Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen

help me, pleaseee

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biết

Đk: \(x\ge6\)

pt\(\Leftrightarrow\sqrt{5x^2+4x}=5\sqrt{x}+\sqrt{x^2-3x-18}\)

\(\Leftrightarrow5x^2+4x=25x+x^2-3x-18+10\sqrt{x\left(x^2-3x-18\right)}\)

\(\Leftrightarrow2x^2-9x+9=5\sqrt{x^3-3x^2-18x}\)

\(\Leftrightarrow4x^4+81x^2+81-36x^3-162x+36x^2=25\left(x^3-3x^2-18x\right)\)

\(\Leftrightarrow4x^4-61x^3+192x^2+288x+81=0\)

\(\Leftrightarrow\left(x-9\right)\left(4x+3\right)\left(x^2-7x-3\right)=0\)

\(\Leftrightarrow\left(4x+3\right)\left(x-9\right)\left(x-\dfrac{7+\sqrt{61}}{2}\right)\left(x-\dfrac{7-\sqrt{61}}{2}\right)=0\)

mà x \(\ge6\) \(\Rightarrow\left\{{}\begin{matrix}4x+3>0\\x-\dfrac{7-\sqrt{61}}{2}>0\end{matrix}\right.\)

\(\Rightarrow\left(x-9\right)\left(x-\dfrac{7+\sqrt{61}}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=\dfrac{7+\sqrt{61}}{2}\end{matrix}\right.\)

Vậy.....

Sau khi bình phương lần thứ nhất, đến:

\(2x^2-9x+9=5\sqrt{x^3-3x^2-18}\)

Thay vì bình phương tiếp lên bậc 4 rất cồng kềnh, em có thể đặt ẩn phụ:

\(\Leftrightarrow2x^2-9x+9=5\sqrt{\left(x+3\right)\left(x^2-6x\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-6x}=a\\\sqrt{x+3}=b\end{matrix}\right.\) ta được:

\(2a^2+3b^2=5ab\)

\(\Leftrightarrow\left(a-b\right)\left(2a-3b\right)=0\)

\(a,PT\Leftrightarrow\left|x+3\right|=3x-6\\ \Leftrightarrow\left[{}\begin{matrix}x+3=3x-6\left(x\ge-3\right)\\x+3=6-3x\left(x< -3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\left(tm\right)\\x=\dfrac{3}{4}\left(ktm\right)\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{9}{2}\\ b,PT\Leftrightarrow\left|x-1\right|=\left|2x-1\right|\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2x-1\\1-x=2x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

\(c,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=25x^2-20x+4\\ \Leftrightarrow25x^2-15x=0\\ \Leftrightarrow5x\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{3}{5}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\\ d,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=2-5x\\ \Leftrightarrow x\in\varnothing\)

vô đây Câu hỏi của Phan hữu Dũng - Toán lớp 9 - Học toán với OnlineMath

Cho mình copy nhé:

Đặt \(\sqrt{3x-2}=a;\sqrt{x-1}=b\left(a,b\ge0\right)\)

\(\Rightarrow\begin{cases}a^2=3x-2\\b^2=x-1\end{cases}\)\(\Rightarrow a^2+b^2=4x-3\)

\(pt\Leftrightarrow a+b=a^2+b^2-6+2ab\)

\(\Leftrightarrow a^2+b^2-6+2ab-a-b=0\)

\(\Leftrightarrow\left(a+b\right)^2-\left(a+b\right)-6=0\)

\(\Leftrightarrow\left(a+b\right)^2+2\left(a+b\right)-3\left(a+b\right)-6=0\)

\(\Leftrightarrow\left(a+b\right)\left(a+b+2\right)-3\left(a+b+2\right)=0\)

\(\Leftrightarrow\left(a+b-3\right)\left(a+b+2\right)=0\)

\(\Leftrightarrow a+b=3\)hoặc\(a+b=-2\)(loại,vì a\(\ge\)0;b\(\ge\)0 =>a+b\(\ge\)0)

• Với a+b=3

\(\Rightarrow\sqrt{3x-2}+\sqrt{x-1}=3\)

\(\Leftrightarrow\sqrt{3x-2}=3-\sqrt{x-1}\)

\(\Rightarrow3x-2=9+x-1-6\sqrt{x-1}\)

\(\Rightarrow2x-10=-6\sqrt{x-1}\)

\(\Rightarrow4x^2-40x+100=36\left(x-1\right)\)

\(\Rightarrow4x^2-76x+1236=0\)

\(\Rightarrow4x^2-8x-68x+136=0\)

\(\Rightarrow4x\left(x-2\right)-68\left(x-2\right)=0\)

\(\Rightarrow\left(4x-68\right)\left(x-2\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=17\left(loai\right)\\x=2\left(TM\right)\end{array}\right.\)

Vậy phương trình đã cho có nghiệm là x=2

giải pt 

ảo ma