Giaỉ phương trình :
a) \(x^3=2\)
b) \(27x^3=-81\)
c) \(\sqrt[3]{3x+1}=4\)
d) \(\sqrt[3]{x-2}+2=x\)
K
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Những câu hỏi liên quan
14 tháng 8 2021
c: Ta có: \(\sqrt{2x}=\sqrt{5}\)
\(\Leftrightarrow2x=5\)
hay \(x=\dfrac{5}{2}\)
d: Ta có: \(\sqrt{3x-1}=4\)
\(\Leftrightarrow3x-1=16\)
\(\Leftrightarrow3x=17\)
hay \(x=\dfrac{17}{3}\)
14 tháng 8 2021
Ta có: \(\sqrt{4\cdot\left(1-x\right)^2}=6\)
\(\Leftrightarrow2\left|x-1\right|=6\)
\(\Leftrightarrow\left|x-1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
HP
31 tháng 8 2021
a, ĐK: \(x\ge2\)
\(\sqrt{2x+1}-\sqrt{x-2}=x+3\)
\(\Leftrightarrow\dfrac{x+3}{\sqrt{2x+1}+\sqrt{x-2}}=x+3\)
\(\Leftrightarrow\left(x+3\right)\left(\dfrac{1}{\sqrt{2x+1}+\sqrt{x-2}}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\left(l\right)\\\sqrt{2x+1}+\sqrt{x-2}=1\left(vn\right)\end{matrix}\right.\)
Phương trình vô nghiệm.
HP
31 tháng 8 2021
b, ĐK: \(x\ge-1\)
\(\sqrt{x+3}+2x\sqrt{x+1}=2x+\sqrt{x^2+4x+3}\)
\(\Leftrightarrow\sqrt{x+3}+2x\sqrt{x+1}=2x+\sqrt{\left(x+3\right)\left(x+1\right)}\)
\(\Leftrightarrow-\sqrt{x+3}\left(\sqrt{x+1}-1\right)+2x\left(\sqrt{x+1}-1\right)=0\)
\(\Leftrightarrow\left(2x-\sqrt{x+3}\right)\left(\sqrt{x+1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+3}=2x\\\sqrt{x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x+3=4x^2\end{matrix}\right.\\x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)
17 tháng 5 2021
b, \(đk:x\ge2\)
Xét x=2 thay vào pt thấy không thỏa mãn => x>2 hay 27x-54>0
\(x^3-11x+36x-18=4\sqrt[4]{27x-54}\)
\(\Leftrightarrow27x^3-297x^2+972x-486=4\sqrt[4]{\left(27x-54\right).81.81.81}\le189+27x\) (cosi với 4 số dương, dấu = xảy ra khi x=5)
\(\Leftrightarrow x^3-11x^2+35x-25\le0\)
\(\Leftrightarrow\left(x-1\right)\left(x-5\right)^2\le0\) (*)
Có \(\left\{{}\begin{matrix}x>2\\\left(x-5\right)^2\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1>0\\\left(x-5\right)^2\ge0\end{matrix}\right.\)\(\Rightarrow\left(x-1\right)\left(x-5\right)^2\ge0\) (2*)
Từ (*) và (2*) ,dấu = xra khi x=5 (thỏa mãn)
Vây pt có nghiệm duy nhất x=5
17 tháng 5 2021
c,Có \(6\sqrt[3]{4x^3+x}=16x^4+5>0\)
\(\Leftrightarrow4x^3+x>0\)
Có: \(16x^4+5=6\sqrt[3]{4x^3+x}\le2\left(4x^3+x+2\right)\) (theo cosi với 3 số dương,dấu = xảy ra khi \(x=\dfrac{1}{2}\))
\(\Leftrightarrow16x^4-8x^3-2x+1\le0\)
\(\Leftrightarrow\left(2x-1\right)^2\left(4x^2+2x+1\right)\le0\) (*)
(tương tự câu b) Dấu = xảy ra khi \(x=\dfrac{1}{2}\)(thỏa mãn)
Vậy....
