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a) \(\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\) (1)
\(\Leftrightarrow9x-7=\sqrt{\left(7x+5\right)\left(7x+5\right)}\)
\(\Leftrightarrow9x-\sqrt{\left(7x+5\right)\left(7x+5\right)}=7\)
\(\Leftrightarrow9x-\sqrt{\left(7x+5\right)^2}=7\)
\(\Leftrightarrow9x-\left|7x+5\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}9x-\left(7x+5\right)=7\left(đk:7x+5\ge0\right)\\9x-\left[-\left(7x+5\right)\right]=7\left(đk:7x+5< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\left(đk:x\ge-\dfrac{5}{7}\right)\\x=\dfrac{1}{8}\left(đk:x< -\dfrac{5}{7}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x\in\varnothing\end{matrix}\right.\)
\(\Leftrightarrow x=6\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{6\right\}\)
b) \(\sqrt{4x-20}+3\sqrt{\dfrac{x+5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\) (2)
\(\Leftrightarrow\sqrt{4\left(x-5\right)}+3\cdot\dfrac{\sqrt{x+5}}{3}-\dfrac{1}{3}\cdot\sqrt{9\left(x-5\right)}=4\)
\(\Leftrightarrow\sqrt{4}\sqrt{x-5}+\sqrt{x+5}-\dfrac{1}{3}\cdot\sqrt{9}\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x+5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x+5}-\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}+\sqrt{x+5}=4\)
\(\Leftrightarrow\sqrt{x-5}=4-\sqrt{x+5}\)
\(\Leftrightarrow x-5=\left(4-\sqrt{x+5}\right)^2\)
\(\Leftrightarrow x-5=16-8\sqrt{x+5}+x+5\)
\(\Leftrightarrow-5=16-8\sqrt{x+5}+5\)
\(\Leftrightarrow-5=21-8\sqrt{x+5}\)
\(\Leftrightarrow8\sqrt{x+5}=21+5\)
\(\Leftrightarrow8\sqrt{x+5}=26\)
\(\Leftrightarrow\sqrt{x+5}=\dfrac{13}{4}\)
\(\Leftrightarrow x+5=\dfrac{169}{16}\)
\(\Leftrightarrow x=\dfrac{169}{16}-5\)
\(\Leftrightarrow x=\dfrac{89}{16}\)
Vậy tập nghiệm phương trình (2) là \(S=\left\{\dfrac{89}{16}\right\}\)
Nick cũ không đi giải lấy nick mới giải làm gì vậy Tuấn Anh Phan Nguyễn ? :D
a: \(\Leftrightarrow\dfrac{2x-3}{x-1}=4\)
=>4x-4=2x-3
=>2x=1
hay x=1/2
b: \(\Leftrightarrow\sqrt{\dfrac{2x-3}{x-1}}=2\)
=>(2x-3)=4x-4
=>4x-4=2x-3
=>2x=1
hay x=1/2(nhận)
c: \(\Leftrightarrow\sqrt{2x+3}\left(\sqrt{2x-3}-2\right)=0\)
=>2x+3=0 hoặc 2x-3=4
=>x=-3/2 hoặc x=7/2
e: \(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
=>căn (x-5)=2
=>x-5=4
hay x=9
a)
ĐKXĐ: \(x> \frac{-5}{7}\)
Ta có: \(\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)
\(\Rightarrow 9x-7=\sqrt{7x+5}.\sqrt{7x+5}=7x+5\)
\(\Rightarrow 2x=12\Rightarrow x=6\) (hoàn toàn thỏa mãn)
Vậy......
b) ĐKXĐ: \(x\geq 5\)
\(\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow \sqrt{4}.\sqrt{x-5}+3\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}\sqrt{9}.\sqrt{x-5}=4\)
\(\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow 2\sqrt{x-5}=4\Rightarrow \sqrt{x-5}=2\Rightarrow x-5=2^2=4\Rightarrow x=9\)
(hoàn toàn thỏa mãn)
Vậy..........
