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a)
ĐKXĐ: \(x> \frac{-5}{7}\)
Ta có: \(\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)
\(\Rightarrow 9x-7=\sqrt{7x+5}.\sqrt{7x+5}=7x+5\)
\(\Rightarrow 2x=12\Rightarrow x=6\) (hoàn toàn thỏa mãn)
Vậy......
b) ĐKXĐ: \(x\geq 5\)
\(\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow \sqrt{4}.\sqrt{x-5}+3\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}\sqrt{9}.\sqrt{x-5}=4\)
\(\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow 2\sqrt{x-5}=4\Rightarrow \sqrt{x-5}=2\Rightarrow x-5=2^2=4\Rightarrow x=9\)
(hoàn toàn thỏa mãn)
Vậy..........
c) ĐK: \(x\in \mathbb{R}\)
Đặt \(\sqrt{6x^2-12x+7}=a(a\geq 0)\Rightarrow 6x^2-12x+7=a^2\)
\(\Rightarrow 6(x^2-2x)=a^2-7\Rightarrow x^2-2x=\frac{a^2-7}{6}\)
Khi đó:
\(2x-x^2+\sqrt{6x^2-12x+7}=0\)
\(\Leftrightarrow \frac{7-a^2}{6}+a=0\)
\(\Leftrightarrow 7-a^2+6a=0\)
\(\Leftrightarrow -a(a+1)+7(a+1)=0\Leftrightarrow (a+1)(7-a)=0\)
\(\Rightarrow \left[\begin{matrix} a=-1\\ a=7\end{matrix}\right.\) \(\Rightarrow a=7\) vì \(a\geq 0\)
\(\Rightarrow 6x^2-12x+7=a^2=49\)
\(\Rightarrow 6x^2-12x-42=0\Leftrightarrow x^2-2x-7=0\)
\(\Leftrightarrow (x-1)^2=8\Rightarrow x=1\pm 2\sqrt{2}\)
(đều 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!
\(a,\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)\(ĐKXĐ:x\ge-\frac{5}{7}\)
\(\Leftrightarrow9x-7=7x+5\)
\(\Leftrightarrow9x-7x=5+7\)
\(\Leftrightarrow2x=12\)
\(\Leftrightarrow x=6\)
\(b,\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow\sqrt{4\left(x-5\right)}+3.\frac{\sqrt{x-5}}{\sqrt{9}}-\frac{1}{3}\sqrt{9\left(x-5\right)}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}\left(2+1-1\right)=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\)
\(\Leftrightarrow x=9\)
a) \(\sqrt{\left(2x-1\right)^2}=3\)
⇔ \(\left|2x-1\right|=3\)
⇔ \(\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\)
⇔ \(\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
b) \(3\sqrt{x}-2\sqrt{9x}+\sqrt{16x}=5\)
ĐKXĐ : \(x\ge0\)
⇔ \(3\sqrt{x}-2\sqrt{3^2x}+\sqrt{4^2x}=5\)
⇔ \(3\sqrt{x}-2\cdot3\sqrt{x}+4\sqrt{x}=5\)
⇔ \(7\sqrt{x}-6\sqrt{x}=5\)
⇔ \(\sqrt{x}=5\)
⇔ \(x=25\)( tm )
c) \(\sqrt{4x+20}-3\sqrt{5+x}+\frac{3}{4}\sqrt{9x+45}=6\)
ĐKXĐ : \(x\ge-5\)
⇔ \(\sqrt{2^2\left(x+5\right)}-3\sqrt{x+5}+\frac{3}{4}\sqrt{3^2\left(x+5\right)}=6\)
⇔ \(2\sqrt{x+5}-3\sqrt{x+5}+\frac{3}{4}\cdot3\sqrt{x+5}=6\)
⇔ \(-\sqrt{x+5}+\frac{9}{4}\sqrt{x+5}=6\)
⇔ \(\frac{5}{4}\sqrt{x+5}=6\)
⇔ \(\sqrt{x+5}=\frac{24}{5}\)
⇔ \(x+5=\frac{576}{25}\)
⇔ \(x=\frac{451}{25}\left(tm\right)\)
a) \(\left|3x+1\right|=\left|x+1\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=x+1\\3x+1=-x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