d) Đk: \(x\ge\dfrac{3}{4}\)
Áp dụng bđt cosi:
\(\sqrt{2x-1}\le\dfrac{2x-1+1}{2}=x\)
\(\Rightarrow\dfrac{1}{\sqrt{2x-1}}\ge\dfrac{1}{x}\) (*)
\(\sqrt[4]{4x-3}\le\dfrac{4x-3+1+1+1}{4}=x\)
\(\dfrac{\Rightarrow1}{\sqrt[4]{4x-3}}\ge\dfrac{1}{x}\) (2*)
Từ (*) và (2*) \(\Rightarrow\dfrac{1}{\sqrt{2x-1}}+\dfrac{1}{\sqrt[4]{4x-3}}\ge\dfrac{2}{x}\)
Dấu = xảy ra khi x=1 (tm)
2 tháng 7 2021
a) ĐKXĐ: \(x\ge0\)
Ta có: \(3\sqrt{18x}-5\sqrt{8x}+4\sqrt{50x}=38\)
\(\Leftrightarrow9\sqrt{2x}-10\sqrt{2x}+20\sqrt{2x}=38\)
\(\Leftrightarrow19\sqrt{2x}=38\)
\(\Leftrightarrow\sqrt{2x}=2\)
\(\Leftrightarrow2x=4\)
hay x=2(thỏa ĐK)
b) ĐKXĐ: \(x\ge0\)
Ta có: \(3\sqrt{12x}-2\sqrt{27x}+4\sqrt{3x}=8\)
\(\Leftrightarrow6\sqrt{3x}-6\sqrt{3x}+4\sqrt{3x}=8\)
\(\Leftrightarrow\sqrt{3x}=2\)
\(\Leftrightarrow3x=4\)
hay \(x=\dfrac{4}{3}\)
c) ĐKXĐ: \(x\ge5\)
Ta có: \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\)
hay x=9
H
2 tháng 7 2021
a)
\(3.3\sqrt{2x}-5.2\sqrt{2x}+4.5.\sqrt{2x}=38\\ \Leftrightarrow19\sqrt{2x}=38\\ \Leftrightarrow\sqrt{2x}=2\\ \Leftrightarrow x=2\)
b)
\(3.2.\sqrt{3x}-2.3.\sqrt{3x}+4.\sqrt{3x}=8\\ \Leftrightarrow4\sqrt{3x}=8\\ \Leftrightarrow\sqrt{3x}=2\\\Leftrightarrow x=\dfrac{2^2}{3}=\dfrac{4}{3} \)
c)
\(\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=9\)
TA
3 tháng 7 2017
a ; \(3x-7\sqrt{x}+4=0
\)
\(3x-3\sqrt{x}-4\sqrt{x}+4=0\)\(\left(\sqrt{x}-1\right)\left(3\sqrt{x}-4\right)=0\)
từ đó suy ra x
18 tháng 11 2021
a, ĐKXĐ: \(x\ge-\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{3}{2}.2\sqrt{1+3x}-\dfrac{5}{3}.3\sqrt{1+3x}-\dfrac{1}{4}.4\sqrt{1+3x}=1\\ \Leftrightarrow3\sqrt{1+3x}-5\sqrt{1+3x}-\sqrt{1+3x}=1\\ \Leftrightarrow-3\sqrt{1+3x}=1\\ \Leftrightarrow\sqrt{1+3x}=-\dfrac{1}{3}\left(vô.lí\right)\)
b, \(\Leftrightarrow\sqrt{\left(x-\dfrac{1}{2}\right)^2}=3\\ \Leftrightarrow\left|x-\dfrac{1}{2}\right|=3\\ \Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=3\\x-\dfrac{1}{2}=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
18 tháng 11 2021
a) ĐKXĐ: \(x\ge-\dfrac{1}{3}\)
\(pt\Leftrightarrow3\sqrt{3x+1}-5\sqrt{3x+1}-\sqrt{3x+1}=1\)
\(\Leftrightarrow-3\sqrt{3x+1}=1\Leftrightarrow\sqrt{3x+1}=-\dfrac{1}{3}\left(VLý\right)\)
Vậy \(S=\varnothing\)
b) \(pt\Leftrightarrow\sqrt{\left(x-\dfrac{1}{2}\right)^2}=3\Leftrightarrow\left|x-\dfrac{1}{2}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=3\\x-\dfrac{1}{2}=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
AP
2
XO
8 tháng 5 2022
\(x^2+1+3x=x\sqrt{x^2+1}+3\sqrt{x^2+1}\)
<=> \(\sqrt{x^2+1}\left(\sqrt{x^2+1}-x\right)-3\left(\sqrt{x^2+1}-x\right)=0\)
\(\Leftrightarrow\left(\sqrt{x^2+1}-x\right)\left(\sqrt{x^2+1}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+1}=x\\\sqrt{x^2+1}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+1=x^2\\x^2=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}∃x̸\\x=\pm\sqrt{8}\end{matrix}\right.\)
8 tháng 5 2022
`x^2 + 3x + 1 = (x + 3) \sqrt{x^2 + 1}`
Nghiệm của pt là `x = +- 2 \sqrt{2}`
a) \(x^3=2\\ x=\Leftrightarrow x=\sqrt[3]{2}\)
b)\(27x^3=-81\\ \Leftrightarrow x^3=-3\\ x=\Leftrightarrow x=\sqrt[3]{-3}\)
c) \(\sqrt[3]{3x+1}=4\\ \Leftrightarrow3x+1=64\\ \Leftrightarrow3x=64-1\\ \Leftrightarrow3x=63\\ \Leftrightarrow x=21\)
d)\(\sqrt[3]{x-2}+2=x\\ \Leftrightarrow\sqrt[3]{x-2}=x-2\\ \Leftrightarrow x-2=\left(x-2\right)^3\\ \Leftrightarrow\left(x-2\right)-\left(x-2\right)^3=0\\ \Leftrightarrow\left(x-2\right)\left[1-\left(x-2\right)^2\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\1-\left(x-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\\x=1\end{matrix}\right.\)
Giải:
a) \(x^3=2\)
\(\Leftrightarrow x^3=\left(\sqrt[3]{2}\right)^3\)
\(\Leftrightarrow x=\sqrt[3]{2}\)
Vậy ...
b) \(27x^3=-81\)
\(\Leftrightarrow x^3=-3\)
\(\Leftrightarrow x^3=\left(\sqrt[3]{-3}\right)^3\)
\(\Leftrightarrow x=\sqrt[3]{-3}\)
Vậy ...
c) \(\sqrt[3]{3x+1}=4\)
\(\Leftrightarrow\sqrt[3]{3x+1}=\sqrt[3]{64}\)
\(\Leftrightarrow3x+1=64\)
\(\Leftrightarrow3x=63\)
\(\Leftrightarrow x=21\)
Vậy ...
d) \(\sqrt[3]{x-2}+2=x\)
\(\Leftrightarrow\sqrt[3]{x-2}=x-2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\\x=1\end{matrix}\right.\)
Vậy ...