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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
7 tháng 10 2021
c: Ta có: \(\sqrt{x-1}+\sqrt{9x-9}-\sqrt{4x-4}=4\)
\(\Leftrightarrow2\sqrt{x-1}=4\)
\(\Leftrightarrow x-1=4\)
hay x=5
e: Ta có: \(\sqrt{4x^2-28x+49}-5=0\)
\(\Leftrightarrow\left|2x-7\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-7=5\\2x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
8 tháng 10 2021
a. ĐKXĐ: $x\in\mathbb{R}$
PT $\Leftrightarrow \sqrt{(x-2)^2}=2-x$
$\Leftrightarrow |x-2|=2-x$
$\Leftrightarrow 2-x\geq 0$
$\Leftrightarrow x\leq 2$
b. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{4}.\sqrt{x-2}-\frac{1}{5}\sqrt{25}.\sqrt{x-2}=3\sqrt{x-2}-1$
$\Leftrightarrow 2\sqrt{x-2}-\sqrt{x-2}=3\sqrt{x-2}-1$
$\Leftrightarrow 1=2\sqrt{x-2}$
$\Leftrightarrow \frac{1}{2}=\sqrt{x-2}$
$\Leftrightarrow \frac{1}{4}=x-2$
$\Leftrightarrow x=\frac{9}{4}$ (tm)
MN
3
20 tháng 7 2019
\(2x+3+\sqrt{4x^2+9x+2}=2\sqrt{x+2}+\sqrt{4x+1}\left(x\ge-\frac{1}{4}\right)\)
\(\Leftrightarrow2\left(x+2\right)-1+\sqrt{\left(x+2\right)\left(4x+1\right)}=2\sqrt{x+2}+\sqrt{4x+1}\)
\(\Leftrightarrow4\left(x+2\right)-2+2\sqrt{x+2}.\sqrt{4x+1}=4\sqrt{x+2}+2\sqrt{4x+1}\)
Đặt \(\hept{\begin{cases}2\sqrt{x+2}=a\left(a\ge0\right)\\\sqrt{4x+1}=b\left(b\ge0\right)\end{cases}\Rightarrow}a^2-b^2=4\left(x+2\right)-4x-1=7\)\(\Leftrightarrow\left(a-b\right)\left(a+b\right)=7\)(1)
\(pt:a^2-2+ab=2a+2b\)
\(\Leftrightarrow a\left(a+b\right)-2\left(a+b\right)=2\)
\(\Leftrightarrow\left(a-2\right)\left(a+b\right)=2\)(2)
Nhân chéo 2 vế của (1) với (2) được
\(7\left(a-2\right)\left(a+b\right)=2\left(a-b\right)\left(a+b\right)\)
\(\Leftrightarrow7\left(a-2\right)=2\left(a-b\right)\left(Do\left(a+b\right)>0\right)\)
\(\Leftrightarrow7a-14=2a-2b\)
\(\Leftrightarrow5a=14-2b\)
\(\Leftrightarrow10\sqrt{x+2}=14-2\sqrt{4x+1}\)
\(\Leftrightarrow5\sqrt{x+2}=7-\sqrt{4x+1}\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{4x+1}\le7\\25\left(x+2\right)=49-14\sqrt{4x+1}+4x+1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}0\le4x+1\le49\\21x=-14\sqrt{4x+1}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}-\frac{1}{4}\le x\le0\\441x^2=196\left(4x+1\right)\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}-\frac{1}{4}\le x\le0\\441x^2-784x-196=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}-\frac{1}{4}\le x\le0\\49\left(9x+2\right)\left(x-2\right)=0\end{cases}}\)
\(\Leftrightarrow x=-\frac{2}{9}\left(TmĐKXĐ\right)\)
Vậy
T
22 tháng 7 2019
Incursion_03 em thử nha, sai thì thôi ạ, em hơi nghiện liên hợp r.
ĐK: x>=-1/4
PT \(\Leftrightarrow2x+\frac{31}{9}+\sqrt{4x^2+9x+2}-\frac{4}{9}=2\sqrt{x+2}-\frac{8}{3}+\sqrt{4x+1}-\frac{1}{3}+3\)
\(\Leftrightarrow2\left(x+\frac{2}{9}\right)+\frac{\left(x+\frac{2}{9}\right)\left(4x+\frac{73}{9}\right)}{\sqrt{4x^2+9x+2}+\frac{4}{9}}=\frac{4\left(x+\frac{2}{9}\right)}{2\sqrt{x+2}+\frac{8}{3}}+\frac{4\left(x+\frac{2}{9}\right)}{\sqrt{4x+1}+\frac{1}{3}}\)
\(\Leftrightarrow\left(x+\frac{2}{9}\right)\left[2+\frac{4x+\frac{73}{9}}{\sqrt{4x^2+9x+2}+\frac{4}{9}}-4\left(\frac{1}{2\sqrt{x+2}+\frac{8}{3}}+\frac{1}{\sqrt{4x+1}+\frac{1}{3}}\right)\right]=0\)
Cái ngoặc to em chịu:( đang suy nghĩ
HP
17 tháng 9 2021
d. \(\sqrt{9x^2+12x+4}=4\)
<=> \(\sqrt{\left(3x+2\right)^2}=4\)
<=> \(|3x+2|=4\)
<=> \(\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
17 tháng 9 2021
c: Ta có: \(\dfrac{5\sqrt{x}-2}{8\sqrt{x}+2.5}=\dfrac{2}{7}\)
\(\Leftrightarrow35\sqrt{x}-14=16\sqrt{x}+5\)
\(\Leftrightarrow x=1\)
NK
6
15 tháng 9 2021
\(\sqrt{4x^2-4x+1}=3-x\left(x\in R\right)\\ \Leftrightarrow\sqrt{\left(2x-1\right)^2}=3-x\\ \Leftrightarrow2x-1=3-x\\ \Leftrightarrow3x=4\Leftrightarrow x=\dfrac{4}{3}\\ \sqrt{9x+9}+\sqrt{x+1}-\sqrt{4x+4}=2\left(x+1\right)\left(x\ge-1\right)\\ \Leftrightarrow\sqrt{x+1}\left(\sqrt{9}+1+\sqrt{4}\right)=2\left(x+1\right)\\ \Leftrightarrow6\sqrt{x+1}=2\left(x+1\right)\\ \Leftrightarrow3\sqrt{x+1}=x+1\\ \Leftrightarrow\sqrt{x+1}\left(3-\sqrt{x+1}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\\sqrt{x+1}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x+1=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=8\left(tm\right)\end{matrix}\right.\)
HP
15 tháng 9 2021
a, ĐK: \(x\in R\)
\(\sqrt{4x^2-4x+1}=3-x\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=3-x\)
\(\Leftrightarrow\left|2x-1\right|=3-x\)
TH1: \(\left\{{}\begin{matrix}2x-1\ge0\\2x-1=3-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x=\dfrac{4}{3}\end{matrix}\right.\Leftrightarrow x=\dfrac{4}{3}\)
TH2: \(\left\{{}\begin{matrix}2x-1< 0\\1-2x=3-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< \dfrac{1}{2}\\x=-2\end{matrix}\right.\Leftrightarrow x=-2\)
