5.

ĐKXĐ: ...

\(\Leftrightarrow3x^2-14x-5+\sqrt{3x+1}-4+1-\sqrt{6-x}=0\)

\(\Leftrightarrow\left(3x+1\right)\left(x-5\right)+\frac{3\left(x-5\right)}{\sqrt{3x+1}+4}+\frac{x-5}{1+\sqrt{6-x}}=0\)

\(\Leftrightarrow\left(x-5\right)\left(3x+1+\frac{3}{\sqrt{3x+1}+4}+\frac{1}{1+\sqrt{6-x}}\right)=0\)

\(\Leftrightarrow x=5\)

6.

ĐKXĐ: \(-4\le x\le4\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\frac{\sqrt{4-x}+2}{\sqrt{x+4}+2}=2\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\sqrt{4-x}+2=2\sqrt{x+4}+4\)

\(\Leftrightarrow2\sqrt{x+4}-\frac{4}{5}+\frac{14}{5}-\sqrt{4-x}=0\)

\(\Leftrightarrow\frac{2\left(x+4-\frac{4}{25}\right)}{\sqrt{x+4}+\frac{2}{5}}+\frac{\frac{196}{25}-4+x}{\frac{14}{5}+\sqrt{4-x}}=0\)

\(\Leftrightarrow\left(x-\frac{96}{25}\right)\left(\frac{2}{\sqrt{x+4}+\frac{2}{5}}+\frac{1}{\frac{14}{5}+\sqrt{4-x}}\right)=0\)

\(\Rightarrow x=\frac{96}{25}\)

1.

Bạn coi lại đề

2.

ĐKXĐ: \(1\le x\le2\)

Nhận thấy \(\sqrt{x+2}+\sqrt{x-1}>0;\forall x\) , nhân 2 vế của pt với nó:

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

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

\(\Leftrightarrow3\sqrt{2-x}+3=\sqrt{x+2}+\sqrt{x-1}\)

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

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

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

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

1)\(\sqrt{2x^2-2x+\frac{1}{2}}=\frac{1}{\sqrt{2}}\left(ĐKXĐ:x^2-x+\frac{1}{4}\ge0\right)\)

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

   \(2x^2-2x=0\)

    \(2x\left(x-1\right)=0\)

            \(\Rightarrow\orbr{\begin{cases}2x=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

2)\(\sqrt{9x-9}-2\sqrt{\frac{x-1}{4}}=6\left(ĐKXĐ:x\ge1\right)\)

    \(\sqrt{9\left(x-1\right)}-2.\frac{\sqrt{x-1}}{2}=6\)

   \(3\sqrt{x-1}-\left(\sqrt{x-1}\right)=6\)

  \(2\sqrt{x-1}=6\)

   \(\sqrt{x-1}=3=\sqrt{9}\)

    \(\Rightarrow x=10\)

4)\(1-3x+\sqrt{x^2-6x+9}=0\)

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

    \(1-3x+x-3=0\)

    \(x=-1\)

5)\(\frac{1}{2}\sqrt{\frac{3x+9}{4}}+\sqrt{x+3}=\sqrt{1-x}\)

    \(\frac{1}{2}.\frac{\sqrt{3x+9}}{2}+\sqrt{x+3}=\sqrt{1-x}\)

    \(\frac{\sqrt{3x+9}}{4}+\sqrt{x+3}=\sqrt{1-x}\)

      \(\frac{\sqrt{3x+9}+4\sqrt{x+3}}{4}=\frac{4\sqrt{1-x}}{4}\)

     \(\Rightarrow\sqrt{3}.\sqrt{x+3}+4\sqrt{x+3}=4\sqrt{1-x}\)

     \(\Rightarrow\left(\sqrt{3}+4\right)\left(\sqrt{x+3}\right)=\sqrt{2-2x}\)

6)\(\sqrt{4x^2-9}.\left(\sqrt{x+1}+1\right)=0\)

    \(\Rightarrow\orbr{\begin{cases}4x^2-9=0\\\sqrt{x+1}+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}4x^2=9\\\sqrt{x+1}=-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{3}{2}\\x=-1\end{cases}}\)

1/ \(\sqrt{x-2}-\sqrt{1-3x}=0\\ đk:\left\{{}\begin{matrix}x-2\ge0\\1-3x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\le\frac{1}{3}\end{matrix}\right.\)

=> pt vô no

2/ \(\sqrt{15-x}+\sqrt{3-x}=6\\ đk\left\{{}\begin{matrix}15-x\ge0\\3-x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le15\\x\le3\end{matrix}\right.\Leftrightarrow x\le3\)

\(pt\Leftrightarrow15-x+3-x+2\sqrt{\left(15-x\right)\left(3-x\right)}=36\)

\(\Leftrightarrow2\sqrt{\left(15-x\right)\left(3-x\right)}=2x+36\)

\(\Leftrightarrow4\left(15-x\right)\left(3-x\right)=\left(2x+18\right)^2\left(đk:x\ge-9\right)\)

\(\Leftrightarrow-144x=144\Leftrightarrow x=-1\left(nhan\right)\)

Câu 1: ĐKXĐ: \(\left\{{}\begin{matrix}x-2\ge0\\1-3x\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge2\\x\le\frac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\) Không tồn tại x thỏa mãn ĐKXĐ \(\Rightarrow\) pt vô nghiệm

Câu 2:

ĐKXĐ: \(x\le3\)

\(\Leftrightarrow15-x+3-x+2\sqrt{\left(15-x\right)\left(3-x\right)}=36\)

\(\Leftrightarrow x+9=\sqrt{x^2-18x+45}\) (\(x\ge-9\))

\(\Leftrightarrow x^2+18x+81=x^2-18x+45\)

\(\Leftrightarrow36x=-36\Rightarrow x=-1\)

Câu 3:

ĐKXĐ: \(x\ge1\)

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

\(\Leftrightarrow x-1=4+x+1+4\sqrt{x+1}\)

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

Phương trình vô nghiệm

6.

