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)

a.\(2\sqrt{12x}-3\sqrt{3x}+4\sqrt{48x}=17\)

=>\(4\sqrt{3x}-3\sqrt{3x}+16\sqrt{3x}=17\)

=>\(17\sqrt{3x}=17\)

=>\(\sqrt{3x}=1\)

=>\(x=\dfrac{1}{3}\)

b.Ta có:\(\sqrt{x^2-6x+9}=1\)

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

=>\(\left|x-3\right|=1\)

Vậy có hai trường hợp:

TH1:\(x-3=1\)

=>\(x=4\)

TH2:\(x-3=-1\)

=>\(x=2\)

a) ĐKXĐ: \(x\ge0\)

Ta có: \(2\sqrt{12x}-3\sqrt{3x}+4\sqrt{48x}=17\)

\(\Leftrightarrow2\cdot2\cdot\sqrt{3x}-3\cdot\sqrt{3x}+4\cdot4\cdot\sqrt{3x}=17\)

\(\Leftrightarrow4\sqrt{3x}-3\sqrt{3x}+16\sqrt{3x}=17\)

\(\Leftrightarrow17\sqrt{3x}=17\)

\(\Leftrightarrow\sqrt{3x}=1\)

\(\Leftrightarrow3x=1\)

hay \(x=\dfrac{1}{3}\)(nhận)

Vậy: \(S=\left\{\dfrac{1}{3}\right\}\)

b) ĐKXĐ: \(x\in R\)

Ta có: \(\sqrt{x^2-6x+9}=1\)

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

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

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

Vậy: S={2;4}

đây là PT sao lại giải BPT?

sorry mih ghi nhầm bn ạ mà chẳng wan trọng lắm đâu bn cứ tập trung mà giải hộ mình cái phương trình ấy

Đk:\(-\sqrt{17}\le x\le\sqrt{17}\)

Khi \(y=\sqrt{17-x^2}\ge0\) thì ta có hpt

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

Đặt \(\hept{\begin{cases}S=x+y\\P=xy\end{cases}}\left(S^2\ge4P\right)\) ta có:

\(hpt\Leftrightarrow\hept{\begin{cases}S+P=9\\S^2-2P=17\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}S=9-P\\S^2-2P=17\end{cases}}\)

\(\Leftrightarrow\left(9-P\right)^2-2P=17\Leftrightarrow\left(P-4\right)\left(P-16\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}P=4\\P=16\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}S=9-P=9-4=5\\S=9-P=9-16=-7\left(loai\right)\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x+y=5\\xy=4\end{cases}}\Rightarrow\orbr{\begin{cases}\hept{\begin{cases}x=4\\y=1\end{cases}}\\\hept{\begin{cases}x=1\\y=4\end{cases}}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x+y=5\\xy=4\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\hept{\begin{cases}x=4\\y=1\end{cases}}\\\begin{cases}x=1\\y=4\end{cases}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=4\\y=1\end{cases};\hept{\begin{cases}x=1\\y=4\end{cases}}}\) (thỏa mãn)

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 \(x=1\)

a) đk: \(1\le x\le5\)

 \(\sqrt[4]{5-x}+\sqrt[4]{x-1}=\sqrt{2}\)

<=> \(\left(\sqrt[4]{5-x}+\sqrt[4]{x-1}\right)^4=\sqrt{2}^4\)

<=> \(5-x+x-1+4\sqrt[4]{5-x}^3.\sqrt[4]{x-1}+6\sqrt[4]{5-x}^2.\sqrt[4]{x-1}^2+4\sqrt[4]{5-x}.\sqrt[4]{x-1}^3=4\)

<=> \(\sqrt[4]{\left(5-x\right)\left(x-1\right)}.\left(2\sqrt[4]{5-x}^2+3\sqrt[4]{5-x}.\sqrt[4]{x-1}+2\sqrt[4]{x-1}^2\right)=0\)

<=> \(\left[{}\begin{matrix}\sqrt[4]{\left(5-x\right)\left(x-1\right)}=0\left(2\right)\\2\sqrt[4]{5-x}^2+3\sqrt[4]{\left(5-x\right)\left(x-1\right)}+2\sqrt[4]{x-1}^2=0\left(1\right)\end{matrix}\right.\)

Giải (2) <=> \(\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\left(tm\right)\)

Giải (1) : Đặt \(\sqrt[4]{5-x}=a;\sqrt[4]{x-1}=b\)(đk : a, b \(\ge\)0)

Khi đó, ta có: \(2a^2+3ab+2b^2=0\)

<=> 2(a² + 3/2ab + 9/16b²) + \(\dfrac{7}{8}b^2=0\)

<=> \(2\left(a+\dfrac{3}{4}b\right)^2+\dfrac{7}{8}b^2=0\)

<=> \(\left\{{}\begin{matrix}a+\dfrac{3}{4}b=0\\b=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}a=0\\b=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}\sqrt[4]{x-1}=0\\\sqrt[4]{5-x}=0\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)(vô lí)

câu a

Học tại nhà - Toán - Bài 110035

b,  ĐK \(x\ge-4\)

PT 

<=> \(\left(x-\sqrt{x+4}\right)+\left(\sqrt{2x^2-10x+17}-2x+3\right)=0\)

<=> \(\frac{x^2-x-4}{x+\sqrt{x+4}}+\frac{-2x^2+2x+8}{\sqrt{2x^2-10x+17}+2x-3}=0\)với \(x+\sqrt{x+4}\ne0\)

<=> \(\frac{x^2-x-4}{x+\sqrt{x+4}}-\frac{2\left(x^2-x-4\right)}{\sqrt{2x^2-10x+17}+2x-3}=0\)

<=> \(\orbr{\begin{cases}x^2-x-4=0\\\frac{1}{x+\sqrt{x+4}}-\frac{2}{\sqrt{2x^2-10x+17}+2x-3}=0\left(2\right)\end{cases}}\)

Giải (2)

=> \(2x+2\sqrt{x+4}=2x-3+\sqrt{2x^2-10x+17}\)

<=> \(\sqrt{2x^2-10x+17}=2\sqrt{x+4}+3\)

<=> \(2x^2-10x+17=4\left(x+4\right)+9+12\sqrt{x+4}\)

<=> \(x^2-7x-4=6\sqrt{x+4}\)

<=> \(\left(x-6\right)^2+5x-40=6\sqrt{6\left(x-6\right)-5x+40}\)

Đặt x-6=a;\(\sqrt{6\left(x-6\right)-5x+40}=b\)

=> \(\hept{\begin{cases}a^2+5x-40=6b\\b^2+5x-40=6a\end{cases}}\)

=> \(a^2-b^2+6\left(a-b\right)=0\)

<=> \(\orbr{\begin{cases}a=b\\a+b+6=0\end{cases}}\)

+ a=b

=> \(x-6=\sqrt{x+4}\)

=> \(\hept{\begin{cases}x\ge6\\x^2-13x+32=0\end{cases}}\)=> \(x=\frac{13+\sqrt{41}}{2}\)

+ a+b+6=0

=> \(x+\sqrt{x+4}=0\)(loại)

Vậy \(S=\left\{\frac{13+\sqrt{41}}{2};\frac{1+\sqrt{17}}{2}\right\}\)

a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)

\(\Leftrightarrow3\sqrt{x+5}=6\)

\(\Leftrightarrow x+5=4\)

hay x=-1

b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=289\)

hay x=290