c) \(\sqrt{9x^2-12x+4}=\sqrt{x^2}\)
\(\Leftrightarrow\sqrt{\left(3x-2\right)^2}=\sqrt{x^2}\)
\(\Leftrightarrow\left|3x-2\right|=\left|x\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=x\\3x-2=-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)
d) \(\sqrt{x^2+4x+4}=\sqrt{4x^2-12x+9}\)
\(\Leftrightarrow\sqrt{\left(x+2\right)^2}=\sqrt{\left(2x-3\right)^2}\)
\(\Leftrightarrow\left|x+2\right|=\left|2x-3\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=2x-3\\x+2=-2x+3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{3}\end{matrix}\right.\)
e) \(\left|x^2-1\right|+\left|x+1\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-1=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow x=-1\)
f) \(\sqrt{x^2-8x+16}+\left|x+2\right|=0\)
\(\Leftrightarrow\sqrt{\left(x-4\right)^2}+\left|x+2\right|=0\)
\(\Leftrightarrow\left|x-4\right|+\left|x+2\right|=0\)
⇒ vô nghiệm
a)\(\sqrt{4x+20}\) +\(\sqrt{x-5}\) -\(\dfrac{1}{3}\)\(\sqrt{9x-45}\)=4 ; ĐKXĐ : x ≥_+ 5
⇔ \(\sqrt{2^2x+2^2.5}\) +\(\sqrt{x-5}\) -\(\dfrac{1}{3}\)\(\sqrt{3^2x-3^2.5}\) =4
⇔ 2\(\sqrt{x+5}\) +\(\sqrt{x-5}\) -\(\dfrac{1}{3}\)3\(\sqrt{x-5}\) =4 ⇔ 2\(\sqrt{x+5}\) +\(\sqrt{x-5}\) -\(\sqrt{x-5}\) =4⇔2\(\sqrt{x+5}\)=4(tm)
⇔\(\sqrt{x+5}\)=2⇔x+5=4 ⇔x=-1
Vậy x=-1
b) \(\sqrt{x^2-36}\) - \(\sqrt{x-6}\) =0 ; ĐKXĐ: x≥_+6
⇔ \(\sqrt{\left(x-6\right)\left(x+6\right)}\) - \(\sqrt{x-6}\) =0 ⇔ \(\sqrt{x-6}\).\(\sqrt{x+6}\) - \(\sqrt{x-6}\) =0
⇔ \(\sqrt{x-6}\)(\(\sqrt{x+6}\) -1 )=0 ⇔\([\) \(\begin{matrix}\sqrt{x-6}&=0\\\sqrt{x+6}-1&=0\end{matrix}\) ⇔ \([\) \(\begin{matrix}x-6&=0\\x+6-1&=0\end{matrix}\) ⇔\([\) \(\begin{matrix}x&=6\left(ktm\right)\\x&=-5\left(tm\right)\end{matrix}\)
Vậy x=-5
c) \(\sqrt{4-x^2}\) -x +2 =0 ; ĐKXĐ: -2≤x≤2
⇔ \(\sqrt{\left(2-x\right)\left(2+x\right)}\) -x+2 =0 ⇔ \(\sqrt{\left(2-x\right)\left(2+x\right)}\) -(x-2)=0
⇔ \(\sqrt{\left(2-x\right)\left(2+x\right)}\) =(x-2) ⇔ (2-x)(2+x)=(x-2)2 ⇔ 4-x2 = x2-4x+4 ⇔ -x2-x2+4x=4-4
⇔-2x2+4x=0 ⇔ -2x(x-2)=0 ⇔ \([\) \(\begin{matrix}-2x&=0\\x-2&=0\end{matrix}\) ⇔\([\) \(\begin{matrix}x&=0\left(tm\right)\\x&=2\left(tm\right)\end{matrix}\)
Vậy S=\(\left\{0;2\right\}\)
d) \(\sqrt{\left(2x-3\right)\left(x-1\right)}-\sqrt{x-1}=0\) ; ĐKXĐ: x≥\(\dfrac{3}{2}\);x ≥ 1
⇔\(\sqrt{2x-3}.\sqrt{x-1}-\sqrt{x-1}=0\) ⇔ \(\sqrt{x-1}.\left(\sqrt{2x-3}-1\right)=0\)
⇔ \(\left[{}\begin{matrix}\sqrt{x-1}=0\\\sqrt{2x-3}-1=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x-1=0\\2x-3-1=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=1\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
Vậy s=\(\left\{1:2\right\}\)
a) x = 21.
