Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\sqrt{25x+75}+3\sqrt{x-2}=2+4\sqrt{x+3}+\sqrt{9x-18}\) (ĐKXĐ : \(x\ge2\) )
\(\Leftrightarrow5\sqrt{x+3}+3\sqrt{x-2}-4\sqrt{x+3}-3\sqrt{x-2}=2\)
\(\Leftrightarrow\sqrt{x+3}=2\)
\(\Leftrightarrow x+3=4\)
\(\Leftrightarrow x=1\) ( Thỏa mãn ĐKXĐ )
c) \(\sqrt{4x+20}+\sqrt{x+5}-\dfrac{1}{3}\sqrt{9x+45}=4\) (ĐKXĐ : \(x\ge-5\) )
\(\Leftrightarrow2\sqrt{x+5}+\sqrt{x+5}-\sqrt{x+5}=4\)
\(\Leftrightarrow2\sqrt{x+5}=4\)
\(\Leftrightarrow\sqrt{x+5}=2\)
\(\Leftrightarrow x+5=4\)
\(\Leftrightarrow x=-1\) ( Thỏa mãn ĐKXĐ )
Vậy.......
Lời giải:
a) ĐK: \(x\geq 0\)
\(4\sqrt{x}-2\sqrt{9x}+\sqrt{16x}=5\)
\(\Leftrightarrow 4\sqrt{x}-2\sqrt{9}.\sqrt{x}+\sqrt{16}.\sqrt{x}=5\)
\(\Leftrightarrow 4\sqrt{x}-6\sqrt{x}+4\sqrt{x}=5\)
\(\Leftrightarrow 2\sqrt{x}=5\Rightarrow \sqrt{x}=\frac{5}{2}\Rightarrow x=\frac{25}{4}\) (thỏa man)
b) ĐK: \(x\geq -5\)
PT \(\Leftrightarrow \sqrt{4}.\sqrt{x+5}-3\sqrt{x+5}+\frac{4}{3}\sqrt{9}.\sqrt{x+5}=6\)
\(\Leftrightarrow 2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow 3\sqrt{x+5}=6\Rightarrow \sqrt{x+5}=2\)
\(\Rightarrow x+5=2^2=4\Rightarrow x=-1\) (thỏa mãn)
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
ĐK:5+x\(\ge0\)\(\Leftrightarrow x\ge-5\)
\(\sqrt{4x+20}-3\sqrt{5+x}=6-\dfrac{4}{3}\sqrt{9x+45}\)
\(\Leftrightarrow\sqrt{4\left(x+5\right)}-3\sqrt{x+5}=6-\dfrac{4}{3}\sqrt{9\left(x+5\right)}\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+\dfrac{4}{3}.3\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\Leftrightarrow\sqrt{x+5}=2\)
\(\Leftrightarrow x+5=4\)
\(\Leftrightarrow x=-1\)(tm)
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
Ta có : \(\sqrt{x-5}-\sqrt{4x-20}-\frac{1}{5}.\sqrt{9x-45}=3\)
\(\Leftrightarrow\sqrt{x-5}+\sqrt{4\left(x-5\right)}-\frac{1}{5}\sqrt{9\left(x-5\right)}=3\)
\(\Leftrightarrow\sqrt{x-5}+2\sqrt{x-5}-\frac{3}{5}\sqrt{x-5}=3\left(^∗\right)\)
Đặt \(\sqrt{x-5}=t,\hept{\begin{cases}t>0\\x\ge5\end{cases}}\)
Từ (*) ta có : \(t+2t+\frac{-3}{5}t=3\)
\(\Leftrightarrow5t+10t-3t=15\)
\(\Leftrightarrow t=\frac{5}{4}\left(t/m\right)\)
\(\Leftrightarrow\sqrt{x-5}=\frac{5}{4}\)
\(\Leftrightarrow x-5=\frac{25}{16}\)
\(\Leftrightarrow x=\frac{105}{16}\)
Nghiệm cuối của phương trình là : \(\left\{\frac{105}{16}\right\}\)
\(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\\ 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ 2\sqrt{x-5}=4\\ \sqrt{x-5}=2\\ x-5=4\\ x=9\)
ĐK:x\(\ge5\)
\(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)
Vậy S={9}