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:

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

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

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

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

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

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

\(\Leftrightarrow...\)

a) \(\sqrt{9-12x+4x^2}=4\Leftrightarrow\sqrt{\left(2x\right)^2-2.2x.3+9}=4\Leftrightarrow\sqrt{\left(2x-3\right)^2}=4\left(1\right)\)Nếu \(x< \dfrac{3}{2}\)

\(\left(1\right)\Leftrightarrow3-2x=4\Leftrightarrow2x=-1\Leftrightarrow x=\dfrac{-1}{2}\)(nhận)

Nếu \(x\ge\dfrac{3}{2}\)

\(\left(1\right)\Leftrightarrow2x-3=4\Leftrightarrow2x=7\Leftrightarrow x=\dfrac{7}{2}\)(nhận)

Vậy S=\(\left\{\dfrac{-1}{2};\dfrac{7}{2}\right\}\)

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

\(\left(1\right)\Leftrightarrow1-x+\left[-\left(x+1\right)\right]=1\Leftrightarrow1-x+\left(-x-1\right)=1\Leftrightarrow1-x-x-1=1\Leftrightarrow-2x=1\Leftrightarrow x=\dfrac{-1}{2}\)(loại)

Nếu -1≤x<1

\(\left(1\right)\Leftrightarrow1-x+x+1=1\Leftrightarrow2=1\)(loại)

Nếu x≥1

\(\left(1\right)\Leftrightarrow x-1+x+1=1\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\)(loại)

Vậy S=∅

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:

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

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

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

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

\(a=2\sqrt{x+1}+\sqrt{4x+1}\)

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

<=> a^2 = -3 + 4a 

.

fsđsđf

Em chỉ mới nghĩ ra được câu a thôi.

a) ĐK: x >1/4

PT<\(\Leftrightarrow\) \(2a^2-\left(4x-1\right)a+2x-1=0\)

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

b)ĐK: x > -1

Đặt \(\sqrt{x+1}=a\). Có hệ:

\(\hept{\begin{cases}x^2-3x+2=a\left(1\right)\\a^2-1=x\left(2\right)\end{cases}}\). Thay (2) vào (1) được:

\(\left(a^2-1\right)^2-3\left(a^2-1\right)+2=a\)

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

Cái ngoặc nhỏ thi dễ nhưng cáu ngoặc sau thì em chịu rồi:(( để mai em nghĩ cách khác

\(Đk:x\ge\dfrac{3}{2}\Rightarrow x>0\)

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

\(\Leftrightarrow2x^3-8x^2+10x-2-2\sqrt{2x-3}=0\)

\(\Leftrightarrow\left(2x^3-8x^2+8x\right)+\left[\left(2x-3\right)-2\sqrt{2x-3}+1\right]=0\)

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

Ta có: \(\left\{{}\begin{matrix}2x\left(x-2\right)^2\ge0\left(x>0\right)\\\left(\sqrt{2x-3}-1\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow2x\left(x-2\right)^2+\left(\sqrt{2x-3}-1\right)^2\ge0\)

Do đó: \(\left\{{}\begin{matrix}2x\left(x-2\right)^2=0\\\left(\sqrt{2x-3}-1\right)^2=0\end{matrix}\right.\Leftrightarrow x=2\)

Thử lại ta có x=2 là nghiệm duy nhất của phương trình đã cho.

x^3-4x^2+5x-1-căn 2x-3=0

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

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

=>\(\left(x-2\right)\left[\left(x-1\right)\left(x-2\right)-\dfrac{2}{\sqrt{2x-3}+1}\right]=0\)

=>x-2=0

=>x=2

ĐLXĐ:\(x\ge-1\)

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

\(\Leftrightarrow\left[\sqrt{x^2+4x+12}-\left(6-3x\right)\right]-\left[\sqrt{x+1}-\left(x-2\right)\right]=0\)

