ĐKXĐ: mọi \(x\)

Ta có \(x^2+4x+7=\left(x+4\right)\sqrt{x^2+7}\)

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

\(\Leftrightarrow\left(x+4\right)\left(\sqrt{x^2+7}-4\right)-x^2-4x+4x-7+16=0\) ( thêm bớt )

\(\Leftrightarrow\left(x+4\right)\left(\sqrt{x^2+7}-4\right)-\left(x^2-9\right)=0\)

\(\Leftrightarrow\left(x+4\right)\dfrac{x^2-9}{\sqrt{x^2+7}+4}-\left(x^2-9\right)=0\)

\(\Leftrightarrow\left(x^2-9\right)\left(\dfrac{x+4}{\sqrt{x^2+7}+4}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-9=0\\\dfrac{x+4}{\sqrt{x^2+7}+4}-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\pm3\\\dfrac{x+4}{\sqrt{x^2+7}+4}=1\left(\text{*}\right)\end{matrix}\right.\)

Giải (*), ta được phương trình

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

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

\(\Leftrightarrow x^2+7=x^2\)

\(\Leftrightarrow7=0\) ( vô lý )

Suy ra phương trình (*) vô nghiệm 

Vậy \(S=\left\{\pm3\right\}\)

a, Ta có : \(\left\{{}\begin{matrix}x^2+y^2=1\\x^2-y^2-x+y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=1\\\left(x-y\right)\left(x+y\right)-\left(x-y\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=1\\\left(x-y\right)\left(x+y-1\right)=0\end{matrix}\right.\)

TH1 : \(x-y=0\Rightarrow x=y\)

- Thay vào PT ( I ) ta được : \(x^2+x^2=2x^2=1\)

\(\Rightarrow x=y=\dfrac{\sqrt{2}}{2}\)

TH2 : \(x+y-1=0\)

- Kết hợp PT ( I ) ta được hệ : \(\left\{{}\begin{matrix}x+y=1\\\left(x+y\right)^2-2xy=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=1\\-2xy=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=1\\xy=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\end{matrix}\right.\)

Vậy hệ phương trình có tập nghiệm là \(S=\left\{\left(\dfrac{\sqrt{2}}{2};\dfrac{\sqrt{2}}{2}\right);\left(1;0\right);\left(0;1\right)\right\}\)

b.

Đặt \(\sqrt{x^2+7}=t>0\)

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

\(\Delta=\left(x+4\right)^2-16x=\left(x-4\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{x+4+x-4}{2}=x\\t=\dfrac{x+4-x+4}{2}=4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2+7}=x\left(x\ge0\right)\\\sqrt{x^2+7}=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+7=x^2\left(vô-nghiệm\right)\\x^2+7=16\end{matrix}\right.\)

\(\Rightarrow x=\pm3\)

1. ĐKXĐ: $x\geq \frac{-3}{5}$

PT $\Leftrightarrow 5x+3=3-\sqrt{2}$

$\Leftrightarrow x=\frac{-\sqrt{2}}{5}$

2. ĐKXĐ: $x\geq \sqrt{7}$ 

PT $\Leftrightarrow (\sqrt{x}-7)(\sqrt{x}+7)=4$

$\Leftrightarrow x-49=4$

$\Leftrightarrow x=53$ (thỏa mãn)

a,\(x+4\sqrt{7-x}\) \(-4\sqrt{x-1}-\sqrt{\left(7-x\right)\left(x-1\right)}-1=0\) (dk \(1\le x\le7\) )

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-1}=4\\\sqrt{x-1}=\sqrt{7-x}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=17\left(l\right)\\x=4\left(tm\right)\end{cases}}}\)

mà sao bạn k làm giúp mình câu b

bạn ơi mình sai đề câu b

b, \(x^2+2x+4=3\sqrt{x^3+4x}\)

Đặt \(\sqrt{x^2+7}=a\left(a>0\right)\)

Khi đó phương trình trở thành :

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

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

\(\Leftrightarrow\left(a-x\right)\left(a-4\right)=0\Leftrightarrow\orbr{\begin{cases}a-x=0\\a-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a=x\\a=4\end{cases}}}\)

+) \(a=x\Rightarrow\sqrt{x^2+7}=x\)( điều kiện bổ sung \(x\ge0\))

\(\Leftrightarrow x^2+7=x^2\Leftrightarrow7=0\)( vô lý ) => loại

+) \(a=4\)( thỏa mãn điều kiện a > 0 )  \(\Rightarrow\sqrt{x^2+7}=4\Leftrightarrow x^2+7=16\)

\(\Leftrightarrow x^2=9\Leftrightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)

Vậy phương trình có tập nghiệm S = { 3 ; -3 }

Tích cho mk nhoa !!!! ~~

P/S: Không cần đặt ẩn phụ cho phí t/g!

\(ĐK:x\inℝ\)

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x^2+7}=x\left(1\right)\\\sqrt{x^2+7}=4\left(2\right)\end{cases}}\)

Giải (1) ta thấy vô nghiệm

\(\left(2\right)\Leftrightarrow x^2+7=16\Leftrightarrow x^2=9\Leftrightarrow x=\pm3\)

Vậy phương trình có tập nghiệm S = {3;-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:

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

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

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

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