\(x^2-2-2\sqrt{4x-7}=0\)

\(\Leftrightarrow\left(4x-7-2\sqrt{4x-7}+1\right)+\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow\left(\sqrt{4x-7}-1\right)^2+\left(x-2\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{4x-7}-1=0\\x-2=0\end{matrix}\right.\)

Tự làm tiếp nhé.

. . .

\(4x^2-5x+1+2\sqrt{x-1}=0\)

\(\Leftrightarrow\left(x-1\right)\left(4x-1\right)+2\sqrt{x-1}=0\)

\(\Leftrightarrow\sqrt{x-1}\left[\left(4x-1\right)\sqrt{x-1}+2\right]=0\)

\(\Rightarrow x=1\)

. . .

\(\sqrt{x^2-4x+4}+\sqrt{x^2-6x+9}=1\)

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

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

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

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

Đến đây lập bảng xét dấu

. . .

\(x^2-x+2=2\sqrt{x^2-x+1}\)

\(\Leftrightarrow\left(\sqrt{x^2-x+1}-1\right)^2=0\)

Tự làm tiếp nhé.

\(\sqrt{3x+1}-\sqrt{6-x}+3x^2-14x-8=0\)

\(\Leftrightarrow\left(\sqrt{3x+1}-4\right)+\left(1-\sqrt{6-x}\right)+\left(3x^2-14-5\right)=0\)

\(\Leftrightarrow\dfrac{3x+1-16}{\sqrt{3x+1}+4}+\dfrac{1-6+x}{1+\sqrt{6-x}}+\left(x-5\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\dfrac{3\left(x-5\right)}{\sqrt{3x+1}+4}+\dfrac{x-5}{1+\sqrt{6-x}}+\left(x-5\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{1+\sqrt{6-x}}+3x+1\right)\left(x-5\right)=0\)

\(\Rightarrow x=5\)

. . .

\(\sqrt{2x^2-4x+5}-x+4=0\)

\(\Leftrightarrow\sqrt{2x^2-4x+5}=x-4\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-4\ge0\\2x^2-4x+5=x^2-8x+16\end{matrix}\right.\)

Tự làm tiếp nhé.

. . .

\(\sqrt{2x+3}+\sqrt{x-1}=\sqrt{x+6}\)

\(\Leftrightarrow\sqrt{2x+3}=\sqrt{x+6}-\sqrt{x-1}\)

\(\Leftrightarrow2x+3=x+6-2\sqrt{\left(x+6\right)\left(x-1\right)}+x-1\)

\(\Leftrightarrow2\sqrt{x^2+5x-6}=2\)

\(\Leftrightarrow x^2+5x-6=1\)

Tự làm tiếp nhé.

. . .

\(x+y+\dfrac{1}{2}=\sqrt{x}+\sqrt{y}\)

\(\Leftrightarrow\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\left(y-\sqrt{y}+\dfrac{1}{4}\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\left(\sqrt{y}-\dfrac{1}{2}\right)^2=0\)

Tự làm tiếp nhé.

- Lô Vỹ Vy Vy Nếu câu hỏi liên quan đến hình học, thì mỗi lần đăng một câu thôi, nếu câu hỏi liên quan đến đại số và số học thì có thể đẳng 3 - 4 câu một lần. Lần sau đừng đăng dày đặc như thế này nữa.

ĐK : | x| \(\ge\sqrt{7}\)

x²  + 4x - 7 = ( x + 4 ) \(\sqrt{x^2-7}\)

\(\Leftrightarrow\left(x^2-7\right)+4x-\left(x+4\right)\sqrt{x^2-7}=0\)

\(\Leftrightarrow\left(x^2-7\right)+4x-x\sqrt{x^2-7}-4\sqrt{x^2-7}=0\)

\(\Leftrightarrow\sqrt{x^2-7}\left(\sqrt{x^2-7}-x\right)-4\left(\sqrt{x^2-7}-x\right)=0\)

\(\Leftrightarrow\left(\sqrt{x^2-7}-x\right)\left(\sqrt{x^2-7}-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x^2-7}-x=0\\\sqrt{x^2-7}-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{x^2-7}=x\\\sqrt{x^2-7}=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x^2-7=x^2\\x^2-7=16\end{cases}}}\)

<=> x² =23 <=> x = \(\pm\sqrt{23}\)( T/m đk)

Có thể đặt \(t=\sqrt{x^2-7}\left(t\ge0\right)\)cho dễ nhìn

a) \(\sqrt{1-4x+4x^2}=5\) 

\(\Leftrightarrow\sqrt{\left(1-2x\right)^2}=5\)

\(\Leftrightarrow\left|1-2x\right|=5\)

\(\Leftrightarrow2x-1=5\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\)

b) \(\sqrt{x^2+6x+9}=3x-1\)

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

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

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

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=2\)

\(a,\sqrt{1-4x+4x^2}=5\\ \Leftrightarrow\sqrt{\left(1-2x\right)^2}=5\\ \Leftrightarrow\left|1-2x\right|=5\)

\(TH_1:x\le\dfrac{1}{2}\)

\(1-2x=5\\ \Leftrightarrow x=-2\left(tm\right)\)

\(TH_2:x\ge\dfrac{1}{2}\)

\(-1+2x=5\\ \Leftrightarrow x=3\left(tm\right)\)

Vậy \(S=\left\{-2;3\right\}\)

\(b,\sqrt{x^2+6x+9}=3x-1\\ \Leftrightarrow\sqrt{\left(x+3\right)^2}=3x-1\\ \Leftrightarrow\left|x+3\right|=3x-1\)

\(TH_1:x\ge-3\\ x+3=3x-1\\ \Leftrightarrow-2x=-4\Leftrightarrow x=2\left(tm\right)\)

\(TH_2:x< 3\\ -x-3=3x-1\\ \Leftrightarrow-4x=2\\ \Leftrightarrow x=-\dfrac{1}{2}\left(tm\right)\)

Vậy \(S=\left\{2;-\dfrac{1}{2}\right\}\)

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

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

_Với x<1 \(\Rightarrow1-x-x+2=3\)

\(\Rightarrow3-2x=3\)

\(\Rightarrow x=0\) (t/m)

_Với \(1\le x< 2\Rightarrow x-1+2-x=3\)

\(\Rightarrow0x=2\) (Vô lý)

_Với x>2 \(\Rightarrow x-1+x-2=3\)

\(\Leftrightarrow2x=6\Leftrightarrow x=3\) (t/m)

Vậy x=0 hoặc x=6.

\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=3\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)

ĐKXĐ: \(x\ge1\)

\(pt\Leftrightarrow\sqrt{\left(2x-1\right)^2}=x-1\Leftrightarrow\left|2x-1\right|=x-1\)

\(\Leftrightarrow2x-1=x-1\left(do.x\ge1\right)\)

\(\Leftrightarrow x=0\left(ktm\right)\)

Vậy \(S=\varnothing\)

ĐK \(x\ge1\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=\dfrac{2}{3}\left(ktm\right)\end{matrix}\right.\)