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b:
ĐKXĐ: x>0
\(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)^2-2-4\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)+6=0\)
\(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}-2\right)^2=0\)
\(\Leftrightarrow x+1-2\sqrt{x}=0\)
=>x=1
b:
ĐKXĐ: x>0
\(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)^2-2-4\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)+6=0\)
\(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}-2\right)^2=0\)
\(\Leftrightarrow x+1-2\sqrt{x}=0\)
=>x=1
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
Lần sau xé nhỏ câu hỏi giùm con nha má, để nhiều thế này thất thu T_T!
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..........
1)\(\sqrt{9\left(x-1\right)}=21\Leftrightarrow3\sqrt{x-1}=21\Leftrightarrow\sqrt{x-1}=7\Leftrightarrow\hept{\begin{cases}7\ge0\\x-1=49\end{cases}\Leftrightarrow x=50}\)
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
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
Nguyễn Thành Trương , mình đang sài latop, nhìn bài của cậu, tớ muốn quẹo cả cổ -.-
Hoài Dung Copy ảnh. Mở paint past vào chỉnh hướng rồi xem :)
a) ta có \(\sqrt{12x^2+12x+19}+\sqrt{20x^2+20x+14}=-4x^2-4x+6\)
\(\Leftrightarrow\sqrt{12\left(x+\dfrac{1}{2}\right)^2+16}+\sqrt{20\left(x+\dfrac{1}{2}\right)^2+9}=-\left(2x+1\right)^2+7\)ta có : \(VT\ge\sqrt{16}+\sqrt{9}=7\) và \(VT\le7\)
\(\Rightarrow VT=VP\) \(\Leftrightarrow x=\dfrac{-1}{2}\) vậy \(x=\dfrac{-1}{2}\)
b) điều kiện \(x>0\)
ta có : \(\left(x+\dfrac{1}{x}\right)-4\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)+6=0\)
\(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)^2-4\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)+4=0\)
\(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}-2\right)^2=0\) \(\Leftrightarrow\sqrt{x}+\dfrac{1}{\sqrt{x}}-2=0\)
\(\Leftrightarrow\sqrt{x}+\dfrac{1}{\sqrt{x}}=2\Leftrightarrow\dfrac{x+\sqrt{x}}{\sqrt{x}}=2\Leftrightarrow x+\sqrt{x}=2\sqrt{x}\)
\(\Leftrightarrow x-\sqrt{x}=0\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(L\right)\\x=1\left(N\right)\end{matrix}\right.\)
vậy \(x=1\)
Mysterious Person giup mk nha