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NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 5 2019
Câu 1:
\(\Leftrightarrow\sqrt{2\left(x+1\right)\left(x+3\right)}+\sqrt{\left(x-1\right)\left(x+1\right)}=2\left(x+1\right)\)
- Với \(x< -1\Rightarrow\left\{{}\begin{matrix}VT\ge0\\VP< 0\end{matrix}\right.\) pt vô nghiệm
- Nhận thấy \(x=-1\) là 1 nghiệm
- Nếu \(x>-1\) kết hợp ĐKXĐ các căn thức ta được \(x\ge1\), pt tương đương:
\(\sqrt{2\left(x+3\right)}+\sqrt{x-1}=2\sqrt{x+1}\)
\(\Leftrightarrow2x+6+x-1+2\sqrt{2\left(x+3\right)\left(x-1\right)}=4x+4\)
\(\Leftrightarrow2\sqrt{2x^2+4x-6}=x-1\)
\(\Leftrightarrow4\left(2x^2+4x-6\right)=\left(x-1\right)^2\)
\(\Leftrightarrow7x^2+18x-25=0\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\frac{25}{7}< 0\left(l\right)\end{matrix}\right.\)
Vậy pt có nghiệm \(x=\pm1\)
Câu 2:
ĐKXĐ: \(x\ge1\)
\(\sqrt{x-1+2\sqrt{x-1}+1}-\sqrt{x-1-2\sqrt{x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}-\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\)
\(\Leftrightarrow\sqrt{x-1}+1-\left|\sqrt{x-1}-1\right|=2\)
- Nếu \(\sqrt{x-1}-1\ge0\Leftrightarrow x\ge2\) pt trở thành:
\(\sqrt{x-1}+1-\sqrt{x-1}+1=2\Leftrightarrow2=2\) (luôn đúng)
- Nếu \(1\le x< 2\) pt trở thành:
\(\sqrt{x-1}+1-1+\sqrt{x-1}=2\Leftrightarrow x=2\left(l\right)\)
Vậy nghiệm của pt là \(x\ge2\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 5 2019
Câu 3:
Bình phương 2 vế ta được:
\(2x^2+2x+5+2\sqrt{\left(x^2+x+4\right)\left(x^2+x+1\right)}=2x^2+2x+9\)
\(\Leftrightarrow\sqrt{\left(x^2+x+4\right)\left(x^2+x+1\right)}=2\)
\(\Leftrightarrow\left(x^2+x+4\right)\left(x^2+x+1\right)=4\)
Đặt \(x^2+x+1=a>0\) pt trở thành:
\(a\left(a+3\right)=4\Leftrightarrow a^2+3a-4=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-4\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x^2+x+1=1\Leftrightarrow x^2+x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Câu 5:
ĐKXĐ: \(x\ge1\)
\(\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1-6\sqrt{x-1}+9}=1\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}-3\right)^2}=1\)
\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|=1\)
Mà \(VT=\left|\sqrt{x-1}-2\right|+\left|3-\sqrt{x-1}\right|\ge\left|\sqrt{x-1}-2+3-\sqrt{x-1}\right|=1\)
\(\Rightarrow VT\ge VP\Rightarrow\) Đẳng thức xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}\sqrt{x-1}-2\ge0\\\sqrt{x-1}-3\le0\end{matrix}\right.\) \(\Rightarrow5\le x\le10\)
Vậy nghiệm của pt là \(5\le x\le10\)
Y
4 tháng 7 2019
Làm hơi tắt xíu, có gì ko hiểu cmt nha :>
\(a.\sqrt{x-1}=3\left(ĐK:x\ge1\right)\Leftrightarrow x-1=9\Leftrightarrow x=10\)
\(b.\sqrt{x^2-4x+4}=2\\ \Leftrightarrow\sqrt{\left(x-2\right)^2}=2\\ \Leftrightarrow\left|x-2\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-2=2\left(x\ge2\right)\\2-x=2\left(x< 2\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)
\(c.\sqrt{25x^2-10x+1}=4x-9\\ \Leftrightarrow\sqrt{\left(5x-1\right)^2}=4x-9\\ \Leftrightarrow\left|5x-1\right|=4x-9\\\Leftrightarrow \left[{}\begin{matrix}5x-1=4x-9\left(x\ge\frac{1}{5}\right)\\1-5x=4x-9\left(x< \frac{1}{5}\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-8\left(ktm\right)\\x=\frac{10}{9}\left(ktm\right)\end{matrix}\right.\)
Y
4 tháng 7 2019
