giải pt: giúp mịch vs ạ :(
\(\sqrt{4x^2-4x+1}=\sqrt{x^2-6x+9}\)
:< xong lúc nào cũng được nhưng nhanh hộ mịch :)))
~Thanks you
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AD
2
H9
HT.Phong (9A5)
CTVHS
27 tháng 6 2023
a) \(\sqrt{1-4x+4x^2}=5\)
\(\Leftrightarrow\sqrt{\left(1-2x\right)^2}=5\)
\(\Leftrightarrow\left|1-2x\right|=5\)
\(\Leftrightarrow2x-1=5\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
b) \(\sqrt{x^2+6x+9}=3x-1\)
\(\Leftrightarrow\sqrt{\left(x+3\right)^2=3x-1}\)
\(\Leftrightarrow\left|x+3\right|=3x-1\)
\(\Leftrightarrow x+3=3x-1\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=2\)
Y
27 tháng 6 2023
\(a,\sqrt{1-4x+4x^2}=5\\ \Leftrightarrow\sqrt{\left(1-2x\right)^2}=5\\ \Leftrightarrow\left|1-2x\right|=5\)
\(TH_1:x\le\dfrac{1}{2}\)
\(1-2x=5\\ \Leftrightarrow x=-2\left(tm\right)\)
\(TH_2:x\ge\dfrac{1}{2}\)
\(-1+2x=5\\ \Leftrightarrow x=3\left(tm\right)\)
Vậy \(S=\left\{-2;3\right\}\)
\(b,\sqrt{x^2+6x+9}=3x-1\\ \Leftrightarrow\sqrt{\left(x+3\right)^2}=3x-1\\ \Leftrightarrow\left|x+3\right|=3x-1\)
\(TH_1:x\ge-3\\ x+3=3x-1\\ \Leftrightarrow-2x=-4\Leftrightarrow x=2\left(tm\right)\)
\(TH_2:x< 3\\ -x-3=3x-1\\ \Leftrightarrow-4x=2\\ \Leftrightarrow x=-\dfrac{1}{2}\left(tm\right)\)
Vậy \(S=\left\{2;-\dfrac{1}{2}\right\}\)
ND
14 tháng 8 2019
\(ĐK:\forall x\in R\)
\(PT\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\sqrt{\left(x-3\right)^2}\)
\(\Leftrightarrow\left|2x-1\right|=x-3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x-3\\1-2x=x-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\frac{4}{3}\end{matrix}\right.\)(t/m)
1 tháng 12 2021
a,ĐKXĐ:\(x\ge2\)
\(4\sqrt{x-2}+\sqrt{9x-18}-\sqrt{\dfrac{x-2}{4}}=26\\ \Leftrightarrow4\sqrt{x-2}+3\sqrt{x-2}-\dfrac{\sqrt{x-2}}{2}=26\\ \Leftrightarrow8\sqrt{x-2}+6\sqrt{x-2}-\sqrt{x-2}=52\\ \Leftrightarrow13\sqrt{x-2}=52\\ \Leftrightarrow\sqrt{x-2}=4\\ \Leftrightarrow x-2=16\\ \Leftrightarrow x=18\left(tm\right)\)
b,ĐKXĐ:\(x\in R\)
\(3x+\sqrt{4x^2-8x+4}=1\\ \Leftrightarrow2\sqrt{x^2-2x+1}=1-3x\\ \Leftrightarrow\left|x-1\right|=\dfrac{1-3x}{2}\\ \Leftrightarrow\left[{}\begin{matrix}x-1=\dfrac{1-3x}{2}\\x-1=\dfrac{3x-1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-2=1-3x\\2x-2=3x-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
c, ĐKXĐ:\(x\ge0\)
\(\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=7\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+1\right)-2\left(2\sqrt{x}+1\right)=7\\ \Leftrightarrow2x+\sqrt{x}-4\sqrt{x}-2=7\\ \Leftrightarrow2x-3\sqrt{x}-9=0\\ \Leftrightarrow\left(2x+3\sqrt{x}\right)-\left(6\sqrt{x}+9\right)=0\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+3\right)-3\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left(\sqrt{x}-3\right)\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\2\sqrt{x}=-3\left(vô.lí\right)\end{matrix}\right.\\ \Leftrightarrow x=9\left(tm\right)\)
25 tháng 9 2020
\(a,\sqrt{1-6x+9x^2}=\sqrt{4x^2+12x+9} \) Đkxđ : x ≥ \(\frac{1}{3}\) ; x ≥ \(\frac{3}{2}\)
⇌ \(\sqrt{\left(1-3x\right)^2}\) = \(\sqrt{\left(2x+3\right)^2}\)
⇌ | 1 - 3x | = | 2x + 3 |
⇌ 1 - 3x = \(\pm\left(2x+3\right)\)
⇒ \(\left[{}\begin{matrix}1-3x=2x+3\\1-3x=-\left(2x+3\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{2}{5}\left(lo\text{ại}\right)\\x=4\left(tm\right)\end{matrix}\right.\)
\(b, \sqrt{x^2}+2x+1+\sqrt{4x^2}=2\) Đkxđ : x ≥ 0
⇌ | x | + 2x + 1 + | 2x| = 2
⇌ x + 2x + 1 + 2x = 2
⇌ 5x = 1
⇌ x = \(\frac{1}{5}\)
NH
0
23 tháng 8 2021
d) \(\sqrt{x^2-6x+9}=2\Leftrightarrow\sqrt{\left(x-3\right)^2}=2\Leftrightarrow x-3=2\Leftrightarrow x=5\)
