ĐKXĐ: ....

Đặt \(x+\sqrt{17-x^2}=a\ge-\sqrt{17}\Rightarrow x\sqrt{17-x^2}=\frac{a^2-17}{2}\)

Phương trình trở thành:

\(a+\frac{a^2-17}{2}=9\Leftrightarrow a^2+2a-35=0\Rightarrow\left[{}\begin{matrix}a=5\\a=-7\left(l\right)\end{matrix}\right.\)

\(\Rightarrow x+\sqrt{17-x^2}=5\)

\(\Leftrightarrow\sqrt{17-x^2}=5-x\)

\(\Leftrightarrow17-x^2=x^2-10x+25\)

\(\Leftrightarrow2x^2-10x+8=0\Rightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)

Lời giải:

ĐKXĐ:......

Ta có: Đặt \(y=\sqrt{17-x^2}\Rightarrow x^2+y^2=17\)

Ta chuyển phương trình về hệ phương trình:

\(\left\{\begin{matrix} x+y+xy=9\\ x^2+y^2=17\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} xy=9-(x+y)\\ (x+y)^2-2xy=17\end{matrix}\right.\)

\(\Rightarrow (x+y)^2-2[9-(x+y)]=17\)

\(\Leftrightarrow (x+y)^2+2(x+y)-35=0\)

\(\Leftrightarrow (x+y-5)(x+y+7)=0\)

Nếu \(x+y=5\Rightarrow xy=9-5=4\)

Theo định lý Viete đảo thì $x,y$ là nghiệm của PT: \(X^2-5X+4=0\)

\(\Rightarrow (x,y)=(1,4)\Leftrightarrow (x,\sqrt{17-x^2})=(1,4)\)

\(\Rightarrow x=1\)

Nếu \(x+y=-7\Rightarrow xy=9-(-7)=16\)

Vì \(x+y<0; y\geq 0\Rightarrow x< 0\Rightarrow xy\leq 0\Leftrightarrow 16\leq 0\) (vô lý nên loại)

Vậy \(x=1\)

Đk:\(-\sqrt{17}\le x\le\sqrt{17}\)

Đặt \(t=x+\sqrt{17-x^2}\left(t>0\right)\)

\(\Rightarrow t^2=17+2x\sqrt{17-x^2}\)

\(\Rightarrow x\sqrt{17-x^2}=\frac{t^2-17}{2}\)

thay vào pt 

\(t+\frac{t^2-17}{2}=9\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}t=-7\left(loai\right)\\t=5\left(tm\right)\end{array}\right.\)

\(\Rightarrow x+\sqrt{17-x^2}=5\)

\(\Leftrightarrow\sqrt{17-x^2}=5-x\)

Với \(x< \sqrt{17}\) bình 2 vế ta có:

\(17-x^2=x^2-10x+25\)

\(\Leftrightarrow2x^2-10x+8=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=4\end{cases}\left(tm\right)}\)

dòng cuối là \(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=4\end{array}\right.\)(thỏa mãn)

â) \(\sqrt{x+9}=7\\ \Rightarrow x+9=49\\ \Rightarrow x=40\)

b) \(\sqrt{x-4}=4-x\\ \Rightarrow x-4=16-8x+x^2\\ \Rightarrow x^2-9x+20=0\\ \Rightarrow\left(x-4\right)\left(x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

c) \(\sqrt{x^2-12x+36}=81\\ \Rightarrow x-6=81\\ \Rightarrow x=87\)

a: Ta có: \(\sqrt{x+9}=7\)

\(\Leftrightarrow x+9=49\)

hay x=40

b: Ta có: \(\sqrt{x-4}=4-x\)

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

\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=5\left(loại\right)\end{matrix}\right.\)

c: Ta có: \(\sqrt{x^2-12x+36}=81\)

\(\Leftrightarrow\left|x-6\right|=81\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=81\\x-6=-81\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=87\\x=-75\end{matrix}\right.\)

a: Ta có: \(\sqrt{4-3x}=8\)

\(\Leftrightarrow4-3x=64\)

\(\Leftrightarrow3x=-60\)

hay x=-20

b: ta có: \(\sqrt{4x-8}-12\sqrt{\dfrac{x-2}{9}}=-1\)

\(\Leftrightarrow2\sqrt{x-2}-12\cdot\dfrac{\sqrt{x-2}}{3}=-1\)

\(\Leftrightarrow x-2=\dfrac{1}{4}\)

hay \(x=\dfrac{9}{4}\)

\(\left\{{}\begin{matrix}8>0\left(luondung\right)\\4-3x=64\end{matrix}\right.\) \(\Leftrightarrow x=-20\left(ktm\right)\)

a:Ta có: \(\sqrt{2x+9}=\sqrt{5-4x}\)

\(\Leftrightarrow2x+9=5-4x\)

\(\Leftrightarrow6x=-4\)

hay \(x=-\dfrac{2}{3}\left(nhận\right)\)

b: Ta có: \(\sqrt{2x-1}=\sqrt{x-1}\)

\(\Leftrightarrow2x-1=x-1\)

hay x=0(loại)

c: Ta có: \(\sqrt{x^2+3x+1}=\sqrt{x+1}\)

