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1)
ĐK: \(x\geq 5\)
PT \(\Leftrightarrow \sqrt{4(x-5)}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9(x-5)}=6\)
\(\Leftrightarrow \sqrt{4}.\sqrt{x-5}+3\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}.\sqrt{9}.\sqrt{x-5}=6\)
\(\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=6\)
\(\Leftrightarrow 2\sqrt{x-5}=6\Rightarrow \sqrt{x-5}=3\Rightarrow x=3^2+5=14\)
2)
ĐK: \(x\geq -1\)
\(\sqrt{x+1}+\sqrt{x+6}=5\)
\(\Leftrightarrow (\sqrt{x+1}-2)+(\sqrt{x+6}-3)=0\)
\(\Leftrightarrow \frac{x+1-2^2}{\sqrt{x+1}+2}+\frac{x+6-3^2}{\sqrt{x+6}+3}=0\)
\(\Leftrightarrow \frac{x-3}{\sqrt{x+1}+2}+\frac{x-3}{\sqrt{x+6}+3}=0\)
\(\Leftrightarrow (x-3)\left(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}\right)=0\)
Vì \(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}>0, \forall x\geq -1\) nên $x-3=0$
\(\Rightarrow x=3\) (thỏa mãn)
Vậy .............
Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
Em xin phép làm bài EZ nhất :)
4,ĐK :\(\forall x\in R\)
Đặt \(x^2+x+2=t\) (\(t\ge\dfrac{7}{4}\))
\(PT\Leftrightarrow\sqrt{t+5}+\sqrt{t}=\sqrt{3t+13}\)
\(\Leftrightarrow2t+5+2\sqrt{t\left(t+5\right)}=3t+13\)
\(\Leftrightarrow t+8=2\sqrt{t^2+5t}\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge-8\\\left(t+8\right)^2=4t^2+20t\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\3t^2+4t-64=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left(t-4\right)\left(3t+16\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left[{}\begin{matrix}t=4\left(tm\right)\\t=-\dfrac{16}{3}\left(l\right)\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow x^2+x+2=4\)\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy ....
giải phương trình$\sqrt{x}+\sqrt{1-x}+2\sqrt{x-x^2}-2\sqrt[4]{x-x^2}=1$√x+√1−x+2√x−x2−24√x−x2=1$\sqrt{x^2+10x+7}=3\sqrt{x+3}+2\sqrt{x+7}-6$√x2+10x+7=3√x+3+2√x+7−6$\sqrt[3]{x+1}+\sqrt[3]{x+2}=1+\sqrt[3]{x-3x+12}$3√x+1+3√x+2=1+3√x−3x+12$\left(4x+2\right)\sqrt{x+8}=3x^2+7x+8$(4x+2)√x+8=3x2+7x+8$x+4\sqrt{5-x}=4\sqrt{x-1}+\sqrt{-x^2+6x-5}+1$x+
ải phương trình
$\sqrt{x}+\sqrt{1-x}+2\sqrt{x-x^2}-2\sqrt[4]{x-x^2}=1$√x+√1−x+2√x−x2−24√x−x2=1
4√5−x=4√x−1+√−x2+6x−5+1