b) x1 = 6; x2 = 7.
c) x = 2.
d) x1 = 1; x2 = 2.
a) \(x=6\)
b) \(x=6\)
c) \(x=2\)
d) \(x=1\) hoặc \(x=2\)
a) x=9
b) S= {6;7}
c) x=2
d) x=2
\(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}\) = 4
⇔\(\sqrt{4\left(x-5\right)}\)+\(\sqrt{x-5}\) - \(\dfrac{1}{3}\)\(\sqrt{9x-45}\) = 4
⇔2\(\sqrt{x-5}\)+\(\sqrt{x-5}\) - \(\dfrac{1}{3}\).3.\(\sqrt{x-5}\) = 4
⇔2\(\sqrt{x-5}\) = 4
Điều kiện xác định: x\ge5x ≥ 5
PT ⇔ \(\sqrt{x-5}\) = 2
⇔\(x\) - 5 = 4
⇔\(x\) = 9
Vậy pt có nghiệm là 9
b)\(\sqrt{x^2-36}\)-\(\sqrt{x-6}\) = 0
⇔(\(\sqrt{x-6}\))2-\(\sqrt{x-6}\)= 0
⇔\(\sqrt{x-6}\).\(\sqrt{x-6}\)-\(\sqrt{x-6}\) = 0
⇔\(\sqrt{x-6}\) = \(\sqrt{x-6}\):\(\sqrt{x-6}\)
⇔\(\sqrt{x}\) = 6
c) \(\sqrt{4x-x^2}\)-\(x+2=0\)
Điều kiện xác định : −2 ≤ x ≤ 2
Với điều kiện đã cho thì VT\ge0,VP\le0VT ≥ 0,VP ≤ 0 nên phương trình đã cho tương đương với
\(\int_{x-2=0}^{\sqrt{4-x^2}=0}\Leftrightarrow\int_{x=2}^{x=2}\)⇔\(x=2\)
Vậy pt có nghiệm là 2
d) \(\sqrt{\left(2x-3\right)\left(x-1\right)}\)\(-\sqrt{x-1}\) = 0
Điều kiện xác định x \(\ge\)\(\dfrac{3}{2}\) ; x \(\ge\)1
\left\{{}\begin{matrix}\left(2x-3\right)\left(x-1\right)\ge0\\x-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3\ge0\\x-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\x\ge1\end{matrix}\right.\Leftrightarrow x\ge\dfrac{3}{2}
\left\{{}\begin{matrix}\left(2x-3\right)\left(x-1\right)\ge0\\x-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3\ge0\\x-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\x\ge1\end{matrix}\right.\Leftrightarrow x\ge\dfrac{3}{2}
\left\{{}\begin{matrix}\left(2x-3\right)\left(x-1\right)\ge0\\x-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3\ge0\\x-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\x\ge1\end{matrix}\right.\Leftrightarrow x\ge\dfrac{3}{7
\2left\{{}\begin{matrix}\left(2x-3\right)\left(x-1\right)\ge0\\x-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3\ge0\\x-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\x\ge1\end{matrix}\right.\Leftrightarrow x\ge\dfrac{3}{21}
\left\{{}\begin{matrix}\left(2x-3\right)\left(x-1\right)\ge0\\x-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3\ge0\\x-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\x\ge1\end{matrix}\right.\Leftrightarrow x\ge\dfrac{3}{2}Với điều kiện trên thì
\(\sqrt{\left(2x-3\right)\left(x-1\right)}\) = \(\sqrt{2x-3}.\sqrt{x-1}\)
⇔\(\sqrt{\left(2x-3\right)\left(x-1\right)}\)\(-\sqrt{x-1}\) = 0
⇔\(\sqrt{2x-3}.\sqrt{x-1}\)\(-\sqrt{x-1}\) = 0
⇔\(\sqrt{x-1}\)\(\left(\sqrt{2x-3-1}\right)\) = 0
\\(\sqrt{x-1}=0\) ⇔ \(x=1\)
left\{{}\begin{matrix}\left(2x-3\right)\left(x-1\right)\ge0\\x-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3\ge0\\x-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\x\ge1\end{matrix}\right.\Leftrightarrow x\ge\dfrac{3}2}\(\sqrt{2x-3}-1\) = 0 ⇔ \(x=2\)
