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![](https://rs.olm.vn/images/avt/0.png?1311)
9) Sửa: \(2\sqrt{8\sqrt{3}}-2\sqrt{5\text{ }\sqrt{3}}-3\sqrt{20\sqrt{3}}\)
\(=2\sqrt{2^2\cdot2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{2^2\cdot5\sqrt{3}}\)
\(=2\cdot2\sqrt{2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\cdot2\sqrt{5\sqrt{3}}\)
\(=4\sqrt{2\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}\)
\(=4\sqrt{2\sqrt{3}}-8\sqrt{5\sqrt{3}}\)
10) \(\sqrt{12x}-\sqrt{48x}-3\sqrt{3x}+27\)
\(=\sqrt{2^2\cdot3x}-\sqrt{4^2\cdot3x}-3\sqrt{3x}+27\)
\(=2\sqrt{3x}-4\sqrt{3x}-3\sqrt{3x}+27\)
\(=-5\sqrt{3x}++27\)
11) \(\sqrt{18x}-5\sqrt{8x}+7\sqrt{18x}+28\)
\(=\sqrt{3^2\cdot2x}-5\sqrt{2^2\cdot2x}+7\sqrt{3^2\cdot2x}+28\)
\(=3\sqrt{2x}-5\cdot2\sqrt{2x}+7\cdot3\sqrt{2x}+28\)
\(=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+28\)
\(=14\sqrt{2x}+28\)
12) \(\sqrt{45a}-\sqrt{20a}+4\sqrt{45a}+\sqrt{a}\)
\(=\sqrt{3^2\cdot5a}-\sqrt{2^2\cdot5a}+4\sqrt{3^2\cdot5a}+\sqrt{a}\)
\(=3\sqrt{5a}-2\sqrt{5a}+4\cdot3\sqrt{5a}+\sqrt{a}\)
\(=3\sqrt{5a}-2\sqrt{5a}+12\sqrt{5a}+\sqrt{a}\)
\(=13\sqrt{5a}+\sqrt{a}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Pt \(\Leftrightarrow\sqrt{\left(x-2\right)^2}=5\Leftrightarrow\left|x-2\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
Vậy...
b)Đk: \(x\ge-1\)
Pt \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}\)
\(\Leftrightarrow4\sqrt{x+1}=16\)\(\Leftrightarrow x+1=16\)\(\Leftrightarrow x=15\) (tm)
Vậy...
\(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\) (a>0)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=a+\sqrt{a}-\left(2\sqrt{a}+1\right)+1=a-\sqrt{a}\)
b) \(A=a-\sqrt{a}=a-2.\dfrac{1}{2}\sqrt{a}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
Dấu "=" xảy ra khi \(\sqrt{a}=\dfrac{1}{2}\Leftrightarrow a=\dfrac{1}{4}\left(tmđk\right)\)
Vậy \(A_{min}=-\dfrac{1}{4}\)
a) \(\sqrt{x^2-4x+4}=5\Rightarrow\sqrt{\left(x-2\right)^2}=5\Rightarrow\left|x-2\right|=5\)
\(\Rightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
b) \(\sqrt{16x+16}-3\sqrt{x+1}+\sqrt{4x+4}=16-\sqrt{x+1}\)
\(\Rightarrow\sqrt{16\left(x+1\right)}-3\sqrt{x+1}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Rightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Rightarrow4\sqrt{x+1}=16\Rightarrow\sqrt{x+1}=4\Rightarrow x=15\)
a) \(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=a+\sqrt{a}-2\sqrt{a}-1+1=a-\sqrt{a}\)
b) Ta có: \(a-\sqrt{a}=\left(\sqrt{a}\right)^2-2.\sqrt{a}.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)
\(=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
\(\Rightarrow A_{min}=-\dfrac{1}{4}\) khi \(a=\dfrac{1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(x=\dfrac{9}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 = \(\dfrac{-4}{6}=\dfrac{-2}{3}\)
![](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)
Bài 1:
a) \(A=\sqrt{8}+\sqrt{18}-\sqrt{32}\)
\(=2\sqrt{2}+3\sqrt{2}-4\sqrt{2}\)
\(=\sqrt{2}\)
b) \(B=\sqrt{9-4\sqrt{5}}-\sqrt{5}\)
\(=\sqrt{4-4\sqrt{5}+5}-\sqrt{5}\)
\(=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{5}\)
\(=\left|2-\sqrt{5}\right|-\sqrt{5}\)
\(=\sqrt{5}-2-\sqrt{5}\)
\(=-2\)
Bài 2:
a) \(\left\{{}\begin{matrix}2x-3y=4\\x+3y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x=6\\x+3y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\2+3y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
Vậy phương trình có nghiệm là: \(\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
b) ĐKXĐ: \(x\ne\pm2\)
Với \(x\ne\pm2\), ta có:
\(\dfrac{10}{x^2-4}+\dfrac{1}{2-x}=1\)
\(\Leftrightarrow\dfrac{10}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}=1\)
\(\Leftrightarrow\dfrac{10-x-2}{x^2-4}=1\)
\(\Leftrightarrow\dfrac{8-x}{x^2-4}=1\)
\(\Rightarrow x^2-4=8-x\)
\(\Leftrightarrow x^2+x-12=0\)
\(\Leftrightarrow x^2-3x+4x-12=0\)
\(\Leftrightarrow x\left(x-3\right)+4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+4=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\) (TM)
Vậy phương trình có tập nghiệm là: S ={3; -4}
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{2\sqrt{x}-9-\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)-\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2\sqrt{x}-9-x+2\sqrt{x}+3-2x+3\sqrt{x}+2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{-3x+7\sqrt{x}-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(A=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\) (ĐK: \(x\ne4;x\ne9;x\ge0\))
\(A=\dfrac{2\sqrt{x}-9}{x-2\sqrt{x}-3\sqrt{x}+6}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(A=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(A=\dfrac{2\sqrt{x}-9-\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)-\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(A=\dfrac{2\sqrt{x}-9-\left(x-3\sqrt{x}+\sqrt{x}-3\right)-\left(2x-4\sqrt{x}+\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(A=\dfrac{2\sqrt{x}-9-x+2\sqrt{x}+3-2x+3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(A=\dfrac{-3x+7\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^2+8=3\sqrt{x^3+8}\)
\(\left(x^2+8\right)^2=\left(3\sqrt{x^2+8}\right)^2\)
\(x^4+16x^2+64=9x^2+72\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
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:
\(A^2=8+2\sqrt{10+2\sqrt{5}}+8-2\sqrt{10+2\sqrt{5}}+2\cdot\sqrt{8^2-\left(2\sqrt{10+2\sqrt{5}}\right)^2}\)
\(=16+2\cdot\sqrt{64-4\cdot\left(10+2\sqrt{5}\right)}\)
\(=16+2\cdot\sqrt{24-8\sqrt{5}}\)
\(=16+2\cdot\sqrt{20-2\cdot2\sqrt{5}\cdot2+4}\)
\(=16+2\cdot\sqrt{\left(2\sqrt{5}-2\right)^2}\)
\(=16+2\cdot\left(2\sqrt{5}-2\right)=12+4\sqrt{5}\)
\(=10+2\cdot\sqrt{10}\cdot\sqrt{2}+2\)
\(=\left(\sqrt{10}+\sqrt{2}\right)^2\)
=>\(A=\sqrt{10}+\sqrt{2}\)