a) Đk: \(\left[{}\begin{matrix}x\le-1\\x\ge1\end{matrix}\right.\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-1}-1=0\\\sqrt{x^2-1}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-1}=1\\x^2-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=2\left(1\right)\\x^2=1\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x=\pm\sqrt{2}\left(N\right)\)

\(\left(2\right)\Leftrightarrow x=\pm1\left(N\right)\)

Kl: \(x=\pm\sqrt{2}\), \(x=\pm1\)

b) Đk: \(\left[{}\begin{matrix}x\le-2\\x\ge2\end{matrix}\right.\)

\(\sqrt{x^2-4}-x+2=0\)

\(\Leftrightarrow\sqrt{x^2-4}=x-2\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-4=x^2-4x+4\\x\ge2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x=8\\x\ge2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\left(N\right)\\x\ge2\end{matrix}\right.\)

kl: x=2

c) \(\sqrt{x^4-8x^2+16}=2-x\)

\(\Leftrightarrow\sqrt{\left(x^2-4\right)^2}=2-x\)

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

Th1: \(x^2-4< 0\Leftrightarrow-2< x< 2\)

(*) \(\Leftrightarrow x^2-4=x-2\Leftrightarrow x^2-x-2=0\Leftrightarrow\left[{}\begin{matrix}x=2\left(L\right)\\x=-1\left(N\right)\end{matrix}\right.\)

Th2: \(x^2-4\ge0\Leftrightarrow\left[{}\begin{matrix}x\le-2\\x\ge2\end{matrix}\right.\)

(*)\(\Leftrightarrow x^2-4=2-x\Leftrightarrow x^2+x-6=0\Leftrightarrow\left[{}\begin{matrix}x=2\left(N\right)\\x=-3\left(N\right)\end{matrix}\right.\)

Kl: x=-3, x=-1,x=2

d) \(\sqrt{9x^2+6x+1}=\sqrt{11-6\sqrt{2}}\)

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

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

Th1: \(3x+1\ge0\Leftrightarrow x\ge-\dfrac{1}{3}\)

(*) \(\Leftrightarrow3x+1=3-\sqrt{2}\Leftrightarrow x=\dfrac{2-\sqrt{2}}{3}\left(N\right)\)

Th2: \(3x+1< 0\Leftrightarrow x< -\dfrac{1}{3}\)

(*) \(\Leftrightarrow3x+1=-3+\sqrt{2}\Leftrightarrow x=\dfrac{-4+\sqrt{2}}{3}\left(N\right)\)

Kl: \(x=\dfrac{2-\sqrt{2}}{3}\), \(x=\dfrac{-4+\sqrt{2}}{3}\)

e) Đk: \(x\ge-\dfrac{3}{2}\)

\(\sqrt{4^2-9}=2\sqrt{2x+3}\) \(\Leftrightarrow\sqrt{7}=2\sqrt{2x+3}\) \(\Leftrightarrow7=8x+12\)

\(\Leftrightarrow8x=-5\Leftrightarrow x=-\dfrac{5}{8}\left(N\right)\)

kl: \(x=-\dfrac{5}{8}\)

f) Đk: x >/ 5

\(\sqrt{4x-20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)

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

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

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

\(\Leftrightarrow x-5=4\)

\(\Leftrightarrow x=9\left(N\right)\)

kl: x=9

1)

ĐK: \(x\geq 5\)

PT \(\Leftrightarrow \sqrt{4(x-5)}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9(x-5)}=6\)

\(\Leftrightarrow \sqrt{4}.\sqrt{x-5}+3\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}.\sqrt{9}.\sqrt{x-5}=6\)

\(\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=6\)

\(\Leftrightarrow 2\sqrt{x-5}=6\Rightarrow \sqrt{x-5}=3\Rightarrow x=3^2+5=14\)

2)

ĐK: \(x\geq -1\)

\(\sqrt{x+1}+\sqrt{x+6}=5\)

\(\Leftrightarrow (\sqrt{x+1}-2)+(\sqrt{x+6}-3)=0\)

\(\Leftrightarrow \frac{x+1-2^2}{\sqrt{x+1}+2}+\frac{x+6-3^2}{\sqrt{x+6}+3}=0\)

\(\Leftrightarrow \frac{x-3}{\sqrt{x+1}+2}+\frac{x-3}{\sqrt{x+6}+3}=0\)

\(\Leftrightarrow (x-3)\left(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}\right)=0\)

Vì \(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}>0, \forall x\geq -1\) nên $x-3=0$

\(\Rightarrow x=3\) (thỏa mãn)