e) đk: \(x\ge2\)\(\sqrt{x^2-3x+2}=\sqrt{x-1}\Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}=\sqrt{x-1}\Leftrightarrow\sqrt{x-2}=1\Leftrightarrow x-2=1\Leftrightarrow x=3\)f) \(\sqrt{4x^2-4x+1}=\sqrt{x^2-6x+9}\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\sqrt{\left(x-3\right)^2}\Leftrightarrow2x-1=x-3\Leftrightarrow x=-2\)
23 tháng 8 2021
c: Ta có: \(\sqrt{x+4\sqrt{x-4}}=2\)
\(\Leftrightarrow\left|\sqrt{x-4}+2\right|=2\)
\(\Leftrightarrow x-4=0\)
hay x=4
20 tháng 10 2018
\(1)\) ĐKXĐ : \(x\ge3\)
\(\sqrt{x^2-4x+3}+\sqrt{x-1}=0\)
\(\Leftrightarrow\)\(\sqrt{\left(x^2-4x+4\right)-1}+\sqrt{x-1}=0\)
\(\Leftrightarrow\)\(\sqrt{\left(x-2\right)^2-1}+\sqrt{x-1}=0\)
\(\Leftrightarrow\)\(\sqrt{\left(x-2-1\right)\left(x-2+1\right)}+\sqrt{x-1}=0\)
\(\Leftrightarrow\)\(\sqrt{\left(x-3\right)\left(x-1\right)}+\sqrt{x-1}=0\)
\(\Leftrightarrow\)\(\sqrt{x-1}\left(\sqrt{x-3}+1\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\sqrt{x-1}=0\\\sqrt{x-3}+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x\in\left\{\varnothing\right\}\end{cases}}}\)
Vậy \(x=1\)
\(2)\)\(\sqrt{x^2-2x+1}-\sqrt{x^2-6x+9}=10\)
\(\Leftrightarrow\)\(\sqrt{\left(x-1\right)^2}-\sqrt{\left(x-3\right)^2}=10\)
\(\Leftrightarrow\)\(\left|x-1\right|-\left|x-3\right|=10\)
+) Với \(\hept{\begin{cases}x-1\ge0\\x-3\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge1\\x\ge3\end{cases}\Leftrightarrow}x\ge3}\) ta có :
\(x-1-x+3=10\)
\(\Leftrightarrow\)\(0=8\) ( loại )
+) Với \(\hept{\begin{cases}x-1< 0\\x-3< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 1\\x< 3\end{cases}\Leftrightarrow}x< 1}\) ta có :
\(1-x+x-3=10\)
\(\Leftrightarrow\)\(0=12\) ( loại )
Vậy không có x thỏa mãn đề bài
Chúc bạn học tốt ~
PS : mới lp 8 sai đừng chửi nhé :v
L
1
4 tháng 9 2023
\(\sqrt{4x^2}=3\left(ĐK:4x^2\ge0\forall x\in R\right)\\ \Leftrightarrow\sqrt{\left(2x\right)^2}=3\\ \Leftrightarrow\left|2x\right|=3\\ \Leftrightarrow\left[{}\begin{matrix}2x=-3\\2x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\left(tm\right)\\x=\dfrac{3}{2}\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{3}{2};\dfrac{3}{2}\right\}\)
\(\sqrt{x^2-6x+9}=2\\ \Leftrightarrow\sqrt{\left(x-3\right)^2}=2\left(ĐK:\left(x-3\right)^2\ge0\forall x\in R\right)\\ \Leftrightarrow\left|x-3\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2+3\\x=-2-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=-5\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left(\pm5\right)\)
\(\sqrt{\left(2x-3\right)^2}=6\left(ĐK:\left(2x-3\right)^2\ge0\forall x\in R\right)\\ \Leftrightarrow\left|2x-3\right|=6\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=3+6\\2x=-6+3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=4,5\left(tm\right)\\x=-1,5\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{4,5;-1,5\right\}\)
\(\sqrt{25x^2}=100\\ \sqrt{\left(5x\right)^2}=100\left(ĐK:\left(5x\right)^2\ge0\forall x\in R\right)\\\Leftrightarrow \left|5x\right|=100\\ \Leftrightarrow\left[{}\begin{matrix}5x=100\\5x=-100\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=20\left(tm\right)\\x=-20\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{\pm20\right\}\)
Ta có: \(\sqrt{4x^2-4x+1}=\sqrt{x^2-6x+9}\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\sqrt{\left(x-3\right)^2}\)
\(\Leftrightarrow2x-1=x-3\)
\(\Leftrightarrow2x-x=-3+1\)
\(\Leftrightarrow x=-2\)
Vậy phương trình có nghiệm x=-2
\(\sqrt{4x^2-4x+1}=\sqrt{x^2-6x+9}\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\sqrt{\left(x-3\right)^2}\)
\(\Leftrightarrow\left|2x-1\right|=\left|x-3\right|\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=x-3\\2x-1=3-x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=\frac{4}{3}\end{cases}}\)