Tìm x biết 6(x-3)(x-4)-6x(x-2)=4
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\(\left(x+2\right)^3-x^2\left(x-6\right)-4=0\\ \Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2-4=0\\ \Leftrightarrow12x-12=0\\ \Leftrightarrow12x=12\\ \Leftrightarrow x=1\)
\(6x^2-\left(2x-3\right)\left(3x+2\right)=1\\ \Leftrightarrow6x^2-\left[3x.\left(2x-3\right)+2.\left(2x-3\right)\right]=1\\ \Leftrightarrow6x^2-\left(6x^2-9x+4x-6\right)=1\\ \Leftrightarrow6x^2-\left(6x^2-5x-6\right)=1\\ \Leftrightarrow6x^2-6x^2+5x+6=1\\ \Leftrightarrow5x=-5\\ \Leftrightarrow x=-1\)
a) \(-\dfrac{2}{5}+\dfrac{5}{6}x=-\dfrac{4}{15}\\ \Leftrightarrow\dfrac{5}{6}x=\dfrac{2}{15}\\ \Leftrightarrow x=\dfrac{4}{25}\)
b) \(\dfrac{2}{3}+\dfrac{7}{4}\div x=\dfrac{5}{6}\\ \Leftrightarrow\dfrac{7}{4}\div x=\dfrac{1}{6}\\ \Leftrightarrow x=\dfrac{7}{24}\)
a: Ta có: \(-\dfrac{2}{5}+\dfrac{5}{6}x=\dfrac{-4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{5}{6}=\dfrac{2}{15}\)
hay \(x=\dfrac{4}{25}\)
b: Ta có: \(\dfrac{7}{4}:x+\dfrac{2}{3}=\dfrac{5}{6}\)
\(\Leftrightarrow\dfrac{7}{4}:x=\dfrac{1}{6}\)
hay \(x=\dfrac{21}{2}\)
6(x-3)(x-4)-6x(x-2)=4
<=>6(x2-7x+12)-6x2+12x=4
<=>6x2-42x+72-6x2+12x-4=0
<=>-30x+68=0
<=>-30x =-68
<=>x =34/15
Ta có :
\(\left(x-5\right)\left(6x+1\right)-\left(2x-3\right)\left(3x+4\right)-x=6\)
\(\Rightarrow\left(6x^2-30x+x-5\right)-\left(6x^2-9x+8x-12\right)-x=6\)
\(\Rightarrow6x^2-29x-5-\left(6x^2-x-12\right)-x=6\)
\(\Rightarrow6x^2-29x-5-6x^2+x+12-x=6\)
\(\Rightarrow\left(6x^2-6x^2\right)+\left(-29x+x-x\right)+\left(12-5\right)=6\)
\(\Rightarrow-29x+7=6\)
\(\Rightarrow-29x=6-7\)
\(\Rightarrow-29x=-1\)
\(\Rightarrow x=\frac{1}{29}\)
Vậy \(x=\frac{1}{29}\)
bài 5:
1: \(\dfrac{12x^3y^2}{18xy^5}=\dfrac{12x^3y^2:6xy^2}{18xy^5:6xy^2}=\dfrac{2x^2}{3y^3}\)
2: \(\dfrac{10xy-5x^2}{2x^2-8y^2}=\dfrac{5x\cdot2y-5x\cdot x}{2\left(x^2-4y^2\right)}\)
\(=\dfrac{5x\left(2y-x\right)}{-2\left(x+2y\right)\left(2y-x\right)}=\dfrac{-5x}{2\left(x+2y\right)}\)
3: \(\dfrac{x^2-xy-x+y}{x^2+xy-x-y}\)
\(=\dfrac{\left(x^2-xy\right)-\left(x-y\right)}{\left(x^2+xy\right)-\left(x+y\right)}\)
\(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)
4: \(\dfrac{\left(x+1\right)\left(x^2-2x+1\right)}{\left(6x^2-6\right)\left(x^3-1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)^2}{6\left(x^2-1\right)\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)}{6\left(x-1\right)\left(x+1\right)\cdot\left(x^2+x+1\right)}\)
\(=\dfrac{1}{6\left(x^2+x+1\right)}\)
5: \(\dfrac{2x^2-7x+3}{1-4x^2}\)
\(=-\dfrac{2x^2-7x+3}{4x^2-1}\)
\(=-\dfrac{2x^2-6x-x+3}{\left(2x-1\right)\left(2x+1\right)}\)
\(=-\dfrac{2x\left(x-3\right)-\left(x-3\right)}{\left(2x-1\right)\left(2x+1\right)}\)
\(=-\dfrac{\left(x-3\right)\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{-x+3}{2x+1}\)
Bài 3:
1: \(9x^3-xy^2\)
\(=x\cdot9x^2-x\cdot y^2\)
\(=x\left(9x^2-y^2\right)\)
\(=x\left(3x-y\right)\left(3x+y\right)\)
2: \(x^2-3xy-6x+18y\)
\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)
\(=x\left(x-3y\right)-6\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-6\right)\)
3: \(x^2-3xy-6x+18y\)
\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)
\(=x\left(x-3y\right)-6\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-6\right)\)
4: \(6xy-x^2+36-9y^2\)
\(=36-\left(x^2-6xy+9y^2\right)\)
\(=36-\left(x-3y\right)^2\)
\(=\left(6-x+3y\right)\left(6+x-3y\right)\)
5: \(x^4-6x^2+5\)
\(=x^4-x^2-5x^2+5\)
\(=x^2\left(x^2-1\right)-5\left(x^2-1\right)\)
\(=\left(x^2-5\right)\left(x^2-1\right)\)
\(=\left(x^2-5\right)\left(x-1\right)\left(x+1\right)\)
6: \(9x^2-6x-y^2+2y\)
\(=\left(9x^2-y^2\right)-\left(6x-2y\right)\)
\(=\left(3x-y\right)\left(3x+y\right)-2\left(3x-y\right)\)
\(=\left(3x-y\right)\left(3x+y-2\right)\)
Bài 1:
a: \(x^3-6x^2+11x-6\)
\(=x^3-x^2-5x^2+5x+6x-6\)
\(=\left(x-1\right)\left(x^2-5x+6\right)\)
\(=\left(x-1\right)\left(x-2\right)\left(x-3\right)\)
b: \(x^3-6x^2-9x+14\)
\(=x^3-7x^2+x^2-7x-2x+14\)
\(=\left(x-7\right)\left(x^2+x-2\right)\)
\(=\left(x-7\right)\left(x+2\right)\left(x-1\right)\)
c: \(x^3+6x^2+11x+6\)
\(=x^3+3x^2+3x^2+9x+2x+6\)
\(=\left(x+3\right)\left(x^2+3x+2\right)\)
\(=\left(x+3\right)\left(x+1\right)\left(x+2\right)\)
6(x-3)(x-4)-6x(x-2)=4
<=>(6x-18)(x-4)-6x2+12=4
<=>6x2-24x-18x+72-6x2+12=4
<=>-30x+72=4
<=>-30x=-68
<=>x=34/15