Vậy .............

a) Ta có: \(\sqrt{4x-8}+5\sqrt{x-2}-\sqrt{9x-18}=20\)       \(\left(ĐK:x\ge2\right)\)

        \(\Leftrightarrow\sqrt{4}.\sqrt{x-2}+5\sqrt{x-2}-\sqrt{9}.\sqrt{x-2}=20\)

        \(\Leftrightarrow2.\sqrt{x-2}+5\sqrt{x-2}-3.\sqrt{x-2}=20\)

        \(\Leftrightarrow4.\sqrt{x-2}=20\)

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

        \(\Leftrightarrow x-2=25\)

        \(\Leftrightarrow x=27\left(TM\right)\)

Vậy \(S=\left\{27\right\}\)

a, PT <=> \(2\sqrt{x-2}+5\sqrt{x-2}-\sqrt{9\left(x-2\right)}=20\)

\(2\sqrt{x-2}+5\sqrt{x-2}-\sqrt{9}\sqrt{x-2}=20\)

\(\left(2+5-3\right)\sqrt{x-2}=20\)

\(4\sqrt{x-2}=20\Leftrightarrow\sqrt{x-2}=5\Leftrightarrow x-2=25\Leftrightarrow x=27\)

a) giải pt ra ta được  : x=-1

b) giải pt ra ta được  : x=2

c)giải pt ra ta được  : x vô ngiệm

d)giải pt ra ta được  : x=vô ngiệm

~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~

~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~

điều kiện -4<=x<=4x<=4

\(a,\sqrt{\left(x+4\right)^2}+\sqrt{\left(x-4\right)^2}\)

\(A=\left|x+4\right|+\left|x-4\right|\)

KẾT HỢP ĐIỀU KIỆN

\(A=x+4+4-x\)

\(A=8\)

\(B=\sqrt{\left(3x\right)^2-6x+1}+\sqrt{\left(2x\right)^2-12x+3^2}\)

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

\(B=\left|3x-1\right|+\left|2x-3\right|\)

\(TH1:x>=\frac{3}{2}\)

\(B=3x-1+2x-3\)

\(B=5x-4\)

\(TH2:\frac{1}{3}< =x< \frac{3}{2}\)

\(B=3x-1-2x+3\)

\(B=x+2\)

\(TH3:x< \frac{1}{3}\)

\(B=-3x+1-2x+3\)

\(B=4-5x\)

câu c và câu d tương tự

câu c tách ra: \(C=\sqrt{\left(\sqrt{x}-3\right)^2}-\sqrt{\left(2\sqrt{x}+1\right)^2}\)

còn câu d tách ra :\(D=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)

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

bạn tự làm nốt câu c, d nha 

Bài 2 :

ĐKXĐ : \(\left\{{}\begin{matrix}x\le3\\x\le5\end{matrix}\right.\)

=> \(x\le3\)

Ta có : \(\sqrt{3-x}+\sqrt{5-x}=2\)

=> \(\sqrt{3-x}=2-\sqrt{5-x}\)

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

=> \(-4\sqrt{5-x}=-6\)

=> \(\sqrt{5-x}=\frac{3}{2}\)

=> \(x=2,75\) ( TM )

Ta có : \(A=\sqrt{3-2,75}-\sqrt{5-2,75}=-1\)

Vậy ...

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Mình làm một vài câu thôi nhé, các câu còn lại tương tự.

Giải:

a) ??? Đề thiếu

b) \(\sqrt{-3x+4}=12\)

\(\Leftrightarrow-3x+4=144\)

\(\Leftrightarrow-3x=140\)

\(\Leftrightarrow x=\dfrac{-140}{3}\)

Vậy ...

c), d), g), h), i), p), q), v), a') Tương tự b)

w), x) Mình đã làm ở đây:

Câu hỏi của Ami Yên - Toán lớp 9 | Học trực tuyến

z) \(\sqrt{16\left(x+1\right)^2}-\sqrt{9\left(x+1\right)^2}=4\)

\(\Leftrightarrow4\left(x+1\right)-3\left(x+1\right)=4\)

\(\Leftrightarrow x+1=4\)

\(\Leftrightarrow x=3\)

Vậy ...

b') \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)

\(\Leftrightarrow3\sqrt{x+1}+2\sqrt{x+1}=\sqrt{x+1}\)

\(\Leftrightarrow3\sqrt{x+1}+2\sqrt{x+1}-\sqrt{x+1}=0\)

\(\Leftrightarrow4\sqrt{x+1}=0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

Vậy ...