\(A=\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}\)

\(=\sqrt{\left(2x-1\right)^2}+\sqrt{\left(2x-3\right)^2}=\left|2x-1\right|+\left|2x-3\right|=\left|2x-1\right|+\left|3-2x\right|\ge\left|2x-1+3-2x\right|=2\)

\(\Rightarrow A\ge2\)

Dấu "=" xảy ra khi \(\left(2x-1\right)\left(3-2x\right)\ge0\)

\(\Leftrightarrow\dfrac{1}{2}\le x\le\dfrac{3}{2}\)

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\(\sqrt{\left(1+2x\right)^2}+\sqrt{\left(2x-3\right)^2}=|1+2x|+|2x-3|=|1+2x|+|3-2x|>=|1+2x+3-2x|=4\)

=>p min=4 

dau "="xay ra  <=>(1-2x)(3-2x)>=0

=>x

Ta có: \(P=\sqrt{1+4x+4x^2}+\sqrt{4x^2-12x+9}\)

    \(\Leftrightarrow P=\sqrt{4x^2+4x+1}+\sqrt{9-12x+4x^2}\)

    \(\Leftrightarrow P=\sqrt{\left(2x+1\right)^2}+\sqrt{\left(3-2x\right)^2}\)

    \(\Leftrightarrow P=\left|2x+1\right|+\left|3-2x\right|\)

Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)cho phương trình \(P,\)ta có:

     \(P=\left|2x+1\right|+\left|3-2x\right|\ge\left|2x+1-3-2x\right|=\left|-2\right|=2\)

     \(\Rightarrow\)\(P_{min}=2\)

Dấu "=" xảy ra khi và chỉ khi: \(\left(2x+1\right).\left(3-2x\right)>0\)

C₁: Các bạn lập bảng xét dấu nha mình làm cách kia cho các bạn dễ hiểu

C₂:

+ \(\hept{\begin{cases}2x+1>0\\3-2x>0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}2x>-1\\-2x>-3\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x>-\frac{1}{2}\\x< \frac{3}{2}\end{cases}}\)\(\Rightarrow\)\(-\frac{1}{2}< x< \frac{3}{2}\)( TM )

+ \(\hept{\begin{cases}2x+1< 0\\3-2x< 0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}2x< -1\\-2x< -3\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x< -\frac{1}{2}\\x>\frac{3}{2}\end{cases}}\)\(\Rightarrow\)\(-\frac{1}{2}>x>\frac{3}{2}\)( L )

     \(\Rightarrow\)\(-\frac{1}{2}< x< \frac{3}{2}\)

Vậy \(P_{min}=2\)\(\Leftrightarrow\)\(-\frac{1}{2}< x< \frac{3}{2}\)

a. ĐKXĐ: $x\geq 2$ hoặc $x=1$

PT $\Leftrightarrow \sqrt{(x-1)(x-2)}=\sqrt{x-1}$

$\Leftrightarrow \sqrt{x-1}(\sqrt{x-2}-1)=0$

\(\Leftrightarrow \left[\begin{matrix} \sqrt{x-1}=0\\ \sqrt{x-2}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1\\ x=3\end{matrix}\right.\) (đều thỏa mãn)

b.

PT $\Leftrightarrow \sqrt{(x-2)^2}=\sqrt{(2x-3)^2}$

$\Leftrightarrow |x-2|=|2x-3|$

\(\Leftrightarrow \left[\begin{matrix} x-2=2x-3\\ x-2=3-2x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1\\ x=\frac{5}{3}\end{matrix}\right.\)

c. ĐKXĐ: $x=2$ hoặc $x\geq 3$

PT $\Leftrightarrow \sqrt{(x-2)(x-3)}=\sqrt{x-2}$

$\Leftrightarrow \sqrt{x-2}(\sqrt{x-3}-1)=0$

\(\Leftrightarrow \left[\begin{matrix} \sqrt{x-2}=0\\ \sqrt{x-3}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=2\\ x=4\end{matrix}\right.\) (đều tm)

d.

PT $\Leftrightarrow \sqrt{(2x-1)^2}=\sqrt{(x-3)^2}$

$\Leftrightarrow |2x-1|=|x-3|$

\(\Leftrightarrow \left[\begin{matrix} 2x-1=x-3\\ 2x-1=3-x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=-2\\ x=\frac{4}{3}\end{matrix}\right.\)

a: Ta có: \(\sqrt{x^2-3x+2}=\sqrt{x-1}\)

\(\Leftrightarrow x^2-3x+2=x-1\)

\(\Leftrightarrow x^2-4x+3=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=3\left(nhận\right)\end{matrix}\right.\)

b: Ta có: \(\sqrt{x^2-4x+4}=\sqrt{4x^2-12x+9}\)

\(\Leftrightarrow\left|x-2\right|=\left|2x-3\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=x-2\\2x-3=-x+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{3}\end{matrix}\right.\)

c: Ta có: \(\sqrt{x^2-5x+6}=\sqrt{x-2}\)

\(\Leftrightarrow x^2-5x+6=x-2\)

\(\Leftrightarrow x^2-6x+8=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)

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

3: Ta có: \(\sqrt{4x+1}=x+1\)

\(\Leftrightarrow x^2+2x+1=4x+1\)

\(\Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

4: Ta có: \(2\sqrt{x-1}+\dfrac{1}{3}\sqrt{9x-9}=15\)

\(\Leftrightarrow3\sqrt{x-1}=15\)

\(\Leftrightarrow x-1=25\)

hay x=26

5: Ta có: \(\sqrt{4x^2-12x+9}=7\)

\(\Leftrightarrow\left|2x-3\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

a

ĐK: \(x\ge1\left(\sqrt{x-1}\ge0\right)\)

\(PT\Leftrightarrow\sqrt{x^2-x-2x+2}=\sqrt{x-1}\\ \Leftrightarrow\sqrt{x\left(x-1\right)-2\left(x-1\right)}=\sqrt{x-1}\\ \Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}=\sqrt{x-1}\\ \Leftrightarrow\left(\sqrt{x-1}\right)\left(\sqrt{x-2}-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=0\\\sqrt{x-2}=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=3\left(nhận\right)\end{matrix}\right.\)

b

ĐK: \(\left\{{}\begin{matrix}x^2-4x+4>0\\4x^2-4x+9>0\end{matrix}\right.\)

PT \(\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(2x-3\right)^2}\)

\(\Leftrightarrow\left|x-2\right|=\left|2x-3\right|\\ \Leftrightarrow\left[{}\begin{matrix}x-2=2x-3\\x-2=3-2x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=\dfrac{5}{3}\left(nhận\right)\end{matrix}\right.\)

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