tìm x biết
\(\sqrt[3]{x+1}\)= \(\sqrt{x-3}\)
K
Khách
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Những câu hỏi liên quan
16 tháng 10 2023
Tìm \(x\) để \(A.B>\dfrac{3}{2}?\)
\(ĐK:x>0;\\ A.B\\ =\dfrac{3+\sqrt{x}}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ =1+\dfrac{1}{\sqrt{x}}\\ A.B>\dfrac{3}{2}\\ \Leftrightarrow1+\dfrac{1}{\sqrt{x}}>\dfrac{3}{2}\\ \Leftrightarrow\dfrac{1}{\sqrt{x}}>\dfrac{3}{2}-1\\ \Leftrightarrow\dfrac{1}{\sqrt{x}}>\dfrac{1}{2}\\ \Leftrightarrow\sqrt{x}< 2\\ \Leftrightarrow0< x< 4\)
Vậy \(0< x< 4\) thì \(A.B>\dfrac{3}{2}\)
31 tháng 10 2021
\(1,\\ a,ĐK:\left\{{}\begin{matrix}x\ge0\\x+5\ge0\end{matrix}\right.\Leftrightarrow x\ge0\\ b,Sửa:B=\left(\sqrt{3}-1\right)^2+\dfrac{24-2\sqrt{3}}{\sqrt{2}-1}\\ B=4-2\sqrt{3}+\dfrac{2\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\\ B=4-2\sqrt{3}+2\sqrt{3}=4\\ 3,\\ =\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{1+\sqrt{x}}\right]\cdot\dfrac{\sqrt{x}-3+2-2\sqrt{x}}{\left(1-\sqrt{x}\right)\left(\sqrt{x}-3\right)}-2\\ =\left(1-\sqrt{x}\right)\cdot\dfrac{-\sqrt{x}-1}{\left(1-\sqrt{x}\right)\left(\sqrt{x}-3\right)}-2\\ =\dfrac{-\sqrt{x}-1}{\sqrt{x}-3}-2=\dfrac{-\sqrt{x}-1-2\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{-3\sqrt{x}+5}{\sqrt{x}-3}\)
TK
25 tháng 5 2021
\(\dfrac{\sqrt{x-1}}{\sqrt{x+3}}=\dfrac{\sqrt{x-2}}{1}\)(Đk x>2;x≠-3)
⇔\(\sqrt{\left(x-2\right)\left(x+3\right)}=\sqrt{x-1}\)
⇔\(\left(x-2\right)\left(x+3\right)=x-1\)
⇔\(x^2+x-6-x+1=0\)
⇔\(x^2-5=0\)
⇔\(x^2=5\)
⇔x=\(\pm\sqrt{5}\)(thỏa điều kiện)
Vậyx=\(\pm\sqrt{5}\)
25 tháng 5 2021
ĐKXĐ:x khác -3; x≥2
quy đồng và khử mẩu 2 vế ta đc:
\(\sqrt{x-1}=\sqrt{x-2}\cdot\sqrt{x+3}\)Bình phương 2 vế ta đc:
x-1=(x-2)*(x+3)<=> x-1=x2+x-6 <=> x2-5=0
<=>\(\left\{{}\begin{matrix}x=\sqrt{5}\left(nhận\right)\\x=-\sqrt{5}\left(loại\right)\end{matrix}\right.\)
vậy x=\(\sqrt{5}\)
26 tháng 8 2021
b: Thay \(x=7-2\sqrt{6}\) vào A, ta được:
\(A=\dfrac{3\cdot\left(\sqrt{6}-1\right)}{-7+2\sqrt{6}-5\left(\sqrt{6}+1\right)-1}\)
\(=\dfrac{3\cdot\left(\sqrt{6}-1\right)}{-8+2\sqrt{6}-5\sqrt{6}-5}\)
\(=\dfrac{-3\sqrt{6}+3}{13+3\sqrt{6}}=\dfrac{93-48\sqrt{6}}{115}\)
24 tháng 11 2021
\(a,\Leftrightarrow x-1=4\Leftrightarrow x=5\\ b,\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\3x+1=4x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\left(tm\right)\end{matrix}\right.\Leftrightarrow x=4\\ c,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=3\\ \Leftrightarrow x+5=9\\ \Leftrightarrow x=4\left(tm\right)\)
\(d,\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{5}+1\right)^2}\\ \Leftrightarrow\left|x-2\right|=\sqrt{5}+1\\ \Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{5}+1\\2-x=\sqrt{5}+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}+3\\x=1-\sqrt{5}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 1 2024
\(P=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}+1}=\dfrac{3}{\sqrt{x}+3}\)
\(P\in Z\Rightarrow\sqrt{x}+3=Ư\left(3\right)=\left\{-3;-1;1;3\right\}\)
Mà \(\sqrt{x}+3\ge3;\forall x\ge0\)
\(\Rightarrow\sqrt{x}+3=3\)
\(\Rightarrow\sqrt{x}=0\Rightarrow x=0\)
H9
HT.Phong (9A5)
CTVHS
25 tháng 7 2023
a) \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left(2x+1\right)^2=6^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
b) \(\sqrt{4x^2-4\sqrt{7}x+7}=\sqrt{7}\)
\(\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)
\(\Leftrightarrow\left(2x-\sqrt{7}\right)^2=\left(\sqrt{7}\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt[]{7}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)
25 tháng 7 2023
a) \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
b) \(pt\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)
\(\Leftrightarrow\left|2x-\sqrt{7}\right|=\sqrt{7}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)
\(đkxđ:x-3\ge0\Leftrightarrow x\ge3\)
\(\sqrt[3]{x+1}=\sqrt{x-3}\)\(\Leftrightarrow\)\(\left(\sqrt[3]{x+1}\right)^6=\left(\sqrt{x-3}\right)^6\)\(\Leftrightarrow\)\(\left(x+1\right)^2=\left(x-3\right)^3\)
\(\Leftrightarrow\)\(x^2+2x+1=x^3-9x^2+27x-27\)
\(\Leftrightarrow x^3-10x^2+25x-28=0\)
\(\Leftrightarrow x^3-7x^2-3x^2+21x+4x-28=0\)
\(\Leftrightarrow x^2\left(x-7\right)-3x\left(x-7\right)+4\left(x-7\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(x^2-3x+4\right)=0\Leftrightarrow\orbr{\begin{cases}x-7=0\\x^2-3x+4=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=7\left(nhận\right)\\x^2-2x.\frac{3}{2}+\frac{9}{4}+\frac{7}{4}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\\left(x-\frac{3}{2}\right)^2+\frac{7}{4}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\\left(x-\frac{3}{2}\right)^2=-\frac{7}{4}\left(vôlí\right)\end{cases}}\)
Vậy \(x=7\)