ĐKXĐ: \(x\ge2\)

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

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

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

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

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

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

4.

ĐKXĐ: \(x\ge4\)

Đặt \(\sqrt{x-4}=t\ge0\Rightarrow x=t^2+4\)

\(\Rightarrow3\left(t^2+4\right)+7t=14t-20\)

\(\Leftrightarrow3t^2-7t+34=0\)

Phương trình vô nghiệm

5.

ĐKXĐ: ...

- Với \(x=0\) ko phải nghiệm

- Với \(x\ne0\Rightarrow\sqrt{x+1}-1\ne0\) , nhân 2 vế của pt cho \(\sqrt{x+1}-1\) và rút gọn ta được:

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

\(\Leftrightarrow2x=4\Rightarrow x=2\)

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 ...

- Câu a có chút thiếu sót, mong thông cảm :)

\(\sqrt{3x-1}\) = 4

a) ĐK: \(x\ge -1\)

Ta có: \(x^2+\sqrt{x+1}=1\)

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

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

\(\Leftrightarrow \sqrt{x+1}[(x-1)\sqrt{x+1}+1]=0\)

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

Với \((1)\Rightarrow x+1=0\Rightarrow x=-1\) (thỏa mãn)

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

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

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

\(\Leftrightarrow x.\frac{x+1+\sqrt{x+1}-1}{\sqrt{x+1}+1}=0\)

\(\Leftrightarrow x.\frac{x+\sqrt{x+1}}{\sqrt{x+1}+1}=0\)

\(\Rightarrow \left[\begin{matrix} x=0\\ x+\sqrt{x+1}=0\end{matrix}\right.\)

Với \(x+\sqrt{x+1}=0\Rightarrow x=-\sqrt{x+1}\Rightarrow \left\{\begin{matrix} x\leq 0\\ x^2=x+1\end{matrix}\right.\Rightarrow x=\frac{1-\sqrt{5}}{2}\)

Vậy \(x=\left\{-1; \frac{1-\sqrt{5}}{2}; 0\right\}\)

b) ĐK: \(-3\leq x\leq 6\)

Ta có: \((\sqrt{3+x}+\sqrt{6-x})^2=3+x+6-x+2\sqrt{(3+x)(6-x)}\)

\(=9+2\sqrt{(3+x)(6-x)}\geq 9\)

\(\Rightarrow \sqrt{3+x}+\sqrt{6-x}\geq 3\) do \(\sqrt{3+x}+\sqrt{6-x}\) không âm.

Dấu "=" xảy ra khi \(\sqrt{(3+x)(6-x)}=0\Leftrightarrow x=-3; x=6\)

Vậy \(x=-3\) or $x=6$

Bài 1: Giải phương trình

a) ĐKXĐ: \(x\ge3\)

Ta có: \(\sqrt{100\cdot\left(x-3\right)}=\sqrt{20}\)

\(\Leftrightarrow\left|100\cdot\left(x-3\right)\right|=\left|20\right|\)

\(\Leftrightarrow100\cdot\left|x-3\right|=20\)

\(\Leftrightarrow\left|x-3\right|=\frac{1}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=\frac{1}{5}\\x-3=-\frac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{16}{5}\left(nhận\right)\\x=\frac{14}{5}\left(loại\right)\end{matrix}\right.\)

Vậy: \(S=\left\{\frac{16}{5}\right\}\)

b) Ta có: \(\sqrt{\left(x-3\right)^2}=7\)

\(\Leftrightarrow\left|x-3\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=7\\x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-4\end{matrix}\right.\)

Vậy: S={10;-4}

c) Ta có: \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{-7}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{\frac{5}{2};\frac{-7}{2}\right\}\)

\(f,\sqrt{x^2-25}-\sqrt{x-5}=0\)

=> \(\sqrt{x^2-25}=\sqrt{x-5}\)

=>\(x^2-25=x-5\)

=>\(x^2-x=25-5=20\)

=>( đến đoạn này mình xin chịu )

\(a,\sqrt{16x}=8\)

=>\(16x=8^2\)

=>\(16x=64\)

=>\(x=64:16=4\)

Vậy \(x\in\left\{4\right\}\)

\(b,\sqrt{x^2}=2x-1\)

=>\(x=2x-1\)

=>\(2x-x=1\)

=>\(x=1\)

Vậy \(x\in\left\{1\right\}\)

\(c,\sqrt{9.\left(x-1\right)}=21\)

=>\(9.\left(x-1\right)=21^2=441\)

=> \(x-1=441:9=49\)

=>\(x=49+1=50\)

Vậy \(x\in\left\{50\right\}\)

\(d,\sqrt{4\left(1-x\right)^2}-6=0\)

=>\(\sqrt{4\left(1-x\right)^2}=0+6=6\)

=> \(4\left(1-x\right)^2=6^2=36\)

=>\(\left(1-x\right)^2=36:4=9\)

=>\(1-x=\sqrt{9}=3\)

=>\(x=1-3=-2\)

Vậy \(x\in\left\{-2\right\}\)

\(g,\sqrt{9\left(2-3x\right)^2}=6\)

=> \(9.\left(2-3x\right)^2=6^2=36\)

=> \(\left(2-3x\right)^2=36:9=4\)

=> \(2-3x=\sqrt{4}=2\)

=>\(3x=2-2=0\)

=>\(x=0:3=0\)

Vậy \(x\in\left\{0\right\}\)

( còn các bài còn lại mình sẽ nghĩ tiếp , HS6-7 làm bài )