\(\sqrt{x^2+16}-\sqrt{x^2+7}=3x-8\)

\(\Leftrightarrow\left(\sqrt{x^2+16}-5\right)+\left(4-\sqrt{x^2+7}\right)=3x-9\)

\(\Leftrightarrow\frac{x^2-9}{\sqrt{x^2+16}+5}+\frac{9-x^2}{\sqrt{x^2+7}+4}=3\left(x-3\right)\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{x+3}{\sqrt{x^2+16}+5}-\frac{x+3}{\sqrt{x^2+7}+4}-3\right)=0\)

\(\Leftrightarrow x=3\)

2:

\(A=\dfrac{x_2-1+x_1-1}{x_1x_2-\left(x_1+x_2\right)+1}\)

\(=\dfrac{3-2}{-7-3+1}=\dfrac{1}{-9}=\dfrac{-1}{9}\)

B=(x1+x2)^2-2x1x2

=3^2-2*(-7)

=9+14=23

C=căn (x1+x2)^2-4x1x2

=căn 3^2-4*(-7)=căn 9+28=căn 27

D=(x1^2+x2^2)^2-2(x1x2)^2

=23^2-2*(-7)^2

=23^2-2*49=431

D=9x1x2+3(x1^2+x2^2)+x1x2

=10x1x2+3*23

=69+10*(-7)=-1

ĐKXĐ: \(x\ge1\).

Phương trình đã cho tương đương:

\(\sqrt{x+3}+\sqrt{x-1}=\dfrac{8}{\sqrt{4x^4-12x^3+9x^2+16}-\left(2x^2-3x\right)}\)

\(\Leftrightarrow\sqrt{x+3}+\sqrt{x-1}=\dfrac{\sqrt{4x^4-12x^3+9x^2+16}+\left(2x^2-3x\right)}{2}\)

\(\Leftrightarrow\sqrt{4x^4-12x^3+9x^2+16}+\left(2x^2-3x\right)-2\sqrt{x+3}-2\sqrt{x-1}=0\)

\(\Leftrightarrow\left(\sqrt{4x^4-12x^3+9x^2+16}-2\sqrt{x+3}\right)+\left(2x^2-3x-2\sqrt{x-1}\right)=0\)

\(\Leftrightarrow\dfrac{4x^4-12x^3+9x^2-4x+4}{\sqrt{4x^4-12x^3+9x^2+16}+2\sqrt{x+3}}+\dfrac{4x^4-12x^3+9x^2-4x+4}{2x^2-3x+2\sqrt{x-1}}=0\)

\(\Leftrightarrow\left(x-2\right)\left(4x^3-4x^2+x-2\right)\left(\dfrac{1}{\sqrt{4x^4-12x^3+9x^2+16}+2\sqrt{x+3}}+\dfrac{1}{2x^2-3x+2\sqrt{x-1}}\right)=0\).

Do \(x\ge1\) nên ta có \(\dfrac{1}{\sqrt{4x^4-12x^3+9x^2+16}+2\sqrt{x+3}}+\dfrac{1}{2x^2-3x+2\sqrt{x-1}}>0\).

Do đó \(\left[{}\begin{matrix}x-2=0\Leftrightarrow x=2\left(TMĐK\right)\\4x^3-4x^2+x-2=0\left(1\right)\end{matrix}\right.\).

Giải phương trình bậc 3 ở (1) ta được \(x=\dfrac{\sqrt[3]{36\sqrt{13}+53\sqrt{6}}}{\sqrt[6]{279936}}+\dfrac{1}{\sqrt[6]{7776}\sqrt[3]{36\sqrt{13}+53\sqrt{6}}}+\dfrac{1}{3}\approx1,157298106\left(TMĐK\right)\).

Vậy...

Vì trong bài làm của mình có một số dòng khá dài nên bạn có thể vào trang cá nhân của mình để đọc tốt hơn!

c.

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

Đặt \(\sqrt{x^2+3}=t>0\)

\(\Rightarrow t^2-\left(3x+1\right)t+2x^2+2x=0\)

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

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{3x+1-x+1}{2}=x+1\\t=\dfrac{3x+1+x-1}{2}=2x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+3}=x+1\left(x\ge-1\right)\\\sqrt{x^2+3}=2x\left(x\ge0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+3=x^2+2x+1\left(x\ge-1\right)\\x^2+3=4x^2\left(x\ge0\right)\end{matrix}\right.\)

\(\Leftrightarrow x=1\)

a.

Đề bài ko chính xác, pt này ko giải được

b.

ĐKXĐ: \(x\ge-\dfrac{7}{2}\)

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

Đặt \(\sqrt{2x+7}=t\ge0\)

\(\Rightarrow t^2-\left(2x+7\right)t+x^2+7x=0\)

\(\Delta=\left(2x+7\right)^2-4\left(x^2+7x\right)=49\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{2x+7-7}{2}=x\\t=\dfrac{2x+7+7}{2}=x+7\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-7=0\left(x\ge0\right)\\x^2+12x+42=0\left(vn\right)\end{matrix}\right.\)

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

\(\left(\sqrt{x^2+16}-5\right)\)\(-3\left(x-3\right)-\left(\sqrt{x^2+7}-4\right)=0\)

\(\Leftrightarrow\frac{\left(\sqrt{x^2+16}-5\right)\left(\sqrt{x^2+16}+5\right)}{\sqrt{x^2+16}+5}\)\(-3\left(x-3\right)-\frac{\left(\sqrt{x^2+7}-4\right)\left(\sqrt{x^2+7}+4\right)}{\sqrt{x^2+7}+4}=0\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{\sqrt{x^2+16}+5}-3-\frac{1}{\sqrt{x^2+7}+4}\right)=0\)

ben trong ngoac bn tu xu li nhe

\(\Rightarrow x=3\)