c) ĐK: \(x\in \mathbb{R}\)
Đặt \(\sqrt{6x^2-12x+7}=a(a\geq 0)\Rightarrow 6x^2-12x+7=a^2\)
\(\Rightarrow 6(x^2-2x)=a^2-7\Rightarrow x^2-2x=\frac{a^2-7}{6}\)
Khi đó:
\(2x-x^2+\sqrt{6x^2-12x+7}=0\)
\(\Leftrightarrow \frac{7-a^2}{6}+a=0\)
\(\Leftrightarrow 7-a^2+6a=0\)
\(\Leftrightarrow -a(a+1)+7(a+1)=0\Leftrightarrow (a+1)(7-a)=0\)
\(\Rightarrow \left[\begin{matrix} a=-1\\ a=7\end{matrix}\right.\) \(\Rightarrow a=7\) vì \(a\geq 0\)
\(\Rightarrow 6x^2-12x+7=a^2=49\)
\(\Rightarrow 6x^2-12x-42=0\Leftrightarrow x^2-2x-7=0\)
\(\Leftrightarrow (x-1)^2=8\Rightarrow x=1\pm 2\sqrt{2}\)
(đều thỏa mãn)
Vậy..........
\(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
<=> x + 1 = 16
<=> x = 15 (nhận)
~ ~ ~
\(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow\sqrt{x+5}=2\)
<=> x + 5 = 4
<=> x = - 1 (nhận)
a, \(\sqrt{9x+9}-4\sqrt{\dfrac{x+1}{4}}=5\) \(x\ge-1\)
\(\Leftrightarrow3\sqrt{x+1}-2\sqrt{x+1}=5\)
\(\Leftrightarrow x+1=25\Leftrightarrow x=24\)
2) "biểu thức"=\(\sqrt{x-5}-4\sqrt{x-5}-\sqrt{x-5}=12\Leftrightarrow4\sqrt{x-5}=12\Leftrightarrow\sqrt{x-5}=3\Leftrightarrow x=14\)
Kl: x=14
3) "biểu thức"=\(4\sqrt{x-1}-3\sqrt{x-1}+\sqrt{x-1}=5\Leftrightarrow2\sqrt{x-1}=5\Leftrightarrow\sqrt{x-1}=\dfrac{5}{2}\Leftrightarrow x=\left(\dfrac{5}{2}\right)^2+1=\dfrac{29}{4}\)
Kl: x=29/4
Sửa đề: \(\sqrt{4x-20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
=>\(2\sqrt{x-5}+\dfrac{3\sqrt{x-5}}{3}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)
=>2 căn x-5=4
=>căn x-5=2
=>x-5=4
=>x=9
\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)
\(ĐK:x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)
\(\Leftrightarrow4x^2-9=4x+12\)
\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)
\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(ĐK:x\ge5\)
\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)
\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)
ĐK:x>=1
\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)
\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)
\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)
\(ĐK:x\ge3\)
\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)
\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)
\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}=0\) (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)
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. ĐKXĐ: $x\in\mathbb{R}$
PT $\Leftrightarrow 4x=\sqrt{(3x+1)^2}$
\(\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ (4x)^2=(3x+1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ (4x-3x-1)(4x+3x+1)=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ (x-1)(7x+1)=0\end{matrix}\right.\Leftrightarrow x=1\)
Vậy $x=1$ là nghiệm của pt.
2. ĐKXĐ: $x\geq -5$
PT $\Leftrightarrow \sqrt{4}.\sqrt{x+5}-3\sqrt{5+x}+\frac{4}{3}.\sqrt{9}.\sqrt{x+5}=0$
$\Leftrightarrow 2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=0$
$\Leftrightarrow 3\sqrt{x+5}=0$
$\Leftrightarrow \sqrt{x+5}=0$
$\Leftrightarrow x=-5$
\(2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(2\sqrt{x-5}=4\)
\(\sqrt{x-5}=2\)
\(\left\{{}\begin{matrix}2>0\left(luondung\right)\\x-5=4\end{matrix}\right.\)\(\Rightarrow x=9\left(tm\right)\)