\(d.\sqrt{x^2+2x+1}=\sqrt{x+1}\left(ĐK:x\ge-1\right)\\ \Leftrightarrow x^2+2x+1=x+1\\ \Leftrightarrow x^2+x=0\Leftrightarrow x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
e. ĐK: \(\left[{}\begin{matrix}x\ge3\\x\le-3\end{matrix}\right.\)
\(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\\ \Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}+\sqrt{\left(x-3\right)^2}=0\\ \Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}+\sqrt{x-3}\right)=0\\ \Leftrightarrow\sqrt{x-3}=0\\ \Leftrightarrow x-3=0\Leftrightarrow x=3\)
Câu cuối chưa nghĩ ra, sorry :<
CM
2
24 tháng 8 2020
\(\sqrt{4x^2-4x+1}=\sqrt{x^2+10x+25}\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\sqrt{\left(x+5\right)^2}\)
\(\Leftrightarrow\left|2x-1\right|=\left|x+5\right|\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=x+5\\2x-1=-\left(x+5\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=x+5\\2x-1=-x-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-\frac{4}{3}\end{cases}}\)
24 tháng 8 2020
a)
\(\sqrt{x+3}+2\sqrt{4\left(x+3\right)}-\frac{1}{3}\sqrt{9\left(x+3\right)}=8\)
\(\sqrt{x+3}+2\cdot2\sqrt{x+3}-\frac{1}{3}\cdot3\sqrt{x+3}=8\)
\(\sqrt{x+3}+4\sqrt{x+3}-\sqrt{x+3}=8\)
\(4\sqrt{x+3}=8\)
\(\sqrt{x+3}=2\)
\(\orbr{\begin{cases}2\ge0\left(llđ\right)\\x+3=2^2\end{cases}}\)
\(x+3=4\)
\(x=1\)
b)
\(\orbr{\begin{cases}x^2+10x+25\ge0\\4x^2-4x+1=x^2+10x+25\end{cases}}\)
\(\orbr{\begin{cases}\left(x+5\right)^2\ge0\left(lld\right)\\3x^2-6x-24=0\end{cases}}\)
\(\orbr{\begin{cases}x=6\\x=-\frac{4}{3}\end{cases}}\)
PA
25 tháng 7 2017
\(x^2-4x-6=\sqrt{2x^2-8x+12}\)
\(\Leftrightarrow\left(x^2+2x\right)-\left(6x+6+\sqrt{2x^2-8x+12}\right)=0\)
\(\Leftrightarrow x\left(x+2\right)-\dfrac{36x^2+72x+36-\left(2x^2-8x+12\right)}{\left(6x+6\right)-\sqrt{2x^2-8x+12}}=0\)
\(\Leftrightarrow x\left(x+2\right)-\dfrac{2\left(17x+6\right)\left(x+2\right)}{\left(6x+6\right)-\sqrt{2x^2-8x+12}}=0\)
\(\Leftrightarrow\left(x+2\right)\left[x-\dfrac{2\left(17x+6\right)}{\left(6x+6\right)-\sqrt{2x^2-8x+12}}\right]=0\)
Pt \(x-\dfrac{2\left(17x+6\right)}{\left(6x+6\right)-\sqrt{2x^2-8x+12}}\) vô nghiệm
=> x + 2 = 0
<=> x = - 2 (nhận)
PA
25 tháng 7 2017
\(\sqrt{x+2-4\sqrt{x-2}}+\sqrt{x+7-6\sqrt{x-2}}=1\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-2}-2\right)^2}+\sqrt{\left(\sqrt{x-2}-3\right)^2}=1\)
\(\Leftrightarrow\left|\sqrt{x-2}-2\right|+\left|\sqrt{x-2}-3\right|=1\)
Ta có:
\(VT=\left|\sqrt{x-2}-2\right|+\left|3-\sqrt{x-2}\right|\ge\left|\sqrt{x-2}-2+3-\sqrt{x-2}\right|=1\)
Dấu "=" xảy ra khi \(\left(\sqrt{x-2}-2\right)\left(3-\sqrt{x-2}\right)\ge0\)
Bảng xét dấu:
Vậy \(6\le x\le11\)
12 tháng 6 2018
a) Ta có: \(\sqrt{x-2}+\sqrt{4x-8}=12\)
\(\Rightarrow\sqrt{x-2}+\sqrt{4}.\sqrt{x-2}=12\)
\(\Rightarrow\sqrt{x-2}\left(2+1\right)=12\)
\(\Rightarrow\sqrt{x-2}=4\)
\(\Rightarrow x-2=4^2\Rightarrow x=18\)
b) \(\sqrt{x-1}-\sqrt{x-4}=1\)
\(\Rightarrow\left(\sqrt{x-1}-\sqrt{x-4}\right)^2=1\)
\(\Rightarrow\left(\sqrt{x-1}\right)^2-2.\sqrt{x-1}.\sqrt{x-4}+\left(\sqrt{x-4}\right)^2=1\)
\(\Rightarrow x-1-2.\sqrt{\left(x-1\right)\left(x-4\right)}+x-4=1\)
\(\Rightarrow2x-2\sqrt{\left(x-1\right)\left(x-4\right)}-6=0\)
\(\Rightarrow2\left(x-\sqrt{\left(x-1\right)\left(x-4\right)}-3\right)=0\)
\(\Rightarrow x-\sqrt{\left(x-1\right)\left(x-4\right)}-3=0\)
....
