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1) \(\sqrt[]{3x+7}-5< 0\)
\(\Leftrightarrow\sqrt[]{3x+7}< 5\)
\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)
\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)
\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)
a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow x+5=4\)
hay x=-1
b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x-1=289\)
hay x=290
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 .............
b, \(đk:x\ge2\)
Xét x=2 thay vào pt thấy không thỏa mãn => x>2 hay 27x-54>0
\(x^3-11x+36x-18=4\sqrt[4]{27x-54}\)
\(\Leftrightarrow27x^3-297x^2+972x-486=4\sqrt[4]{\left(27x-54\right).81.81.81}\le189+27x\) (cosi với 4 số dương, dấu = xảy ra khi x=5)
\(\Leftrightarrow x^3-11x^2+35x-25\le0\)
\(\Leftrightarrow\left(x-1\right)\left(x-5\right)^2\le0\) (*)
Có \(\left\{{}\begin{matrix}x>2\\\left(x-5\right)^2\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1>0\\\left(x-5\right)^2\ge0\end{matrix}\right.\)\(\Rightarrow\left(x-1\right)\left(x-5\right)^2\ge0\) (2*)
Từ (*) và (2*) ,dấu = xra khi x=5 (thỏa mãn)
Vây pt có nghiệm duy nhất x=5
c,Có \(6\sqrt[3]{4x^3+x}=16x^4+5>0\)
\(\Leftrightarrow4x^3+x>0\)
Có: \(16x^4+5=6\sqrt[3]{4x^3+x}\le2\left(4x^3+x+2\right)\) (theo cosi với 3 số dương,dấu = xảy ra khi \(x=\dfrac{1}{2}\))
\(\Leftrightarrow16x^4-8x^3-2x+1\le0\)
\(\Leftrightarrow\left(2x-1\right)^2\left(4x^2+2x+1\right)\le0\) (*)
(tương tự câu b) Dấu = xảy ra khi \(x=\dfrac{1}{2}\)(thỏa mãn)
Vậy....
d) Đk: \(x\ge\dfrac{3}{4}\)
Áp dụng bđt cosi:
\(\sqrt{2x-1}\le\dfrac{2x-1+1}{2}=x\)
\(\Rightarrow\dfrac{1}{\sqrt{2x-1}}\ge\dfrac{1}{x}\) (*)
\(\sqrt[4]{4x-3}\le\dfrac{4x-3+1+1+1}{4}=x\)
\(\dfrac{\Rightarrow1}{\sqrt[4]{4x-3}}\ge\dfrac{1}{x}\) (2*)
Từ (*) và (2*) \(\Rightarrow\dfrac{1}{\sqrt{2x-1}}+\dfrac{1}{\sqrt[4]{4x-3}}\ge\dfrac{2}{x}\)
Dấu = xảy ra khi x=1 (tm)
a.
ĐKXĐ: \(x>0\)
\(\sqrt{x\left(x+3\right)}+2\sqrt{x+2}=2x+\sqrt{\dfrac{\left(x+2\right)\left(x+3\right)}{x}}\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{x}-\sqrt{x+3}\right)+\sqrt{\dfrac{x+2}{x}}\left(\sqrt{x+3}-2\sqrt{x}\right)=0\)
\(\Leftrightarrow\sqrt{x}\left(\dfrac{4x-x-3}{2\sqrt{x}+\sqrt{x+3}}\right)-\sqrt{\dfrac{x+2}{x}}\left(\dfrac{4x-x-3}{\sqrt{x+3}+2\sqrt{x}}\right)=0\)
\(\Leftrightarrow\dfrac{3\left(x-1\right)}{2\sqrt{x}+\sqrt{x+3}}\left(\sqrt{x}-\sqrt{\dfrac{x+2}{x}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{x+2}{x}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-1\left(loại\right)\end{matrix}\right.\)
b.
ĐKXĐ: \(x\ge-\dfrac{1}{2};x\ne1-\sqrt{2}\)
\(x+2+x\sqrt{2x+1}=x\sqrt{x+2}+\sqrt{\left(x+2\right)\left(2x+1\right)}\)
\(\Leftrightarrow\sqrt{x+2}\left(\sqrt{2x+1}-\sqrt{x+2}\right)-x\left(\sqrt{2x+1}-\sqrt{x+2}\right)=0\)
\(\Leftrightarrow\left(\sqrt{2x+1}-\sqrt{x+2}\right)\left(\sqrt{x+2}-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x+1}=\sqrt{x+2}\\\sqrt{x+2}=x\left(x\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=x+2\\x^2-x-2=0\left(x\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-1\left(loại\right)\end{matrix}\right.\)
1.
ĐKXĐ: \(x< 5\)
\(\Leftrightarrow\sqrt{\dfrac{42}{5-x}}-3+\sqrt{\dfrac{60}{7-x}}-3=0\)
\(\Leftrightarrow\dfrac{\dfrac{42}{5-x}-9}{\sqrt{\dfrac{42}{5-x}}+3}+\dfrac{\dfrac{60}{7-x}-9}{\sqrt{\dfrac{60}{7-x}}+3}=0\)
\(\Leftrightarrow\dfrac{9x-3}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{9x-3}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}=0\)
\(\Leftrightarrow\left(9x-3\right)\left(\dfrac{1}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{1}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}\right)=0\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
b.
ĐKXĐ: \(x\ge2\)
\(\sqrt{\left(x-2\right)\left(x-1\right)}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}-\sqrt{x-2}+\sqrt{x+3}-\sqrt{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x-2}-\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{x-2}-\sqrt{x+3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-2=x+3\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow x=2\)