- Giải phương trình sau: 5/3 * sqrt(9x - 18) - 1/2 * sqrt(16) overline 6x-32 -15=0
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![](https://rs.olm.vn/images/avt/0.png?1311)
b: Ta có: \(\sqrt{x^2-6x+9}-\dfrac{\sqrt{6}+\sqrt{3}}{\sqrt{2}+1}=0\)
\(\Leftrightarrow x^2-6x+9=3\)
\(\Leftrightarrow x^2-6x+6=0\)
\(\text{Δ}=\left(-6\right)^2-4\cdot1\cdot6=36-24=12\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{6-2\sqrt{3}}{2}=3-\sqrt{3}\\x_2=3+\sqrt{3}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\sqrt{1-8x+16x^2}=\dfrac{1}{3}\)
\(\Leftrightarrow\sqrt{1^2-2\cdot4x\cdot1+\left(4x\right)^2}=\dfrac{1}{3}\)
\(\Leftrightarrow\sqrt{\left(4x-1\right)^2}=\dfrac{1}{3}\)
\(\Leftrightarrow\left|4x-1\right|=\dfrac{1}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-1=\dfrac{1}{3}\left(ĐK:x\ge\dfrac{1}{4}\right)\\4x-1=\dfrac{1}{3}\left(ĐK:x< \dfrac{1}{4}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{4}{3}\\4x=\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\left(tm\right)\\x=\dfrac{1}{6}\left(tm\right)\end{matrix}\right.\)
b) \(\sqrt{16x-32}+\sqrt{25x-50}=18+\sqrt{9x-18}\) (ĐK: \(x\ge2\))
\(\Leftrightarrow\sqrt{16\left(x-2\right)}+\sqrt{25\left(x-2\right)}=18+\sqrt{9\left(x-2\right)}\)
\(\Leftrightarrow4\sqrt{x-2}+5\sqrt{x-2}=18+3\sqrt{x-2}\)
\(\Leftrightarrow6\sqrt{x-2}=18\)
\(\Leftrightarrow\sqrt{x-2}=3\)
\(\Leftrightarrow x-2=9\)
\(\Leftrightarrow x=9+2\)
\(\Leftrightarrow x=11\left(tm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2: ĐKXĐ: x>=0
\(\sqrt{3x}-2\sqrt{12x}+\dfrac{1}{3}\cdot\sqrt{27x}=-4\)
=>\(\sqrt{3x}-2\cdot2\sqrt{3x}+\dfrac{1}{3}\cdot3\sqrt{3x}=-4\)
=>\(\sqrt{3x}-4\sqrt{3x}+\sqrt{3x}=-4\)
=>\(-2\sqrt{3x}=-4\)
=>\(\sqrt{3x}=2\)
=>3x=4
=>\(x=\dfrac{4}{3}\left(nhận\right)\)
3:
ĐKXĐ: x>=0
\(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18}=0\)
=>\(3\sqrt{2x}+5\cdot2\sqrt{2x}-20-3\sqrt{2}=0\)
=>\(13\sqrt{2x}=20+3\sqrt{2}\)
=>\(\sqrt{2x}=\dfrac{20+3\sqrt{2}}{13}\)
=>\(2x=\dfrac{418+120\sqrt{2}}{169}\)
=>\(x=\dfrac{209+60\sqrt{2}}{169}\left(nhận\right)\)
4: ĐKXĐ: x>=-1
\(\sqrt{16x+16}-\sqrt{9x+9}=1\)
=>\(4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>\(\sqrt{x+1}=1\)
=>x+1=1
=>x=0(nhận)
5: ĐKXĐ: x<=1/3
\(\sqrt{4\left(1-3x\right)}+\sqrt{9\left(1-3x\right)}=10\)
=>\(2\sqrt{1-3x}+3\sqrt{1-3x}=10\)
=>\(5\sqrt{1-3x}=10\)
=>\(\sqrt{1-3x}=2\)
=>1-3x=4
=>3x=1-4=-3
=>x=-3/3=-1(nhận)
6: ĐKXĐ: x>=3
\(\dfrac{2}{3}\sqrt{x-3}+\dfrac{1}{6}\sqrt{x-3}-\sqrt{x-3}=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}\cdot\left(\dfrac{2}{3}+\dfrac{1}{6}-1\right)=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}\cdot\dfrac{-1}{6}=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}=\dfrac{2}{3}:\dfrac{1}{6}=\dfrac{2}{3}\cdot6=\dfrac{12}{3}=4\)
=>x-3=16
=>x=19(nhận)
![](https://rs.olm.vn/images/avt/0.png?1311)
f) Ta có: \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)
\(\Leftrightarrow4\left|x+1\right|-3\left|x+1\right|=4\)
