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\(\dfrac{1}{2}.\sqrt{108}-10.\sqrt{\dfrac{1}{5}}-\sqrt{147}+\sqrt{20}\)
\(=\dfrac{1}{2}.\sqrt{6^2.3}-10.\dfrac{\sqrt{5}}{5}-\sqrt{7^2.3}+\sqrt{2^2.5}\)
\(=3\sqrt{3}-2\sqrt{5}+2\sqrt{5}-7\sqrt{3}\)
\(=-4\sqrt{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)
\(\Leftrightarrow2\cdot3\sqrt{x-3}-\dfrac{1}{5}\cdot5\sqrt{x-3}-\dfrac{1}{7}\cdot7\cdot\sqrt{x-3}=20\)
\(\Leftrightarrow4\sqrt{x-3}=20\)
\(\Leftrightarrow x-3=25\)
hay x=28
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=3\cdot7\sqrt{3}+2\cdot6\sqrt{3}-4\cdot4\sqrt{3}-11\sqrt{3}\)
\(=21\sqrt{3}+12\sqrt{3}-16\sqrt{3}-11\sqrt{3}\)
\(=6\sqrt{3}\)
b: \(=\sqrt{\left(3-\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=3-\sqrt{5}+\sqrt{5}-1\)
=2
c: \(=\left(4-\sqrt{3}\right)\sqrt{\left(4+\sqrt{3}\right)^2}\)
\(=\left(4-\sqrt{3}\right)\left(4+\sqrt{3}\right)\)
=16-3
=13
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2\sqrt{9x-27}-\frac{1}{5}\sqrt{25x-75}-\frac{1}{7}\sqrt{49x-147}=20\)
\(< =>2\sqrt{9\left(x-3\right)}-\frac{1}{5}\sqrt{25\left(x-3\right)}-\frac{1}{7}\sqrt{49\left(x-3\right)}=20\)
\(< =>2\cdot3\sqrt{\left(x-3\right)}-\frac{1}{5}.5\sqrt{\left(x-3\right)}-\frac{1}{7}.7\sqrt{\left(x-3\right)}=20\) \(đk:x\ge0\)
\(< =>6\sqrt{\left(x-3\right)}-\sqrt{\left(x-3\right)}-\sqrt{\left(x-3\right)}=20\)
\(< =>\sqrt{\left(x-3\right)}\left(6-1-1\right)=20\)
\(< =>4\sqrt{\left(x-3\right)}=20\)
\(< =>\sqrt{\left(x-3\right)}=5\)
\(< =>x-3=25\)
\(< =>x=28\left(tm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)ĐK:x\ge-1\\ \Leftrightarrow x+1=2\sqrt{x+1}\\ \Leftrightarrow x^2+2x+1=4x+4\\ \Leftrightarrow x^2+2x-4x+1-4=0\\ \Leftrightarrow x^2-2x-3=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{3;-1\right\}\)
\(b)ĐK:x\ge2\\ \Leftrightarrow2x-4=\sqrt{x-2}\\ \Leftrightarrow4x^2-16x+16=x-2\\ \Leftrightarrow4x^2-16x-x+16+2=0\\ \Leftrightarrow4x^2-17x+18=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{9}{4};2\right\}\)
\(c)ĐK:x\ge3\\ \Leftrightarrow2\sqrt{9\left(x-3\right)}-\dfrac{1}{5}\sqrt{25\left(x-3\right)}-\dfrac{1}{7}\sqrt{49\left(x-3\right)}=20\\ \Leftrightarrow2.3\sqrt{x-3}-\dfrac{1}{5}\cdot5\sqrt{x-3}-\dfrac{1}{7}\cdot7\sqrt{x-3}=20\\ \Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\\ \Leftrightarrow4\sqrt{x-3}=20\\ \Leftrightarrow\sqrt{x-3}=5\\ \Leftrightarrow x-3=25\\ \Leftrightarrow x=25+3\\ \Leftrightarrow x=28\left(tm\right)\)
