Câu hỏi của Tran Mai Ngoc - Toán lớp 7 - Học toán với OnlineMath

\(ĐKXĐ:x\ne1,x\ne3,x\ne8,x\ne20\)

\(\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}=-\dfrac{3}{4}\)

\(\Rightarrow\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{2\left(x-8\right)\cdot\left(x-20\right)+5\left(x-1\right)\cdot\left(x-20\right)+12\left(x-1\right)\cdot\left(x-3\right)-\left(x-1\right)\cdot\left(x-3\right)\cdot\left(x-8\right)}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{\left(2x-16\right)\cdot\left(x-20\right)+\left(5x-5\right)\cdot\left(x-20\right)+\left(12x-12\right)\cdot\left(x-3\right)-\left(x^2-3x-x+3\right)\cdot\left(x-8\right)}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{2x^2-40x-16x+320+5x^2-100x-5x+100+12x^2-36x-12x+36-\left(x^2-4x+3\right)\cdot\left(x-8\right)}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{2x^2-40x-16x+320+5x^2-100x-5x+100+12x^2-36x-\left(x^3-8x^2-4x^2+32x+3x-24\right)}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}\)

\(\Leftrightarrow\dfrac{2x^2-40x-16x+320+5x^2-100x-5x+100+12x^2-36x-12x+36-\left(x^3-12x^2+35x-24\right)}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{2x^2-40x-16x+320+5x^2-100x-5x+100+12x^2-36x-12x+36-x^3+12x^2-35x+24}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{31x^2-244x+480-x^3}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{-x^3+31x^2-244x+480}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{-x^3+3x^2+28x^2-84x-160x+480}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{-x^2\cdot\left(x-3\right)+28x\cdot\left(x-3\right)-160\left(x-3\right)}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{-\left(x-3\right)\left(x^2-28x+160\right)}{\left(x-1\right)\left(x-3\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{-1\left(x^2-8x-20x+160\right)}{\left(x-1\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{-1\left(x^2-8x-20x+160\right)}{\left(x-1\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{-1\left(x\cdot\left(x-8\right)-20\left(x-8\right)\right)}{\left(x-1\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{-1\left(x-20\right)\left(x-8\right)}{\left(x-1\right)\left(x-8\right)\left(x-20\right)}=-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{-1}{x-1}=-\dfrac{3}{4}\)

\(\Leftrightarrow-\dfrac{1}{x-1}=-\dfrac{3}{4}\)

\(\Leftrightarrow-4=-3\left(x-1\right)\)

\(\Leftrightarrow-4=-3\left(x-1\right)\)

\(\Leftrightarrow-4=-3x+3\)

\(\Leftrightarrow3x=3+4\)

\(\Leftrightarrow3x=7\)

\(\Rightarrow x=\dfrac{7}{3}\)

Vậy \(x=\dfrac{7}{3}\)

Đúng rồi có ý kiến: tự làm 2 GP, chép 1 GP (hoặc vẫn đc tick và hiển thị câu trả lời do hoc24 chọn nhưng k có GP nào) Vì nếu chép người hỏi cx cần biết nó có đúng không? >>> vẫn cần một cái tick từ GV.

P/s: nếu ý kiến này thành sự thật thì Bình Trần Thị lỗ nặng

Giờ tui vẫn còn cóp rải rác

c) \(2\left(x-4\right)^2-\left(x-4\right)=0\)

\(\Rightarrow\left(x-4\right)\cdot\left[2\left(x-4\right)-1\right]=0\)

\(\Leftrightarrow\left(x-4\right)-\left(2x-8-1\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(2x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\2x-9=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{9}{2}\end{matrix}\right.\)

Vậy \(x_1=4;x_2=\dfrac{9}{2}\)

d) \(3x\left(x-1\right)+\left(x-1\right)^2=0\)

\(\Rightarrow\left(x-1\right)\left(3+x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2+x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2+x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2+x=0\end{matrix}\right.\)

Vậy \(x_1=-2;x_2=1\)

d) \(5x\left(x-9\right)^2-\left(9-x\right)^3=0\)

\(\Rightarrow5x\left(x^2-18x+81\right)-\left(729-243x+27x^2-x^3\right)=0\)

\(\Leftrightarrow5x^3-90x^2+405x-729+243x-27x^2+x^3=0\)

\(\Leftrightarrow6x^3-117x^2+648x-729=0\)

\(\Leftrightarrow3\left(2x^3-39x^2+216x-243\right)=0\)

\(\Leftrightarrow3\left(2x^3-18x^2-21x^2+189x+27x-243\right)=0\)

\(\Leftrightarrow3\left(2x^2\cdot\left(x-9\right)-21\cdot\left(x-9\right)+27\left(x-9\right)\right)=0\)

\(\Leftrightarrow3\left(x-9\right)\left(2x^2-21x+27\right)=0\)

\(\Leftrightarrow3\left(x-9\right)\cdot\left(2x^2-38-18x+27\right)=0\)

\(\Leftrightarrow3\left(x-9\right)\cdot\left(x\cdot\left(2x-3\right)-9\left(2x-3\right)\right)=0\)

\(\Leftrightarrow3\left(x-9\right)\cdot\left(x-9\right)\cdot\left(2x-3\right)=0\)

\(\Leftrightarrow3\left(x-9\right)^2\cdot\left(2x-3\right)=0\)

\(\Leftrightarrow\left(x-9\right)^2\cdot\left(2x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-9\right)^2=0\\2x-3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=9\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy \(x_1=\dfrac{3}{2};x_2=9\)

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\(\sqrt{x+6}=x\)

\(\Rightarrow x+6=x^2\)

\(\Leftrightarrow x+6-x^2=x^2-x^2\)

\(\Leftrightarrow-x^2+x+6=0\)

\(\Leftrightarrow\left(-x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x-2=0\\x-3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-2\left(loại\right)\\x=3\left(nhận\right)\end{matrix}\right.\)

Vậy \(x=3\)