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a)...ghi lại đề...
\(\Leftrightarrow\sqrt{x^2-x-2x+2}=\sqrt{x-1}\)
\(\Leftrightarrow\sqrt{x\left(x-1\right)-2\left(x-1\right)}=\sqrt{x-1}\)
\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}=\sqrt{x-1}\)
\(\Leftrightarrow\sqrt{x-2}\cdot\sqrt{x-1}=\sqrt{x-1}\)
\(\Leftrightarrow\sqrt{x-2}=\frac{\sqrt{x-1}}{\sqrt{x-1}}=1\)
\(\Leftrightarrow\sqrt{x-2}^2=1^2\)
\(\Leftrightarrow x-2=1\)(Vì \(x-2\ge0\Leftrightarrow x\ge2\))
\(\Leftrightarrow x=3\)
\(\)
\(a,\sqrt{x^2-3x+2}=\sqrt{x-1}\)
\(\Rightarrow x^2-3x+2=x-1\)
\(\Rightarrow x^2-4x+3=0\)
\(\Rightarrow x^2-x-3x+3=0\)
\(\Rightarrow\left(x-3\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)
Vậy..........
a)
ĐK: $x\geq 2$
PT \(\Leftrightarrow \sqrt{(x-1)(x-2)}=\sqrt{x-1}\)
\(\Leftrightarrow \sqrt{x-1}(\sqrt{x-2}-1)=0\)
\(\Rightarrow \left[\begin{matrix} \sqrt{x-1}=0\\ \sqrt{x-2}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1(\text{loại vì x}\geq 2)\\ \sqrt{x-2}=1\end{matrix}\right.\)
\(\Rightarrow x=1^2+2=3\) là nghiệm duy nhất thỏa mãn
b)
ĐK: $x\in\mathbb{R}$
Bình phương 2 vế:
\(\Rightarrow x^2-4x+4=4x^2-12x+9\)
\(\Leftrightarrow (x-2)^2=(2x-3)^2\)
\(\Leftrightarrow (x-2)^2-(2x-3)^2=0\Leftrightarrow (x-2-2x+3)(x-2+2x-3)=0\)
\(\Leftrightarrow (-x+1)(3x-5)=0\Rightarrow \left[\begin{matrix} x=1\\ x=\frac{5}{3}\end{matrix}\right.\) (đều thỏa mãn)
Vậy..........
c)
ĐKXĐ: $x\geq 3$
PT \(\Leftrightarrow \sqrt{(x-2)(x-3)}=\sqrt{x-2}\)
\(\Leftrightarrow (x-2)(x-3)=x-2\) (bình phương 2 vế không âm)
\(\Leftrightarrow (x-2)(x-3-1)=0\)
\(\Rightarrow \left[\begin{matrix} x-2=0\\ x-4=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=2(\text{loại vì x}\geq 3)\\ x=4\end{matrix}\right.\)
Vậy $x=4$
d)
ĐK: $x\in\mathbb{R}$
PT \(\Leftrightarrow 4x^2-4x+1=x^2-6x+9\) (bình phương 2 vế không âm)
\(\Leftrightarrow (2x-1)^2=(x-3)^2\Leftrightarrow (2x-1)^2-(x-3)^2=0\)
\(\Leftrightarrow (2x-1-x+3)(2x-1+x-3)=0\)
\(\Leftrightarrow (x+2)(3x-4)=0\Rightarrow \left[\begin{matrix} x+2=0\\ 3x-4=0\end{matrix}\right.\)
\(\Leftrightarrow \left[\begin{matrix} x=-2\\ x=\frac{4}{3}\end{matrix}\right.\) (đều thỏa mãn)
Vậy.........
a) \(\Leftrightarrow\sqrt{\left(x+3\right)^2}=4\)
\(\Leftrightarrow\left|x+3\right|=4\) \(\Leftrightarrow\left[{}\begin{matrix}x+3=4\\x+3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\) ( TM )
b) \(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=5x+3\)
\(\Leftrightarrow\left|2x-1\right|=5x+3\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x+3\ge0\\\left[{}\begin{matrix}2x-1=5x+3\\2x-1=-5x-3\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\frac{3}{5}\\\left[{}\begin{matrix}x=-\frac{4}{3}\left(KTM\right)\\x=-\frac{2}{7}\left(TM\right)\end{matrix}\right.\end{matrix}\right.\)
a \(\sqrt{x^2+6x+9}=4\Leftrightarrow\sqrt{\left(x+3\right)^2=4}\)
\(\Leftrightarrow x+3=4\)
\(\Rightarrow x=1\)
c)\(C=5+\sqrt{-4x^2-4x}\)
\(C=5+\sqrt{1-\left(4x^2+4x+1\right)}\)
\(C=5+\sqrt{1-\left(2x+1\right)^2}\)
Ta có: \(-\left(2x+1\right)^2\le0\)
\(\sqrt{1-\left(2x+1\right)^2}\le1\)
\(\sqrt{1-\left(2x+1\right)^2}+5\le6\Leftrightarrow C\le6\)
Vậy \(C_{max}=6\) khi \(2x+1=0\Leftrightarrow x=-\frac{1}{2}\)
f) \(F=\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}\)
