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7 tháng 2 2021
a, ĐKXĐ : \(x\ge\dfrac{1}{2}\)
PT <=> 2x - 1 = 5
<=> x = 3 ( TM )
Vậy ...
b, ĐKXĐ : \(x\ge5\)
PT <=> x - 5 = 9
<=> x = 14 ( TM )
Vậy ...
c, PT <=> \(\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy ...
d, PT<=> \(\left|x-3\right|=3-x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=x-3\\x-3=3-x\end{matrix}\right.\)
Vậy phương trình có vô số nghiệm với mọi x \(x\le3\)
e, ĐKXĐ : \(-\dfrac{5}{2}\le x\le1\)
PT <=> 2x + 5 = 1 - x
<=> 3x = -4
<=> \(x=-\dfrac{4}{3}\left(TM\right)\)
Vậy ...
f ĐKXĐ : \(\left[{}\begin{matrix}x\le0\\1\le x\le3\end{matrix}\right.\)
PT <=> \(x^2-x=3-x\)
\(\Leftrightarrow x=\pm\sqrt{3}\) ( TM )
Vậy ...
7 tháng 2 2021
a) \(\sqrt{2x-1}=\sqrt{5}\) (x \(\ge\dfrac{1}{2}\))
<=> 2x - 1 = 5
<=> x = 3 (tmđk)
Vậy S = \(\left\{3\right\}\)
b) \(\sqrt{x-5}=3\) (x\(\ge5\))
<=> x - 5 = 9
<=> x = 4 (ko tmđk)
Vậy x \(\in\varnothing\)
c) \(\sqrt{4x^2+4x+1}=6\) (x \(\in R\))
<=> \(\sqrt{\left(2x+1\right)^2}=6\)
<=> |2x + 1| = 6
<=> \(\left[{}\begin{matrix}\text{2x + 1=6}\\\text{2x + 1}=-6\end{matrix}\right.< =>\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-7}{2}\end{matrix}\right.\)(tmđk)
Vậy S = \(\left\{\dfrac{5}{2};\dfrac{-7}{2}\right\}\)
28 tháng 6 2021
a)ĐK:\(\begin{cases}25x^2-9 \ge 0\\5x+3 \ge 0\\\end{cases}\)
`<=>` \(\begin{cases}(5x-3)(5x+3) \ge 0\\5x+3 \ge 0\\\end{cases}\)
`<=>` \(\begin{cases}\left[ \begin{array}{l}x\ge \dfrac35\\x \le -\dfrac35\end{array} \right.\\\end{cases}\)
`<=>` \(\left[ \begin{array}{l}x=-\dfrac35\\x \ge \dfrac35\end{array} \right.\)
`pt<=>\sqrt{5x+3}(\sqrt{5x-3}-2)=0`
`<=>` \(\left[ \begin{array}{l}5x+3=0\\\sqrt{5x-3}=2\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=-\dfrac35\\5x-3=4\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=-\dfrac35\\x=7/5\end{array} \right.\)
`b)sqrt{x-3}/sqrt{2x+1}=2`
ĐK:\(\begin{cases}x-3 \ge 0\\2x+1>0\\\end{cases}\)
`<=>x>=3`
`pt<=>sqrt{x-3}=2sqrt{2x+1}`
`<=>x-3=8x+4`
`<=>7x=7`
`<=>x=1(l)`
`c)sqrt{x^2-2x+1}+sqrt{x^2-4x+4}=3`
`<=>sqrt{(x-1)^2}+sqrt{(x-2)^2}=3`
`<=>|x-1|+|x-2|=3`
`**x>=2`
`pt<=>x-1+x-2=3`
`<=>2x=6`
`<=>x=3(tm)`
`**x<=1`
`pt<=>1-x+2-x=3`
`<=>3-x=3`
`<=>x=0(tm)`
`**1<=x<=2`
`pt<=>x-1+2-x=3`
`<=>=-1=3` vô lý
Vậy `S={0,3}`
1 tháng 10 2021
\(\left(\sqrt{x+3}-\sqrt{x+1}\right)\left(x^2+\sqrt{x^2+4x+3}\right)=2x\left(đk:x\ge0\right)\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x+3}-\sqrt{x+1}\right)\left(\sqrt{x+3}+\sqrt{x+1}\right)\left(x^2+\sqrt{\left(x+1\right)\left(x+3\right)}\right)}{\sqrt{x+3}+\sqrt{x+1}}=2x\)