Vậy pt có nghiệm là (1;2)
a)\(\sqrt{4x-20}\)+\(\sqrt{x-5}\)-\(\dfrac{1}{3}\)\(\sqrt{9x-45}\)=4\(\Leftrightarrow\)2\(\sqrt{x-5}\)+\(\sqrt{x-5}\)-\(\dfrac{1}{3}\).3\(\sqrt{x-5}\)=4\(\Leftrightarrow\)2\(\sqrt{x-5}\)=4\(\Leftrightarrow\)\(\sqrt{x-5}\)=2\(\Leftrightarrow\)x-5=4\(\Leftrightarrow\)x=9
b)\(\sqrt{x^2-36}\)-\(\sqrt{x-6}\)=0\(\Leftrightarrow\)\(\sqrt{\left(x-6\right)\left(x+6\right)}\)-\(\sqrt{x-6}\)=0\(\Leftrightarrow\)x+6\(\sqrt{x-6}\)-\(\sqrt{x-6}\)=0\(\Leftrightarrow\)x+5\(\sqrt{x-6}\)=0\(\Leftrightarrow\)x+5=0,\(\sqrt{x-6}\)=0\(\Leftrightarrow\)x=-5,x-6=0\(\Leftrightarrow\)x=5,x=6
c)\(\sqrt{4-x^2}\)-x+2=0\(\Leftrightarrow\)\(\sqrt{\left(2-x\right)\left(2+x\right)}\)-x+2=0\(\Leftrightarrow\)2+x\(\sqrt{2-x}\)-x+2=0\(\Leftrightarrow\)4\(\sqrt{2-x}\)=0\(\Leftrightarrow\)4\(\sqrt{2-x}\)=0\(\)\(\Leftrightarrow\)2-x=0\(\Leftrightarrow\)x=2
d)\(\sqrt{\left(2x-3\right)\left(x-1\right)}\)-\(\sqrt{x-1}\)=0\(\Leftrightarrow\)2x-3\(\sqrt{x-1}\)-\(\sqrt{x-1}\)=0\(\Leftrightarrow\)(2x-3-1)\(\sqrt{x-1}\)=0\(\Leftrightarrow\)2x-4=0,x-1=0\(\Leftrightarrow\)x=2,x=1\(\Rightarrow\)x=2
A)đkxđ x>=5
X=9(tm)
Vậy pt có 1 nghiệm là S={9}
B)x=6
x=7
Vậy pt có 2 nghiệm S={6;7}
C)đkxđ -2=<x=<2
x=2(tm)
Vậy pt có 1 nghiệm là S={2}
D)đkxđ x>=3/2
x=2(tm)
Vậy pt có nghiệm là S={2}
A)đkxđ x>=5
X=9(tm)
Vậy pt có 1 nghiệm là S={9}
B)x=6
x=7
Vậy pt có 2 nghiệm S={6;7}
C)đkxđ -2=<x=<2
x=2(tm)
Vậy pt có 1 nghiệm là S={2}
D)đkxđ x>=3/2
x=2(tm)
Vậy pt có nghiệm là S={2}
a) \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\)
<=> \(\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\)
<=> \(2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
<=> 2\(\sqrt{x-5}\) =4 đkxđ: x≥5
<=> \(\sqrt{x-5}=2\)
<=> x-5=4
<=> x=9 (tm)
vậy S= { 9}
b) \(\sqrt{x^2-36}-\sqrt{x-6}=0\)
<=> \(\sqrt{\left(x-6\right)\left(x+6\right)}-\sqrt{x-6}=0\)
<=> \(\sqrt{x-6}.\sqrt{x+6}-\sqrt{x-6}=0\)
<=> \(\sqrt{x-6}.\sqrt{x+6-1}=0\)
<=> \(\sqrt{x-6}.\sqrt{x+5}=0\)
<=> \(\sqrt{x-6}=0\) hoặc \(\sqrt{x+5}=0\)
<=> x = 6 (ktm) hoặc x = 5 (tm)
vậy S= {5}
c) \(\sqrt{4-x^2}-x+2=0\) đkxđ: -2≤0≤2
<=> \(\sqrt{\left(2-x\right)\left(2+x\right)}-x+2=0\)
<=> \(\left(\sqrt{\left(2-x\right)\left(2+x\right)}\right)^2=\left(x-2\right)^2\)
<=> \(\left(2-x\right)\left(2+x\right)=x-2\)
<=> \(4-x^2=x^2-4x+4\)
<=> \(-2x^2+4x=0\)
<=> -2x(x-2)=0
<=> -2x=0 hoặc x-2=0
<=> x=0(tm) hoặc x=2 (tm)
vậy S={0;2}
d) \(\sqrt{\left(2x-3\right)\left(x-1\right)}-\sqrt{x-1}=0\) đkxđ: x≥3/2 ; x≥1
<=> \(\sqrt{2x-3}.\sqrt{x-1}-\sqrt{x-1}=0\)
<=> \(\sqrt{x-1}.\sqrt{2x-3-1}=0\)