12 tháng 7 2018
1) \(\sqrt{3-x}=3x-5\)
\(\Leftrightarrow\left(\sqrt{3-x}\right)^2=\left(3x-5\right)^2\)
\(\Leftrightarrow3-x=9^2-30x+25\)
\(\Rightarrow x=\frac{11}{9};x=2\)
2) \(x-\sqrt{4x-3}\)
\(\Leftrightarrow x-\sqrt{4x-3}-x=2x-x\)
\(\Leftrightarrow-\sqrt{4-x}=2-x\)
\(\Leftrightarrow\left(-\sqrt{4x-3}\right)^2=\left(2-x\right)^2\)
\(\Leftrightarrow4x-3=4-4x+x^2\)
\(\Rightarrow x=1;x=7\)
4) \(\sqrt{x+1}=x-1\)
\(\Leftrightarrow\left(\sqrt{x+1}\right)^2=\left(x-1\right)^2\)
\(\Leftrightarrow x+1=x^2-2x+1\)
\(\Leftrightarrow x=3;x=0\)
\(\Rightarrow x=3;x=0\)
5) \(\sqrt{x^2-1}=x+1\)
\(\Leftrightarrow\left(\sqrt{x^2-1}\right)^2=\left(x+1\right)^2\)
\(\Leftrightarrow x^2-1=x^2+2x+1\)
\(\Rightarrow x=-1\)
6) \(\sqrt{x^2-4x+3}=x-2\)
\(\Leftrightarrow\left(\sqrt{x^2-4x+3}\right)^2=\left(x-2\right)^2\)
\(\Leftrightarrow x^2-4x+3=x^2-4x+4\)
\(\Leftrightarrow x=3;x=4\)
\(\Rightarrow x=3;x=4\)
7) \(\sqrt{x^2-1}=x-1\)
\(\Leftrightarrow\left(\sqrt{x^2-1}\right)^2=\left(x-1\right)^2\)
\(\Leftrightarrow x^2-1=x^2-2x+1\)
\(\Rightarrow x=1\)
8) \(x-2\sqrt{x-1}=16\)
\(\Leftrightarrow x-2\sqrt{x-1}-x=16-x\)
\(\Leftrightarrow-2\sqrt{x-1}=16-x\)
\(\Leftrightarrow\left(-2\sqrt{x-1}\right)^2=\left(16-x\right)^2\)
\(\Leftrightarrow4x-4=256-32x+x^2\)
\(\Leftrightarrow x=26;x=10\)
\(\Rightarrow x=26;x=10\)
9) \(\sqrt{5-x^2}=x-1\)
\(\Leftrightarrow\left(\sqrt{5-x^2}\right)^2=\left(x+1\right)^2\)
\(\Leftrightarrow5-x^2=x^2-2x+1\)
\(\Leftrightarrow x=2;x=-1\)
\(\Rightarrow x=2;x=-1\)
10) \(x-\sqrt{4x-3}=2\)
\(\Leftrightarrow x-\sqrt{4x-3}-x=2-x\)
\(\Leftrightarrow-\sqrt{4x-3}=2-x\)
\(\Leftrightarrow\left(-\sqrt{4x-3}\right)^2=\left(2-x\right)^2\)
\(\Leftrightarrow4x-3=4-4x+x^2\)
\(\Leftrightarrow x=7;x=1\)
\(\Rightarrow x=1;x=7\)
Mk ko chắc