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AH
Akai Haruma
Giáo viên
3 tháng 8 2021
a. ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow -5x-5\sqrt{x}+12\sqrt{x}+12=0$
$\Leftrightarrow -5\sqrt{x}(\sqrt{x}+1)+12(\sqrt{x}+1)=0$
$\Leftrightarrow (\sqrt{x}+1)(12-5\sqrt{x})=0$
Dễ thấy $\sqrt{x}+1>1$ với mọi $x\geq 0$ nên $12-5\sqrt{x}=0$
$\Leftrightarrow \sqrt{x}=\frac{12}{5}$
$\Leftrightarrow x=5,76$ (thỏa mãn)
AH
Akai Haruma
Giáo viên
3 tháng 8 2021
b. ĐKXĐ: $x^2\geq 5$
PT $\Leftrightarrow \frac{1}{3}\sqrt{4}.\sqrt{x^2-5}+2\sqrt{\frac{1}{9}}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$
$\Leftrightarrow \frac{2}{3}\sqrt{x^2-5}+\frac{2}{3}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$
$\Leftrightarrow -\frac{5}{3}\sqrt{x^2-5}=0$
$\Leftrightarrow \sqrt{x^2-5}=0$
$\Leftrightarrow x=\pm \sqrt{5}$
24 tháng 7 2023
=>\(5\cdot\dfrac{3\sqrt{x-3}}{5}-7\cdot\dfrac{2\sqrt{x-3}}{3}-7\cdot\sqrt{x^2-9}+18\cdot\sqrt{\dfrac{9}{81}\left(x^2-9\right)}=0\)
=>\(3\cdot\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}=7\cdot\sqrt{x^2-9}-18\cdot\dfrac{3}{9}\cdot\sqrt{x^2-9}\)
=>\(-\dfrac{5}{3}\sqrt{x-3}=\sqrt{x^2-9}\)
=>\(\sqrt{x-3}\left(\sqrt{x+3}+\dfrac{5}{3}\right)=0\)
=>x-3=0
=>x=3
22 tháng 7 2023
\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)
\(ĐK:x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)
\(\Leftrightarrow4x^2-9=4x+12\)
\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)
\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(ĐK:x\ge5\)
\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)
22 tháng 7 2023
\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)
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 \(\dfrac{5}{3}+\sqrt{x+3}>0\))
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)
15 tháng 10 2023
a: ĐKXĐ: x-5>=0
=>x>=5
\(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\cdot\sqrt{9x-45}=4\)
=>\(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)
=>\(2\sqrt{x-5}=4\)
=>x-5=4
=>x=9(nhận)
b: ĐKXĐ: x-1>=0
=>x>=1
\(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}=4\)
=>\(\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}=4\)
=>\(-2\sqrt{x-1}=4\)
=>\(\sqrt{x-1}=-2\)(vô lý)
Vậy: Phương trình vô nghiệm
c: ĐKXĐ: x-2>=0
=>x>=2
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot\sqrt{9x-18}+6\cdot\sqrt{\dfrac{x-2}{81}}=-4\)
=>\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot3\sqrt{x-2}+6\cdot\dfrac{\sqrt{x-2}}{9}=-4\)
=>\(\sqrt{x-2}\left(\dfrac{1}{3}-2+\dfrac{2}{3}\right)=-4\)
=>\(-\sqrt{x-2}=-4\)
=>x-2=16
=>x=18(nhận)
d: ĐKXĐ: x+3>=0
=>x>=-3
\(\sqrt{9x+27}+4\sqrt{x+3}-\dfrac{3}{4}\cdot\sqrt{16x+48}=0\)
=>\(3\sqrt{x+3}+4\sqrt{x+3}-\dfrac{3}{4}\cdot4\sqrt{x+3}=0\)
=>\(4\sqrt{x+3}=0\)
=>x+3=0
=>x=-3(nhận)
NV
15 tháng 10 2023
a) \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\)
= \(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\)
= \(2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
= \(2\sqrt{x-5}=4\)
= \(\sqrt{x-5}=2\)
