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????
xin lỗi nha !
mình mới học lớp 3
mà bài này khó nắm
\(\Leftrightarrow\sqrt[3]{3x+1}+\sqrt[3]{5-x}=\sqrt[3]{4x-3}+\sqrt[3]{9-2x}\)
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{3x+1}=a\\\sqrt[3]{5-x}=b\\\sqrt[3]{4x-3}=c\\\sqrt[3]{9-2x}=d\end{matrix}\right.\)
Ta được: \(\left\{{}\begin{matrix}a+b=c+d\\a^3+b^3=c^3+d^3\end{matrix}\right.\)
TH1:
Nếu \(a+b=c+d=0\Leftrightarrow\sqrt[3]{3x+1}+\sqrt[3]{5-x}=\sqrt[3]{4x-3}+\sqrt[3]{9-2x}=0\)
\(\Rightarrow\left\{{}\begin{matrix}3x+1=-\left(5-x\right)\\4x-3=-\left(9-2x\right)\end{matrix}\right.\) \(\Rightarrow x=-3\)
TH2: nếu \(a+b=c+d\ne0\)
\(a+b=c+d\Leftrightarrow\left(a+b\right)^3=\left(c+d\right)^3\)
\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=c^3+d^3+3cd\left(c+d\right)\)
\(\Leftrightarrow ab\left(a+b\right)=cd\left(c+d\right)\) (do \(a^3+b^3=c^3+d^3\))
\(\Leftrightarrow ab=cd\) (do \(a+b=c+d\ne0\))
\(\Leftrightarrow\sqrt[3]{\left(3x+1\right)\left(5-x\right)}=\sqrt[3]{\left(4x-3\right)\left(9-2x\right)}\)
\(\Leftrightarrow\left(3x+1\right)\left(5-x\right)=\left(4x-3\right)\left(9-2x\right)\)
\(\Leftrightarrow5x^2-28x+32=0\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{8}{5}\end{matrix}\right.\)
Vậy \(x=\left\{-3;4;\dfrac{8}{5}\right\}\)
Cái cuối này căn bậc 2 hay căn bậc 3 em? Căn bậc 2 thì hơi nghi ngờ về khả năng giải được của pt này.
a, \(\dfrac{1}{2}\sqrt{x-5}-\sqrt{4x-20+3}=0\left(dkxd:x\ge5\right)\)
\(< =>\dfrac{\sqrt{x-5}}{2}=\sqrt{4x-17}\)
\(< =>\dfrac{x-5}{4}=4x-17\)
\(< =>x-5=16x-68\)
\(< =>15x=68-5=63\)
\(< =>x=\dfrac{63}{15}=\dfrac{21}{5}\)(ktm)
b, \(\sqrt{2x+1}-2\sqrt{x}+1=0\left(dkxd:x\ge0\right)\)
\(< =>\sqrt{2x+1}+1=2\sqrt{x}\)
\(< =>2x+1+1+2\sqrt{2x+1}=4x\)
\(< =>2x-2\sqrt{2x+1}-2=0\)
\(< =>2x+1-2\sqrt{2x+1}+1-4=0\)
\(< =>\left(\sqrt{2x+1}-1\right)^2=4\)
\(< =>\left\{{}\begin{matrix}\sqrt{2x+1}-1=2\\\sqrt{2x+1}-1=-2\end{matrix}\right.\)
\(< =>\left\{{}\begin{matrix}\sqrt{2x+1}=3\\\sqrt{2x+1}=-1\left(loai\right)\end{matrix}\right.\)
\(< =>2x+1=9< =>2x=8< =>x=4\)(tmdk)
\(2x-3\sqrt{2x-1}-5=0\)
<=> \(-3\sqrt{2x-1}-5=0-2x\)
<=> \(-3\sqrt{2x-1}-5=-2x\)
<=> \(-3\sqrt{2x-1}=-2x+5\)
<=> \(\left(-3\sqrt{2x-1}\right)^2=\left(-2x+5\right)^2\)
<=> 18x - 9 = 4x2 - 20 + 25
<=> \(\orbr{\begin{cases}x=\frac{17}{2}\left(\text{thỏa mãn}\right)\\x=1\left(\text{loại}\right)\end{cases}}\)
=> x = 17/2
a: ĐKXĐ: \(\left\{{}\begin{matrix}2x-3>=0\\x-1>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{3}{2}\\x>1\end{matrix}\right.\Leftrightarrow x>=\dfrac{3}{2}\)
\(\dfrac{\sqrt{2x-3}}{\sqrt{x-1}}=2\)
=>\(\sqrt{\dfrac{2x-3}{x-1}}=2\)
=>\(\dfrac{2x-3}{x-1}=4\)
=>4(x-1)=2x-3
=>4x-4=2x-3
=>4x-2x=-3+4
=>2x=1
=>\(x=\dfrac{1}{2}\left(loại\right)\)
b: ĐKXĐ: 2x+15>=0
=>x>=-15/2
\(x+\sqrt{2x+15}=0\)
=>\(\sqrt{2x+5}=-x\)
=>\(\left\{{}\begin{matrix}-x>=0\\\left(-x\right)^2=2x+5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{15}{2}< =x< =0\\x^2-2x-5=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-\dfrac{15}{2}< =x< =0\\\left(x-1\right)^2=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{15}{2}< =x< =0\\\left[{}\begin{matrix}x-1=\sqrt{6}\\x-1=-\sqrt{6}\end{matrix}\right.\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-\dfrac{15}{2}< =x< =0\\\left[{}\begin{matrix}x=\sqrt{6}+1\left(loại\right)\\x=-\sqrt{6}+1\left(nhận\right)\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\sqrt[3]{2x+4}=\dfrac{2x-1+5}{\sqrt[3]{\left(2x-1\right)^2}-\sqrt[3]{5\left(2x-1\right)}+\sqrt[3]{25}}\)
\(\Leftrightarrow\sqrt[3]{2x+4}=\dfrac{2x+4}{\sqrt[3]{\left(2x-1\right)^2}-\sqrt[3]{10x-5}+\sqrt[3]{25}}\)
=>\(\sqrt[3]{2x+4}\left(\dfrac{\sqrt[3]{\left(2x+4\right)^2}}{\sqrt[3]{\left(2x-1\right)^2}-\sqrt[3]{10x-5}+\sqrt[3]{25}}-1\right)=0\)
=>2x+4=0
=>x=-2
ĐKXĐ: \(x\ge0\)
\(pt\Leftrightarrow\left(2\sqrt{x}-3\right)\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{3}{2}\\\sqrt{x}=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
\(2x-5\sqrt{x}+3=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{9}{4}\end{matrix}\right.\)