Vậy \(S=\left\{28\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
đề =
\(2\sqrt{9\left(x-3\right)}-\frac{1}{5}\sqrt{25\left(x-3\right)}-\frac{1}{7}\sqrt{49\left(x-3\right)}=20\)
=>\(6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)
=>\(4\sqrt{x-3}=20\)
=>\(\sqrt{x-3}=5\)
=>\(x-3=25\)
=>\(x=28\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a)sqrt{1-4x+4x^2}+5=x-2`
`<=>\sqrt{(2x-1)^2}=x-2-5`
`<=>|2x-1|=x-7(x>=7)`
`<=>[(2x-1=x-7),(2x-1=7-x):}`
`<=>[(x=-6(ktm)),(3x=8):}`
`<=>x=8/3(ktm)`
Vậy PTVN
`b)3sqrt{12+4x}+4/7sqrt{147+49x}=3/2sqrt{48+16x}+4(x>=-3)`
`<=>6sqrt{x+3}+4sqrt{x+3}=6sqrt{x+3}+4`
`<=>4sqrt{x+3}=4`
`<=>sqrt{x+3}=1<=>x+3=1`
`<=>x=-2(tm)`
Vậy `S={-2}`
a) \(\sqrt{1-4x+4x^2}+5=x-2\Leftrightarrow\sqrt{\left(1-2x\right)^2}+5=x-2\Leftrightarrow\left|1-2x\right|=x-7\left(1\right)\)TH1: \(1-2x\ge0\Leftrightarrow x\le\dfrac{1}{2}\)
\(\left(1\right)\Leftrightarrow1-2x=x-7\Leftrightarrow3x=8\Leftrightarrow x=\dfrac{8}{3}\)(không thỏa đk)
TH2: \(1-2x< 0\Leftrightarrow x>\dfrac{1}{2}\)
\(\left(1\right)\Leftrightarrow2x-1=x-7\Leftrightarrow x=-6\)(không thỏa đk)
Vậy \(S=\varnothing\)
b) \(3\sqrt{12+4x}+\dfrac{4}{7}\sqrt{147+49x}=\dfrac{3}{2}\sqrt{48+16x}+4\Leftrightarrow6\sqrt{3+x}+4\sqrt{3+x}=6\sqrt{3+x}+4\Leftrightarrow4\sqrt{3+x}=4\Leftrightarrow\sqrt{3+x}=1\Leftrightarrow3+x=1\Leftrightarrow x=-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì \(7^n+147\) là số chính phương
=> Đặt: \(7^n+147\) với a là số nguyên khi đó ta có:
\(7^n+147=a^2\)không mất tính tổng quát g/s a nguyên dương
mà: n là số tự nhiên nên \(7^n⋮7\); \(147=7^2.3⋮7\)=> \(a^2⋮7\)=> \(a⋮7\)=> \(a^2⋮7^2\)
=> \(7^n⋮7^2\)=> n \(\ge\)2
+) Với n = 2k khi đó: \(k\ge1\)
Ta có: \(7^{2k}+147=a^2\)
<=> \(\left(a-7^k\right)\left(a+7^k\right)=147\)
Vì: \(\hept{\begin{cases}0< a-7^k< a+7^k\\a-7^k;a+7^k⋮7\end{cases}}\)
Do đó: \(\hept{\begin{cases}a+7^k=21\\a-7^k=7\end{cases}}\Leftrightarrow7^k=7\Leftrightarrow k=1\)=> n = 2
Thử lại thỏa mãn
+) Với n = 2k + 1 ta có:
\(7^{2k+1}:4\) dư -1
\(147\): 4 dư 3
=> \(7^{2k+1}+147\) chia 4 dư 2
mà số chính phương chia 4 bằng 0 hoặc 1
=> Loại
Vậy: n = 2