Giải phương trình
\(4x^2+\sqrt{2x+9}=9\)
\(\left(x+1\right)\left(x+3\right)=5\sqrt{5x+11}\)
\(\sqrt{4x+9}+3\left(2x+1\right)=2x^2\)
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ND
Nguyễn Đức Trí
VIP
1 tháng 9 2023
1) \(\sqrt[]{9\left(x-1\right)}=21\)
\(\Leftrightarrow9\left(x-1\right)=21^2\)
\(\Leftrightarrow9\left(x-1\right)=441\)
\(\Leftrightarrow x-1=49\Leftrightarrow x=50\)
2) \(\sqrt[]{1-x}+\sqrt[]{4-4x}-\dfrac{1}{3}\sqrt[]{16-16x}+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}+\sqrt[]{4\left(1-x\right)}-\dfrac{1}{3}\sqrt[]{16\left(1-x\right)}+5=0\)
\(\)\(\Leftrightarrow\sqrt[]{1-x}+2\sqrt[]{1-x}-\dfrac{4}{3}\sqrt[]{1-x}+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}\left(1+3-\dfrac{4}{3}\right)+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}.\dfrac{8}{3}=-5\)
\(\Leftrightarrow\sqrt[]{1-x}=-\dfrac{15}{8}\)
mà \(\sqrt[]{1-x}\ge0\)
\(\Leftrightarrow pt.vô.nghiệm\)
3) \(\sqrt[]{2x}-\sqrt[]{50}=0\)
\(\Leftrightarrow\sqrt[]{2x}=\sqrt[]{50}\)
\(\Leftrightarrow2x=50\Leftrightarrow x=25\)
H9
HT.Phong (9A5)
CTVHS
1 tháng 9 2023
1) \(\sqrt{9\left(x-1\right)}=21\) (ĐK: \(x\ge1\))
\(\Leftrightarrow3\sqrt{x-1}=21\)
\(\Leftrightarrow\sqrt{x-1}=7\)
\(\Leftrightarrow x-1=49\)
\(\Leftrightarrow x=49+1\)
\(\Leftrightarrow x=50\left(tm\right)\)
2) \(\sqrt{1-x}+\sqrt{4-4x}-\dfrac{1}{3}\sqrt{16-16x}+5=0\) (ĐK: \(x\le1\))
\(\Leftrightarrow\sqrt{1-x}+2\sqrt{1-x}-\dfrac{4}{3}\sqrt{1-x}+5=0\)
\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}+5=0\)
\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}=-5\) (vô lý)
Phương trình vô nghiệm
3) \(\sqrt{2x}-\sqrt{50}=0\) (ĐK: \(x\ge0\))
\(\Leftrightarrow\sqrt{2x}=\sqrt{50}\)
\(\Leftrightarrow2x=50\)
\(\Leftrightarrow x=\dfrac{50}{2}\)
\(\Leftrightarrow x=25\left(tm\right)\)
4) \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\left(ĐK:x\ge-\dfrac{1}{2}\right)\\2x+1=-6\left(ĐK:x< -\dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\left(tm\right)\\x=-\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)
5) \(\sqrt{\left(x-3\right)^2}=3-x\)
\(\Leftrightarrow\left|x-3\right|=3-x\)
\(\Leftrightarrow x-3=3-x\)
\(\Leftrightarrow x+x=3+3\)
\(\Leftrightarrow x=\dfrac{6}{2}\)
\(\Leftrightarrow x=3\)
16 tháng 6 2021
a) ĐK: \(x\ge3\)
PT \(\Leftrightarrow\sqrt{\left(x-3\right)\left(x-2\right)}-\sqrt{x-2}+\sqrt{x+1}-\sqrt{\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-3}-1\right)+\sqrt{x+1}\left(1-\sqrt{x-3}\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-\sqrt{x+1}\right)\left(\sqrt{x-3}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=\sqrt{x+1}\\\sqrt{x-3}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x-2=x+1\\x-3=1\end{matrix}\right.\) \(\Leftrightarrow x=4\) (Thỏa mãn)
Vậy ...
17 tháng 12 2023
a: ĐKXĐ: \(x\in R\)
\(\sqrt{\left(2x+3\right)^2}=5\)
=>|2x+3|=5
=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
b: ĐKXĐ: \(x\in R\)
\(\sqrt{9\left(x-2\right)^2}=18\)
=>\(\sqrt{9}\cdot\sqrt{\left(x-2\right)^2}=18\)
=>\(3\cdot\left|x-2\right|=18\)
=>\(\left|x-2\right|=6\)
=>\(\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
c: ĐKXĐ: x>=2
\(\sqrt{9x-18}-\sqrt{4x-8}+3\sqrt{x-2}=40\)
=>\(3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)
=>\(4\sqrt{x-2}=40\)
=>\(\sqrt{x-2}=10\)
=>x-2=100
=>x=102(nhận)
d: ĐKXĐ: \(x\in R\)
\(\sqrt{4\left(x-3\right)^2}=8\)
=>\(\sqrt{\left(2x-6\right)^2}=8\)
=>|2x-6|=8
=>\(\left[{}\begin{matrix}2x-6=8\\2x-6=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=14\\2x=-2\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)
e: ĐKXĐ: \(x\in R\)
\(\sqrt{4x^2+12x+9}=5\)
=>\(\sqrt{\left(2x\right)^2+2\cdot2x\cdot3+3^2}=5\)
=>\(\sqrt{\left(2x+3\right)^2}=5\)
=>|2x+3|=5
=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
f: ĐKXĐ:x>=6/5
\(\sqrt{5x-6}-3=0\)
=>\(\sqrt{5x-6}=3\)
=>\(5x-6=3^2=9\)
=>5x=6+9=15
=>x=15/5=3(nhận)