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LS
Lê Song Phương
CTVHS
8 tháng 10 2023
a) đkxđ \(x\ge1\)
pt đã cho \(\Leftrightarrow\left(\sqrt{2x-1}-3\right)+\left(\sqrt{x-1}-2\right)=0\)
\(\Leftrightarrow\dfrac{2x-10}{\sqrt{2x-1}+3}+\dfrac{x-5}{\sqrt{x-1}+2}=0\)
\(\Leftrightarrow\left(x-5\right)\left(\dfrac{2}{\sqrt{2x-1}+3}+\dfrac{1}{\sqrt{x-1}+2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\left(nhận\right)\\\dfrac{2}{\sqrt{2x-1}+3}+\dfrac{1}{\sqrt{x-1}+3}=0\end{matrix}\right.\)
Hiển nhiên pt thứ 2 vô nghiệm vì \(VT>0\) với mọi \(x\ge1\). Do đó pt đã cho có nghiệm duy nhất là \(x=5\)
b) đkxđ: \(x\ge-3\)
Để ý rằng \(x^2+2x+7=\left(x^2+1\right)+\left(2x+6\right)=\left(x^2+1\right)+2\left(x+3\right)\) nên nếu ta đặt \(\sqrt{x^2+1}=u\left(u\ge1\right)\) và \(\sqrt{x+3}=v\left(v\ge0\right)\) thì pt đã chot rở thành:
\(u^2+2v^2=3uv\)
\(\Leftrightarrow\left(u-v\right)\left(u-2v\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}u=v\\u=2v\end{matrix}\right.\)
Nếu \(u=v\) thì \(\sqrt{x^2+1}=\sqrt{x+3}\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge-3\\x^2+1=x+3\end{matrix}\right.\)
Mà \(x^2+1=x+3\) \(\Leftrightarrow x^2-x-2=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\) (nhận)
Nếu \(u=2v\) thì \(\sqrt{x^2+1}=2\sqrt{x+3}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-3\\x^2+1=4x+12\end{matrix}\right.\)
mà \(x^2+1=4x+12\)\(\Leftrightarrow x^2-4x-11=0\)
\(\Leftrightarrow x=2\pm\sqrt{15}\) (nhận)
Vậy pt đã cho có tập nghiệm \(S=\left\{2;-1;2\pm\sqrt{15}\right\}\)
8 tháng 10 2023
a) \(\sqrt{2x-1}+\sqrt{x-1}=5\) (ĐK: \(x\ge1\))
\(\Leftrightarrow\left(\sqrt{2x-1}+\sqrt{x-1}\right)^2=5^2\)
\(\Leftrightarrow2x-1+x-1+2\sqrt{\left(2x-1\right)\left(x-1\right)}=25\)
\(\Leftrightarrow3x-2+2\sqrt{\left(2x-1\right)\left(x-1\right)}=25\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)\left(x-1\right)}=\dfrac{27-3x}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{27-3x}{2}\ge0\\\left(2x-1\right)\left(x-1\right)=\left(\dfrac{27-3x}{2}\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}27-3x\ge0\\2x^2-2x-x+1=\dfrac{729-162x+9x^2}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x\le27\\8x^2-12x+4=9x^2-162x+729\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le9\\x^2-150x+725=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le9\\\left[{}\begin{matrix}x-5=0\\x-145=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le9\\\left[{}\begin{matrix}x=5\left(tm\right)\\x=145\left(ktm\right)\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow x=5\)
1 tháng 10 2021
Tham khảo:
1) Giải phương trình : \(11\sqrt{5-x}+8\sqrt{2x-1}=24+3\sqrt{\left(5-x\right)\left(2x-1\right)}\) - Hoc24
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 1 2021
1.
\(\Leftrightarrow\left(2x+1\right)\sqrt{2x^2+4x+5}-\left(2x+1\right)\left(x+3\right)+x^2-2x-4=0\)
\(\Leftrightarrow\left(2x+1\right)\left(\sqrt{2x^2+4x+5}-\left(x+3\right)\right)+x^2-2x-4=0\)
\(\Leftrightarrow\dfrac{\left(2x+1\right)\left(x^2-2x-4\right)}{\sqrt{2x^2+4x+5}+x+3}+x^2-2x-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\\dfrac{2x+1}{\sqrt{2x^2+4x+5}+x+3}+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2x+1+\sqrt{2x^2+4x+5}+x+3=0\)
\(\Leftrightarrow\sqrt{2x^2+4x+5}=-3x-4\) \(\left(x\le-\dfrac{4}{3}\right)\)
\(\Leftrightarrow2x^2+4x+5=9x^2+24x+16\)
\(\Leftrightarrow7x^2+20x+11=0\)
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 1 2021
2.
