 3 căn x ra 2 căn x +căn x rồi đặt nhân tử 2 căn x ra
rút gọn được \(\frac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
a, \(5\sqrt{2x^2+3x+9}=2x^2+3x+3\) (*)
Đặt \(2x^2+3x=a\left(a\ge-9\right)\)
=> \(5\sqrt{a+9}=a+3\)
<=> \(25\left(a+9\right)=a^2+6a+9\)
<=> \(25a+225=a^2+6a+9\)
<=> \(0=a^2+6a+9-25a-225=a^2-19a-216\)
<=> 0= \(a^2-27a+8a-216\)
<=> \(\left(a-27\right)\left(a+8\right)=0\)
=> \(\left[{}\begin{matrix}a=27\\a=-8\end{matrix}\right.\) <=>\(\left[{}\begin{matrix}2x^2+3x=27\\2x^2+3x=-8\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}2x^2+3x-27=0\\2x^2+3x+8=0\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}\left(x-3\right)\left(2x+9\right)=0\\2\left(x^2+2.\frac{3}{4}+\frac{9}{16}\right)+\frac{55}{8}=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=3\left(tm\right)\\x=-\frac{9}{2}\left(tm\right)\\2\left(x+\frac{3}{4}\right)^2=-\frac{55}{8}\left(ktm\right)\end{matrix}\right.\)
Vậy pt (*) có tập nghiệm \(S=\left\{3,-\frac{9}{2}\right\}\)
b, \(9-\sqrt{81-7x^3}=\frac{x^3}{2}\left(đk:x\le\sqrt[3]{\frac{81}{7}}\right)\)(*)
<=> \(\sqrt{81-7x^3}=9-\frac{x^3}{2}\)
<=>\(81-7x^3=\left(9-\frac{x^3}{2}\right)^2=81-9x^3+\frac{x^6}{4}\)
<=> \(-7x^3+9x^3-\frac{x^6}{4}=0\) <=> \(2x^3-\frac{x^6}{4}=0\)<=> \(8x^3-x^6=0\)
<=> \(x^3\left(8-x^2\right)=0\)
=> \(\left[{}\begin{matrix}x=0\\8=x^2\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}x=0\left(tm\right)\\x=\pm2\sqrt{2}\left(ktm\right)\end{matrix}\right.\)
Vậy pt (*) có nghiệm x=0
d,\(\sqrt{9x-2x^2}-9x+2x^2+6=0\) (*) (đk: \(0\le x\le\frac{1}{2}\))
<=> \(\sqrt{9x-2x^2}-\left(9x-2x^2\right)+6=0\)
Đặt \(\sqrt{9x-2x^2}=a\left(a\ge0\right)\)
Có \(a-a^2+6=0\)
<=> \(a^2-a-6=0\) <=> \(a^2-3x+2x-6=0\)
<=> \(\left(a-3\right)\left(a+2\right)=0\)
=> \(a-3=0\) (vì a+2>0 vs mọi \(a\ge0\))
<=> a=3 <=>\(\sqrt{9x-2x^2}=3\) <=> \(9x-2x^2=9\)
<=> 0=\(2x^2-9x+9\) <=> \(2x^2-6x-3x+9=0\) <=>\(\left(2x-3\right)\left(x-3\right)=0\)
=> \(\left[{}\begin{matrix}2x=3\\x=3\end{matrix}\right.< =>\left[{}\begin{matrix}x=\frac{3}{2}\\x=3\end{matrix}\right.\)(t/m)
Vậy pt (*) có tập nghiệm \(S=\left\{\frac{3}{2},3\right\}\)
a/ ĐKXĐ: \(x\ge2\)
\(\Leftrightarrow2\sqrt{\left(x-2\right)\left(x+2\right)}-6\sqrt{x-2}+\sqrt{x+2}-3=0\)
\(\Leftrightarrow2\sqrt{x-2}\left(\sqrt{x+2}-3\right)+\sqrt{x+2}-3=0\)
\(\Leftrightarrow\left(2\sqrt{x-2}+1\right)\left(\sqrt{x+2}-3\right)=0\)
\(\Leftrightarrow\sqrt{x+2}-3=0\Rightarrow x=11\)
b/ ĐKXĐ: ....