\(\Leftrightarrow\left|x+1\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
g) Ta có: \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)
\(\Leftrightarrow5\sqrt{x+1}-\sqrt{x+1}=0\)
\(\Leftrightarrow x+1=0\)
hay x=-1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Đk:x\ge2\\ PT\Leftrightarrow\dfrac{10\sqrt{x-2}-\sqrt{x-2}+1}{2}=6\sqrt{x-2}\\ \Leftrightarrow9\sqrt{x-2}+1=12\sqrt{x-2}\\ \Leftrightarrow\sqrt{x-2}=\dfrac{1}{3}\Leftrightarrow x-2=\dfrac{1}{9}\\ \Leftrightarrow x=\dfrac{19}{9}\left(tm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{5}\sqrt[]{25x+50}-5\sqrt[]{x+2}+\sqrt[]{9x+18}+9=0\)
\(\Leftrightarrow\dfrac{1}{5}\sqrt[]{25\left(x+2\right)}-5\sqrt[]{x+2}+\sqrt[]{9\left(x+2\right)}+9=0\)
\(\Leftrightarrow\dfrac{1}{5}.5\sqrt[]{x+2}-5\sqrt[]{x+2}+3\sqrt[]{x+2}+9=0\)
\(\Leftrightarrow\sqrt[]{x+2}-5\sqrt[]{x+2}+3\sqrt[]{x+2}+9=0\)
\(\Leftrightarrow\sqrt[]{x+2}\left(1-5+3\right)+9=0\)
\(\Leftrightarrow-\sqrt[]{x+2}+9=0\)
\(\Leftrightarrow\sqrt[]{x+2}=9\)
\(\Leftrightarrow x+2=81\)
\(\Leftrightarrow x=79\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
a, \(2\sqrt{18}-4\sqrt{50}-3\sqrt{32}=6\sqrt{2}-20\sqrt{2}-12\sqrt{2}=-2\sqrt{2}\)
b, \(\sqrt{14-6\sqrt{5}}+\sqrt{14+6\sqrt{5}}=\sqrt{\left(\sqrt{5}-3\right)^2}+\sqrt{\left(\sqrt{5}+3\right)^2}\)
\(=\left|\sqrt{5}-3\right|+\left|\sqrt{5}+3\right|\)
\(=-\sqrt{5}+3+\sqrt{5}+3=6\)
c, \(\dfrac{\sqrt{10}+10}{1+\sqrt{10}}-\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}=\dfrac{\sqrt{10}\left(1+\sqrt{10}\right)}{1+\sqrt{10}}-\dfrac{\sqrt{10}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{5}-\sqrt{2}}\)
\(=\sqrt{10}-\sqrt{10}=0\)
2.
ĐK: \(x\in R\)
\(\sqrt{9x^2-30x+25}=5\)
\(\Leftrightarrow\sqrt{\left(3x-5\right)^2}=5\)
\(\Leftrightarrow\left|3x-5\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=5\\3x-5=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10}{3}\\x=0\end{matrix}\right.\)
Vậy ...
Nếu em thấy các câu hỏi do lag mà bị gửi đúp (tức là rất nhiều câu hỏi giống nhau xuất hiện cùng 1 chỗ) thì xóa giúp mình nhé cho đỡ vướng. Nhưng nhớ để lại 1 câu. Cảm ơn em.
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{2}{2}\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\left(tm\right)\)
b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))
\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=16\)
\(\Leftrightarrow x=15\left(tm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(4\sqrt{4x-8}-2\sqrt{9x-18}+\sqrt{16x-32}=5\)
\(\rightarrow4.2\sqrt{x-2}-2.3\sqrt{x-2}+4\sqrt{x-2}=5\)
\(\rightarrow\sqrt{x-2}\left(8-6+4\right)=5\)
\(\rightarrow6\sqrt{x-2}=5\)
\(\rightarrow\sqrt{x-2}=\frac{5}{6}\)
\(\rightarrow x-2=\frac{25}{36}\)
\(\Rightarrow x=\frac{97}{36}\)
b)\(\sqrt{x^2+6x+9}-2=7\)
\(\rightarrow\sqrt{\left(x+3\right)^2}=9\)
\(\rightarrow x+3=9\)
\(\Rightarrow x=6\)
Nhớ tick mik nha
à ừ, nhầm chút!! thông cảm nha! có gì bạn tham khảo bài của Mo Nguyễn Văn