Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: ĐKXĐ: x>=3
Sửa đề: \(\sqrt{4x-12}-\sqrt{9x-27}+\sqrt{\dfrac{25x-75}{4}}-3=0\)
=>\(2\sqrt{x-3}-3\sqrt{x-3}+\dfrac{5}{2}\sqrt{x-3}-3=0\)
=>\(\dfrac{3}{2}\sqrt{x-3}=3\)
=>\(\sqrt{x-3}=2\)
=>x-3=4
=>x=7(nhận)
b: ĐKXĐ: x>=0
\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< =-\dfrac{3}{4}\)
=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{3}{4}< =0\)
=>\(\dfrac{4\sqrt{x}-8+3\sqrt{x}+3}{4\left(\sqrt{x}+1\right)}< =0\)
=>\(7\sqrt{x}-5< =0\)
=>\(\sqrt{x}< =\dfrac{5}{7}\)
=>0<=x<=25/49
c: ĐKXĐ: x>=5
\(\sqrt{9x-45}-14\sqrt{\dfrac{x-5}{49}}+\dfrac{1}{4}\sqrt{4x-20}=3\)
=>\(3\sqrt{x-5}-14\cdot\dfrac{\sqrt{x-5}}{7}+\dfrac{1}{4}\cdot2\cdot\sqrt{x-5}=3\)
=>\(\dfrac{3}{2}\sqrt{x-5}=3\)
=>\(\sqrt{x-5}=2\)
=>x-5=4
=>x=9(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
1
a,A=\(\left(\sqrt{45}-\sqrt{20}+\sqrt{5}\right):\sqrt{6}\)
A=\(\left(3\sqrt{5}-2\sqrt{5}+\sqrt{5}\right):\sqrt{6}\)
A=\(2\sqrt{5}:\sqrt{6}=\dfrac{2\sqrt{5}}{\sqrt{6}}=\dfrac{\sqrt{30}}{3}\)
b, B=\(\dfrac{\sqrt{10}-\sqrt{15}}{\sqrt{8}-\sqrt{12}}=\dfrac{\sqrt{5.2}-\sqrt{5.3}}{\sqrt{4.2}-\sqrt{4.3}}=\dfrac{\sqrt{5}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{4}\left(\sqrt{2}-\sqrt{3}\right)}\)
B=\(\dfrac{\sqrt{5}}{2}\)
Câu 1)
a) \(\left(\sqrt{45}-\sqrt{20}+\sqrt{5}\right):\sqrt{6}=\left(\sqrt{9.5}-\sqrt{4.5}+\sqrt{5}\right):\sqrt{6}=\left(3\sqrt{5}-2\sqrt{5}+\sqrt{5}\right):\sqrt{6}=\dfrac{2\sqrt{5}}{\sqrt{6}}=\dfrac{\sqrt{30}}{3}\)
b) \(\dfrac{\sqrt{10}-\sqrt{15}}{\sqrt{8}-\sqrt{12}}=\dfrac{\sqrt{5}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{4}\left(\sqrt{2}-\sqrt{3}\right)}=\dfrac{\sqrt{5}}{\sqrt{4}}=\dfrac{\sqrt{5}}{2}\)
Câu 2)
ĐK: x\(\ge5\)
\(\sqrt{x-5}+\sqrt{4x-20}-\dfrac{1}{5}\sqrt{9x-45}=3\Leftrightarrow\sqrt{x-5}+\sqrt{4\left(x-5\right)}-\dfrac{1}{5}\sqrt{9\left(x-5\right)}=3\Leftrightarrow\sqrt{x-5}+2\sqrt{x-5}-\dfrac{3}{5}\sqrt{x-5}=3\Leftrightarrow\dfrac{12}{5}\sqrt{x-5}=3\Leftrightarrow\sqrt{x-5}=\dfrac{5}{4}\Leftrightarrow x-5=\dfrac{25}{16}\Leftrightarrow x=\dfrac{105}{16}\left(tm\right)\)
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.......
a, \(\sqrt{x-5} = 3 \)
<=> x - 5 = 9
<=> x = 14.
b, \(\sqrt{4-5x}=12\)
<=> 4 - 5x = 144
<=> 5x = -140
<=> x = -28.
c, \(\sqrt{x^{2}-6x+9}=3\)
<=> x2 - 6x + 9 = 9
<=> (x - 3)2 = 9
TH1:
x - 3 = 3
<=> x = 6.
TH2:
x - 3 = -3
<=> x = 0
d, \(\sqrt{4x+20}+\sqrt{x+5}-\dfrac{1}{3} \sqrt{9x+45}=4\)
<=> \(2\sqrt{x+5}+\sqrt{x+5}-\sqrt{x+5} = 4\)
<=> \(2\sqrt{x+5}\)= 4
<=> \(\sqrt{x+5}\) = 2
<=> x + 5 = 4
<=> x = -1.
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 : \(\left\{\frac{105}{16}\right\}\)
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: \(\Leftrightarrow\dfrac{2x-3}{x-1}=4\)
=>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
\(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\\ 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ 2\sqrt{x-5}=4\\ \sqrt{x-5}=2\\ x-5=4\\ x=9\)
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}
`sqrt{x-5}+2sqrt{4x-20}-1/2sqrt{9x-45}=12`
Điều kiện:`x>=5`
`pt<=>sqrt{x-5}+2sqrt{4(x-5)}-1/2sqrt{9(x-5)}=12`
`<=>sqrt{x-5}+4sqrt{x-5}-3/2sqrt{x-5}=12`
`<=>7/2sqrt{x-5}=12`
`<=>sqrt{x-5}=24/7`
`<=>x-5=576/49`
`<=>x=821/49(Tmđk)`
Vậy `S={821/49}.`
Ta có: \(\sqrt{x-5}+2\sqrt{4x-20}-\dfrac{1}{3}\sqrt{9x-45}=12\)
\(\Leftrightarrow4\sqrt{x-5}=12\)
\(\Leftrightarrow x-5=9\)
hay x=14