\(\sqrt{4x^2-4x+1}=\sqrt{x^2+10x+25}\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}2x-1=x+5\\2x-1=-\left(x+5\right)\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=x+5\\2x-1=-x-5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-\frac{4}{3}\end{cases}}\)

a) 

\(\sqrt{x+3}+2\sqrt{4\left(x+3\right)}-\frac{1}{3}\sqrt{9\left(x+3\right)}=8\)  

\(\sqrt{x+3}+2\cdot2\sqrt{x+3}-\frac{1}{3}\cdot3\sqrt{x+3}=8\)    

\(\sqrt{x+3}+4\sqrt{x+3}-\sqrt{x+3}=8\)    

\(4\sqrt{x+3}=8\)          

\(\sqrt{x+3}=2\) 

\(\orbr{\begin{cases}2\ge0\left(llđ\right)\\x+3=2^2\end{cases}}\) 

\(x+3=4\) 

\(x=1\) 

b) 

\(\orbr{\begin{cases}x^2+10x+25\ge0\\4x^2-4x+1=x^2+10x+25\end{cases}}\) 

\(\orbr{\begin{cases}\left(x+5\right)^2\ge0\left(lld\right)\\3x^2-6x-24=0\end{cases}}\) 

\(\orbr{\begin{cases}x=6\\x=-\frac{4}{3}\end{cases}}\)        

Câu 1 :

Xét điều kiện:\(\hept{\begin{cases}x\ge5\\x\le1\end{cases}}\)(Vô lý) 

Vậy pt vô nghiệm

Câu 2 : 

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

Vậy x=-1

Câu 3 : 

\(\sqrt{3x^2-4x+3}=1-2x\)\(\Leftrightarrow3x^2-4x+3=1+4x^2-4x\)

\(\Leftrightarrow x^2=2\Leftrightarrow x=\sqrt{2}\)

Câu 4 : 

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

\(\Leftrightarrow x=15\)

a,\(\Leftrightarrow\left(4x-1\right)^2\left(x^2+1\right)=4\left(x^2-x+1\right)^2\)

\(\Leftrightarrow\left(16x^2-8x+1\right)\left(x^2+1\right)=4\left(x^4+x^2+1-2x^3+2x^2-2x\right)\)

\(\Leftrightarrow16x^4+17x^2-8x^3-8x+1=4x^4+12x^2+4-8x^3-8x\)

\(\Leftrightarrow12x^4+5x^2-3=0\left(1\right)\)

Dat \(x^2=t\left(t\ge0\right)\)

\(\left(1\right)\Leftrightarrow12t^2+5t-3=0\)

\(\Delta=25-4.12.\left(-3\right)=169>0\)

Suy ra PT co hai nghiem phan biet

\(t_1=\frac{1}{3};t_2=-\frac{3}{4}\)

\(x=\frac{1}{\sqrt{3}}\)

1) Ta có: \(\left(\sqrt{12}-6\sqrt{3}+\sqrt{24}\right)\cdot\sqrt{6}-\left(\frac{5}{2}\sqrt{2}+12\right)\)

\(=\left(2\sqrt{3}-6\sqrt{3}+2\sqrt{6}\right)\cdot\sqrt{6}-\left(\sqrt{\frac{25}{4}\cdot2}+12\right)\)

\(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\sqrt{6}-\left(\sqrt{\frac{50}{4}}+12\right)\)

\(=-12\sqrt{2}+12-\frac{5\sqrt{2}}{2}-12\)

\(=\frac{-24\sqrt{2}-5\sqrt{2}}{2}\)

\(=\frac{-29\sqrt{2}}{2}\)

2) Ta có: \(\frac{26}{2\sqrt{3}+5}-\frac{4}{\sqrt{3}-2}\)

\(=\frac{26\left(5-2\sqrt{3}\right)}{\left(5+2\sqrt{3}\right)\left(5-2\sqrt{3}\right)}+\frac{4}{2-\sqrt{3}}\)

\(=\frac{26\left(5-2\sqrt{3}\right)}{25-12}+\frac{4\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)

\(=2\left(5-2\sqrt{3}\right)+4\left(2+\sqrt{3}\right)\)

\(=10-4\sqrt{3}+8+4\sqrt{3}\)

\(=18\)

3) ĐK để phương trình có nghiệm là: x≥0

Ta có: \(\sqrt{x^2-6x+9}=2x\)

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

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

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

Vậy: S={1}

4) ĐK để phương trình có nghiệm là: \(x\ge\frac{1}{2}\)

Ta có: \(\sqrt{4x^2+1}=2x-1\)

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

\(\Leftrightarrow4x^2+1=4x^2-4x+1\)

\(\Leftrightarrow4x^2+1-4x^2+4x-1=0\)

\(\Leftrightarrow4x=0\)

hay x=0(loại)

Vậy: S=∅

ĐKXĐ: \(x\ge-3\)

Đặt \(\sqrt{\frac{x+3}{2}}=a+1\ge0\Rightarrow x+3=2a^2+4a+2\)

Ta được hệ: \(\left\{{}\begin{matrix}2x^2+4x-a=1\\2a^2+4a-x=1\end{matrix}\right.\)

Trừ vế cho vế:

\(2\left(x^2-a^2\right)+4\left(x-a\right)+\left(x-a\right)=0\)

\(\Leftrightarrow\left(x-a\right)\left(2x+2a+5\right)=0\)

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

ta có \(2x^2+4x=\sqrt{\frac{x+3}{2}}\)

⇔4x⁴+16x³+16x²=\(\frac{x+3}{2}\)

⇔x+3=8x⁴+32x³+32x²

⇔x+3-8x⁴-32x³-32x²=0

⇔10x-9x+3-8x⁴-12x³-20x³+4x²-30x²-6x²=0

⇔(-6x²-9x+3)+(-8x⁴-12x³+4x²)+(-20x³-30x²+10x)

⇔-3(2x²+3x-1)-4x²(2x²+3x-1)-10x(2x²+3x-1)

⇔-(2x²+3x-1)(4x²+10x+3)

⇔\(\left[{}\begin{matrix}2x^2+3x-1=0\\4x^2+10+3=0\end{matrix}\right.\)

1. 2x²+3x-1=0

⇔x²+\(\frac{3}{2}\)x-\(\frac{1}{2}\)=0

⇔(x+\(\frac{3}{4}\))²=\(\frac{17}{16}\)

⇔\(x=\left\{{}\begin{matrix}\frac{-3+\sqrt{17}}{4}\\\frac{-3-\sqrt{17}}{4}\end{matrix}\right.\)

2.tương tự

x= \(\left\{{}\begin{matrix}\frac{-5-\sqrt{13}}{4}\\\frac{-5+\sqrt{13}}{4}\end{matrix}\right.\)

thử lại nghiệm thì chỉ có \(\frac{-3+\sqrt{17}}{4}\) và\(\frac{-5-\sqrt{13}}{4}\)thỏa mãn

⇒x=\(\frac{-3+\sqrt{17}}{4}\) và x=\(\frac{-5-\sqrt{13}}{4}\)

hơi dài