ĐKXĐ: ...
\(\Leftrightarrow2x\sqrt{2x+7}+7\sqrt{2x+7}=x^2+2x+7+7x\)
\(\Leftrightarrow\left(x^2-2x\sqrt{2x+7}+2x+7\right)+7\left(x-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)^2+7\left(x-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)\left(x+7-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2x+7}\\x+7=\sqrt{2x+7}\end{matrix}\right.\)
\(\Leftrightarrow...\)
PA
1
NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 3 2022
ĐKXĐ: \(2\le x\le5\)
\(\left(\sqrt{2x-4}-\sqrt{5-x}\right)\sqrt{3x-3}=3x-9\)
\(\Leftrightarrow\dfrac{\left(3x-9\right)\sqrt{3x-3}}{\sqrt{2x-4}+\sqrt{5-x}}=3x-9\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-9=0\Rightarrow x=3\\\dfrac{\sqrt{3x-3}}{\sqrt{2x-4}+\sqrt{5-x}}=1\left(1\right)\end{matrix}\right.\)
Xét (1):
\(\Leftrightarrow\sqrt{3x-3}=\sqrt{2x-4}+\sqrt{5-x}\)
\(\Leftrightarrow3x-3=x+1+2\sqrt{\left(2x-4\right)\left(5-x\right)}\)
\(\Leftrightarrow x-2=\sqrt{\left(2x-4\right)\left(5-x\right)}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\\left(x-2\right)^2=\left(2x-4\right)\left(5-x\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\\left(x-2\right)\left(3x-12\right)=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
Vậy pt có 3 nghiệm \(x=\left\{2;3;4\right\}\)
TQ
11 tháng 6 2021
a) \(\sqrt{7+\sqrt{2x}=3+\sqrt{5}}\) (x≥0) Đặt \(\sqrt{2x}\) = a ( a>0 )
Khi đó pt :
<=> 7+a =3 + \(\sqrt{5}\)
<=> 4+a = \(\sqrt{5}\)
<=> (4+a)\(^2\) = 5
<=> 16 + 8a + a\(^2\) = 5
<=>a\(^2\) + 8a+ 11 = 0
<=> a = -4 + \(\sqrt{5}\) (Loại) và a = -4-\(\sqrt{5}\)(Loại)
Vậy Pt vô nghiệm.
b) \(\sqrt{3x^2-4x}\) = 2x-3
<=> 3x\(^2\)- 4x = 4x\(^2\)-12x + 9
<=> x\(^2\)-8x+9 = 0
<=> x=1 , x=9
Vậy S={1;9}
c\(\dfrac{\left(7-x\right)\sqrt{7-x}+\left(x-5\right)\sqrt{x-5}}{\sqrt{7-x}+\sqrt{x-5}}\) = 2
<=> \(\dfrac{\left(\sqrt{7-x}\right)^3+\left(\sqrt{x-5}\right)^3}{\sqrt{7-x}+\sqrt{x-5}}=2\)
<=> \(\dfrac{\left(\sqrt{7-x}+\sqrt{x-5}\right)\left(7-x-\sqrt{\left(7-x\right)\left(x-5\right)}+x-5\right)}{\sqrt{7-x}+\sqrt{x-5}}=2\)
<=> \(\sqrt{\left(7-x\right)\left(x-5\right)}=0\)
<=> x=7,x=5
Vậy x=5 hoặc x=7
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) \(\sqrt[]{2x}-\sqrt[]{50}=0\)
\(\Leftrightarrow\sqrt[]{2x}=\sqrt[]{50}\)
\(\Leftrightarrow2x=50\Leftrightarrow x=25\)
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) \(\sqrt{\left(x-3\right)^2}=3-x\)
\(\Leftrightarrow\left|x-3\right|=3-x\)
\(\Leftrightarrow x-3=3-x\)
\(\Leftrightarrow x+x=3+3\)
\(\Leftrightarrow x=\dfrac{6}{2}\)
\(\Leftrightarrow x=3\)
30 tháng 8 2021
a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow x+5=4\)
hay x=-1
b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x-1=289\)
hay x=290
23 tháng 10 2021
\(1,PT\Leftrightarrow2x-1=5\Leftrightarrow x=3\\ 2,\Leftrightarrow x-5=9\Leftrightarrow x=14\\ 3,ĐK:x\ge1\\ PT\Leftrightarrow3\sqrt{x-1}=21\Leftrightarrow\sqrt{x-1}=7\Leftrightarrow x=50\left(tm\right)\\ 4,\Leftrightarrow x=\dfrac{\sqrt{50}}{\sqrt{2}}=\dfrac{5\sqrt{2}}{\sqrt{2}}=5\)
