ĐK: \(x>-1\)

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

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

.Làm nốt. 

~Ko chắc~

À quên: Đặt \(a=\sqrt{x^2-2x+3}\ge\sqrt{2}\)

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

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

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

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

<=>x+1=0

<=>x=-1

Bài 1 :

Mình nghĩ phải sửa đề ntn :

\(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

\(\Leftrightarrow\left[2\left(2x+7\right)\right]^2-\left[3\left(x+3\right)\right]^2=0\)

\(\Leftrightarrow\left[2\left(2x+7\right)-3\left(x+3\right)\right]\left[2\left(2x+7\right)+3\left(x+3\right)\right]=0\)

\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\7x+23=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=\frac{-23}{7}\end{cases}}}\)

Vậy....

b) \(A=\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

Đặt \(q=x^2+x+1\)ta có :

\(A=q\left(q+1\right)-12\)

\(A=q^2+q-12\)

\(A=q^2+4q-3q-12\)

\(A=q\left(q+4\right)-3\left(q+4\right)\)

\(A=\left(q+4\right)\left(q-3\right)\)

Thay \(q=x^2+x+1\)ta có :

\(A=\left(x^2+x+1+4\right)\left(x^2+x+1-3\right)\)

\(A=\left(x^2+x+5\right)\left(x^2+x-2\right)\)

\(A=\left(x^2+x+5\right)\left(x^2+2x-x-2\right)\)

\(A=\left(x^2+x+5\right)\left[x\left(x+2\right)-\left(x+2\right)\right]\)

\(A=\left(x^2+x+5\right)\left(x+2\right)\left(x-1\right)\)

Cảm ơn ạ><

Cách 2 của câu 2:

Đặt \(x^2+x+2=t\)

Ta có: \(A=\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

\(=t\left(t-1\right)-12=t^2-t-12\)

\(=\left(t-4\right)\left(t+3\right)=\left(x^2+x-2\right)\left(x^2+x+5\right)\)

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

ĐK  \(x\ge0\)

Đặt \(x=a,x+1=b\)

\(PT\Leftrightarrow a^4+b^4=\left(a+b\right)^4\)

<=> 4a³b+6a²b²+4ab³=0

<=> ab(2a²+3ab+2b²)=0

=>ab=0 (vì 2a²+3ab+2b²>0)

=>\(\orbr{\begin{cases}a=0\\b=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

Vậy.............................

vào câu hỏi tương tự nhé bạn

Đặt \(\dfrac{x}{\sqrt{4x-1}}=a\)

Theo đề, ta có phương trình:

a+1/a=2

\(\Leftrightarrow a+\dfrac{1}{a}=2\)

\(\Leftrightarrow\dfrac{a^2+1-2a}{a}=0\)

=>a=1

=>\(x=\sqrt{4x-1}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2=4x-1\\x>=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=3\\x>=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow x\in\left\{2+\sqrt{3};2-\sqrt{3}\right\}\)

Đặt \(u=\sqrt{x+1};t=\sqrt{1-x};\text{đ}k:-1\le x\le1\)

Phương trình trở thành:

\(u+2u^2=-t^2+t+3ut\Leftrightarrow\left(u-t\right)^2+u\left(u-t\right)+\left(u-t\right)=0\)

\(\Leftrightarrow\left(u-t\right)\left(2u-t+1\right)=0\Leftrightarrow\orbr{\begin{cases}u=t\\2u+1=t\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x+1}=\sqrt{1-x}\\2\sqrt{x+1}+1=\sqrt{1-x}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{-24}{25}\end{cases}}}\)

mình dùng cách khác nhé :((

\(\sqrt{x+1}+2\left(x+1\right)=x-1+\sqrt{1-x}+3\sqrt{1-x^2}\left(đk:-1\le x\le1\right)\)

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

\(< =>\frac{x}{\sqrt{x+1}+1}+2x-1-x+1=-\frac{x}{\sqrt{1-x}+1}+\frac{9\left(1-x^2-1\right)}{3\sqrt{1-x^2}+3}\)

\(< =>\frac{x}{\sqrt{x+1}+1}+x+\frac{x}{\sqrt{1-x}+1}+\frac{9x^2}{3\sqrt{1-x^2}+3}=0\)

\(< =>x\left(\frac{1}{\sqrt{x+1}+1}+1+\frac{1}{\sqrt{1+x}+1}+\frac{9x}{3\sqrt{1-x^2}+3}\right)=0< =>x=0\)

rồi đến đây dùng đk đánh giá cái ngoặc khác 0 là ok