<=> \(\sqrt{x-1}=0\) hoặc \(\sqrt{2x-4}=0\)
<=> x=1(tm) hoặc x=2 (tm)
vậy S={1;2}
Đkxd:x≥5
√4x-20 + √x-5 -1÷3 √9x-45=4
=√4(x-5)+√x-5-1÷3.√9(x-5)=4
=2√x-5+√x-5-1÷3.3,√x-5=4
=2√x-5=4
=√x-5=2
=x-5=4
=x=9(nhận)
Vậy S={5}
b)√x^2-36-√x-6=0
=√(x-6)(x+6) -√x-6=0
=(x-6)( x+6-1)=0
ĐKXĐ:x≥6
X-6=0
x=6(nhận)
X+6-1=0
x+6=1=0
X=-5(loại)
Vậy S={6}
c)√4-x^2 -x+2=0. ĐKXĐ: -2≥x≥2
=4√2+x .√2-x-x+2=0
=√x+2 .(√2x-2 - √x+2)=0
√x+2=0.or (√2x+2-√x+2) =0
√x+2=0
X=2(nhận)
(√2x+2-√x+2)=0
4-x^2=0
X=4(loại)
Vậy S={2}
d)√(2x+3)(x-1) -√x-1 =0
ĐKXĐ:x≥3÷2
Pt=√2x-3 . √x-1 -√x-1=0
=√x-1(√2x-3 -1)=0
√x-1 =0 or √2x-3 -1=0
√x-1=0
X=1(loại)
√2x-3 -1=0
X=2(nhận)
Vậy S={2}
b , căn x^2 - 36 - căn x - 6 = 0
<=> căn (x - 6 )( x + 6 ) - căn x - 6 = 0
<=> căn x - 6 . căn x + 6 - căn x - 6 = 0
<=> ( căn x - 6 ).( căn x + 6 - 1 )
<=> căn x - 6 = 0 <=> x - 6 = 0
căn x + 6 - 1 = 0 x + 6 = 1
<=> x - 6 = 0 <=> x = 6 ( TMĐK )
x = 1 - 6 x = - 5 ( KTMĐK)
a)\(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\left(1\right)đk:x\ge5\left(1\right)\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}.3\sqrt{x-5}=4 \Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2 \Rightarrow x-5=4\Leftrightarrow x=9\left(tm đkxđ\right)\)
vậy: S={9}
b)đkxđ:
\(\left\{{}\begin{matrix}\left(x-6\right)\left(x+6\right)\ge0\\x-6\ge0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}x-6\ge0\\x+6\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-6\le0\\x+6\le0\end{matrix}\right.\end{matrix}\right.\\x\ge6\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge6\\x\ge-6\end{matrix}\right.\\\left\{{}\begin{matrix}x\le6\\x\le-6\end{matrix}\right.\end{matrix}\right.\\x\ge6\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge6\\x\le-6\end{matrix}\right.\\x\ge6\end{matrix}\right.\)\(\Leftrightarrow\)\(x\ge6\)
Ta có: \(\sqrt{x^2-36}-\sqrt{x-6}=0\)
\(\Leftrightarrow\sqrt{\left(x-6\right)\left(x+6\right)}-\sqrt{x-6}=0\)
\(\Leftrightarrow\sqrt{x-6}\left(\sqrt{x+6}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-6}=0\\\sqrt{x+6}-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-6}=0\\\sqrt{x+6}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-6=0\\x+6=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\left(tm đkxđ\right)\\x=-5\left(ktm đkxđ\right)\end{matrix}\right.\)
vậy: S={6}
c) Đkxđ: \(4-x^2\ge0\Leftrightarrow\left(2-x\right)\left(2+x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2-x\ge0\\2+x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}2-x\le0\\2+x\le0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le2\\x\ge-2\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge2\\x\le-2\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow-2\le x\le2\)
Ta có: \(\sqrt{4-x^2}-x+2=0\)