\(\Leftrightarrow\dfrac{\left(x+3-x-1\right)\left(x^2+\sqrt{\left(x+1\right)\left(x+3\right)}\right)}{\sqrt{x+3}+\sqrt{x+1}}=2x\)
\(\Leftrightarrow\dfrac{x^2+\sqrt{\left(x+1\right)\left(x+3\right)}}{\sqrt{x+3}+\sqrt{x+1}}=x\)
\(\Leftrightarrow x\sqrt{x+3}+x\sqrt{x+1}-x^2-\sqrt{\left(x+1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\sqrt{x+3}\left(x-\sqrt{x+1}\right)-x\left(x-\sqrt{x+1}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{x+1}\right)\left(\sqrt{x+3}-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{x+1}\\x=\sqrt{x+3}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-1=0\\x^2-x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\left(tm\right)\\x=\dfrac{1-\sqrt{5}}{2}\left(ktm\right)\\x=\dfrac{1+\sqrt{13}}{2}\left(tm\right)\\x=\dfrac{1-\sqrt{13}}{2}\left(ktm\right)\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 2 2021
Do \(x^6-x^3+x^2-x+1=\left(x^3-\dfrac{1}{2}\right)^2+\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}>0\) ; \(\forall x\) nên BPT tương đương:
\(\sqrt{13}-\sqrt{2x^2-2x+5}-\sqrt{2x^2-4x+4}\ge0\)
\(\Leftrightarrow\sqrt{4x^2-4x+10}+\sqrt{4x^2-8x+8}\le\sqrt{26}\) (1)
Ta có:
\(VT=\sqrt{\left(2x-1\right)^2+3^2}+\sqrt{\left(2-2x\right)^2+2^2}\ge\sqrt{\left(2x-1+2-2x\right)^2+\left(3+2\right)^2}=\sqrt{26}\) (2)
\(\Rightarrow\left(1\right);\left(2\right)\Rightarrow\sqrt{4x^2-4x+10}+\sqrt{4x^2-8x+8}=\sqrt{26}\)
Dấu "=" xảy ra khi và chỉ khi \(2\left(2x-1\right)=3\left(2-2x\right)\Leftrightarrow x=\dfrac{4}{5}\)
Vậy BPT có nghiệm duy nhất \(x=\dfrac{4}{5}\)
H9
HT.Phong (9A5)
CTVHS
5 tháng 9 2023
1) \(\sqrt{x^2+1}=\sqrt{5}\)
\(\Leftrightarrow x^2+1=5\)
\(\Leftrightarrow x^2=5-1\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow x^2=2^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
2) \(\sqrt{2x-1}=\sqrt{3}\) (ĐK: \(x\ge\dfrac{1}{2}\))
\(\Leftrightarrow2x-1=3\)
\(\Leftrightarrow2x=3+1\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=\dfrac{4}{2}\)
\(\Leftrightarrow x=2\left(tm\right)\)
3) \(\sqrt{43-x}=x-1\) (ĐK: \(x\le43\))
\(\Leftrightarrow43-x=\left(x-1\right)^2\)
\(\Leftrightarrow x^2-2x+1=43-x\)
\(\Leftrightarrow x^2-x-42=0\)
\(\Leftrightarrow\left(x-7\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\left(tm\right)\\x=-6\left(tm\right)\end{matrix}\right.\)
4) \(x-\sqrt{4x-3}=2\) (ĐK: \(x\ge\dfrac{3}{4}\))
