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1/ Đk : \(2x^2-6x-1\ge0\Leftrightarrow\left\{{}\begin{matrix}x\le\frac{3-\sqrt{11}}{2}\\x\ge\frac{3+\sqrt{11}}{2}\end{matrix}\right.\)
Bình phương 2 vế của phương trình, ta có :
\(4x^4+36x^2+1-24x^3-4x^2+12x-4x-5=0\)
\(\Leftrightarrow4x^4-24x^3+32x^2+8x-4=0\)
\(\left[{}\begin{matrix}x=1-\sqrt{2}\left(TM\right)\\x=2-\sqrt{3}\left(l\right)\\x=\sqrt{2}+1\left(l\right)\\x=\sqrt{3}+2\left(TM\right)\end{matrix}\right.\)
Vậy ....
\(\Leftrightarrow\sqrt[3]{3x-5}=\left(2x-3\right)^3-x+2\)
\(\Leftrightarrow3x-5+\sqrt[3]{3x-5}=\left(2x-3\right)^3+2x-3\)
Đặt \(\left\{{}\begin{matrix}2x-3=a\\\sqrt[3]{3x-5}=b\end{matrix}\right.\)
\(\Rightarrow a^3+a=b^3+b\)
\(\Leftrightarrow a^3-b^3+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+b^2+ab+1\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left[\left(a+\frac{b}{2}\right)^2+\frac{3b^2}{4}+1\right]=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow2x-3=\sqrt[3]{3x-5}\)
\(\Leftrightarrow\left(2x-3\right)^3=3x-5\)
\(\Leftrightarrow8x^3-36x^2+51x-22=0\)
\(\Leftrightarrow\left(x-2\right)\left(8x^2-20x+11\right)=0\)
\(\Leftrightarrow...\)
a,\(\sqrt{\left(3x-1\right)^2}=5=>|3x-1|=5=>\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
b, \(\sqrt{4x^2-4x+1}=3=\sqrt{\left(2x-1\right)^2}=3=>\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
c, \(\sqrt{x^2-6x+9}+3x=4=>|x-3|=4-3x\)
TH1: \(|x-3|=x-3< =>x\ge3=>x-3=4-3x=>x=1,75\left(ktm\right)\)
TH2 \(|x-3|=3-x< =>x< 3=>3-x=4-3x=>x=0,5\left(tm\right)\)
Vậy x=0,5...
d, đk \(x\ge-1\)
=>pt đã cho \(< =>9\sqrt{x+1}-6\sqrt{x+1}+4\sqrt{x+1}=12\)
\(=>7\sqrt{x+1}=12=>x+1=\dfrac{144}{49}=>x=\dfrac{95}{49}\left(tm\right)\)
a) Ta có: \(\sqrt{\left(3x-1\right)^2}=5\)
\(\Leftrightarrow\left|3x-1\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=6\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
b) Ta có: \(\sqrt{4x^2-4x+1}=3\)
\(\Leftrightarrow\left|2x-1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
c) Ta có: \(\sqrt{x^2-6x+9}+3x=4\)
\(\Leftrightarrow\left|x-3\right|=4-3x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=4-23x\left(x\ge3\right)\\x-3=23x-4\left(x< 3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+23x=4+3\\x-23x=4+3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{24}\left(loại\right)\\x=\dfrac{-4}{22}=\dfrac{-2}{11}\left(loại\right)\end{matrix}\right.\)
\(ĐKXĐ:x\ge\frac{1}{2}\)
Áp dụng BĐT AM - GM cho các số dương ta có :
\(\sqrt{2x-1}=\sqrt{1.\left(2x-1\right)}\le\frac{1+2x-1}{2}=x\)
\(\sqrt[4]{4x-3}=\sqrt[4]{1.1.1.\left(4x-3\right)}\le\frac{1+1+1+4x-3}{4}=x\)
\(\sqrt[6]{6x-5}=\sqrt[6]{1.1.1.1.1.\left(6x-5\right)}\le\frac{1+1+1+1+1+6x-5}{6}=x\)
\(\Rightarrow\sqrt{2x-1}+\sqrt[4]{4x-3}+\sqrt[6]{6x-5}\le3x\)
Dấu "=" xảy ra \(\Leftrightarrow x=1\) ( Thỏa mãn ĐKXĐ )
Vậy pt có nghiệm duy nhất \(x=1\)
a: Ta có: \(\sqrt{\left(x-3\right)^2}=3-x\)
\(\Leftrightarrow\left|x-3\right|=3-x\)
\(\Leftrightarrow x-3\le0\)
hay \(x\le3\)
b: Ta có: \(\sqrt{4x^2-20x+25}+2x=5\)
\(\Leftrightarrow\left|2x-5\right|=5-2x\)
\(\Leftrightarrow2x-5\le0\)
hay \(x\le\dfrac{5}{2}\)
đặt \(\sqrt{3x+1}=a\)
=> pt <=> 4x^2 +a +6=a^2 +12x
chuyển hết nt sang vế phải để vt =0 ptđttnt có ntc=a+2x-3
câu 2 đặt \(\sqrt[3]{3x-5}=2y-3\) rồi làm tt như bài trên lớp
sau khi chuyển cậu có pt a62-4x^2-a+12x-6=0
=> a^2+2ax-3a-2ax-4x^2+6x+2a+4x-6=0
<=> (a+2x-3)(a-2x+2)=0