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36B
37C
38D
39B
40D
41A
42B
43B
44A
45B
46B
47A
48C
50B
51B
52B
53D
54C
55D
56C
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ĐKXĐ: \(x\ge1\)
\(\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1-6\sqrt{x-1}+9}=0\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(3-\sqrt{x-1}\right)^2}=0\)
\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\left|3-\sqrt{x-1}\right|=0\)
Do \(\left|\sqrt{x-1}-2\right|+\left|3-\sqrt{x-1}\right|\ge\left|\sqrt{x-1}-2+3-\sqrt{x-1}\right|=1>0\) với mọi x thuộc TXĐ
\(\Rightarrow\) Phương trình đã cho vô nghiệm
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a: Khi x=2 thì (1) sẽ là:
4-2(m+2)+m+1=0
=>m+5-2m-4=0
=>1-m=0
=>m=1
x1+x2=m+1=3
=>x2=3-2=1
b: Δ=(m+2)^2-4(m+1)
=m^2+4m+4-4m-4=m^2>=0
=>Phương trình luôn có hai nghiệm
P=(x1+x2)^2-4x1x1+3x1x2
=(x1+x2)^2-x1x2
=(m+2)^2-m-1
=m^2+4m+4-m-1
=m^2+3m+3
=(m+3/2)^2+3/4>=3/4
Dấu = xảy ra khi m=-3/2
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1) \(\sqrt{2x-5}=7\)
\(\left(\sqrt{2x-5}\right)^2=7^2\)
\(2x-5=49\)
\(2x=54\)
\(x=27\)
2) \(3+\sqrt{x-2}=4\)
\(\sqrt{x-2}=1\)
\(\left(\sqrt{x-2}\right)^2=1^2\)
\(x-2=1\)
\(x=3\)
1) \(\sqrt{2x-5}=7\left(đk:x\ge\dfrac{5}{2}\right)\)
\(\Leftrightarrow2x-5=49\Leftrightarrow2x=54\Leftrightarrow x=27\left(tm\right)\)
2) \(3+\sqrt{x-2}=4\left(đk:x\ge2\right)\)
\(\Leftrightarrow\sqrt{x-2}=1\Leftrightarrow x-2=1\Leftrightarrow x=3\)
3) \(\Leftrightarrow\sqrt{\left(x-1\right)^2}=1\Leftrightarrow\left|x-1\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)
4) \(\Leftrightarrow\sqrt{\left(x-2\right)^2}=1\Leftrightarrow\left|x-2\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
5) \(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\sqrt{\left(x+4\right)^2}\)
\(\Leftrightarrow\left|2x-1\right|=\left|x+4\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x+4\\2x-1=-x-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
6) \(ĐK:x\ge-2\)
\(\Leftrightarrow5\sqrt{x+2}-3\sqrt{x+2}-\sqrt{x+2}=\sqrt{x+7}\)
\(\Leftrightarrow\sqrt{x+2}=\sqrt{x+7}\)
\(\Leftrightarrow x+2=x+7\Leftrightarrow2=7\left(VLý\right)\)
Vậy \(S=\varnothing\)
7) \(ĐK:x\ge-1\)
\(\Leftrightarrow5\sqrt{2x+1}+3\sqrt{x+1}=4\sqrt{x+1}+4\sqrt{2x+1}\)
\(\Leftrightarrow\sqrt{2x+1}=\sqrt{x+1}\)
\(\Leftrightarrow2x+1=x+1\Leftrightarrow x=0\left(tm\right)\)