a,    \(\sqrt{5+\sqrt{x-1}}\)=6-x

=>bình phương lên => trục \(\sqrt{x-1}\)với x-6 => có nhân tử chung

c,    đat \(\sqrt{x^2+7x+7}\)=a => pt 3a²+2a-5=0 => giờ thì đơn giản rồi

b, mk k bít lm

a/ ĐKXĐ: \(x\ge\frac{1}{2}\)

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

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

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

\(\Rightarrow x=1\)

2/ ĐKXĐ:\(\left[{}\begin{matrix}x=0\\x\ge2\\x\le-3\end{matrix}\right.\)

- Nhận thấy \(x=0\) là 1 nghiệm

- Với \(x\ge2\):

\(\Leftrightarrow\sqrt{x-1}+\sqrt{x-2}=2\sqrt{x+3}=\sqrt{4x+12}\)

Ta có \(VT\le\sqrt{2\left(x-1+x-2\right)}=\sqrt{4x-6}< \sqrt{4x+12}\)

\(\Rightarrow VT< VP\Rightarrow\) pt vô nghiệm

- Với \(x\le-3\)

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

\(\Leftrightarrow3-2x+2\sqrt{x^2-3x+2}=-4x-12\)

\(\Leftrightarrow2\sqrt{x^2-3x+2}=-2x-15\) (\(x\le-\frac{15}{2}\))

\(\Leftrightarrow4x^2-12x+8=4x^2+60x+225\)

\(\Rightarrow x=-\frac{217}{72}\left(l\right)\)

Vậy pt có nghiệm duy nhất \(x=0\)

Bài 3: ĐKXĐ: \(-3\le x\le6\)

Đặt \(\sqrt{3+x}+\sqrt{6-x}=t\) \(\Rightarrow3\le t\le3\sqrt{2}\)

\(t^2=9+2\sqrt{\left(3+x\right)\left(6-x\right)}\Rightarrow-\sqrt{\left(3+x\right)\left(6-x\right)}=\frac{9-t^2}{2}\)

Phương trình trở thành:

\(t+\frac{9-t^2}{2}=m\Leftrightarrow-t^2+2t+9=2m\) (2)

a/ Với \(m=3\Rightarrow t^2-2t-3=0\Rightarrow\left[{}\begin{matrix}t=-1\left(l\right)\\t=3\end{matrix}\right.\)

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

\(\Leftrightarrow2\sqrt{\left(3+x\right)\left(6-x\right)}=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=6\end{matrix}\right.\)

b/ Xét hàm \(f\left(t\right)=-t^2+2t+9\) trên \(\left[3;3\sqrt{2}\right]\)

\(-\frac{b}{2a}=1< 3\Rightarrow\) hàm số nghịch biến trên \(\left[3;3\sqrt{2}\right]\)

\(f\left(3\right)=6\) ; \(f\left(3\sqrt{2}\right)=6\sqrt{2}-9\)

\(\Rightarrow6\sqrt{2}-9\le2m\le6\Rightarrow\frac{6\sqrt{2}-9}{2}\le m\le3\)

Bài 4 làm tương tự bài 3

ĐKXĐ:...

a. Đặt \(\left\{{}\begin{matrix}\sqrt{2x^2+4x+16}=a>0\\\sqrt{x+70}=b\ge0\end{matrix}\right.\)

\(\Rightarrow6x^2+10x-92=3a^2-2b^2\)

Pt trở thành:

\(3a^2-2b^2+ab=0\)

\(\Leftrightarrow\left(a+b\right)\left(3a-2b\right)=0\)

\(\Leftrightarrow3a=2b\)

\(\Leftrightarrow9\left(2x^2+4x+16\right)=4\left(x+70\right)\)

\(\Leftrightarrow...\)

b. ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{1-x}=b\ge0\end{matrix}\right.\)

Phương trình trở thành:

\(a^2+2+ab=3a+b\)

\(\Leftrightarrow a^2-3a+2+ab-b=0\)

\(\Leftrightarrow\left(a-1\right)\left(a-2\right)+b\left(a-1\right)=0\)

\(\Leftrightarrow\left(a-1\right)\left(a+b-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a+b=2\end{matrix}\right.\)

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

\(\Leftrightarrow...\)

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

Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{x-2}=b\ge0\end{matrix}\right.\)

Ta có: \(a^2-b^2=x+1-x+2=3\left(1\right)\)

pt \(\Leftrightarrow\left(a-b\right)\left(1+ab\right)=3\left(2\right)\)

Từ (1) và (2) suy ra:

\(\left(a-b\right)\left(1+ab\right)=a^2-b^2\)

\(\Leftrightarrow\left(a-b\right)\left(1-a\right)\left(1-b\right)=0\) \(\Rightarrow a=b=1\)

* Với \(a=1\Rightarrow\sqrt{x+1}=1\Leftrightarrow x=0\left(l\right)\)

*Với \(b=1\Rightarrow\sqrt{x-2}=1\Leftrightarrow x=3\left(n\right)\)

Vậy pt chỉ có một nghiệm là x=3

1.

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\\dfrac{2x+1}{\sqrt{2x^2+4x+5}+x+3}+1=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2x+1+\sqrt{2x^2+4x+5}+x+3=0\)

\(\Leftrightarrow\sqrt{2x^2+4x+5}=-3x-4\) \(\left(x\le-\dfrac{4}{3}\right)\)

\(\Leftrightarrow2x^2+4x+5=9x^2+24x+16\)

\(\Leftrightarrow7x^2+20x+11=0\)

2.

ĐKXĐ: ...

\(\Leftrightarrow2x\sqrt{2x+7}+7\sqrt{2x+7}=x^2+2x+7+7x\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2x+7}\\x+7=\sqrt{2x+7}\end{matrix}\right.\)

\(\Leftrightarrow...\)

3.

ĐKXĐ: ...

Từ pt dưới:

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)+3x-3y=3x^2+3y^2+1+1\)

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

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

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

\(\Leftrightarrow y=x-2\)

Thế vào pt trên:

\(x^2-2x+3=2\sqrt{5x-2}+\sqrt{7x-1}\)

\(\Leftrightarrow x^2-5x+2+2\left(x-\sqrt{5x-2}\right)+\left(x+1-\sqrt{7x-1}\right)=0\)

\(\Leftrightarrow x^2-5x+2+\dfrac{2\left(x^2-5x+2\right)}{x+\sqrt{5x-2}}+\dfrac{x^2-5x+2}{x+1+\sqrt{7x-1}}=0\)

\(\Leftrightarrow x^2-5x+2=0\)