\(\Leftrightarrow x^2+3x=x\)

\(\Leftrightarrow x\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-2\left(loại\right)\end{matrix}\right.\)

a. \(\sqrt{2x+9}=\sqrt{5-4x}\)

<=> 2x + 9 = 5 - 4x 

<=> 2x + 4x = 5 - 9

<=> 6x = -4

<=> x = \(\dfrac{-4}{6}=\dfrac{-2}{3}\)

d: Ta có: \(\sqrt{2x^2-3}=\sqrt{4x-3}\)

\(\Leftrightarrow2x^2-3=4x-3\)

\(\Leftrightarrow2x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

2:

ĐKXĐ: x>=3

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

=>\(\left|\sqrt{x-3}+\sqrt{3}\right|+\left|\sqrt{x-3}-\sqrt{3}\right|=2\sqrt{3}\)

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

\(\Leftrightarrow\sqrt{x-3}+\left|\sqrt{x-3}-\sqrt{3}\right|=\sqrt{3}\)(1)

TH1: x>=6

(1) trở thành \(\sqrt{x-3}+\sqrt{x-3}-\sqrt{3}=\sqrt{3}\)

=>\(2\sqrt{x-3}=2\sqrt{3}\)

=>x-3=3

=>x=6(nhận)

TH2: 3<=x<6

Phương trình (1) sẽ là;

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

=>\(\sqrt{3}=\sqrt{3}\)(luôn đúng)

1:

\(A^2=8+2\sqrt{10+2\sqrt{5}}+8-2\sqrt{10+2\sqrt{5}}+2\cdot\sqrt{8^2-\left(2\sqrt{10+2\sqrt{5}}\right)^2}\)

\(=16+2\cdot\sqrt{64-4\cdot\left(10+2\sqrt{5}\right)}\)

\(=16+2\cdot\sqrt{24-8\sqrt{5}}\)

\(=16+2\cdot\sqrt{20-2\cdot2\sqrt{5}\cdot2+4}\)

\(=16+2\cdot\sqrt{\left(2\sqrt{5}-2\right)^2}\)

\(=16+2\cdot\left(2\sqrt{5}-2\right)=12+4\sqrt{5}\)

\(=10+2\cdot\sqrt{10}\cdot\sqrt{2}+2\)

\(=\left(\sqrt{10}+\sqrt{2}\right)^2\)

=>\(A=\sqrt{10}+\sqrt{2}\)

a) Ta có: \(\sqrt{\left(x+1\right)^2}=3\)

\(\Leftrightarrow\left|x+1\right|=3\)

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

b) Ta có: \(3\sqrt{4x+4}-\sqrt{9x-9}-8\sqrt{\dfrac{x+1}{16}}=5\)

\(\Leftrightarrow6\sqrt{x+1}-3\sqrt{x-3}-2\sqrt{x+1}=5\)

\(\Leftrightarrow4\sqrt{x+1}=5+3\sqrt{x-3}\)

\(\Leftrightarrow16\left(x+1\right)=25+30\sqrt{x-3}+9\left(x-3\right)\)

\(\Leftrightarrow16x+16=25+9x-27+30\sqrt{x-3}\)

\(\Leftrightarrow30\sqrt{x-3}=16x+16+2-9x\)

\(\Leftrightarrow30\sqrt{x-3}=7x+18\)

\(\Leftrightarrow x-3=\left(\dfrac{7x+18}{30}\right)^2\)

\(\Leftrightarrow x-3=\dfrac{49x^2}{900}+\dfrac{7}{25}x+\dfrac{9}{25}\)

\(\Leftrightarrow\dfrac{49}{900}x^2-\dfrac{18}{25}x+\dfrac{84}{25}=0\)

\(\Delta=\left(-\dfrac{18}{25}\right)^2-4\cdot\dfrac{49}{900}\cdot\dfrac{84}{25}=-\dfrac{16}{75}< 0\)

Vậy: Phương trình vô nghiệm

a)Pt\(\Leftrightarrow\left|x+1\right|=3\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

b)Đk:\(x\ge-1\)

Sửa đề: \(3\sqrt{4x+4}-\sqrt{9x+9}-8\sqrt{\dfrac{x+1}{16}}=5\)

Pt \(\Leftrightarrow6\sqrt{x+1}-3\sqrt{x+1}-2\sqrt{x+1}=5\)

\(\Leftrightarrow\sqrt{x+1}=5\)

\(\Leftrightarrow x=24\left(tm\right)\)

a.  \(\sqrt{\left(x+1\right)^2}\)  \(=3\)

⇔   \(\left|x+1\right|=3\)

⇔    \(\left|x\right|=2\)

⇒     \(x=2\) và  \(x=-2\)

a) \(\sqrt{x^2-x-4}=\sqrt{x-1}\)

\(x^2-x-4=x-1\)

\(x^2-x-4-x+1=0\)

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

\(\left(x^2-2.x.1+1^2\right)-6=0\)

\(\left(x-1\right)^2=6\)

⇒\(\left\{{}\begin{matrix}x-1=6\\x-1=-6\end{matrix}\right.\)         ⇒\(\left\{{}\begin{matrix}x=7\\x=-5\end{matrix}\right.\)