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Thực ra cũng EZ thôi :
\(\frac{6}{x^2-9}-1+\frac{4}{x^2-11}-1-\frac{7}{x^2-8}+1-\frac{3}{x^2-12}+1=0=>\)
\(\frac{15-x^2}{x^2-9}+\frac{15-x^2}{x^2-11}-\frac{15-x^2}{x^2-8}-\frac{15-x^2}{x^2-12}=0\)
=> \(\left(15-x^2\right)\left(\frac{1}{x^2-9}+\frac{1}{x^2-11}-\frac{1}{x^2-8}-\frac{1}{x^2-12}\right)=0\)
=>\(15-x^2=0=>x=\pm\sqrt{15}\)
Hình như còn nghiệm , any body help me ?
Câu 2/
Điều kiện xác định b tự làm nhé:
\(\frac{6}{x^2-9}+\frac{4}{x^2-11}-\frac{7}{x^2-8}-\frac{3}{x^2-12}=0\)
\(\Leftrightarrow x^4-25x^2+150=0\)
\(\Leftrightarrow\left(x^2-10\right)\left(x^2-15\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=10\\x^2=15\end{cases}}\)
Tới đây b làm tiếp nhé.
a. ĐK: \(\frac{2x-1}{y+2}\ge0\)
Áp dụng bđt Cô-si ta có: \(\sqrt{\frac{y+2}{2x-1}}+\sqrt{\frac{2x-1}{y+2}}\ge2\)
\(\)Dấu bằng xảy ra khi \(\frac{y+2}{2x-1}=1\Rightarrow y+2=2x-1\Rightarrow y=2x-3\)
Kết hợp với pt (1) ta tìm được x = -1, y = -5 (tmđk)
b. \(pt\Leftrightarrow\left(\frac{6}{x^2-9}-1\right)+\left(\frac{4}{x^2-11}-1\right)-\left(\frac{7}{x^2-8}-1\right)-\left(\frac{3}{x^2-12}-1\right)=0\)
\(\Leftrightarrow\left(15-x^2\right)\left(\frac{1}{x^2-9}+\frac{1}{x^2-11}+\frac{1}{x^2-8}+\frac{1}{x^2-12}\right)=0\)
\(\Leftrightarrow x^2-15=0\Leftrightarrow\orbr{\begin{cases}x=\sqrt{15}\\x=-\sqrt{15}\end{cases}}\)
B4
a) \(\frac{9}{\sqrt{3}}=\frac{9\cdot\sqrt{3}}{\sqrt{3}\cdot\sqrt{3}}=\frac{9\sqrt{3}}{3}=3\sqrt{3}\)
b)\(\frac{3}{\sqrt{5}-\sqrt{2}}=\frac{3\left(\sqrt{5}+\sqrt{2}\right)}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}=\frac{3\left(\sqrt{5}+\sqrt{2}\right)}{3}=\sqrt{5}+\sqrt{2}\)
c)\(\frac{\sqrt{2}+1}{\sqrt{2}-1}=\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\frac{\left(\sqrt{2}+1\right)^2}{1}=\left(\sqrt{2}+1\right)^2\)
d)\(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{1}=14\)
B3
a)\(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\) \(đk:x\ge1\)
\(\frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\sqrt{x-1}\cdot\left(\frac{1}{2}-\frac{9}{2}+3\right)=-17\)
\(\sqrt{x-1}\cdot\left(-1\right)=-17\)
\(\sqrt{x-1}=17\)
\(\left[{}\begin{matrix}x-1=289\left(tm\right)\\x-1=-289\left(ktm\right)\end{matrix}\right.\)
\(x=290\left(tm\right)\)
Lời giải:
a) ĐK: $x\geq 2$
PT $\Leftrightarrow \sqrt{(x-2)(x+2)}-3\sqrt{x-2}=0$
$\Leftrightarrow \sqrt{x-2}(\sqrt{x+2}-3)=0$
\(\Rightarrow \left[\begin{matrix} \sqrt{x-2}=0\\ \sqrt{x+2}-3=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=2\\ x=7\end{matrix}\right.\) (thỏa mãn)
Vậy..........
b) ĐK: $x\geq 0$
PT $\Leftrightarrow (\sqrt{x}-3)^2=0$
$\Leftrightarrow \sqrt{x}-3=0$
$\Leftrightarrow x=9$ (thỏa mãn)
c) ĐK: $x\geq 3$
PT $\Leftrightarrow \sqrt{9(x-3)}+\sqrt{x-3}-\frac{1}{2}\sqrt{4(x-3)}=7$
$\Leftrightarrow 3\sqrt{x-3}+\sqrt{x-3}-\sqrt{x-3}=7$
$\Leftrightarrow 3\sqrt{x-3}=7$
$\Leftrightarrow x-3=(\frac{7}{3})^2$
$\Rightarrow x=\frac{76}{9}$
d)
ĐK: $x\geq \frac{-1}{2}$
PT $\Leftrightarrow 3\sqrt{4(2x+1)}-\frac{1}{3}\sqrt{9(2x+1)}-\frac{1}{2}\sqrt{25(2x+1)}+\sqrt{\frac{1}{4}(2x+1)}=6$
$\Leftrightarrow 6\sqrt{2x+1}-\sqrt{2x+1}-\frac{5}{2}\sqrt{2x+1}+\frac{1}{2}\sqrt{2x+1}=6$
$\Leftrightarrow 3\sqrt{2x+1}=6$
$\Leftrightarrow \sqrt{2x+1}=2$
$\Rightarrow x=\frac{3}{2}$ (thỏa mãn)
ĐKXĐ: \(\left\{{}\begin{matrix}x^2\ne9\\x^2\ne11\\x^2\ne8\\x^2\ne12\end{matrix}\right.\Leftrightarrow x\notin\left\{3;-3;\sqrt{11};-\sqrt{11};2\sqrt{2};-2\sqrt{2};2\sqrt{3};-2\sqrt{3}\right\}\)
Đặt \(x^2-11=a\)(Điều kiện: \(a\notin\left\{-2;0;-3;1\right\}\))
PT\(\Leftrightarrow\frac{6}{a+2}+\frac{4}{a}-\frac{7}{a+3}-\frac{3}{a-1}=0\)
\(\Leftrightarrow\frac{6}{a+2}-1+\frac{4}{a}-1+\frac{-7}{a+3}+1+\frac{-3}{a-1}+1=0\)
\(\Leftrightarrow\frac{6-a-2}{a+2}+\frac{4-a}{a}+\frac{-7+a+3}{a+3}+\frac{-3+a-1}{a-1}=0\)
\(\Leftrightarrow-\frac{a-4}{a+2}-\frac{a-4}{a}+\frac{a-4}{a+3}+\frac{a-4}{a-1}=0\)
\(\Leftrightarrow\left(a-4\right)\left(-\frac{1}{a+2}-\frac{1}{a}+\frac{1}{a+3}+\frac{1}{a-1}\right)=0\)
\(\Leftrightarrow a-4=0\)
hay a=4
\(\Leftrightarrow x^2-11=4\)
\(\Leftrightarrow x^2=15\)
hay \(x=\pm\sqrt{15}\)