\(\Leftrightarrow\frac{x^2+4x+12-36+36x-9x^2}{\sqrt{x^2+4x+12}+2-3x}-\frac{x+1-x^2+4x-4}{\sqrt{x+1}+x+2}=0\)

\(\Leftrightarrow\frac{-8x^2+40x-24}{\sqrt{x^2+4x+12}+2-3x}-\frac{-x^2+5x-3}{\sqrt{x+1}+x-2}=0\)

\(\Leftrightarrow\frac{8\left(-x^2+5x-3\right)}{\sqrt{x^2+4x+12}+2-3x}-\frac{-x^2+5x-3}{\sqrt{x+1}+x-2}=0\)

\(\Leftrightarrow\left(-x^2+5x-3\right)\left[\frac{8}{\sqrt{x^2+4x+12}+2-3x}-\frac{1}{\sqrt{x+1}+x-2}\right]=0\)

TH1:\(-x^2+5x-3=0\Rightarrow\orbr{\begin{cases}x=\frac{5+\sqrt{13}}{2}\\x=\frac{5-\sqrt{13}}{2}\end{cases}}\)

TH2:........ ( chắc vô nghiệm )

phần mẫu phải là \(\sqrt{x^2+4x+12}+6-3x\) chứ :vv Hơi lỗi nhưng cảm ơn nhé !!

\(x^2+4x+12=\left(x+1\right)^2+2\left(x+1\right)+9\)

Đặt \(\sqrt{x+1}=a\ge0\).

PT \(\Leftrightarrow\sqrt{a^4+2a^2+9}=2a^2+a-6\)

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

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

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

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

\(\Leftrightarrow\left(x-2\right)\left[\dfrac{x-2}{\sqrt{x-2}\left(\sqrt{x-1}+1\right)}-\dfrac{x+3}{\sqrt{x+3}\left(\sqrt{x-1}-1\right)}\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left[\dfrac{\sqrt{x-2}}{\sqrt{x-1}+1}-\dfrac{\sqrt{x+3}}{\sqrt{x-1}-1}\right]=0\)

Pt \(\dfrac{\sqrt{x-2}}{\sqrt{x-1}+1}-\dfrac{\sqrt{x+3}}{\sqrt{x-1}-1}=0\) vô n_o

(vì \(\dfrac{\sqrt{x-2}}{\sqrt{x-1}+1}< \dfrac{\sqrt{x+3}}{\sqrt{x-1}-1}\forall x\ge2\Rightarrow VT< 0\))

=> x - 2 = 0

<=> x = 2 (nhận)

\(\sqrt{4x+1}-\sqrt{3x-2}=\dfrac{x+3}{5}\)

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

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

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

TH1:

x + 3 = 0

<=> x = - 3 (loại)

TH2:

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

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

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

\(\Leftrightarrow\dfrac{4x+1-9}{\sqrt{4x+1}+3}+\dfrac{3x-2-4}{\sqrt{3x-2}+2}=0\)

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

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

Pt \(\dfrac{4}{\sqrt{4x+1}+3}+\dfrac{3}{\sqrt{3x-2}+2}>0\forall x\ge\dfrac{2}{3}\) => vô n_o

=> x - 2 = 0

<=> x = 2 (nhận)

~ ~ ~

Vậy x = 2

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

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

\(\Leftrightarrow\sqrt{2\left(x+3\right)\left(x+1\right)}-\dfrac{\left(4x^2+8x+4\right)-\left(x^2-1\right)}{\sqrt{x^2-1}+2x+2}=0\)

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

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

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

TH1

x + 1 = 0

<=> x = - 1 (loại)

TH2

\(2\sqrt{x+3}-\dfrac{3x+5}{\sqrt{x-1}+2\sqrt{x+1}}=0\)

mà \(2\sqrt{x+3}=\dfrac{4x+12}{2\sqrt{x+3}}>\dfrac{3x+5}{\sqrt{x-1}+2\sqrt{x+1}}\forall x\ge1\)

=> VT > 0

=> vô n_o

~ ~ ~

Vậy pt vô n_o