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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
a: Ta có: \(\sqrt{\left(x-3\right)^2}=3-x\)
\(\Leftrightarrow\left|x-3\right|=3-x\)
\(\Leftrightarrow x-3\le0\)
hay \(x\le3\)
b: Ta có: \(\sqrt{4x^2-20x+25}+2x=5\)
\(\Leftrightarrow\left|2x-5\right|=5-2x\)
\(\Leftrightarrow2x-5\le0\)
hay \(x\le\dfrac{5}{2}\)
f) Ta có: \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)
\(\Leftrightarrow4\left|x+1\right|-3\left|x+1\right|=4\)
\(\Leftrightarrow\left|x+1\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
g) Ta có: \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)
\(\Leftrightarrow5\sqrt{x+1}-\sqrt{x+1}=0\)
\(\Leftrightarrow x+1=0\)
hay x=-1
a.\(\sqrt{\left(x-3\right)^2}=3-x\)
\(\Leftrightarrow x-3=3-x\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
b.\(\sqrt{4x^2-20x+25}+2x=5\)
\(\Leftrightarrow\sqrt{\left(2x-5\right)^2}=5-2x\)
\(\Leftrightarrow2x-5=5-2x\)
\(\Leftrightarrow4x=10\)
\(\Leftrightarrow x=\dfrac{5}{2}\)
c.
d.\(\sqrt{x^2-\dfrac{1}{2}x+\dfrac{1}{16}}=\dfrac{1}{4}-x\)
\(\Leftrightarrow\sqrt{\left(x-\dfrac{1}{4}\right)^2}=\dfrac{1}{4}-x\)
\(\Leftrightarrow x-\dfrac{1}{4}=\dfrac{1}{4}-x\)
\(\Leftrightarrow x=\dfrac{1}{4}\)
a: =>|x-3|=3-x
=>x-3<=0
hay x<=3
b: =>|2x-5|=-2x+5
=>2x-5<=0
=>x<=5/2
c: =>|căn x-1-1|=căn x-1-1
=>căn x-1-1>=0
=>căn x-1>=1
=>x-1>=1
hay x>=2
các biểu thức trong căn pt hết về HĐT rồi phá ra là done
a. ĐKXĐ: $x\geq 1$
PT $\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{3}{2}.\sqrt{9}.\sqrt{x-1}+24.\sqrt{\frac{1}{64}}.\sqrt{x-1}=-17$
$\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17$
$\Leftrightarrow -\sqrt{x-1}=-17$
$\Leftrightarrow \sqrt{x-1}=17$
$\Leftrightarrow x-1=289$
$\Leftrightarrow x=290$
b. ĐKXĐ: $x\geq \frac{1}{2}$
PT $\Leftrightarrow \sqrt{9}.\sqrt{2x-1}-0,5\sqrt{2x-1}+\frac{1}{2}.\sqrt{25}.\sqrt{2x-1}+\sqrt{49}.\sqrt{2x-1}=24$
$\Leftrightarrow 3\sqrt{2x-1}-0,5\sqrt{2x-1}+2,5\sqrt{2x-1}+7\sqrt{2x-1}=24$
$\Leftrightarrow 12\sqrt{2x-1}=24$
$\Leftrihgtarrow \sqrt{2x-1}=2$
$\Leftrightarrow x=2,5$ (tm)
c. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{36}.\sqrt{x-2}-15\sqrt{\frac{1}{25}}\sqrt{x-2}=4(5+\sqrt{x-2})$
$\Leftrightarrow 6\sqrt{x-2}-3\sqrt{x-2}=20+4\sqrt{x-2}$
$\Leftrightarrow \sqrt{x-2}=-20< 0$ (vô lý)
Vậy pt vô nghiệm
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) \(\sqrt{\left(x-3\right)^2}=3\Leftrightarrow\left|x-3\right|=3\) \(\Leftrightarrow\left[{}\begin{matrix}x-3=3\\x-3=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\left(N\right)\\x=0\left(N\right)\end{matrix}\right.\)
b) \(\sqrt{4x^2-20x+25}+2x=5\Leftrightarrow\left|2x-5\right|+2x-5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-5\ge0\\2x-5+2x-5=0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-5\le0\\5-2x+2x-5=0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\4x-10=0\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{5}{2}\\0x=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\x=\dfrac{10}{4}\left(N\right)\end{matrix}\right.\\x\le\dfrac{5}{2}\end{matrix}\right.\) ** 10/4 = 5/2 rồi**
Kl: x \< 5/2
c) \(\sqrt{1-12x+36x^2}=5\Leftrightarrow\left|1-6x\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}1-6x=5\\1-6x=-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\left(N\right)\\x=1\left(N\right)\end{matrix}\right.\)
Kl: x=-2/3, x=1
d) Đk: x >/ 1
\(\sqrt{x+2\sqrt{x-1}}=2\Leftrightarrow\left|\sqrt{x-1}+1\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}+1=2\left(1\right)\\\sqrt{x-1}+2=-2\left(VN\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{x-1}=1\Leftrightarrow x=2\)(N)
Kl: x=2
e) Đk: x >/ 1
\(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}-1\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}\ge1\\\left|\sqrt{x-1}-1\right|=\sqrt{x-1}-1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{x-1}-1=\sqrt{x-1}-1\) (luôn đúng)
kl: x >/ 1
f) \(\sqrt{x^2-\dfrac{1}{2}x+\dfrac{1}{16}}=\dfrac{1}{4}-x\) \(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{4}\\\left|\dfrac{1}{4}-x\right|=\dfrac{1}{4}-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{4}\\\dfrac{1}{4}-x=\dfrac{1}{4}-x\end{matrix}\right.\)
(luôn đúng)
Kl: x \< 1/4
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