= \(\left|x-5\right|=4\)
=> \(x-5=\pm4\)
\(x=\pm4+5\)
\(x=9;x=1\)
Vậy x=9; x=1
31 tháng 8 2020
1. \(x^3-6x^2+10x-4=0\)
<=> \(\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(2x-4\right)=0\)
<=> \(\left(x-2\right)\left(x^2-4x+2\right)=0\)
<=> \(\orbr{\begin{cases}x=2\\x^2-4x+2=0\left(1\right)\end{cases}}\)
Giải pt (1): \(\Delta=\left(-4\right)^2-4.2=8>0\)
=> pt (1) có 2 nghiệm: \(x_1=\frac{4+\sqrt{8}}{2}=2+\sqrt{2}\)
\(x_2=\frac{4-\sqrt{8}}{2}=2-\sqrt{2}\)
31 tháng 8 2020
1) Ta có: \(x^3-6x^2+10x-4=0\)
\(\Leftrightarrow\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(2x-4\right)=0\)
\(\Leftrightarrow x^2\left(x-2\right)-4x\left(x-2\right)+2\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^2-4x+2\right)\left(x-2\right)=0\)
+ \(x-2=0\)\(\Leftrightarrow\)\(x=2\)\(\left(TM\right)\)
+ \(x^2-4x+2=0\)\(\Leftrightarrow\)\(\left(x^2-4x+4\right)-2=0\)
\(\Leftrightarrow\)\(\left(x-2\right)^2=2\)
\(\Leftrightarrow\)\(x-2=\pm\sqrt{2}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2+\sqrt{2}\approx3,4142\left(TM\right)\\x=2-\sqrt{2}\approx0,5858\left(TM\right)\end{cases}}\)
Vậy \(S=\left\{0,5858;2;3,4142\right\}\)
5 tháng 7 2021
a) Ta có: \(\sqrt{25x+75}+3\sqrt{x-2}=2\sqrt{x-2}+\sqrt{9x-18}\)
\(\Leftrightarrow5\sqrt{x+3}+3\sqrt{x-2}=2\sqrt{x-2}+3\sqrt{x-2}\)
\(\Leftrightarrow\sqrt{25x+75}=\sqrt{4x-8}\)
\(\Leftrightarrow25x-4x=-8-75\)
\(\Leftrightarrow21x=-83\)
hay \(x=-\dfrac{83}{21}\)
b) Ta có: \(\sqrt{\left(2x-1\right)^2}=4\)
\(\Leftrightarrow\left|2x-1\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
c) Ta có: \(\sqrt{\left(2x+1\right)^2}=3x-5\)
\(\Leftrightarrow\left|2x+1\right|=3x-5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=3x-5\left(x\ge-\dfrac{1}{2}\right)\\2x+1=5-3x\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3x=-5-1\\2x+3x=5-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(nhận\right)\\x=\dfrac{4}{5}\left(loại\right)\end{matrix}\right.\)
d) Ta có: \(\sqrt{4x-12}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)
\(\Leftrightarrow2\sqrt{x-3}-2\sqrt{x-2}=3\sqrt{x-2}+8\)
\(\Leftrightarrow2\sqrt{x-3}-5\sqrt{x-2}=8\)
\(\Leftrightarrow4\left(x-3\right)+25\left(x-2\right)-20\sqrt{x^2-5x+6}=8\)
\(\Leftrightarrow4x-12+25x-50-8=20\sqrt{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow20\sqrt{\left(x-2\right)\left(x-3\right)}=29x-70\)
\(\Leftrightarrow x^2-5x+6=\dfrac{\left(29x-70\right)^2}{400}\)
\(\Leftrightarrow x^2-5x+6=\dfrac{841}{400}x^2-\dfrac{203}{20}x+\dfrac{49}{4}\)
\(\Leftrightarrow\dfrac{-441}{400}x^2+\dfrac{103}{20}x-\dfrac{25}{4}=0\)
\(\Delta=\left(\dfrac{103}{20}\right)^2-4\cdot\dfrac{-441}{400}\cdot\dfrac{-25}{4}=-\dfrac{26}{25}\)(Vô lý)
vậy: Phương trình vô nghiệm
TH
1
18 tháng 11 2018
cộng tất cả lại ==> đưa về dạng bình phương ==> pt có nghiệm duy nhất
\(\left(0;0;0\right)\)
Sửa đề: \(\sqrt{9x^2+18}+2\sqrt{x^2+2}-\sqrt{16x^2+32}-12-5=0\)
\(\Leftrightarrow3\sqrt{x^2+2}+2\sqrt{x^2+2}-4\sqrt{x^2+2}=17\)
=>\(\sqrt{x^2+2}=17\)
=>x^2+2=289
=>x^2=287
=>\(x=\pm\sqrt{287}\)