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\(\Leftrightarrow12x^2-12x+20-10x+17=0\)
\(\Leftrightarrow12x^2-22x+37=0\)
\(\text{Δ}=\left(-22\right)^2-4\cdot12\cdot37=-1292< 0\)
Do đó: Phương trình vô nghiệm
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\(12x^2-4\left(3x-5\right)=10x-17\)
\(\Leftrightarrow12x^2-12x+20=10x-17\)
\(\Leftrightarrow12x^2-12x-10x=-17-20\)
\(\Leftrightarrow-12x^2-22x=-37\)
![](https://rs.olm.vn/images/avt/0.png?1311)
làm khuyến mại 1 câu;
a) = 12x2 -12x2 +20x -10x +17 =0
10x = -17
x = -17/10
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a) 5.(x^2-3x+1)+x.(1-5x)=x-2
\(\Leftrightarrow5x^2-15x+5+x-5x^2=x-2\)
\(\Leftrightarrow-14x-x=-2-5\)
\(\Leftrightarrow-15x=-7\)
\(\Leftrightarrow x=\frac{7}{15}\)
b\(,3x.\left(\frac{4}{3}+1\right)-4x\left(x-2\right)=10\)
\(\Leftrightarrow4x+3x-4x^2+8x-10=0\)
\(\Leftrightarrow-4x^2+15x-10=0\)
Đề sai???
\(c,12x^2-4x\left(3x-5\right)=10x-17\)
\(\Leftrightarrow12x^2-12x^2+20x-10x=-17\)
\(\Leftrightarrow10x=-17\)
\(\Leftrightarrow x=-\frac{17}{10}\)
\(d,4x\left(x-5\right)-7x\left(x-4\right)+3x^2=12\)
\(\Leftrightarrow4x^2-20x-7x^2+28x+3x^2=12\)
\(\Leftrightarrow8x=12\)
\(\Leftrightarrow x=\frac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,5\left(x^2-3x+1\right)+x\left(1-5x\right)=x-2\)
\(\Leftrightarrow5x^2-15x+5+x-5x^2-x+2=0\)
\(\Leftrightarrow-15x+7=0\)
\(\Leftrightarrow-15x=-7\)
\(\Leftrightarrow x=-\frac{7}{-15}\)
\(\Leftrightarrow x=\frac{7}{15}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
c: Ta có: \(x^3-12x^2+48x-64=0\)
\(\Leftrightarrow x-4=0\)
hay x=4
![](https://rs.olm.vn/images/avt/0.png?1311)
c: Ta có: \(x^3-12x^2+48x-64=0\)
\(\Leftrightarrow x-4=0\)
hay x=4
\(\Leftrightarrow12x^2-12x+20-10x+17=0\)
\(\Leftrightarrow12x^2-22x+37=0\)
\(\Leftrightarrow12\left(x^2-\frac{11}{6}x+\frac{37}{12}\right)=0\)
\(\Leftrightarrow12\left(x^2-2x\cdot\frac{11}{12}+\frac{121}{144}-\frac{121}{144}+\frac{37}{12}\right)=0\)
\(\Leftrightarrow12\left[\left(x-\frac{11}{12}\right)^2+\frac{323}{144}\right]=0\)
\(\Leftrightarrow12\left(x-\frac{11}{12}\right)^2+\frac{323}{12}=0\) ( vô nghiệm )