\(\Leftrightarrow\sqrt{4x-3}=x-2\)
\(\Leftrightarrow4x-3=\left(x-2\right)^2\)
\(\Leftrightarrow x^2-4x+4=4x-3\)
\(\Leftrightarrow x^2-8x+7=0\)
\(\Leftrightarrow\left(x-7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
5) \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}=\dfrac{1}{2}\) (ĐK: \(x\ge0\))
\(\Leftrightarrow\sqrt{x}+3=2\sqrt{x}+2\)
\(\Leftrightarrow2\sqrt{x}-\sqrt{x}=3-2\)
\(\Leftrightarrow\sqrt{x}=1\)
\(\Leftrightarrow x=1^2\)
\(\Leftrightarrow x=1\left(tm\right)\)
GH
5 tháng 9 2023
1)
\(\sqrt{x^2+1}=\sqrt{5}\\ \Leftrightarrow x^2+1=5\\ \Leftrightarrow x^2=5-1=4\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy PT có nghiệm `x=2` hoặc `x=-2`
2)
ĐKXĐ: \(x\ge\dfrac{1}{2}\)
\(\sqrt{2x-1}=\sqrt{3}\\ \Leftrightarrow2x-1=3\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\left(tm\right)\)
Vậy PT có nghiệm `x=2`
3)
\(ĐKXĐ:x\le43\)
PT trở thành:
\(43-x=\left(x-1\right)^2=x^2-2x+1\\ \Leftrightarrow43-x-x^2+2x-1=0\\ \Leftrightarrow-x^2+x+42=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-6\left(tm\right)\\x=7\left(tm\right)\end{matrix}\right.\)
Vậy PT có nghiệm `x=-6` hoặc `x=7`
4)
ĐKXĐ: \(x\ge\dfrac{3}{4}\)
PT trở thành:
\(\sqrt{4x-3}=x-2\\ \Leftrightarrow4x-3=\left(x-2\right)^2=x^2-4x+4\\ \Leftrightarrow4x-3-x^2+4x-4=0\\ \Leftrightarrow-x^2+8x-7=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=7\left(tm\right)\end{matrix}\right.\)
Vậy PT có nghiệm \(x=1\) hoặc \(x=7\)
5)
ĐKXĐ: \(x\ge0\)
PT trở thành:
\(\sqrt{x+3}=2\sqrt{x}+2\\ \Leftrightarrow x+3=\left(2\sqrt{x}+2\right)^2=4x+8\sqrt{x}+4\\ \Leftrightarrow x+3-4x-8\sqrt{x}-4=0\\ \Leftrightarrow-3x-8\sqrt{x}-1=0\left(1\right)\)
Đặt \(\sqrt{x}=t\left(t\ge0\right)\)
Khi đó:
(1)\(\Leftrightarrow3t^2+8t+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{-4+\sqrt{13}}{3}\left(loại\right)\\t=\dfrac{-4-\sqrt{13}}{3}\left(loại\right)\end{matrix}\right.\)
Vậy PT vô nghiệm.
Đặt \(\left\{{}\begin{matrix}\sqrt{x+3}=a\\\sqrt{x+1}=b\end{matrix}\right.\left(a,b\ge0\right)\)
\(PT\Leftrightarrow a+2xb-2x-ab=0\\ \Leftrightarrow2x\left(b-1\right)-a\left(b-1\right)=0\\ \Leftrightarrow\left(2x-a\right)\left(b-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x=a\\b=1\end{matrix}\right.\)
Với \(2x=a\Leftrightarrow x+3=4x^2\left(x\ge0\right)\Leftrightarrow x=1\left(tm\right)\)
Với \(b=1\Leftrightarrow x+1=1\Leftrightarrow x=0\left(tm\right)\)
Vậy PT có nghiệm \(x\in\left\{0;1\right\}\)
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