\(\Leftrightarrow\sqrt{\left(2-x\right)\left(2+x\right)}+\left(\sqrt{2-x}\right)^2=0\)
\(\Leftrightarrow\sqrt{2-x}\left(\sqrt{2+x}+\sqrt{2-x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2-x}=0\\\sqrt{2+x}+\sqrt{2-x}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2-x}=0\\\left\{{}\begin{matrix}\sqrt{2-x}=0\\\sqrt{2+x}=0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2-x=0\\\left\{{}\begin{matrix}2-x=0\\2+x=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\left\{{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow x=2\left(tm đkxđ\right) \)
Vậy: S={2}
d) đkxđ:\(\left\{{}\begin{matrix}\left(2x-3\right)\left(x-1\right)\ge0\\x-1\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-3\ge0\\x-1\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-3\le0\\x-1\le0\end{matrix}\right.\end{matrix}\right.\\x\ge1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\x\ge1\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{3}{2}\\x\le1\end{matrix}\right.\end{matrix}\right.\\x\ge1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge\dfrac{3}{2}\\x\le1\end{matrix}\right.\\x\ge1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x\ge\dfrac{3}{2}\\x=1\end{matrix}\right.\)
Ta có: \(\sqrt{\left(2x-3\right)\left(x-1\right)}-\sqrt{x-1}=0\Leftrightarrow\sqrt{x-1}\left(\sqrt{2x-3}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=0\\\sqrt{2x-3}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=0\\\sqrt{2x-3}=1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2x-3=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm đkxđ\right)\\x=\dfrac{3}{2}\left(tm đkxđ\right)\end{matrix}\right.\)
\(Vay:S=\left\{1;\dfrac{3}{2}\right\}\)
a,\(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\left(x\ge5\right)\Rightarrow2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}.3\sqrt{x-5}=4\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\Leftrightarrow x-5=4\Leftrightarrow x=9\left(TMĐKXĐ\right)\)
b,\(\sqrt{x^2-36}-\sqrt{x-6}=0\left(x\ge6\right)\Rightarrow\sqrt{x-6}\left(\sqrt{x+6}-1\right)=0\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-6}=0\\\sqrt{x+6}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x+6=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(TMĐKXĐ\right)\\x=-5\left(KTMĐKCĐ\right)\end{matrix}\right.\)
c,\(\sqrt{\left(2x-3\right)\left(x-1\right)}-\sqrt{x-1}=0\left(x\ge\dfrac{3}{2}\right)\Rightarrow\sqrt{x-1}\left(\sqrt{2x-3}-1\right)=0\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=0\\\sqrt{2x-3}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x-3=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(KTMĐKXĐ\right)\\x=2\left(TMĐKXĐ\right)\end{matrix}\right.\)