Đặt \(\left\{{}\begin{matrix}\sqrt{x-2016}=a>0\\\sqrt{y-2017}=b>0\\\sqrt{z-2018}=a>0\end{matrix}\right.\)
\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{4}-\frac{a-1}{a^2}+\frac{1}{4}-\frac{b-1}{b^2}+\frac{1}{4}-\frac{c-1}{c^2}=0\)
\(\Leftrightarrow\frac{\left(a-2\right)^2}{a^2}+\frac{\left(b-2\right)^2}{b^2}+\frac{\left(c-2\right)^2}{c^2}=0\)
\(\Leftrightarrow a=b=c=2\Rightarrow\left\{{}\begin{matrix}x=2020\\y=2021\\z=2022\end{matrix}\right.\)
a/ ĐK: \(x\ge0\)
\(\Leftrightarrow\sqrt{3+x}=x^2-3\)
Đặt \(\sqrt{3+x}=a>0\Rightarrow3=a^2-x\) pt trở thành:
\(a=x^2-\left(a^2-x\right)\)
\(\Leftrightarrow x^2-a^2+x-a=0\)
\(\Leftrightarrow\left(x-a\right)\left(x+a+1\right)=0\)
\(\Leftrightarrow x=a\) (do \(x\ge0;a>0\))
\(\Leftrightarrow\sqrt{3+x}=x\Leftrightarrow x^2-x-3=0\)
d/ ĐKXĐ: ...
\(\sqrt{6x^2+1}=\sqrt{2x-3}+x^2\)
\(\Leftrightarrow\sqrt{2x-3}-1+x^2+1-\sqrt{6x^2+1}\)
\(\Leftrightarrow\frac{2\left(x-2\right)}{\sqrt{2x-3}+1}+\frac{x^4+2x^2+1-6x^2-1}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}=0\)
\(\Leftrightarrow\frac{2\left(x-2\right)}{\sqrt{2x-3}+1}+\frac{x^2\left(x+2\right)\left(x-2\right)}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{2}{\sqrt{2x-3}+1}+\frac{x^2\left(x+2\right)}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}\right)=0\)
\(\Leftrightarrow x=2\) (phần trong ngoặc luôn dương với mọi \(x\ge\frac{3}{2}\))
a) \(\left\{{}\begin{matrix}x\ge0\\-\sqrt{x+7}< 0\\-5x-4\ne0\\-3x+2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x+7>0\\-5x\ne4\\-3x\ne-2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x>-7\\x\ne\frac{-4}{5}\\x\ne\frac{2}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne\frac{2}{3}\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x\ge0\\x+4\ne0\\x-2\ge0\\-2x-10\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne-4\\x\ge2\\-2x\ne10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ne-5\end{matrix}\right.\Leftrightarrow x\ge2\)
c) \(\left\{{}\begin{matrix}x\ge0\\-x-3\ne0\\2x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne-3\\x\ne-\frac{3}{2}\end{matrix}\right.\Leftrightarrow x\ge0\)
d) \(\left\{{}\begin{matrix}2x-7\ge0\\x\ge0\\3x-4\ne0\\x-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{7}{2}\\x\ge0\\x\ne\frac{4}{3}\\x\ne3\end{matrix}\right.\Leftrightarrow x\ge\frac{7}{2}\)
ĐKXĐ : x\(\ge0\)
ADBĐT BCS ta được
\(\left(\frac{x^2}{3}+4\right)\left(3+1\right)\ge\left(x+2\right)^2\)
\(\Rightarrow4\sqrt{\frac{x^2}{3}+4}\ge2x+4\)(do x\(\ge0\)) (1)
Do x\(\ge0\)nên ADBĐT Cauchy ta được:
\(\sqrt{6x}\le\frac{x+6}{2}\)\(\Rightarrow1+\frac{3x}{2}+\sqrt{6x}\le1+\frac{3x}{2}+\frac{x+6}{2}=1+\frac{4x+6}{2}=2x+4\)(2)
Từ (1) và (2) \(\Rightarrow4\sqrt{\frac{x^2}{3}+4}\ge1+\frac{3x}{2}+\sqrt{6x}\)
Dấu = xảy ra \(\Leftrightarrow x=6\)(thỏa mãn ĐKXĐ)
3) ĐKXĐ \(-1\le x\le1\)
Khi đó phương trình đã cho \(\Leftrightarrow4\left(\sqrt{1+x}+\sqrt{1-x}\right)=8-x^2\)
\(\Leftrightarrow\hept{\begin{cases}16\left(2+2\sqrt{1-x^2}\right)=\left(7+1-x^2\right)\left(2\right)\\8-x^2\ge0\end{cases}}\)
Đặt \(\sqrt{1-x^2}=a\ge0\)
Khi đó phương trình (2) trở thành:
\(\hept{\begin{cases}16\left(2+2a\right)=\left(7+a^2\right)\\x^2\le8\end{cases}}\)
\(\Leftrightarrow a^4+14a^2+49=32+32a\)
\(\Leftrightarrow a^4+14a^2-32a+17=0\)
\(\Leftrightarrow a^4-2a^2+1+16a^2-32a+16=0\)
\(\Leftrightarrow\left(a^2-1\right)^2+16\left(a-1\right)^2=0\)
\(\Leftrightarrow a=1\)
hay \(\sqrt{1-x^2}=1\)
\(\Leftrightarrow x=0\)(thỏa mãn)
em ms hok lớp 1