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\(\sqrt{x^2+16}-\sqrt{x^2+7}=3x-8\)
\(\Leftrightarrow\left(\sqrt{x^2+16}-5\right)+\left(4-\sqrt{x^2+7}\right)=3x-9\)
\(\Leftrightarrow\frac{x^2-9}{\sqrt{x^2+16}+5}+\frac{9-x^2}{\sqrt{x^2+7}+4}=3\left(x-3\right)\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{x+3}{\sqrt{x^2+16}+5}-\frac{x+3}{\sqrt{x^2+7}+4}-3\right)=0\)
\(\Leftrightarrow x=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Vì \(3x-8=\sqrt{x^2+16}-\sqrt{x^2+7}>0\Rightarrow x>\frac{8}{3}\Rightarrow 3x-4>0\)
PT \(\Leftrightarrow 3x-8+\sqrt{x^2+7}-\sqrt{x^2+16}=0\)
\(\Leftrightarrow [(3x-4)-\sqrt{x^2+16}]+(\sqrt{x^2+7}-4)=0\)
\(\Leftrightarrow \frac{(3x-4)^2-(x^2+16)}{3x-4+\sqrt{x^2+16}}+\frac{x^2+7-16}{\sqrt{x^2+7}+4}=0\) (liên hợp)
\(\Leftrightarrow \frac{8x^2-24x}{3x-4+\sqrt{x^2+16}}+\frac{x^2-9}{\sqrt{x^2+7}+4}=0\)
\(\Leftrightarrow (x-3)\left(\frac{8x}{3x-4+\sqrt{x^2+16}}+\frac{x+3}{\sqrt{x^2+7}+4}\right)=0\)
Với mọi \(x>\frac{8}{3}\) ta dễ thấy \(\frac{8x}{3x-4+\sqrt{x^2+16}}+\frac{x+3}{\sqrt{x^2+7}+4}>0\)
Do đó $x-3=0$ hay $x=3$ là nghiệm duy nhất của pt.
Với \(\forall x>\dfrac{8}{3}\) thì \(3x>4\) mà \(x^2>0\) \(\Rightarrow3x-4+\sqrt{x^2+16}>0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\sqrt{x^2+16}-5\right)\)\(-3\left(x-3\right)-\left(\sqrt{x^2+7}-4\right)=0\)
\(\Leftrightarrow\frac{\left(\sqrt{x^2+16}-5\right)\left(\sqrt{x^2+16}+5\right)}{\sqrt{x^2+16}+5}\)\(-3\left(x-3\right)-\frac{\left(\sqrt{x^2+7}-4\right)\left(\sqrt{x^2+7}+4\right)}{\sqrt{x^2+7}+4}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{\sqrt{x^2+16}+5}-3-\frac{1}{\sqrt{x^2+7}+4}\right)=0\)
ben trong ngoac bn tu xu li nhe
\(\Rightarrow x=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3x-2=\sqrt[]{x^2+15}-\sqrt[]{x^2+8}=\dfrac{7}{\sqrt[]{x^2+15}+\sqrt[]{x^2+8}}>0\)
\(\Rightarrow x>\dfrac{2}{3}\)
\(\sqrt[]{x^2+15}-4=3x-3+\sqrt[]{x^2+8}-3\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{\sqrt[]{x^2+15}+4}=3\left(x-1\right)+\dfrac{\left(x-1\right)\left(x+1\right)}{\sqrt[]{x^2+8}+3}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\dfrac{x+1}{\sqrt[]{x^2+15}+4}=3+\dfrac{x+1}{\sqrt[]{x^2+8}+3}\left(1\right)\end{matrix}\right.\)
Do \(x>\dfrac{2}{3}\Rightarrow x+1>0\Rightarrow\dfrac{x+1}{\sqrt[]{x^2+15}+4}< \dfrac{x+1}{\sqrt[]{x^2+8}+3}\)
\(\Rightarrow\) (1) vô nghiệm hay pt có nghiệm duy nhất \(x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b)đk:\(x\ge\dfrac{1}{2}\)
Có: \(\sqrt{2x^2-1}\le\dfrac{2x^2-1+1}{2}=x^2\)
\(x\sqrt{2x-1}=\sqrt{\left(2x^2-x\right)x}\le\dfrac{2x^2-x+x}{2}=x^2\)
=>\(\sqrt{2x^2-1}+x\sqrt{2x-1}\le2x^2\)
Dấu = xảy ra\(\Leftrightarrow x=1\)
Vậy....
c) đk: \(x\ge0\)
\(\Leftrightarrow\sqrt{x}=\sqrt{x+9}-\dfrac{2\sqrt{2}}{\sqrt{x+1}}\)
\(\Rightarrow x=x+9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)
\(\Leftrightarrow0=9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)
Đặt \(a=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\left(a>0\right)\)
\(\Leftrightarrow\dfrac{a^2-2}{2}=\dfrac{8}{x+1}\)
pttt \(9+\dfrac{a^2-2}{2}-4a=0\) \(\Leftrightarrow a=4\) (TM)
\(\Rightarrow4=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\) \(\Leftrightarrow16=\dfrac{2\left(x+9\right)}{x+1}\) \(\Leftrightarrow x=\dfrac{1}{7}\) (TM)
Vậy ...
a)ĐKXĐ: x≥-1/3; x≤6
<=>\(\dfrac{3x-15}{\sqrt{3x+1}+4}+\dfrac{x-5}{\sqrt{x-6}+1}+\left(x-5\right)\cdot\left(3x+1\right)=0\Leftrightarrow\left(x-5\right)\cdot\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{\sqrt{x-6}+1}+3x+1\right)=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)(nhận)
(vì x≥-1/3 nên3x+1≥0 )
\(\sqrt{x^2+16}+8=3x+\sqrt{x^2+7}\)
\(\Leftrightarrow3x+\sqrt{x^2+7}-\sqrt{x^2+16}-8=0\)
\(\Leftrightarrow\frac{\left(\sqrt{x^2+7}-4\right)\left(\sqrt{x^2+7}+4\right)}{\sqrt{x^2+7}-4}-\frac{\left(\sqrt{x^2+16}+5\right)\left(\sqrt{x^2+16}-5\right)}{\sqrt{x^2+16}-5}+3x-9=0\)
\(\Leftrightarrow\frac{x^2+7-16}{\sqrt{x^2+7}-4}-\frac{x^2+16-25}{\sqrt{x^2+16}-5}+3\left(x-3\right)=0\)
\(\Leftrightarrow\frac{x^2-9}{\sqrt{x^2+7}-4}-\frac{x^2-9}{\sqrt{x^2+16}-5}+3\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{x+3}{\sqrt{x^2+7}-4}-\frac{x+3}{\sqrt{x^2+16}-5}+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\\frac{x+3}{\sqrt{x^2+7}-4}-\frac{x+3}{\sqrt{x^2+16}-5}+3=0\end{cases}}\)
:v thoi