\(F=\sqrt{\left(2x-1\right)^2}+\sqrt{\left(2x-3\right)^2}\)
\(F=\left|2x-1\right|+\left|3-2x\right|\ge\left|2x+1+3-2x\right|=4\)
\(F_{min}=4\) khi \(\left(2x-1\right)\left(3-2x\right)\ge0\Leftrightarrow\frac{1}{2}\le x\le\frac{3}{2}\)
Mấy còn lại tương tự =)))
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) \(\sqrt{9-12x+4x^2}=4+x\Leftrightarrow\sqrt{\left(3-2x\right)^2}=4+x\)
\(\Leftrightarrow\left|3-2x\right|=4+x\)
th1: \(3-2x\ge0\Leftrightarrow2x\le3\Leftrightarrow\Leftrightarrow x\le\dfrac{3}{2}\)
\(\Rightarrow\left|3-2x\right|=4+x\Leftrightarrow3-2x=4+x\Leftrightarrow3x=-1\Leftrightarrow x=\dfrac{-1}{3}\left(tmđk\right)\)
th2: \(3-2x< 0\Leftrightarrow2x>3\Leftrightarrow x>\dfrac{3}{2}\)
\(\Rightarrow\left|3-2x\right|=4+x\Leftrightarrow2x-3=4+x\Leftrightarrow x=7\left(tmđk\right)\)
vậy \(x=\dfrac{-1}{3};x=7\)
b) \(\sqrt{4-4x+x^2}=\left(x-1\right)^2+x-6\)
\(\Leftrightarrow\sqrt{\left(2-x\right)^2}=x^2-2x+1+x-6\)
\(\Leftrightarrow\left|2-x\right|=x^2-x-5\)
th1: \(2-x\ge0\Leftrightarrow x\le2\)
\(\Rightarrow\left|2-x\right|=x^2-x-5\Leftrightarrow2-x=x^2-x-5\)
\(\Leftrightarrow x^2=7\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{7}\left(loại\right)\\x=-\sqrt{7}\left(tmđk\right)\end{matrix}\right.\)
th2: \(2-x< 0\Leftrightarrow x>2\)
\(\Rightarrow\left|2-x\right|=x^2-x-5\Leftrightarrow x-2=x^2-x-5\)
\(\Leftrightarrow x^2-2x-3=0\Leftrightarrow x^2+x-3x-3=0\)
\(\Leftrightarrow x\left(x+1\right)-3\left(x+1\right)=0\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\left(tmđk\right)\\x=-1\left(loại\right)\end{matrix}\right.\)
vậy \(x=-\sqrt{7};x=3\)
a) \(\sqrt{9-12x+4x^2}=4+x\)
\(\Leftrightarrow\sqrt{\left(3-2x\right)^2}=4+x\)
\(\Leftrightarrow\left|3-2x\right|=4+x\)
\(\Leftrightarrow\left[{}\begin{matrix}3-2x=4+x\\3-2x=-4-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=7\end{matrix}\right.\)
Vậy \(x_1=-\dfrac{1}{3};x_2=7\).
b) \(\sqrt{4-4x+x^2}=\left(x-1\right)^2+x-6\)
\(\Leftrightarrow\sqrt{\left(2-x\right)^2}=x^2-2x+1+x-6\)
\(\Leftrightarrow\left|2-x\right|=x^2-x-5\)
\(\Leftrightarrow\left[{}\begin{matrix}2-x=x^2-x-5\\2-x=-x^2+x+5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=7\\x^2=2x+3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\left(l\right)\\x=-\sqrt{7}\\x=3\\x=-1\left(l\right)\end{matrix}\right.\)
Vậy \(x_1=-\sqrt{7};x_2=3\).
1
ĐK: \(x\in R\)
\(\sqrt{x^2-4x+4}=\sqrt{4x^2-12+9}\\ \Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(2x-3\right)^2}\\ \Leftrightarrow\left|x-2\right|=\left|2x-3\right|\\ \Leftrightarrow\left[{}\begin{matrix}x-2=2x-3\\2-x=2x-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{3}\end{matrix}\right.\)
2
ĐK: \(\left\{{}\begin{matrix}x+2\sqrt{x-1}\ge0\\x-1\ge0\end{matrix}\right.\Leftrightarrow x\ge1\)
Đặt \(t=\sqrt{x-1}\left(t\ge0\right)\Rightarrow t^2=x-1\Rightarrow x=t^2+1\)
\(\sqrt{x+2\sqrt{x-1}}=2\\ \Leftrightarrow\sqrt{t^2+2t+1}=2\\ \Leftrightarrow\sqrt{\left(t+1\right)^2}=2\left(1\right)\)
Do có \(t\ge0\) nên \(\left(1\right)\Leftrightarrow t+1=2\Leftrightarrow t=2-1=1\)
\(\Rightarrow x=t^2+1=1^2+1=2\) (thỏa mãn)
1: =>|2x-3|=|x-2|
=>2x-3=x-2 hoặc 2x-3=-x+2
=>x=1 hoặc 3x=5
=>x=5/3 hoặc x=1
2: \(\Leftrightarrow\left|\sqrt{x-1}+1\right|=2\)
=>căn x-1+1=2
=>căn x-1=1
=>x-1=1
=>x=2